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THE COMPENDIOUS MEASURER; BEING A BRIEF, YET COMPREHENSIVE, TREATISE ON MENSURATION AND PRACTICAL GEOMETRY.

WITH AN INTRODUCTION TO DECIMAL AND DUODECIMAL ARITHMETIC.

ADAPTED TO THE USE OF SCHOOLS AND PRACTICE.

By CHARLES HUTTON, LL.D. and F.R.S. &c.

LONDON▪ PRINTED FOR G. C. J. AND J. ROBINSON, AND R. BALDWIN, IN PATERNOSTER-ROW; AND G. AND J. WILKIE, ST. PAUL'S CHURCH-YARD.

M.DCC.LXXXVI.

PREFACE.

[]

SOME years ſince I publiſhed a complete Treatiſe on Menſuration, both in Theory and Practice; in which the Elements of that Science are demonſtrated, and the Rules applied to the various practical pur⯑poſes of life. That work has been well received by the Public, and honoured with the high approbation of the more learned Mathematicians.

It has, however, been often repreſented to me, by Tutors and others, that the great ſize and price of that work, as well as the very ſcientific manner in which it is formed, prevent it from being ſo gene⯑rally uſeful in ſchools, and to practical meaſurers, as a more compendious and familiar little book might be, which they could put into the hands of their pu⯑pils, as a work containing all the practical rules of that art, in a form proper for them to copy from, and unmixed with ſuch geometrical and algebraical de⯑monſtrations as occur in the large work.

In compliance, therefore, with ſuch repreſentations, I have drawn up this Compendium of Menſuration, Practical Geometry, and Arithmetic, expreſsly with the view of accommodating it to practical matters, and to the uſe of ſchools. I have, therefore, here brought together all the moſt uſeful rules and pre⯑cepts; have arranged them in an orderly manner, proper for the pupil to copy; and delivered them in plain and familiar language. An example, worked out at full length, is ſet down to each rule, together with drawings or repreſentations of the geometrical figures proper to illuſtrate each problem; and then are ſubjoined ſome more queſtions to each rule, as examples propoſed for the practice of the learner, with [iv]the anſwer ſet down, by which he may know when his work is right.

The introduction to Decimal and Duodecimal Arithmetic, will be found uſeful, by going over thoſe branches before entering on the Menſuration, that the learner may be very ready and expert in numeral calculations.

The Practical Geometry contains a great num⯑ber of geometrical conſtructions and operations; by the practice of which the learner will acquire the free and eaſy uſe of his inſtruments, and ſo become prepared for making the drawings that are uſeful for illuſtrating the various branches of menſuration fol⯑lowing.

The Menſuration itſelf next ſucceeds, and is di⯑vided into various parts; firſt, Menſuration in gene⯑ral, and then as applied to the ſeveral practical uſes in life. The whole being arranged in ſuch order as the learner may properly take in ſucceſſion; or diſtin⯑guiſhed into ſeveral branches, of which he may ſelect and ſtudy any peculiar ones that may be more to his purpoſe than the reſt, when he either has not leiſure or inducement to go over the whole in a regular gra⯑dation. And notwithſtanding the compendious ſize of the book, and the great number of practical branches here treated, it will be found that each one is much more full and complete than the firſt appear⯑ance of ſo ſmall a form may promiſe to admit of. However, if farther ſatisfaction be deſired by any one, either concerning the ſcience in general, or the demonſtrations of the rules, or the more curious and copious diſplay of properties, he may apply to my large treatiſe before mentioned, where he will find every part delivered in the moſt ample form.

CONTENTS.

[]

ARITHMETIC.

DECIMAL Fractions Page 1
Addition of Decimals Page 4
Subtraction of Decimals Page 5
Multiplication of Decimals Page 5
Diviſion of Decimals Page 8
Reduction of Decimals Page 11
Circulating Decimals Page 17
Reduction of Repetends Page 17
Addition of Repetends Page 19
Subtraction of Repetends Page 20
Multiplication of Repetends Page 20
Diviſion of Repetends Page 22
Involution, or Raiſing of Powers Page 24
Evolution, or Extraction of Roots Page 29
To extract the Square Root Page 31
Ditto of Vulgar Fractions or Mixt Numbers Page 35
To find a Mean Proprotional Page 37
To extract the Cube Root Page 40
To find two Mean Proportionals Page 43
To extract any Root whatever Page 46
Ditto of Vulgar Fractions or Mixt Numbers Page 48
Duodecimals, or Croſs Multiplication Page 49

PRACTICAL GEOMETRY.

Definitions Page 56
Problems Page 64
To divide a Line into two equal Parts Page 64
To divide an Angle into two equal Parts Page 64
To divide a Right Angle into three equal Parts Page 64
To draw one Line parallel to another Page 65
To raiſe a Perpendicular on a given Line Page 65
To let fall a Perpendicular on a given Line Page 67
To divide a Line into any Number of equal Parts Page 68
To divide a Line in a given Proportion Page 68
To make one Angle equal to another Page 69
To make an Angle of any Number of Degrees Page 69
To meaſure a given Angle Page 70
To find the Center of a Circle Page 70
To draw a Circle through three Points Page 70
To draw a Tangent to a Circle Page 71
To find a third Proportional to two Lines Page 71
To find a Fourth Proportional to three Lines Page 72
To find a Mean Proportional between two Lines Page 72
To make an Equilateral Triangle Page 72
To make a Triangle with three given Lines Page 73
To make a Square Page 73
To make a Rectangle Page 73
To make a Pentagon Page 74
To make a Hexagon Page 75
To make an Octagon Page 76
To make any regular Polygon Page 76
To inſcribe a Polygon in a Circle Page 77
To deſcribe a Polygon about a Circle Page 78
To find the Center of a Polygon Page 79
To inſcribe a Circle in a Triangle Page 79
To deſcribe a Circle about a Triangle Page 80
To deſcribe a Circle within or about a Square Page 80
To deſcribe a Square or Octagon in or about a Circle Page 80
To inſcribe a Trigon, or a Hexagon, in a Circle Page 81
To inſcribe a Dodecagon in a Circle Page 81
[vii]To inſcribe a Pentagon or a Decagon in a Circle Page 82
To divide a Circle into 12 equal Parts Page 82
To draw a Line equal to the Circumf. of a Circle Page 83
To draw a Right Line equal to any Arc of a Circle Page 83
To divide a Circle into any Number of equal Parts Page 84
To make one Triangle ſimilar to another Page 84
To make a Figure ſimilar to a given one Page 85
To reduce a Figure from one Scale to another Page 85
To make a Triangle equal to a Trapezium Page 86
To make a Triangle equal to a Pentagon Page 86
To make a Triangle equal to a Circle Page 87
To make a Parallelogram equal to a Triangle Page 87
To make a Square equal to a Rectangle Page 88
To make a Square equal to the ſum of two Squares Page 88
To make a Square equal to the diff. of two Squares Page 89
To make a Square equal to any Number of Squares Page 89
To make plane Diagonal Scales Page 90
Geometrical Remarks Page 92

MENSURATION OF SUPERFICIES Page 93

To find the Area of a Parallelogram Page 94
To find the Area of a Triangle Page 96
To find any Side of a Right-angled Triangle Page 98
To find the Area of a Trapezoid Page 100
To find the Area of a Trapezium Page 101
To find the Area of any irregular Figure Page 104
To find the Area of a regular Polygon Page 105
To find the Radius, or Chord, &c. of an Arc Page 107
To find the Diameter and Circumference of a Circle Page 108
To find the Length of any Circular Arc Page 110
To find the Area of a Circle Page 111
To find the Area of a Circular Sector Page 112
To find the Area of a Circular Segment Page 113
To find the Area of a Circular Zone Page 117
To find the Area of a Circular Ring Page 119
To meaſure long irregular Figures Page 120

MENSURATION OF SOLIDS Page 121

Definitions Page 121
To find the Solidity of a Cube Page 125
To find the Solidity of a Parallelopipedon Page 125
To find the Solidity of any Priſm Page 126
To find the Convex Surface of a Cylinder Page 127
To find the Convex Surface of a Cone Page 128
To find the Surface of a Conical Fruſtum Page 129
To find the Solidity of any Pyramid or Cone Page 130
To find the Solidity of the Fruſt. of a Cone or Pyr. Page 131
To find the Solidity of a Wedge Page 134
To find the Solidity of a Priſmoid Page 134
To find the Surface of a Sphere or Globe Page 136
To find the Solidity of a Sphere or Globe Page 137
To find the Solidity of a Spherical Segment Page 138
To find the Solidity of a Spherical Zone Page 138
To find the Surface of a Circular Spindle Page 139
To find the Solidity of a Circular Spindle Page 141
To find the Solidity of the Zone of a Circ. Spindle Page 142
To find the Surface or Solidity of any regular Body Page 145
To find the Surface of a Cylindrical Ring Page 147
To find the Solidity of a Cylindrical Ring Page 147

OF THE CARPENTER'S RULE Page 149

To multiply Numbers by the Rule Page 150
To divide by the Rule Page 150
To Square any Number Page 151
To extract the Square Root Page 151
To find a Mean Proportional Page 151
To find a Third Proportional Page 152
To find a Fourth Proportional Page 152

TIMBER MEASURING Page 153

To meaſure Board or Plank Page 153
To meaſure Squared Timber Page 154
To meaſure Round Timber Page 156

ARTIFICERS WORKS Page 160

Bricklayers Work Page 161
Maſons Work Page 164
Carpenters Work Page 166
Slaters and Tilers Work Page 169
Plaſterers Work Page 170
Painters Work Page 171
Glaziers Work Page 172
Pavers Work Page 174
Plumbers Work Page 175

VAULTED AND ARCHED ROOFS Page 176

To find the Surface of a Vaulted Roof Page 177
To find the Solid Content of a Vaulted Roof Page 177
To find the Surface of a Dome Page 178
To find the Solid Content of a Dome Page 179
To find the Superficies of a Saloon Page 180
To find the Solid Content of a Saloon Page 181
To find the Concave Surface of a Groin Page 182
To find the Solid Content of a Groin Page 183

LAND SURVEYING Page 185

Of the Surveying Chain Page 185
Of the Plain Table Page 187
Of the Theodolite Page 190
Of the Croſs Page 191
Of the Field-Book Page 193

THE PRACTICE OF SURVEYING Page 195

To meaſure a Line Page 195
To meaſure Angles and Bearings Page 197
To meaſure offsets Page 198
To meaſure a Triangular Field Page 199
To meaſure a Four-ſided Field Page 200
To ſurvey any Field by the Chain only Page 202
To ſurvey a Field with the Plain Table Page 203
[x]To ſurvey a Field with the Theodolite Page 205
To ſurvey a Field having crooked Fences Page 207
To ſurvey a Field by two Stations Page 208
To ſurvey a large Eſtate Page 210
To ſurvey a County, or a large Tract Page 213
To ſurvey a Town or City Page 216

OF PLANNING, CASTING-UP, AND DIVIDING Page 218

To lay down the Plan of any Survey Page 218
To caſt up the Contents of Fields Page 220
To transfer a Plan to another Paper Page 226

CONIC SECTIONS AND THEIR SOLIDS Page 229

Definitions Page 229
To deſcribe an Ellipſe Page 231
To find the Tranſverſe, Conj. &c. in an Ellipſe Page 232
To find the Circumference of an Ellipſe Page 236
To find the Area of an Ellipſe Page 237
To find the Area of an Elliptic Segment Page 237
To deſcribe a Parabola Page 238
To find a Parabolic Abſciſs or Ordinate Page 239
To find the Length of a Parabolic Curve Page 239
To find the Area of a Parabola Page 240
To find the Area of a Parabolic Fruſtum Page 241
To conſtruct or deſcribe an Hyperbola Page 242
To find the Tranſverſe, Conj. and in an Hyperbola Page 242
To find the Length of an Hyperbolic Curve Page 246
To find the Area of an Hyperbola Page 248
To find the Solidity of a Spheroid Page 250
To find the Solidity of a Spheroidal Segment Page 250
To find the Solidity of a Spheroidal Fruſtum Page 253
To find the Solidity of an Elliptic Spindle Page 255
To find the Fruſtum or Seg. of an Ellip. Spindle Page 258
To find the Solidity of a Paraboloid Page 260
To find the Solidity of a Paraboloidal Fruſtum Page 261
To find the Solidity of a Parabolic Spindle Page 261
To find the Fruſtum of a Parabolic Spindle Page 262

OF GAUGING Page 265

Of the Gauging Rule Page 266
The Uſe of the Gauging Rule Page 267
Of the Diagonal Rod Page 269
Of Caſks as divided into Varieties Page 270
To find the Content of a Caſk of the firſt Form Page 271
To find the Content of a Caſk of the ſecond Form Page 273
To find the Content of a Caſk of the third Form Page 274
To find the Content of a Caſk of the fourth Form Page 275
To find the Content of a Caſk by four Dimenſions Page 277
To find the Content of a Caſk by three Dimenſions Page 278

OF ULLAGING CASKS Page 280

To find the Ullage by the Sliding Rule Page 280
To Ullage a ſtanding Caſk by the Pen Page 281
To Ullage a lying Caſk by the Pen Page 281

OF SPECIFIC GRAVITY Page 283

A Table of Specific Gravities Page 283
To find the Magnitude of a Body from its Weight Page 284
To find the Weight of a Body from its Magnitude Page 285
To find the Specific Gravity of a Body Page 286
To find the Quantity of Ingredients in a Compound Page 287

WEIGHT AND DIMENSIONS OF BALLS Page 289

To find the Weight of an Iron Ball Page 289
To find the Weight of a Leaden Ball Page 290
To find the Diameter of an Iron Ball Page 291
To find the Diameter of a Leaden Ball Page 293
To find the Weight of an Iron Shell Page 294
To find how much Powder will fill a Shell Page 295
To find the Powder to fill a Rectangular Box Page 296
To find the Powder to fill a Cylinder Page 296
To find the Shell from the Weight of Powder Page 297
To find a Cube to any Weight of Powder Page 298
To find a Cylinder to any Weight of Powder Page 299

OF PILING BALLS AND SHELLS Page 301

To find the Number of Balls in a Triangular Pile Page 301
To find the Number of Balls in a Square Pile Page 302
To find the Number in a Rectangular Pile Page 303
To find the Number in an Incomplete Pile Page 304

OF DISTANCES BY THE VELOCITY OF SOUND Page 306
MISCELLANEOUS QUESTIONS Page 308
A TABLE OF CIRCULAR SEGMENTS Page 317
THE USE OF THE TABLE OF SEGMENTS Page 322

[]

INTRODUCTION.

DECIMAL FRACTIONS.

A Decimal is a fraction whoſe denominator is an unit, or 1, with ſome number of ciphers an⯑nexed; as 1/10 or 45/10000.

Decimals are written down without their denomi⯑nators, the numerators being ſo diſtinguiſhed as to ſhew what the denominators are; which is done, by ſeparating, by a point, ſo many of the right-hand figures from the reſt as there are ciphers in the deno⯑minator; the figures on the left ſide of the point be⯑ing integers, and thoſe on the right decimals.

Thus, 0.5 is underſtood to be 5/10 or ½
And 0.25 is underſtood to be 25/100 or ¼
And 0.75 is underſtood to be 75/100 or ¾
And 1.3 is underſtood to be 13/10 or 1 3/10
And 24.6 is underſtood to be 24 6/10

But when there is not a ſufficient number of figures in the numerator, ciphers are prefixed to ſupply the defect.

So, .02 is 2/100 or 1/50
And .0015 is 15/10000 or 3/2000

So that the denominator of a decimal is a 1, with as many ciphers as there are figures in the decimal.

A finite decimal, is that which ends at a certain number of places. But an infinite decimal, that which no where ends, but is underſtood to be indefinitely continued.

A repeating decimal, has one figure, or ſeveral figures continually repeated. As 20.2433 &c. which is a ſingle or ſimple repetend. And 20.2424 &c. or 20.246246 &c. which are compound re⯑petends; and are otherwiſe called circulates, or cir⯑culating decimals. A point is ſet over a ſingle re⯑petend, and a point over the firſt and laſt figures of a circulating decimal.

The firſt place, next after the decimal mark, is 10th parts, the ſecond is 100th parts, the third is 1000th parts, and ſo on, decreaſing towards the right by 10ths, or increaſing towards the left by 10ths, the ſame as whole or integer numbers do. As in the following Scale of Notation

&c.
Millions
Hundreds of thouſands
Tens of thouſands
Thouſands
Hundreds
Tens
Units
Tenth parts
Hundredth parts
Thouſand parts
Ten thouſand parts
Hundred thouſand parts
Millionth parts
&c.

[3]Ciphers on the right of decimals do not alter their value:

For .5 or 5/10 that is ½
And .50 or 50/100 that is ½
And .500 or 500/1000 that is ½

&c. are all of equal value.

But ciphers before decimal figures, and after the ſeparating point, diminiſh the value in a ten-fold proportion for every cipher.

So, .5 is 5/10 or ½
But .05 is 5/100 or 1/20
And .005 is 5/1000 or 1/200

And ſo on.

So that, in any mixed or fractional number, if the ſeparating point be moved

one, two, three, &c. places to the right hand, every figure will be
10, 100, 1000, &c. times greater than before.

But if the point be moved towards the left hand, then every figure will be diminiſhed in the ſame manner, or the whole quantity will be divided by 10, 100, 1000, &c.

ADDITION OF DECIMALS.

[4]

SET the numbers under each other according to the value of their places, as in whole numbers, or ſo that the decimal points ſtand directly below each other. Then add as in whole numbers, placing the decimal point in the ſum ſtraight below the other points.

EXAMPLES.
(1)	(2)	(3)
276	7530	312.09
39.213	16.201	3.5711
72014.9	3.0142	4195.6
417.	957.13	71.498
5032.	6.72819	9739.215
2214.298	.03014	179.
79993.411	8513.10353	14500.9741

Ex. 4. What is the ſum of .014, .9816, .32, .15914, 72913, and .0047?

Ex. 5. What is the ſum of 27.148, 918.73, 14016, 294304, .7138, and 221.7?

Ex. 6. Required the ſum of 312.984, 21.3918, 2700.42, 3.153, 27.2, and 581.06.

SUBTRACTION OF DECIMALS.

[5]

SET the leſs number under the greater in the ſame manner as in addition. Then ſubtract as in whole numbers, and place the decimal point in the remain⯑der ſtraight below the other points.

EXAMPLES.
(1)	(2)	(3)
From .9173	2.73	214.81
take .2138	1.9185	4.90142
rem. .7035	0.8115	209.90858

Ex. 4. What is the difference between 91.713 and 407?

Ex. 5. What is the difference between 2714 and .916?

Ex. 6. What is the difference between 16.37 and 800.135?

MULTIPLICATION OF DECIMALS.

SET down the factors under each other, and mul⯑tiply them as in whole numbers; and from the pro⯑duct, towards the right hand, point off as many figures for decimals, as there are decimal places in both factors together.

But if there be not as many figures in the product as there ought to be decimals, prefix the proper number of ciphers to ſupply the defect.

[6]

EXAMPLES.
(1)	(2)	(3)
520.3	91.78	.217
.417	.381	.0431
36421	9178	217
5203	73424	651
20812	27534	868
216.9651	34.96818	.0093527

Ex. 4. What is the product of 51.6 and 21?

Ex. 5. What is the product of 314 and .029?

Ex. 6. What is the product of .051 and .0091?

Note, When decimals are to be multiplied by 10, or 100, or 1000, &c. that is by 1 with any number of ciphers, it is done by only moving the decimal point as many places farther to the right hand, as there are ciphers in the ſaid multiplier: ſub⯑joining ciphers if there be not ſo many figures.

EXAMPLES.

1. The product of 51.3 and 10 is
2. The product of 2.714 and 100 is
3. The product of .9163 and 1000 is
4. The product of 21.31 and 10000 is

CONTRACTION.

When the product would contain ſeveral more decimals than are neceſſary for the purpoſe in hand, the work may be much contracted thus, retaining only the proper number of decimals.

Set the units figure of the multiplier ſtraight un⯑der ſuch decimal place of the multiplicand as you [7]intend the laſt of your product ſhall be, writing the other figures of the multiplier in an inverted order: then in multiplying reject all the figures in the mul⯑tiplicand which are on the right of the figure you are multiplying by; ſetting the products down ſo, that their right hand figures fall ſtraight below each other; and carrying to ſuch right-hand figures from the product of the two preceding figures in the multipli⯑cand thus, viz. 1 from 5 to 15, 2 from 15 to 25, 3 from 25 to 35, &c. incluſively; and the ſum of the lines will be the product to the number of decimals required, and will commonly be to the neareſt unit in the laſt figure.

EXAMPLES.

1. Multiply 27.14986 by 92.41035, ſo as to re⯑tain only four places of decimals in the product.

Contracted.	Common way.
27.14986			27.14986
53014.29			92.41035
24434874	13	574930
542997	81	44958
108599	2714	986
2715	108599	44
81	542997	2
14	24434874
2508.9280	2508.9280	650510

2. Multiply 480.14936 by 2.72416, retaining four decimals in the product.

3. Multiply 2490.3048 by .573286, retaining five decimals in the product.

4. Multiply 325.701428 by .7218393, retain⯑ing three decimals in the product.

DIVISION OF DECIMALS.

[8]

DIVIDE as in whole numbers. And to know how many decimals to point off in the quotient, obſerve the following rules:

1. There muſt be as many decimals in the divi⯑dend, as in both the diviſor and quotient together; therefore, point off for decimals in the quotient, as many figures, as the decimal places in the dividend exceed thoſe in the diviſor.

2. If the figures in the quotient are not ſo many as the rule requires, ſupply the defect by pre⯑fixing ciphers.

3. If the decimal places in the diviſor be more than thoſe in the dividend, add ciphers as decimals to the dividend, till the number of decimals in the dividend be equal to thoſe in the diviſor, and the quotient will be integers till all theſe decimals are uſed. And, in caſe of a remainder after all the figures of the dividend are uſed, and more figures are wanted in the quotient, annex ciphers to the remainder, to continue the diviſion as far as neceſ⯑ſary.

4. The firſt figure of the quotient will poſſeſs the ſame place, of integers or decimals, as that figure of the dividend which ſtands over the units place of the firſt product.

EXAMPLES.
1. Divide 3424.6056 by 43.6.	2. Divide 3877875 by .675.
43.6) 3424.6056 (78.546	.675) 3877875.000 (5745000
3726	5028
2380	3037
2005	3375
2616	···· 000

3. Divide .0081892 by .347.
4. Divide 7.13 by .18.
5. Divide 3.15 by 375
6. Divide 109 by .215

CONTRACTIONS.

1. If the diviſor be an integer with any number of ciphers at the end; cut them off, and remove the decimal point in the dividend ſo many places farther to the left as there were ciphers cut off, prefixing ciphers if need be; then proceed as before.

EXAMPLES.
1. Divide 953 by 21000.	2. Divide 41020 by 32000
21.000) .953	32.000) 41.020
7.) .31766	8) 10.255
.04538 &c.	1.281875
Here I firſt divide by 3, and then by 7, becauſe 3 times 7 is 21.	Here I firſt divide by 4, and then by 8, becauſe 4 times 8 is 32.
3. Divide 45.5 by 2170.	4. Divide 61 by 79000.

2. Whence, if the diviſor be 1 with ciphers, the quotient will be the ſame figures with the divi⯑dend, having the decimal point ſo many places far⯑ther to the left as there are ciphers in the diviſor.

EXAMPLES.
217▪3 ÷ 100=2.173	419 by 10 =
5.16 by 1000 =	.21 by 1000 =

3. When the number of figures in the diviſor is great, the diviſion at large will be very troubleſome, but may be contracted thus:

Having, by the 4th general rule, found what place of decimals or integers the firſt figure of the quotient [10]will poſſeſs; conſider how many figures of the quo⯑tient will ſerve the preſent purpoſe; then take the ſame number of the left-hand figures of the diviſor, and as many of the dividend figures as will contain them (leſs than 10 times); by theſe find the firſt fi⯑gure of the quotient; and for each following figure divide the laſt remainder by the diviſor, wanting one figure to the right more than before, but obſerving what muſt be carried to the firſt product for ſuch omitted figures, as in the ſecond contraction of Mul⯑tiplication; and continue the operation till the divi⯑ſor is exhauſted.

When there are not ſo many figures in the diviſor as are required to be in the quotient, begin the divi⯑ſion with all the figures as uſual, and continue it till the number of figures in the diviſor and thoſe re⯑maining to be found in the quotient be equal, after which uſe the contraction.

EXAMPLES.

1. Divide 2508.92806 by 92.41035, ſo as to have four decimals in the quotient.—In this caſe the quotient will contain ſix figures. Hence

92.4103,5)	2508.928,06	(27.1498
	667021
	13849
	4608
	912
	80
	6

2. Divide 4109.2351 by 230.409 ſo that the quotient may contain four decimals.

3. Divide 37.10438 by 5713.96 that the quo⯑tient may contain five decimals.

[11]4. Divide 913.08 by 2137.2 that the quotient may contain three decimals.

REDUCTION OF DECIMALS.

1. To reduce a vulgar fraction to a decimal. Divide the numerator, with as many decimal ciphers annexed, as may be neceſſary, by the denominator; and the quotient will be the decimal ſought.

EXAMPLES.
1. Reduce 1/99 to a decimal.	2. Reduce 1/75 to a decimal.
9) 1.000000	5) 1.0
11) 0.111111	5) 0.20
0.010101 &c.=1/99	3) 0.04
	0.01333 &c.=1/75
Here divide by 9 and 11,	Here divide by 5, 5, and 3,
becauſe 9 times 11 is 99.	becauſe 5 x 5 x 3=75.
And the decimal value of 1/99	And the decimal value of
is the circulate .01.	1/75 is the repetend .013.

OTHER EXAMPLES.

½=.5	1/5=.2	⅛=.125
1/3=.3	1/6=.16	1/9=.1
¼=.25	1/7=.142857	1/10=.1
2/7 =	14/191 =	21/758 =

[12]So that whenever we meet with the repetend .3, in any operation, we may ſubſtitute .⅓ for it; in like manner we may take ⅔ for .6, and ⅙ for .16, and 1/9 for .1, and 9/9 or 1 for .9, &c.

Note, When a great many figures are required in the decimal, and the denominator of the given fraction is a prime number greater than 11, the ope⯑ration will be beſt performed as follows.

Suppoſe, for inſtance, I would find the reciprocal of the prime number 29, or the value of the frac⯑tion 1/29 in decimal numbers. I divide 1.000 by 29, in the common way, ſo far as to find two or three of the firſt figures, or till the remainder becomes a ſingle figure, and then I aſſume the ſupplement to complete the quotient. Thus I ſhall have 1/29=0.03448 8/29 for the complete quotient; which equa⯑tion if I multiply by the numerator 8, it will give 8/29=0.27584 64/29 or rather 8/29=0.27586 6/29. I ſub⯑ſtitute this inſtead of the fraction in the firſt equa⯑tion, and I ſhall have 1/29=0.0344827586 6/29. Again I multiply this equation by 6, and it will give 6/29=0.2068965517 7/29, and then by ſubſtitution 1/29=0.03448275862068965517 7/29. Again, I multiply this equation by 7, and it becomes 7/29=0.24137931034482758620 10/29, and then by ſubſtitu⯑tion 1/29=0.03448275862068965517241379310344 82758620 10/29, where every operation will at leaſt double the number of figures found by the preced⯑ing operation. And this will be an eaſy expedient for converting diviſion into multiplication in all caſes. For this reciprocal of the diviſor being thus found, it may be multiplied by the dividend to produce the quotient.

[13]II. To reduce a decimal to a vulgar fraction.

Under the figures of the given decimal write its proper denominator; which fraction, abbreviated as much as it can be, will be the vulgar fraction ſought.

EXAMPLES.

So .5=5/10=½
And .25=25/100=¼
And .75=75/100=¾
And .6=6/10=3/5
And .625=625/1000=⅝
And .5625=5625/10000=9/10

III. To find the value of a decimal, in the lower denominations.

Multiply the given decimal by the number of parts in the next lower denomination; from the pro⯑duct cut off as many decimals as are in the given number.

Multiply theſe by the parts in the next lower de⯑nomination again, cutting off the ſame number of decimals as before.

And proceed in the ſame manner to the loweſt denomination; then the ſeveral integer parts cut off on the left hand will give the value of the decimal propoſed.

[14]

EXAMPLES.
1. For the value of .3914 l.	2. For the value of
	.2139 lb. avoir.
.3914	.2139
20	16
s 7.8280	12834
12	2139
d 9.9360	oz 3.4224
4	16
q 3.7440	dr 6.7584
Anſ. 7s 9¾ d	Anſ. 3 oz 6 dr

OTHER EXAMPLES.
Queſtions.	Anſwers.
1. — .775 l	16s 6 d
2. — .625 s	0 7½ d
3. — .8635 l	17 s 3 d
4. — .0125 lb troy	3 dwts
5. — .4694 lb troy	5 oz 12 dwts 15 gr
6. — .625 cwt	2 qr 14 lb
7. — .009943 mile	17 yd 1 f 6 in almoſt
8. — .6875 yd cloth	2 qr 3 nl
9. — .3375 acr	1 rd 14 pl
10. — .2083 hhd wine	13 gal
11. — .40625 qt corn	3 bu 1 pk
12. — .42857 month	1 wk 5 da nearly

[15]IV. To bring quantities to decimals of higher denominations.

CASE I.

If a ſingle integer or decimal be propoſed, re⯑duce it to the higher denomination, by dividing as in reduction of whole numbers.

EXAMPLES.
1. Reduce 9d to the de⯑cimal of a pound	2. Reduce 1dwt to the decimal of a lb.
12 9 d	20 1 dwt
20 0.75 s	12 0.05 oz
Anſ. 0.0375 l	Anſ. 0.00416 lb

Queſtions.	Anſwers.
3. Reduce .26 d to 1 ſterl	.001083l
4. Reduce 7 drams to lb avoird	.02734375 lb
5. Reduce 2.15 lb to a cwt	.019196 cwt
6. Reduce 24 yds to a mile	.013636 mile
7. Reduce .056 pole to an acre	.00035 acre
8. Reduce 1.2 pint to hd wine	.00238 hd
9. Reduce 14 min. to a day	.009722 day
10. Reduce .21 pint to a peck	.013125 peck

CASE II.

A compound number may be reduced to a ſuperior name by reducing each of its parts, and taking the ſum of the decimals; the beſt way to do which is thus:

[16]Write the given numbers under each other, pro⯑ceeding orderly from the leaſt to the greateſt name, for dividends; draw a perpendicular line on the left of theſe, and on the left of it write oppoſite to each dividend ſuch a number, for a diviſor, as will reduce it to the next ſuperior name; then begin with the upper diviſion, and affix the quotient of each to the next dividend, as a decimal part of it, before it be divided, and the laſt ſum will be the anſwer.

EXAMPLES.
1. Reduce 3l 12s 6¾d to the denomination of l.	2. Reduce 5 oz 12 dwt 16 gr to the denom. of lb.
4 3	24 16
12 6.75	20 12.66
20 12.5625
Anſ. 3.628125	12 5.633
	Anſ. 0.4694

Queſtions.	Anſwers.
3. Reduce 19 l 17 s 3¼ d to l	19.8635416 l.
4. Reduce 15 s 6 d to l	.775 l
5. Reduce 7½ d to a ſhil	.625 s
6. Reduce 3 cwt 2 qr 14 lb to cwt	3.625 cwt
7. Reduce 17 yd 1ſt 6 in to a mile	.00994318 mil.
8. Reduce 2 qr 3nls to a yard	.6875 yd
9. Reduce 13ac 1.10 14pol to acres	13.3375 acr
10. Reduce 13 gal 1 pint to hd wine	.2083 hd
11. Reduce 3 buſh 1 pec to a qr	.40625 qr
12. Reduce 3. mo 1 we 5 da to mon	3.42857 mon.

CIRCULATING DECIMALS.

[17]

IT has already been obſerved, that when an in⯑finite decimal repeats always one figure, it is a ſingle repetend; and when more than one, a compound repetend, or a circulate: alſo that a point is ſet over a ſingle repetend, and a point over the firſt and laſt figures of a circulate.

It may farther be obſerved, that when other deci⯑mal figures precede a repetend, in any number, it is called a mixed number or quantity, as .23, or .104123: otherwiſe it is a pure repetend as .3 and .123.

Similar repetends begin at the ſame place, and conſiſt of the ſame number of figures: as .3 and 6, or 1.341 and 2.156.

Diſſimilar repetends begin at different places, and conſiſt of an unequal number of figures.

Similar and conterminous repetends, begin and end at the ſame place, as 2.9104 and .0613.

REDUCTION OF REPEATING DECIMALS.

CASE I. To reduce a ſingle Repetend to a vulgar Fraction.

Make the given decimal the numerator; and for a denominator take as many nines as there are recur⯑ring places in the given repetend.

[18]If one or more of the left-hand places, in the gi⯑ven decimal, be ciphers, annex as many ciphers to the right-hand of the nines in the denominator.

EXAMPLES.

1 So .3=3/9=⅓. 4 And 2.63=2 63/99=2 7/11.

2 And .05=5/90=1/18. 5 And .0594405=594405/9999990=17/28.

3 And .123=123/999=41/333. 6 And .769230=769230/999999=10/13.

CASE II. To reduce a mixed Repetend to a vulgar Fraction.

To as many nines as there are figures in the repe⯑tend, annex as many ciphers as there are finite places, for the denominator of the vulgar fraction.

Multiply the nines in the denominator by the finite part of the decimal, and to the product add the repeating part, for the numerator.

Or find the vulgar fraction as before anſwering to the repetend, then join it to the finite part, and reduce them to a common denominator.

EXAMPLES.

1. So [...]
2. And [...]
3. And [...]
4. And [...]

ADDITION OF REPETENDS.

[19]

MAKE every line to begin and end at the ſame place, by extending the repetends, and filling up the vacancies with the proper figures and ciphers. Then add as in common numbers; only increaſe the ſum of the right-hand row, or laſt row of the repetends, by as many units as the firſt row of repetends con⯑tains nines. And the ſum will circulate at the ſame places as the other lines.

EXAMPLES.
(1)	(2)
39.6548=39.65480	91.367=91.3570
81.046=81.04666	72.38=72.3888
42.35=42.35555	7.21=7.2111
9.837=9.83777	4.2965=4.2965
Sum 172.89480	Sum 175.2535
(3)	(4)
9.814=9.81481481	2.41=2.41
1.5=1.50000000	13.215=13.21515151
87.26=87.26666666	5.8=5 8
0.83=0.83333333	27.096=27.09696969
124.09=124.09090909	0.913=0.91391391
Sum 223.50572390	Sum 49.43603512

SUBTRACTION OF REPETENDS.

[20]

MAKE the repetends to begin and end together, as in addition. Then ſubtract as uſual; only, if the repetend of the number to be ſubtracted exceed the repetend of the other number, make the laſt figure of the remainder 1 leſs than it otherwiſe would be.

EXAMPLES.
(1)	(2)
76.32=76.3222	89.576=89.5760
54.7617=54.7617	12.5846=12.5846
Diff. 21.5604	Diff. 76.9913
(3)	(4)
29.21=29.212121	87.4161=87.41614
3.561=3.561561	0.532=0.53232
Diff. 25.650559	Diff. 86.88381

MULTIPLICATION OF REPETENDS.

1. WHEN a repetend is to be multiplied by a finite number: Multiply as in common numbers; only obſerve what muſt be carried from the beginning of the repetend to the end of it. And make all the [21]lines begin and end together when they are to be added.

2. In multiplying a finite decimal by a ſingle repetend; multiply by the repetend, and divide by .9 or 9/10.

3. In more complex caſes, reduce the repetends to vulgar fractions; then divide theſe, and reduce the quotient to a decimal, if neceſſary.

EXAMPLES.
(1)	(2)	(3)
716.2935	2.104	3.028
.27	1.2	17
50140548	4208	21202
143258711	21044	30288
193.399260	2.5253	51.491

(4)
27.1241	Or 3.6=3 6/9=3⅔=11/3.
3.6	Then 27.1241
	11
9) 1627446
	3) 298.3651
1808273	99.45503
813723
99.45503

[22](5)

Mult. 1.206 by 3.5.

1.206=1.26/99=1.2 2/33=1 63/330=398/330, and 3.5=35/9=32/9; then 398/330 x 32/9=12736/2970=4.2882154.

DIVISION OF REPETENDS.

1. IF the dividend only be a repetend, divide as in common numbers, bringing down always the recurring figures, till the quotient become as exact as requiſite.

2. And if the diviſor only be a repetend, it will be beſt to change it into its equivalent vulgar frac⯑tion, then multiply by its numerator, and divide by its denominator.

3. But if both diviſor and dividend be repe⯑tends, change them both to vulgar fractions.

EXAMPLES.
(1)	(2)
1.2) 2.5253	8) 27.912
2.104	3.489027
(3)	(4)
17) 51.491 (3.028	27) 193.39926
49	9) 64.46642
151
151	7.162935

[23] [...]

[...]

7. Divide 4.2882154 by 1.206.

Here 1.206=1.26/99=1.22/33=398/330,

And 4.2882154=4.2882154/999999=42882112/9999990.

Then [...]

Or rather thus: Having found [...], then [...]

INVOLUTION; OR RAISING OF POWERS.

[24]

A POWER is a number produced by multiplying any given number continually by itſelf a certain number of times.

Any number is called the firſt power of itſelf; if it be multiplied by itſelf, the product is called the ſecond power, and ſometimes the ſquare; if this be multiplied by the firſt power again, the product is called the third power, and ſometimes the cube; and if this be multiplied by the firſt power again, the product is called the fourth power, &c. that is, the power is denominated from the number which exceeds the number of multiplications by 1.

Thus: 3 is the firſt power of 3.

3 x 3=9 is the ſecond power of 3.

3 x 3 x 3=27 is the third power of 3.

3 x 3 x 3 x 3=81 is the fourth power of 3.

&c. &c.

And in this manner may be calculated the follow⯑ing table.

[]

TABLE of the Firſt Twelve Powers of Numbers.
1ſt power	1	2	3	4	5	6	7	8	9
2d power	1	4	9	16	25	36	49	64	81
3d power	1	8	27	64	125	216	343	512	729
4th power	1	16	81	256	625	1296	2401	4096	6561
5th power	1	32	243	1024	3125	7776	16807	32768	59049
6th power	1	64	729	4096	15625	46656	117649	262144	531441
7th power	1	128	218-	16384	78125	279936	823543	2097152	4782969
8th power	1	256	6561	65536	390625	1679616	5764801	16777216	43046721
9th power	1	512	19683	262144	1953125	10077696	40353607	134217728	387420489
10th power	1	1024	59049	1048576	9765625	60466176	282475249	1073741824	3486784401
11th power	1	2048	177147	4194304	48828125	362797056	1977326743	8589934592	31381059609
[...]th power	1	4096	531441	16777216	244140625	2176782336	13841287201	68719476736	282420536481

[26]The number which exceeds the multiplications by 1, is called the index or exponent of the power: ſo the index of the firſt power is 1, that of the ſe⯑cond power is 2, that of the third is 3, and ſo on.

Powers are commonly denoted by writing their indices above the firſt power: ſo the ſecond power of 3 may be denoted thus, 3²; the third power thus, 3³; the fourth power thus 3⁴; &c. and the 6th pow⯑er of 503, thus, 503⁶.

Involution is the finding of powers; to do which, from their definition there evidently comes this.

RULE.

Multiply the given number, or firſt power, con⯑tinually by itſelf, till the number of multiplications be 1 leſs than the index of the power to be found, and the laſt product will be the power required.

Note 1. Becauſe fractions are multiplied by taking the products of their numerators and of their denominators, they will be involved by raiſing each of their terms to the power required. And if a mixt number be propoſed, either reduce it to an im⯑proper fraction, or reduce the vulgar fraction to a decimal, and proceed by the rule.

2. The raiſing of powers will be ſometimes ſhor⯑tened by working according to this obſervation, viz. whatever two or more powers are multiplied toge⯑ther, their product is the power whoſe index is the ſum of the indices of the factors; or if a power be multiplied by itſelf, the product will be the power whoſe index is double of that which is multiplied; ſo if I would find the ſixth power, I might multiply the given number twice by itſelf for the third power, [27]then the third power into itſelf would give the ſixth power; or if I would find the ſeventh power; I might firſt find the third and fourth, and their pro⯑duct would be the ſeventh; or laſtly, if I would find the eighth power, I might firſt find the ſecond, then the ſecond into itſelf would be the fourth, and this into itſelf would be the eighth.

[...]

[28] [...]

Ex. 5. The ſquare of ⅔ is ⅔ x ⅔=4/9.

Ex. 6. The cube of 5/9 is 5/9 x 5/9 x 5/9=125/729.

Ex. 7. The ſquare of 3⅖ or 17/5 is 17/5 x 17/5=289/25=11 14/25=11.56.

EVOLUTION; OR EXTRACTION OF ROOTS.

[29]

The root of any given number, or power, is ſuch a number, as being multiplied by itſelf a certain number of times, will produce the power; and it is denominated the firſt, ſecond, third, fourth, &c. root, reſpectively, as the number of multiplications made of it to produce the given power is 0, 1, 2, 3, &c. that is, the name of the root is taken from the number which exceeds the multiplications by 1, like the name of the power in involution.

The index of the root, like that of the power in involution, is 1 more than the number of the multi⯑plications neceſſary to produce the power or given number. So 2 is the index of the ſecond or ſquare root; and 3 the index of the 3d or cubic root; and 4 the index of the 4th root; and ſo on.

Roots are ſometimes denoted by writing √ before the power, with the index of the root againſt it: ſo the third root of 50 is 3√50, and the ſecond root of it is √50, the index 2 being omitted, which index is always underſtood when a root is named or written without one, but if the power be expreſſed by ſeveral numbers with the ſign + or −, &c. between them, then a line is drawn from the top of the ſign of the root or radical ſign, over all the parts of it; ſo the third root of 47−15 is [...]. and ſometimes roots are deſigned like powers, with the reciprocal or [30]the index of the root above the given number. So the root of 3 is 3^½, the root of 50 is 50^½, and the third root of it is 50⅓, alſo the third root of 47—15 is [...]. And this method of notation has juſtly prevailed in the modern algebra; becauſe ſuch roots, being conſidered as fractional powers, need no other directions for any operations to be made with them, than thoſe for integral powers.

A number is called a complete power of any kind, when its root of the ſame kind can be accu⯑rately extracted; but if not, the number is called an imperfect power, and its root a ſurd or irrational quantity. So 4 is a complete power of the ſecond kind, its root being 2; but an imperfect power of the third kind, its third root being a ſurd quantity, which cannot be accurately extracted.

Evolution is the finding of the roots of numbers, either accurately, or in decimals to any propoſed de⯑gree of accuracy.

The power is firſt to be prepared for extraction, or evolution, by dividing it, by means of points or commas, from the place of units, to the left hand in integers, and to the right in decimal fractions, into periods, containing each as many places of figures as are denoted by the index of the root, if the power contain a complete number of ſuch periods; that is each period to have two figures for the ſquare root, three for the cube root, four for the fourth root, and ſo on. And when the laſt period in decimals is not complete, ciphers are added to compleat it.

Note, The root will contain juſt as many places of figures as there are periods or points in [31]the given power; and they will be integers, or de⯑cimals reſpectively, as the periods are ſo from which they are found, or to which they correſpond; that is, there will be as many integer or decimal figures in the root, as there are periods of integers or deci⯑mals in the given number.

TO EXTRACT THE SQUARE ROOT.

1. Having divided the given number into periods of two figures each, find, from the table of powers in page 25, or otherwiſe, a ſquare number either equal to, or the next leſs than the firſt period, which ſubtract from it, and place the root of the ſquare on the right of the given number, after the manner of a quotient in diviſion, for the firſt figure of the root re⯑quired.

2. To the remainder annex the ſecond period for a dividend; and on the left thereof write the double of the root already found, after the manner of a di⯑viſor.

3. Find how often the diviſor is contained in the dividend, wanting its laſt figure on the right hand; place that number for the next figure in the quotient, and on the right of the diviſor, as alſo below the ſame.

4. Multiply the whole increaſed diviſor by it, placing the product below the dividend, which ſub⯑tract from it, and to the remainder bring down the next period, for a new dividend; to which, as before, find a diviſor by doubling the figures already found in the root; and from theſe find the next figure of the root, as in the laſt article; and continue the [32]operation ſtill in the ſame manner till all the periods be uſed, or as far as you pleaſe.

Note. Inſtead of doubling the root to find the new diviſors, you may add the laſt diviſor to the figure below it.

To prove the work, multiply the root by itſelf, and to the product add the remainder, and the ſum will be the given number.

Ex. 1. To extract the root of 17.3056.

Having divided the given number into three periods, namely 17, and 30, and 56, I find that 16 is the next ſquare to 17, the firſt period, which ſet below, and ſub⯑tracting, 1 remains, to which bring down 30, the next pe⯑riod, makes 130 for a divi⯑dend. Then 4, the root of [...] 16, is ſet on the right of the given number for the firſt figure of the root, and its double, or 8, on the left of the dividend for the firſt figure of the diviſor; which being once contained in 13, the dividend wanting its laſt figure, gives 1 for the next figure of the root, which 1 is accordingly ſet in the root, making 4.1, and in the diviſor, making 81, as alſo below the ſame. Theſe multiplied make alſo 81, ſet below the dividend, and ſubtracting, we have 49 remaining, to which the laſt period 56 being brought down, we have 4956 for the new dividend. Then, for a new diviſor, either double the root 4.1, or elſe, which is eaſier, to the laſt diviſor add the figure 1 ſtanding below it, and either way gives [33]82 for the firſt part of the new diviſor. This 82 is 6 times contained in 495, and therefore 6 is the next figure, to ſet in the root and in the diviſor, as alſo below the ſame; which being then multiplied by it, gives 4956, the ſame as the dividend; there⯑fore nothing remains, and 4.16, is the root of 17.3056, as required.

Ex. 2. For the Root of 2025. [...]

Ex. 3. For the Root of .000729. [...]

Note, When all the periods of the given num⯑ber are brought down and uſed, and more figures are required to be found, the operation may be con⯑tinued by adding as many periods of ciphers as we pleaſe, namely, bringing always two ciphers at once to each dividend. And when the root is to be ex⯑tracted to a great number of places, the work may be much abbreviated thus: having proceeded in the extraction after the common method till you have found one more than half the required number of figures in the root, the reſt may be found by divid⯑ing the laſt remainder by its correſponding diviſor, annexing a cipher to every dividual, as in diviſion of decimals; or rather, without annexing ciphers, by omitting continually the right-hand figure of the diviſor, after the manner of the third contraction in diviſion of decimals in page 10.

[34]So the operation for the root of 2, to 12 or 13 places, may be thus.

EXAMPLE 4.

[...]

Here, having found the firſt ſeven figures 1.414213 by the common extraction, by adding always periods of ciphers, the laſt ſix figures 562373 are found by the method of contracted diviſion in decimals, without adding ciphers to the remainder, only [35]pointing off a figure at each time from the laſt diviſor. And the ſame for the two following examples.

Ex. 5. For the root of 3.

[...]

Ex. 6. For the root of 5.

[...]

In like manner may be found the following Roots.

The root of 6 is 2.449490
The root of 7 is 2.645751
The root of 10 is 3.162278
The root of 11 is 3.316625

RULES for the Square Roots of Vulgar Fractions and Mixt Numbers.

Firſt prepare all vulgar fractions, by reducing them to their leaſt terms, both for this and all other roots. Then

1. Take the root of the numerator and of the denominator for the reſpective terms of the root [36]required. And this is the beſt way if the denomi⯑nator be a complete power. But if it be not,

2. Multiply the numerator and denominator to⯑gether; take the root of the product; this root be⯑ing made the numerator to the denominator of the given fraction, or made the denominator to the nu⯑merator of it, will form the fractional root required.

That is [...]And this rule will ſerve whether the root be finite or infinite. Or,

3. Reduce the vulgar fraction to a decimal, and extract its root.

4. Mixt numbers may be either reduced to im⯑proper fractions, and extracted by the firſt or ſecond rule; or the vulgar fraction may be reduced to a decimal, then joined to the integer, and the root of the whole extracted.

Ex. 1. [...]is ⅚

Ex. 2. [...]or [...]is 3/7;

Ex. 3. For the root of 9/12 [...]

[37]Ex. 4. For the root of 5/12 [...]

TO FIND A MEAN PROPORTIONAL.

There are various uſes of the ſquare root, one of which is to find a mean proportional between any two numbers, which is performed thus: Multiply the two given numbers together, then extract the ſquare root out of their product, and it will be the mean proportional ſought.

Ex. 1. To find a Mean Proportional between 3 and 12.

Here 3 x 12=36.

And √36 is 6, the mean proportional ſought.

For 3∶6∷6∶12.

[38]Ex. 2. To find a Mean between 2 and 5.

Here 2 x 5=10 (3.162278 the mean req^d.

[...]

Note, By means of the ſquare root alſo we rea⯑dily find the 4th root, or the 8th root, or the 16th root, &c. that is, the root of any power whoſe index is ſome power of the number 2: namely, by extract⯑ing ſo often the ſquare root as is denoted by the index of that power of 2; that is, two extractions for the 4th root, three for the 8th root, and ſo on.

[39]Thus for the 4th root of 97.41.

[...]

So that the 4th root of 97.41 is 3.14159999, which expreſſes the circumference of a circle whoſe diameter is 1 nearly.

OTHER EXAMPLES.

The 4th root of 21035.8 is 12.0431407.
The 4th root of - 2 is 1.259921.

TO EXTRACT THE CUBE ROOT.

[40]

RULE I.

1. Point the given number into periods of three places each, beginning at units: and there will be as many integral places in the root, as there are points over the integers in the given number.

2. Seek the greateſt cube in the left-hand pe⯑riod, write the root in the quotient, and the cube under the period; from which ſubtract it, and to the remainder bring down the next period: Call this the reſolvend, under which draw a line.

3. Under the reſolvend, write the triple ſquare of the root, ſo that units in the latter ſtand under the place of hundreds in the former; under the tri⯑ple ſquare of the root, write the triple root, remov⯑ed one place to the right; and the ſum of theſe two lines call a diviſor; under which draw a line.

4. Seek how often this diviſor may be had in the reſolvend, its right-hand place excepted, and write the reſult in the quotient.

5. Under the diviſor, write the product of the triple ſquare of the root by the laſt quotient figure, ſetting units place of this line, under that of tens in the diviſor; under this line, write the product of the triple root by the ſquare of the laſt quotient figure, let this line be removed one place beyond the right of the former; and under this line, removed one place forward to the right, write the cube of the laſt quotient figure; the ſum of theſe three lines call the ſubtrahend, under which draw a line.

[41]6. Subtract the ſubtrahend from the reſolvend; to the remainder bring down the next period for a new reſolvend; the diviſor to this, muſt be the triple ſquare of all the quotient added to the triple thereof, &c. as in the third article, &c.

EXAMPLE I.

What is the cube root of 48228544?

[...]

[42]If the work of this example be well conſidered, and compared with the foregoing rule, it will be eaſy to conceive how any other example of the like nature may be wrought; and here obſerve that when the cube root is extracted to more than two places, there is a neceſſity of doing ſome work up⯑on a ſpare piece of paper, in order to come at the root's triple ſquare, and the product of the triple root by the ſquare of the quotient figure, &c.

In this example, the given number is a cube number, and therefore at the end of the operation there remained nothing; for 364 multiplied by 364, the product multiplied by 364 gives 48228544, the given number.

But if the number given be not a cube number; then, to the laſt remainder always bring down three ciphers, and work anew for a decimal fraction if needful.

MORE EXAMPLES.

What is the cube root of

389017	Anſwers.	73
1092727		103
27054036008		3002
219365327791		6031
122615327232		4968

Theſe examples are all performed in the ſame manner as the foregoing one.

TO FIND TWO MEAN PROPORTIONALS.

[43]

There are many uſes of the cube root; one is to find two mean proportionals between two given numbers, which is performed thus:

Divide the greater extreme by the leſs, and the cube root of the quotient multiplied by the leſs extreme, gives the leſs mean. Multiply the ſaid cube root by the leſs mean, and the product is the greater mean proportional.

Note, This is only underſtood of thoſe numbers that are in continued geometric proportion.

EXAMPLE. I.

What are the two mean proportionals between 4 and 108?

108 divided by 4 gives 27, whoſe cube root is 3; and the leſs extreme 4, multiplied thereby, gives 12 for the leſs mean; and 12 multiplied by the ſaid root 3, gives 36 for the greater mean.

For 4 is to 12, as 36 to 108.

EXAMPLE II.

Find the two geometric means between 8 and 1728?

Now 8) 1728 (216, whoſe cube root is 6. And 6 times 8 is 48, the leſs mean, and 6 times 48 is 288 the greater mean.

For 8 is to 48 as 288 to 1728.

[44]If the rule already given for the cube root be thought too tedious, the following one will be found much more ready for uſe.

RULE II. TO EXTRACT THE CUBE ROOT.

1. By trials take the neareſt rational cube to the given cube or number, and call it the aſſumed cube.

2. Then the ſum of the given number and double the aſſumed cube will be to the ſum of the aſſumed cube and double the given number, as the root of the aſſumed cube, is to the root required, nearly. Or as the firſt ſum is to the difference of the given and aſſumed cube, ſo is the aſſumed root, to the difference of the roots nearly.

3. Again, by uſing, in like manner, the cube of the root laſt found as a new aſſumed cube, ano⯑ther root will be obtained ſtill nearer. And ſo on as far as we pleaſe; uſing always the cube of the laſt-found root, for the aſſumed cube.

[45]EXAMPLE. To find the cube root of 21034.8.

Here we ſoon find that the root lies between 20 and 30, and then between 27 and 28. Taking therefore 27, its cube is 19683 the aſſumed cube. Then [...]

[46]Again, for a ſecond operation, the cube of this root is 21035.318645155823, and the proceſs by the latter method will be thus: [...]

TO EXTRACT ANY ROOT WHATEVER.

Let G be the given power or number, n the index of the power, A the aſſumed power, r its root, R the required root of G.

Then as the ſum of n+1 times A and n — 1 times G, is to the ſum of n+1 times G and n — 1 times A, ſo is the aſſumed root r, to the required root R.

Or, as half the ſaid ſum of n+1 times A and n — 1 times G, is to the difference between the given and aſſumed powers, ſo is the aſſumed root r, to the difference between the true and aſſumed roots: which difference added or ſubtracted, gives the true root nearly.

That is, [...].

Or, [...].

[47]And the operation may be repeated as often as we pleaſe, by uſing always the laſt found root for the aſſumed root, and its nth power for the aſſumed power A.

EXAMPLE. To extract the 5th Root of 21035.8.

Here it appears that the 5th root is between 7.3 and 7.4. Taking 7.3, its 5th power is 20730.71593. Hence then we have [...].

[48]OTHER EXAMPLES.

1. What is the 3d root of 2? Anſ. 1.259921.
2. What is the 4th root of 2? Anſ. 1.189207.
3. What is the 4th root of 97.41? Anſ. 3.141599.
4. What is the 5th root of 2? Anſ. 1.148699.
5. What is the 6th root of 21035.8? Anſ. 5.254037.
6. What is the 6th root of 2? Anſ. 1.122462.
7. What is the 7th root of 21035.8? Anſ. 4.145392.
8. What is the 7th root of 2? Anſ. 1.104089.
9. What is the 8th root of 21035.8? Anſ. 3.470323.
10. What is the 8th root of 2? Anſ. 1.090508.
11. What is the 9th root of 21035.8? Anſ. 3.022239.
12. What is the 9th root of 2? Anſ. 1.080059.

GENERAL RULES for extracting any Root out of a Vulgar Fraction or Mixt Number.

1. If the given fraction have a finite root of the kind required, it is beſt to extract the root out of the numerator and denominator, for the terms of the root required.

2. But if the fraction be not a complete power, it may be thrown into a decimal, and then extract⯑ed. Or,

3. Take either of the terms of the given frac⯑tion for the correſponding term of the root; and for the other term of the root, extract the required root of the product, ariſing from the multiplication of ſuch a power of the firſt aſſigned term of the root whoſe index is leſs by 1 than that of the given power, by the other term of the given number.

[49]This rule will do when the root is either finite or infinite.

This is [...].

4. Mixt numbers may be reduced either to improper fractions or decimals, and then extracted.

EXAMPLES.

1. What is the cube root of 8/27? Anſ. ⅔.
2. What is the fourth root of 80/405? Anſ. ⅔.
3. What is the cube root of ½? Anſ. .7937005.
4. What is the cube root of 2 10/27? Anſ. 4/3 or 1⅓.
5. What is the third root of 7⅕? Anſ. 1.930979.

DUODECIMALS: OR CROSS MULTIPLICATION.

DUODECIMALS are ſo called becauſe they de⯑creaſe by twelves, from the place of ſeet towards the right-hand. Inches are ſometimes called primes, and are marked thus′; the next diviſion after inches are called parts, or ſeconds, and are marked thus″; the next are thirds, and marked thus‴; and ſo on.

[50]This rule is otherwiſe called Croſs Multiplica⯑tion, becauſe the factors are ſometimes multiplied croſs ways. And it is commonly uſed by workmen and artificers in caſting up the contents of their work; though a much better way would be by a Decimal Scale.

RULE I.

1. Under the multiplicand write the ſame names or denominations of the multiplier; that is, feet under feet, inches under inches, parts under parts, &c.

2. Multiply each term in the multiplicand, beginning at the loweſt, by the feet in the multi⯑plier, and write each reſult under its reſpective term, obſerving to carry an unit for every 12, from each lower denomination to its next ſuperior.

3. In the ſame manner multiply every term in the multiplicand by the inches in the multiplier, and ſet the reſult of each term one place removed to the right of thoſe in the multiplicand.

4. Proceed in like manner with the ſeconds, and all the reſt of the denominations, if there be any more; and the ſum of all the lines will be the pro⯑duct required.

Or the denominations of the particular products will be as follows:

Feet by feet, give feet.
Feet by primes, give primes.
Feet by ſeconds, give ſeconds.
&c.

Primes by primes, give ſeconds.
Primes by ſeconds, give thirds.
Primes by thirds, give fourths.
&c.

Seconds by ſeconds, give fourths.
Seconds by thirds, give fifths.
Seconds by fourths, give ſixths.
&c.

Thirds by thirds, give ſixths.
Thirds by fourths, give ſevenths.
Thirds by fifths, give eighths.
&c.

In general thus:

When feet are concerned, the product is of the ſame denomination with the term multiplying the feet.

When feet are not concerned, the name of the product will be expreſſed by the ſum of the indices of the two factors.

Ex. 1. Multiply [...]

RULE II.

[52]

When the feet in the multiplicand are expreſſed by a large number.

Multiply firſt by the feet of the multiplier, as before. Then, inſtead of multiplying by the inches and parts, &c. proceed as in the Rule of Practice, by taking ſuch aliquot parts of the multiplicand as correſpond with the inches and ſeconds, &c. of the multiplier. Then the ſum of them all will be the product required.

Ex. 2. Multiply [...]

RULE III.

If the feet in both the multiplicand and multiplier be large numbers.

Multiply the feet only into each other: then, for the inches and ſeconds in the multiplier, take parts of the multiplicand; and for the inches and ſeconds of the multiplicand, take aliquot parts of the feet only in the multiplier. And the ſum of all will be the product.

[53]Ex. 3 Multiply [...]

OTHER EXAMPLES.

[...]

MENSURATION.

[]

MENSURATION is the meaſuring and eſti⯑mating the magnitude and dimenſions of bodies and figures: and it is either angular, lineal, ſuperfi⯑cial, or ſolid, according to the objects it is concerned with. It is accordingly treated in ſeveral parts: as 1ſt, Practical Geometry, which treats of the definitions and conſtruction of geometrical figures; 2d, Trigonometry, which teaches the calculation and conſtruction of triangles, or three-ſided figures, and, by application, of other figures depending on them; 3d, Superficial Menſuration, or the meaſuring the ſurfaces of bodies; 4th, Solid Menſuration, or meaſuring the capacities or ſolid contents of bodies. Beſides theſe general heads, there are ſeveral other ſubordinate diviſions, as alſo the application of each to the practical concerns of life. Of each of which in their order: excepting Trigonometry, which is fully treated of in my large book on Menſuration.

PRACTICAL GEOMETRY.

[]

DEFINITIONS.

1. A POINT has no parts nor dimenſions, neither length, breadth, nor thickneſs.

2. A line is length, without breadth or thickneſs.

[diagram]

3. A ſurface, or ſuperſicies, is an extenſion, or a figure, of two dimenſions, length and breadth; but without thickneſs.

[diagram]

4. A body or ſolid is a figure of three dimenſions, namely, length, breadth, and thickneſs.

Hence ſurfaces are the extremi⯑ties of ſolids; lines the extremities of ſurfaces; and points the ex⯑tremities of lines.

[diagram]

5. Lines are either right, or curved, or mixed of theſe two.

[diagram]

[57]6. A right line, or ſtraight line, lies all in the ſame direction, be⯑tween its extremities; and is the ſhorteſt diſtance between two points.

[diagram]

7. A curve continually changes its direction, between its extreme points.

[diagram]

8. Lines are either parallel, ob⯑lique, perpendicular, or tangential.

[diagram]

9. Parallel lines are always at the ſame diſtance; and never meet, though ever ſo far produced.

[diagram]

10. Oblique right lines change their diſtance, and would meet, if produced, on the ſide of the leaſt diſtance.

[diagram]

11. One line is perpendicular to another, when it inclines not more on the one ſide than on the other.

[diagram]

12. One line is tangential, or a tangent to another, when it touches it, without cutting, when both are produced.

[diagram]

13. An angle is the inclination, or opening, of two lines, having different directions, and meeting in a point.

[diagram]

14. Angles are right or oblique, acute or obtuſe.

15. A right angle, is that which is made by one line perpendicular to another. Or when the angles on each ſide are equal to one an⯑other, they are right angles.

[diagram]

16. An oblique angle is that which is made by two oblique lines; and is either leſs or greater than a right angle.

[58]17. An acute angle is leſs than a right angle.

[diagram]

18. An obtuſe angle is greater than a right angle.

[diagram]

19. Superficies are either plane or curved.

20. A plane, or plane ſuperficies, is that with which a right line may, every way, coincide.—But if not, it is curved.

21. Plane figures are bounded either by right lines or curves.

22. Plane figures bounded by right lines, have names according to the number of their ſides, or of their angles; for they have as many ſides as angles; the leaſt number being three.

23. A figure of three ſides and angles, is called a triangle. And it receives particular denominations from the relations of its ſides and angles.

24. An equilateral triangle, is that whoſe three ſides are all equal.

[diagram]

25. An iſoſeeles triangle, is that which has two ſides equal.

[diagram]

26. A ſcalene triangle, is that whoſe three ſides are all unequal.

[diagram]

[59]27. A right-angled triangle, is that which has one right angle.

[diagram]

28. Other triangles are oblique-angled, and are either obtuſe or acute.

29. An obtuſe-angled triangle has one obtuſe angle.

[diagram]

30. An acute-angled triangle has all its three angles acute.

[diagram]

31. A figure of four ſides and angles, is called a quadrangle, or a quadrilateral.

32. A parallelogram is a quadri⯑lateral which has both its pairs of oppoſite ſides parallel. And it takes the following particular names.

33. A rectangle is a parallelo⯑gram, having all its angles right ones.

[diagram]

34. A ſquare is an equilateral rectangle; having all its ſides equal, and all its angles right ones.

[diagram]

35. A rhomboid is an oblique-angled parallelogram.

[diagram]

36. A rhombus is an equilateral rhomboid; having all its ſides equal, but its angles oblique.

[diagram]

[60]37. A trapezium is a quadrilate⯑ral which hath not both its pairs of oppoſite ſides parallel.

[diagram]

38. A trapeziod hath only one pair of oppoſite ſides parallel.

[diagram]

39. A diagonal is a right line joining any two oppoſite angles of a quadrilateral.

[diagram]

40. Plane figures having more than four ſides are, in general, called polygons: and they receive other particular names according to the number of their ſides or angles.

41. A pentagon is a polygon of five ſides; a hexagon hath ſix ſides; a heptagon, ſeven; an octagon, eight; a nonagon, nine; a decagon, ten; an undecagon, eleven; and a dodecagon hath twelve ſides.

42. A regular polygon hath all its ſides and all its angles equal.—If they are not both equal, the polygon is irregular.

43. An equilateral triangle is alſo a regular figure of three ſides, and the ſquare is one of four: the former being alſo called a trigon, and the latter a tetragon.

44. A circle is a plane figure bounded by a curve line, called the circumference, which is every where equi-diſtant from a certain point within, called its center.

[diagram]

N. B. The circumference itſelf is often called a circle.

[61]45. The radius of a circle, is a right line drawn from the center to the circumference.

[diagram]

46. The diameter of a circle, is a right line drawn through the cen⯑ter, and terminating in the circum⯑ference on both ſides.

[diagram]

47. An arc of a circle, is any part of the circumference.

[diagram]

48. A chord is a right line joining the extremities of an arc.

[diagram]

49. A ſegment is any part of a circle bounded by an arc and its chord.

[diagram]

50. A ſemicircle is half the circle, or a ſegment cut off by a diameter.

[diagram]

[62]51. A ſector is any part of a circle, bounded by an arc, and two radii drawn to its extremities.

[diagram]

52. A quadrant, or quarter of a circle, is a ſector having a quarter of the circumference for its arc, and its two radii are perpendicular to each other.

[diagram]

53. The height or altitude of a figure, is a perpendicular let fall from an angle, or its vertex, to the oppoſite ſide, called the baſe.

[diagram]

54. In a right-angled triangle, the ſide oppoſite the right angle, is called the hypotenuſe; and the other two ſides the legs, or ſome⯑times the baſe and perpendicular.

55. When an angle is denoted by three letters, of which one ſtands at the angular point, and the other two on the two ſides, that which ſtands at the angular point is read in the middle.

[diagram]

56. The circumference of every circle is ſuppoſed to be divided into 360 equal parts, called degrees; and each degree into 60 minutes, each minute into 60 ſeconds, and ſo on. Hence a ſemicircle contains 180 de⯑grees, and a quadrant 90 degrees.

[63]57. The meaſure of a right-lined angle, is an arc of any circle con⯑tained between the two lines which form that angle, the angular point being the center; and it is eſtimated by the number of degrees contained in that arc. Hence a right-angle is an angle of 90 degrees.

[diagram]

The definition of ſolids, or bodies, will be given afterwards, when we come to treat of the menſuration of ſolids.

PROBLEMS.

[64]

PROBLEM I. To divide a given Line AB into two Equal Parts,

From the centers A and B, with any radius greater than half AB, deſcribe arcs cutting each other in m and n. Draw the line mCn, and it will cut the given line into two equal parts in the middle point C.

[diagram]

PROBLEM II. To divide a given Angle ABC into two Equal Parts.

From the center B, with any ra⯑dius, deſcribe the arc AC. From A and C, with one and the ſame ra⯑dius, deſcribe arcs interſecting in m. Draw the line Bm, and it will bi⯑ſect the angle as required.

[diagram]

PROBLEM III. To divide a Right Angle ABC into three Equal Parts.

From the center B, with any ra⯑dius, deſcribe the arc AC. From the center A, with the ſame radius, croſs the arc AC in n. And with the cen⯑ter C, and the ſame radius, cut the arc AC in m. Then through the points m and n draw nm and Bn, and they will triſect the angle as re⯑quired.

[diagram]

PROBLEM IV. To draw a Line Parallel to a given Line AB.

[65]

CASE I. When the Parallel Line is to be at a given Diſtance C.

From any two points m and n, in the line AB, with a radius equal to C, deſcribe the arcs r and o: —Draw CD to touch theſe arcs, without cutting them, and it will be the parallel required.

[diagram]

CASE 2. When the Parallel Line is to paſs through a given Point C.

From any point m, in the line AB, with the radius mC, deſcribe the arc Cn.—From the centre C with the ſame radius, deſcribe the arc mr.—Take the arc Cn in the compaſſes, and apply it from m to r.—Through C and r draw DE, the parallel required.

[diagram]

N.B. This problem is more eaſily effected with a parallel ruler.

PROBLEM V. To erect a Perpendicular from a given Point A in a given Line BC.

CASE 1. When the Point is near the middle of the Line.

On each ſide of the point A take any two equal diſtances Am, An,— From the centers m, n, with any radius greater than Am or An, de⯑ſcribe two arcs cutting in r.— Through A and r draw the line Ar and it will be the perpendicular as required.

[diagram]

[66]CASE 2. When the Point is too near the end of the Line.

With the center C, and any radius, deſcribe the arc mns.—From the point m, with the ſame radius, turn the compaſſes twice over on the arc at n and s.—Again, with the centers n and s, deſcribe arcs inter⯑ſecting in r.—Then draw Ar, and it will be perpendicular as required.

[diagram]

Another Method.

From any point m as a center, with the radius or diſtance mA, de⯑ſcribe an arc cutting the given line in n and A.—Through n and m draw a right line cutting the arc in r.—Laſtly, draw Ar, and it will be the perpendicular as required.

[diagram]

Another Method.

From any plane ſcale of equal parts ſet off Am equal to 4 parts.— With center A, and radius of 3 parts, deſcribe an arc.—And with center m, and radius of 5 parts, croſs it at n.—Draw An for the per⯑pendicular required.

[diagram]

Or any other numbers in the ſame proportion as 3, 4, 5, will do the ſame.

PROBLEM VI. From a given Point A, out of a given Line BC, to let fall a Perpendicular.

[67]

CASE 1. When the Point is nearly oppoſite the Middle of the Line.

With the center A, and any ra⯑dius, deſcribe an arc cutting BC in m and n.—With the centers m and n, and the ſame, or any other ra⯑dius, deſcribe arcs interſecting in r.—Draw ADr, for the perpendicu⯑lar required.

[diagram]

CASE 2. When the Point is nearly oppoſite the End of the Line.

From A draw any line Am to meet BC, in any point m.—Biſect Am at n, and with the center n, and radius An or mn, deſcribe an arc, cutting BC in D.—Draw AD the perpendicular as required.

[diagram]

Another Method.

From B or any point in BC, as a center, deſcribe an arc through the point A.—From any other center m in BC, deſcribe another arc through A, and cutting the former arc again in n.—Through A and n draw the line ADn; and AD will be the perpendicular as re⯑quired.

[diagram]

[68] N. B. Perpendiculars may be more readily raiſed and let fall, in practice, by means of a ſquare, or other fit inſtrument.

PROBLEM VII. To divide a given Line AB into any propoſed Number of Equal Parts.

From A draw any line AC at random, and from B draw BD parallel to it.—On each of theſe lines, beginning at A and B, ſet off as many equal parts, of any length, as AB is to be divided into.—Join the oppoſite points of diviſion by the lines A5, 14, 23, &c. and they will divide AB as required.

[diagram]

PROBLEM VIII. To divide a given Line AB in the ſame Proportion as another Line CD is divided.

From A draw any line AE equal to CD, and upon it trans⯑fer the diviſions of the line CD. —Join BE, and parallel to it draw the lines 11, 22, 33, &c. and they will divide AB as re⯑quired.

[diagram]

PROBLEM IX. At a given Point A, in a given Line AB, to make an Angle equal to a given Angle C.

[69]

With the center C, and any radius, deſcribe an arc mn.—With the center A, and the ſame radius, de⯑ſcribe the arc rs.—Take the diſtance mn between the compaſſes, and ap⯑ply it from r to s.—Then a line drawn through A and s, will make the angle A equal to the angle C as required.

[diagram]

PROBLEM X. At a given Point A, in a given Line AB, to make an Angle of any propoſed Number of Degrees.

With the center A, and radius equal to 60 degrees taken from a ſeale of chords, deſcribe an arc, cutting AB in m.—Then take be⯑tween the compaſſes, the propoſed number of degrees from the ſame ſcale of chords, and apply them from m to n.—Through the point n draw An, and it will make the angle A of the number of degrees propoſed.

[diagram]

Note. Angles of more than 90 degrees are uſually taken off at twice.

Or the angle may be made with any divided arch or inſtrument, by laying the center to the point A, and its radius along AB; then make a mark n at the pro⯑poſed number of degrees, through which draw the line An as before.

PROBLEM XI. To meaſure a given Angle A. (See the laſt figure.)

[70]

Deſcribe the arc mn with the chord of 60 degrees, as in the laſt Problem.—Take the arc mn between the compaſſes, and that extent, applied to the chords, will ſhew the degrees in the given angle.

Note. When the diſtance mn exceeds 90 degrees, it muſt be taken off at twice, as before.

Or the angle may be meaſured by applying the ra⯑dius of a graduated arc, of any inſtrument, to AB, as in the laſt problem; and then noting the degrees cut off by the other leg An of the angle.

PROBLEM XII. To find the Center of a Circle.

Draw any chord AB; and biſect it perpendicularly with CD, which will be a diameter.— Biſect CD in the point o, which will be the center.

[diagram]

PROBLEM XIII. To deſcribe the Circumference of a Circle through thre [...] given Points, A, B, C.

From the middle point B draw chords to the other two points.—Biſect theſe chords per⯑pendicularly by lines meeting in o, which will be the center.— Then from the center o, at the diſtance oA, or oB, or oC, deſcribe the circle.

[diagram]

[71] Note. In the ſame manner may the center of an arc of a circle be found.

PROBLEM XIV. Through a given Point A, to draw a Tangent to a given Circle.

CASE 1. When A is in the Circumference of the Circle.

From the given point A, draw Ao to the center of the circle.—Then through A draw BC perpendicular to Ao, and it will be the tangent as re⯑quired.

[diagram]

CASE 2. Where A is out of the Circumference.

From the given point A, draw Ao to the center, which biſect in the point m.—With the center m, and radius mA or mo, deſcribe an arc cutting the given circle in n.—Through the points A and n, draw the tangent BC:

[diagram]

PROBLEM XV. To find a Third Proportional to two given Lines AB, AC.

Place the two given lines, AB, AC, making any angle at A, and join BC.—In AB take AD equal to AC, and draw DE parallel to BC. So ſhall AE be the third proportional to AB and AC. That is, AB∶AC∷AC∶AE.

[diagram]

PROBLEM XVI. To find a Fourth Proportional to three given Lines, AB, AC, AD.

[72]

Place two of them, AB, AC, making any angle at A, and join BC. Place AD on AB, and draw DE parallel to BC. So ſhall AE be the fourth propor⯑tional required.

That is AB∶AC∷AD∶AE.

[diagram]

PROBLEM XVII. To find a Mean Proportional between two given Lines, AB, BC.

Join AB and BC in one ſtraight line AC, and biſect it in the point o.—With the cen⯑ter o, and radius oA or oC, deſcribe a ſemicircle.—Erect the perpendicular BD, and it will be the mean proportional required.

That is AB∶BD∷BD∶BC.

[diagram]

PROBLEM XVIII. To make an Equilateral Triangle on a given Line AB.

From the centers A and B, with the radius AB, deſcribe arcs, interſecting in C.—Draw AC and BC, and it is done.

Note. An iſoſceles triangle may be made in the ſame man⯑ner, taking for the radius the given length of one of the equal ſides.

[diagram]

PROBLEM XIX. To make a Triangle with three given Lines AB, AC, BC.

[75]

With the center A and radius AC, deſcribe an arc.—With the center B, and radius BC, de⯑ſcribe another arc, cutting the former in C.—Draw AC and BC, and ABC is the triangle required.

[diagram]

PROBLEM XX. To make a Square upon a given Line AB.

Draw BC perpendicular and equal to AB. From A and C with the radius AB, deſcribe arcs interſecting in D.—Draw AD and CD, and it is done.

[diagram]

Another Way.

On the centers A and B, with the radius AB, deſcribe arcs croſſ⯑ing at o.—Biſect Ao in n.—With center o, and radius on, croſs the two arcs in C and D.—Then draw AC, BD, CD.

[diagram]

PROBLEM XXI. To deſcribe a Rectangle, or a Parallelogram, of a given Length and Breadth.

Place BC perpendicular to AB.—With center A, and ra⯑dius AC, deſcribe an arc.— With center C, and radius AB, deſcribe another are, cutting the former in D.—Draw AD and CD, and it is done.

[diagram]

[74] Note. In the ſame manner is deſcribed any oblique parallelogram, only drawing BC, making the given oblique angle with AB, inſtead of perpendicular to it.

PROBLEM XXII. To make a regular Pentagon on a given Line AB.

Make Bm perpendicular and equal to half AB.—Draw Am, and produce it till mn be equal to Bm.—With centers A and B, and radius Bn, deſcribe arcs interſecting in o, which will be the center of the circumſcribing circle.—Then with the center o, and the ſame radius, deſcribe the circle; and about the cir⯑cumference of it apply AB the proper number of times.

[diagram]

Another Method.

Make Bm perpendi⯑cular and equal to AB. —Biſect AB in n; then with the center n, and radius nm, croſs AB produced in o.—With the centers A and B, and radius Ao, deſcribe arcs interſecting in D, the oppoſite angle of the pentagon.—Laſtly, with center D, and radius AB, croſs thoſe arcs again in C and E, the other two angles of the figure.—Then draw the lines from angle to angle, to complete the figure.

[diagram]

[75] A third Method, nearly true.

On the centers A and B, with the radius AB, deſcribe two circles in⯑terſecting in m and n. —With the ſame ra⯑dius, and the center m, deſcribe rAOBS, and draw mn cutting it in o.—Draw rOC and SOB, which will give two angles of the pen⯑tagon.—Laſtly, with radius AB, and centers C and E, deſcribe arcs interſecting in D, the other angle of the pen⯑tagon nearly.

[diagram]

PROBLEM XXIII. To make a Hexagon on a given Line AB.

With the radius AB, and the centers A and B, deſcribe arcs interſecting in o.—With the ſame radius, and center o, deſcribe a circle, which will circumſcribe the hexagon.— Then apply the line AB ſix times round the circumference, marking out the angular points, which connect with right lines.

[diagram]

PROBLEM XXIV. To make an Octagon on a given Line AB.

[76]

Erect AF and BE per⯑pendicular to AB. — Pro⯑duce AB both ways, and biſect the angles mAF and nBE with the lines AH and BC, each equal to AB.— Draw CD and HG parallel to AF or BE, and each equal to AB.—With radius AB, and centers G and D, croſs AF and BE in F and E.—Then join GF, FE, ED, and it is done.

[diagram]

PROBLEM XXV. To make any regular Polygon on a given Line AB.

Draw Ao and Bo making the angles A and B each equal to half the angle of the polygon—With the cen⯑ter o and radius oA, de⯑ſcribe a circle.—Then apply the line AB continually round the circumference the proper number of times, and it is done.

[diagram]

Note. The angle of any polygon, of which the angles oAB and oBA are each one half, is found thus: divide the whole 360 degrees by the number of [77]ſides, and the quotient will be the angle at the center o; then ſubtract that from 180 degrees, and the re⯑mainder will be the angle of the polygon, and is double of oAB or of oBA. And thus you will find the fol⯑lowing table, containing the degrees in the angle o at the center, and the angle of the polygon, for all the regular figures from 3 to 12 ſides.

No. of ſides	Name of the polygon	Angle o at the center	Angle of the polyg	Angle oAB or oBA
3	Trigon	120°	60°	30°
4	Tetragon	90	90	45
5	Pentagon	72	108	54
6	Hexagon	60	120	60
7	Heptagon	51 3/7	128 4/7	64 2/7
8	Octagon	45	135	67½
9	Nonagon	40	140	70
10	Decagon	36	144	72
11	Undecagon	32 8/11	147 3/11	73 7/11
12	Dodecagon	30	150	75

PROBLEM XXVI. In a given Circle to inſcribe any regular Polygon; or, to divide the Circumference into any number of equal Parts. (See the laſt figure.)

At the center o make an angle equal to the angle at the center of the polygon, as contained in the third column of the above table of polygons.—Then the diſtance AB will be one ſide of the polygon; which being carried round the circumference the proper num⯑ber of times, will complete the figure.—Or, the arc AB will be one of the equal parts of the circumfer⯑ence.

[78] Another Method, nearly true.

Draw the diameter AB, which divide into as many equal parts as the figure has ſides.—With the radius AB, and centers A and B, deſcribe arcs croſſing at n; from whence draw nC through the ſecond diviſion on the diameter; ſo ſhall AC be a ſide of the poly⯑gon nearly.

[diagram]

Another Method, ſtill nearer.

Divide the diameter AB into as many equal parts as the figure has ſides, as before.— From the center o raiſe the perpendicular om, which pro⯑duce till mn be three-fourths of the radius om.—From n draw nc through the ſecond diviſion of the diameter, and the line AC will be the ſide of the polygon ſtill nearer than before; or the arc AC one of the equal parts into which the circum⯑ference is to be divided.

[diagram]

PROBLEM XXVII. About a given Circle to circumſcribe any Polygon.

Find the points m, n, p, &c. as in the laſt problem, to which draw radii mo, no, &c. to the center of the circle.—Then through theſe points, m, n, &c. and perpendicular to theſe radii draw the ſides of the polygon.

[diagram]

PROBLEM XXVIII. To find the Center of a given Polygon, or the Center of its inſcribed or circumſcribed Circle.

[79]

Biſect any two ſides with the perpendiculars mo, no; and their interſection will be the center.—Then with the center o, and the diſtance om, de⯑ſcribe the inſcribed circle; or with the diſtance to one of the angles, as A, deſcribe the cir⯑cumſcribing circle.

[diagram]

Note. This method will alſo circumſcribe a circle about any given oblique triangle.

PROBLEM XXIX. In any given Triangle to inſcribe a Circle.

Biſect any two of the angles with the lines Ao, BO, and o will be the center of the circle. —Then with the center o, and radius the neareſt diſtance to any one of the ſides, deſcribe the circle.

[diagram]

PROBLEM XXX. About any given Triangle to circumſcribe a Circle.

[80]

Biſect any two of the ſides AB, BC, with the perpendi⯑culars mo, no.—With the center C, and diſtance to any one of the angles, deſcribe the circle.

[diagram]

PROBLEM XXXI. In, or about, a given Square, to deſcribe a Circle.

Draw the two diagonals of the ſquare, and their inter⯑ſection o will be the center of both the circles.—Then with that center, and the neareſt diſtance to one ſide, deſcribe the inner circle; and with the diſtance to one angle, deſcribe the outer circle.

[diagram]

PROBLEM XXXII. In, or about, a given Circle, to deſcribe a Square, or an Octagon.

Draw two diameters AB, CD, perpendicular to each other.—Then connect their ex⯑tremities, and they will give the inſcribed ſquare ACBD.— Alſo through their extremities draw tangents parallel to them, and they will form the outer ſquare mnop.

[diagram]

[81] Note. If any quadrant, as AC, be biſected in q, it will give one-eighth of the circumference, or the ſide of the octagon.

PROBLEM XXXIII. In a given Circle, to inſcribe a Trigon, a Hexagon, or a Dodecagon.

The radius is the ſide of the hexagon. Therefore from any point A in the circum⯑ference, with the diſtance of the radius, deſcribe the arc BOF. Then is AB the ſide of the hexagon; and there⯑fore carrying it ſix times round will form the hexagon, or divide the circumference into ſix equal parts, each containing 60 degrees.—The ſecond of theſe C, will give AC the ſide of the trigon or equilateral triangle; and the arc AC one third of the circumference, or 120 degrees.—Alſo the half of AB, or An is one-12th of the circumference, or 30 degrees, and gives the ſide of the dodecagon.

[diagram]

Note. If tangents to the circle be drawn through all the angular points of any inſcribed figure, they will form the ſides of a like circumſcribing figure.

PROBLEM XXXIV. In a given Circle to inſcribe a Pentagon, or a Decagon.

[82]

Draw the two diameters AP, mn perpendicular to each other, and biſect the ra⯑dius on at q.—With the center q and diſtance qA, de⯑ſcribe the arc Ar; and with the center A, and radius Ar, deſcribe the arc rB. Then is AB one-fifth of the circum⯑ference; and AB carried five times over will form the pen⯑tagon.—Alſo the arc AB biſect⯑ed in s, will give As the tenth part of the circumference, or the ſide of the decagon.

[diagram]

Note. Tangents being drawn through the angular points, will form the circumſcribing pentagon or deca⯑gon.

PROBLEM XXXV. To divide the Circumference of a given Circle into 12 equal Parts, each of 30 Degrees.
¶ Or to inſcribe a Dodecagon by another Method.

Draw two diameters 17 and 410 perpendicular to each other.—Then with the radius of the circle, and the four extremities 1, 4, 7, 10, as centers, deſcribe ares through the center of the circle; and they will cut the circumference in the points required, dividing it into 12 equal parts, at the points marked with the numbers.

[diagram]

PROBLEM XXXVI. To draw a right Line equal to the Circumference of a given Circle.

[83]

[diagram]

Take III 1 equal to 3 times the diameter and [...]part more; and it will be equal to the circumference, very nearly.

PROBLEM XXXVII. To find a Right Line equal to any given Are AB of a Circle.

Through the point A and the center draw Am, making mn equal to ¾ of the radius no.—Alſo draw the indefinite tangent AP perpendicular to it.—Then through m and B, draw mP; ſo ſhall AP be equal to the arc AB very nearly.

[diagram]

Otherwiſe.

Divide the chord AB into 4 equal parts.—Set one part AC on the arc from B to D.— Draw CD, and the double of it will be nearly equal to the arc ADB.

[diagram]

PROBLEM XXXVIII. To divide a given Circle into any propoſed Number of Parts by equal Lines, ſo that thoſe Parts ſhall be mutu⯑ally equal, both in Area and Perimeter.

[84]

Divide the diameter AB into the propoſed number of equal parts at the points a, b, c, &c.—Then on Aa, Ab, Ac, &c. as diameters, deſcribe ſemicircles on one ſide of the diameter AB; and on Bd, Bc, Bb, &c. deſcribe ſemicircles on the other ſide of the diameter. So ſhall the correſponding joining ſemicircles divide the given circle in the man⯑ner propoſed. And in like manner we may proceed when the ſpaces are to be in any given proportion.—As to the perimeters, they are always equal, whatever the pro⯑portion of the ſpaces is.

[diagram]

PROBLEM XXXIX. To make a Triangle ſimilar to a given Triangle ABC.

Let AB be one ſide of the required triangle. Make the angle a equal to the angle A, and the angle b equal to the angle B; then the triangle abc will be ſimilar to ABC as propoſed.

[diagram]

Note. If ab be equal to AB, the triangles will alſo be equal, as well as ſimilar.

[diagram]

PROBLEM XL. To make a Figure ſimilar to any other given Figure ABCDE.

[85]

From any angle A draw diagonals to the other angles. —Take Ab a ſide of the figure required. Then draw be parallel to BC, and cd to CD, and de to DE, &c.

[diagram]

Otherwiſe.

Make the angles at a, b, e, reſpectively equal to the angles at A, B, E, and the lines will interſect in the cor⯑ners of the figure required.

[diagram]

PROBLEM XLI. To reduce a Complex Figure from one Scale to another, mechanically by means of Squares.

[diagram]

[86]

[diagram]

Divide the given figure, by croſs lines, into ſquares, as ſmall as may be thought neceſſary.—Then divide another paper into the ſame number of ſquares, and either greater, equal, or leſs, in the given proportion. —This done, obſerve what ſquares the ſeveral parts of the given figure are in, and draw, with a pencil, ſimi⯑lar parts in the correſponding ſquares of the new figure. And ſo proceed till the whole is copied.

PROBLEM XLII. To make a Triangle equal to a given Trapezium ABCD.

Draw the diagonal DB, and CE parallel to it, meet⯑ing AB produced in E.—Join DE; ſo ſhall the triangle ADE be equal to the trape⯑zium ABCD.

[diagram]

PROBLEM XLIII. To make a Triangle equal to the figure ABCDEA.

Draw the diagonals DA, DB, and the lines EF, CG parallel to them, meeting the baſe AB, both ways produced, in F and G.—Join DF, DG; and DFG will be the triangle required equal to the given figure ABCDE.

[diagram]

[87] Note. Nearly in the ſame manner may a triangle be made equal to any right-lined figure whatever.

PROBLEM XLIV. To make a Triangle equal to a given Circle.

Draw any ra⯑dius Ao, and the tangent AB per⯑pendicular to it. —On which take AB equal to the circumference of the circle by Problem XXXVI.—Join BO, ſo ſhall ABo be the triangle required, equal to the given circle.

[diagram]

PROBLEM XLV. To make a Rectangle, or a Parallelogram equal to a given Triangle ABC.

Biſect the baſe AB in m.— Through C draw Cno pa⯑rallel to AB.—Through m and B draw mn and Bo pa⯑rallel to each other, and either perpendicular to AB, or making any angle with it. And the rectangle or paralle⯑logram mnoB will be equal to the triangle, as required.

[diagram]

PROBLEM XLVI. To make a Square equal to a given Rectangle ABCD.

[88]

Produce one ſide, AB, till BE be equal to the other ſide BC.—Biſect AE in o; on which as a center, with radius Ao deſcribe a ſemi⯑circle, and produce BC to meet it at F.—On BF make the ſquare BFGH, and it will be equal to the rectangle ABCD, as required.

[diagram]

*⁎* Thus the circle and all right-lined figures, have been reduced to equivalent ſquares.

PROBLEM XLVII. To make a Square equal to two given Squares P and Q.

Set two ſides AB, BC, of the given ſquares, perpen⯑dicular to each other.—Join their extremities AC; ſo ſhall the ſquare Q, con⯑ſtructed on AC, be equal to the two P and Q taken to⯑gether.

[diagram]

Note. Circles, or any other ſimilar figures, are added in the ſame manner. For, if AB and BC be the diameters of two circles, AC will be the diameter of a circle equal to both the other two. And if AB and BC be the like ſides of any two ſimilar figures, then AC will be the like ſide of another ſimilar figure equal to both the two former, and upon which the third figure may be conſtructed by problem XL.

PROBLEM XLVIII. To make a Square equal to the Difference between two given Squares PR. (See the laſt figure).

[89]

On the ſide AC of the greater ſquare, as a diameter, deſcribe a ſemicircle; in which apply AB the ſide of the leſs ſquare.—Join BC, and it will be the ſide of a ſquare equal to the difference between the two P and R, as required.

PROBLEM XLIX. To make a Square equal to the Sum of any Number of Squares taken together.

Draw two indefinite lines Am, An, perpendicular to each other at the point A. On the one of theſe ſet off AB the ſide of one of the given ſquares, and on the other AC the ſide of another of them. Join BC, and it will be the ſide of a ſquare equal to the two together.— Then take AD equal to BC, and AE equal to the ſide of the third given ſquare. So ſhall DE be the ſide of a ſquare equal to the ſum of the three given ſquares. —And ſo on continually, always ſetting more ſides of the given ſquares on the line An, and the ſides of the ſucceſſive ſums on the other line Am.

[diagram]

Note. And thus any number of any ſort of figures may be added together.

PROBLEM L. To make plane Diagonal Scales.

[90]

[diagram]

Draw any line as AB of any convenient length. Divide it into 11 equal parts ^*. Complete theſe into rectangles of a convenient height, by drawing parallel and perpendicular lines. Divide the altitude into 10 equal parts, if it be for a decimal ſcale for common numbers, or into 12 equal parts, if it be for feet and inches; and through theſe points of diviſion draw as many parallel lines, the whole length of the ſcale.— Then divide the length of the firſt diviſion AC into 10 equal parts, both above and below; and connect theſe points of diviſion by diagonal lines, and the ſcale is finiſhed, after being numbered as you pleaſe.

Note. Theſe diagonal ſcales ſerve to take off dimen⯑ſions or numbers of three figures. If the firſt large di⯑viſions be units; the ſecond ſet of diviſions along AC, will be 10th parts; and the diviſions in the altitude, along AD, will be 100th parts. If CD be tens, AC will be units, and AD will be 10th parts. If CB be hundreds, AC will be tens, and AD units. If CB be thouſands, AC will be hundreds, and AD will be tens. And ſo on, each ſet of diviſions being tenth parts of the former ones.

For example, ſuppoſe it were required to take off 243 from the ſcale. Fix one foot of the compaſſes at 2 of the greateſt diviſions, at the bottom of the ſcale, and [91]extend the other to 4 of the ſecond diviſions, along the bottom; then, for the 3, ſlide up both points of the compaſſes by a parallel motion, till they fall upon the third longitudinal line; and in that poſition extend the ſecond point of the compaſſes to the fourth diagonal line, and you have the extent of three figures as required.

Or if you have any line to meaſure the length of. —Take it between the compaſſes, and applying it to ſcale, ſuppoſe it fall between 3 and 4 of the large diviſions; or, more nearly, that it is 3 of the large diviſions, or 3 hundreds, and between 5 and 6 of the ſecond diviſions, or 5 tens or 50, and a little more. Slide up the points of the compaſſes by a parallel motion, keeping one foot always on the vertical diviſion of 3 hundred, till the other point fall ex⯑actly on one of the diagonal lines, which ſuppoſe to be 8, which is 8 units. Which ſhews that the length of the line, propoſed to be meaſured, is 358.

Figure 1. PLANE SCALES FOR TWO FIGURES.

The above are three other forms of ſcales, the firſt of which is a decimal ſcale, for taking off common numbers conſiſting of two figures. The other two are duodecimal ſcales, and ſerve for feet and inches, &c.

[92]Theſe, and other ſcales, engraven on ivory, are fitteſt for practical uſe. And the moſt convenient form of a plane ſcale of equal diviſions, is on the very edge of the ivory, made thin at the edge for laying along any line, and then marking on the paper oppoſite any diviſion required: which is better than taking lengths off a ſcale with compaſſes.

REMARKS.

Note 1. That in a circle, the half chord DC, is a mean proportional between the ſeg⯑ments AD, DB of the diame⯑ter AB perpendicular to it. That is AD: DC∷DC∶DB.

[diagram]

2. The chord AC is a mean proportional between AD and the diameter AB. And the chord BC a mean proportional between DB and AB.

That is, AD∶AC∷AC∶AB, and BD∶BC∷BC∶AB.

3. The angle ACB, in a ſemicircle, is always a right

4. The ſquare of the hypotenuſe of a right-angled triangle, is equal to the ſquares of both the ſides.

That is, AC²=AD²+DC², and BC²=BD²+DC², and AB²=AC²+BC².

5. Triangles that have all the three angles of the one, reſpectively equal to all the three of the other, are called equiangular triangles, or ſimilar triangles.

6. In ſimilar triangles, the like ſides, or ſides oppoſite the equal angles, are proportional.

7. The areas, or ſpaces, of ſimilar triangles, are to each other, as the ſquares of their like ſides.

MENSURATION OF SUPERFICIES.

[]

THE area of any figure, is the meaſure of its ſur⯑face, or the ſpace contained within the bounds of the ſurface, without any regard to thickneſs.

The area is eſtimated by the number of ſquares con⯑tained in the ſurface, the ſide of thoſe ſquares being either an inch, a foot, a yard, &c. And hence the area is ſaid to be ſo many ſquare inches, or ſquare feet, or ſquare yards, &c.

Our ordinary lineal meaſures, or meaſures of length, are as in the firſt table here below; and the annexed table of ſquare meaſures, is taken from it, by ſquaring the ſeveral numbers.

Lineal Meaſures.		Square Meaſures.
12 inches	1 foot	144 inches	1 foot
3 fect	1 yard	9 feet	1 yard
6 feet	1 fathom	36 feet	1 fathom
16½ feet, or	1 pole	272¼ feet	1 pole
5½ yards	or rod	or 30¼ yds.	or rod
40 poles	1 furlong	1600 poles	1 furlong
8 furlongs	1 mile	64 furlongs.	1 mile

PROBLEM I. To find the Area of a Parallelogram; whether it be: Square, a Rectangle, a Rhombus, or a Rhomboid.

[94]

Multiply the length by the breadth, or perpendicular height, and the product will be the area.

EXAMPLES.

1. To find the area of a ſquare, whoſe ſide is 6 inches, or 6 feet, &c.

[...]

[diagram]

2. To find the area of a rectangle, whoſe length is 9 and breadth 4 inches, or feet, &c.

[...]

[diagram]

[95]3. To find the area of a rhombus, whoſe length is 6.20 chains, and perpendicular height 5.45.

[...] Anſ. 3 acres, 1 rood, 20 perches.

[diagram]

Note. Here the ſquare chains are divided by 10 to bring them to acres, becauſe ten ſquare chains make an acre. Alſo the decimals of an acre are multiplied by 4 for roods, and theſe by 40 for perches, becauſe 4 roods make 1 acre, and 40 perches 1 rood.

4. To find the area of the rhomboid, whoſe length is 12 feet 3 inches, and breadth 5 feet 4 inches.

[...]

[diagram]

[96]5. To find the area of a ſquare, whoſe ſide is 35.2; chains.

Anſ. 124 ac 1 ro 1 perch.

6. To find the area of a parallelogram, whoſe length is 12.25 chains, and breadth 8.5 chains.

Anſ. 10 ac 1 ro 26p.

7. To find the area of a rectangular board, whoſe length is 12.5 feet, and breadth 9 inches.

Anſ. 9⅜ feet.

8. To find the ſquare yards of painting in a rhom⯑boid, whoſe length is 37 feet, and breadth 5¼ feet.

Anſw. 21 7/12 ſquare yards.

PROBLEM II. To find the Area of a Triangle.

Rule 1. Multiply the baſe by the perpendicular height, and half the product will be the area.

Rule 2. When the three ſides only are given: Add the three ſides together, and take half the ſum; from the half ſum ſubtract each ſide ſeparately; multiply the half ſum and the three remainders continually to⯑gether; and the ſquare root of the laſt product will be the area of the triangle.

EXAMPLES.

1. Required the area of the triangle, whoſe baſe is 6.25 chains, and perpendicular height 5.20 chains.

[97] [...]

Anſ. 1 ac 2 ro 20 perches.

2. To find the number of ſquare yards in the triangle, whoſe three ſides are 13, 14, 15 feet.

[...]

[diagram]

Anſ. 91/3 ſquare yards

[98]3. How many ſquare yards in a right-angled triangle, whoſe baſe is 40, and perpendicular 30 feet?

Anſ. 66⅔ ſquare yards.

4. To find the area of the triangle, whoſe three ſider are 20, 30, 40 chains.

Anſ. 29 ac or 7 per.

5. How many ſquare yards contains the triangle, whoſe baſe is 49 feet, and height 25¼ feet?

Anſ. 68 53/72 or 68.7361.

6. How many acres &c. in the triangle, whoſe three ſides are 380, 420, 765 yards?

Anſ. 9ac or 38 per.

7. To find the area of the triangle, whoſe baſe is 18 feet 4 inches, and height 11 feet 10 inches.

Anſ. 216 feet 11 inches 4″.

8. How many acres &c. contains the triangle, whoſe three ſides are 49.00, 50.25, 25.69 chains?

Anſ. 61 ac 11 39.68 per.

PROBLEM III. To find one Side of a Right-angled Triangle, having the other two Sides given.

The ſquare of the hypotenuſe is equal to both the ſquares of the two legs. Therefore,

1. To find the hypotenuſe; add the ſquares of the two legs together, and extract the ſquare root of the ſum.

2. To find one leg; ſubſtract the ſquare of the other leg from the ſquare of the hypotenuſe, and extract the root of the difference.

EXAMPLES.

1. Required the hypotenuſe of a right-angled triangle, whoſe baſe is 40, and perpendicular 30.

[99] [...]

[diagram]

2. What is the perpendicular of a right-angled triangle, whoſe baſe AB is 56, and hypotenuſe AC 65?

[...]

3. Required the length of a ſealing ladder to reach the top of a wall, whoſe height is 28 feet, the breadth of the ditch before it being 45 feet.

Anſ. 53 feet.

4. To find the length of a ſhoar, which, ſtrurting 12 feet from the upright of a building, may ſupport a j [...]mb 20 feet from the ground.

Anſ. 23.32380 feet.

5. A line of 320 feet will reach from the top of a precipice, ſtanding cloſe by the ſide of a brook, to the oppoſite bank: required the breadth of the brook; the height of the precipice being 103 feet.

Anſ. 302.9703 feet.

[100]6. A ladder of 50 feet long, being placed in a ſtreet, reached a window 28 feet from the ground on one ſide; and by turning the ladder over, without removing the foot, it touched a moulding 36 feet high on the other ſide: required the breadth of the ſtreet.

Anſ. 76.1233335 feet.

PROBLEM IV. To find the Area of a Trapezoid.

Multiply the ſum of the two parallel ſides by the per⯑pendicular diſtance between them, and half the product will be the area.

EXAMPLES.

1. In a trapezoid, the parallel ſides arc AB 7.5, and DC 12.25, and the perpendicular diſtance AP or cn is 15.4 chains: required the area.

[...]

[diagram]

Anſ. 15 ac or 33.2 perches.

2. How many ſquare feet contains the plank, whoſe length is 12 feet 6 inches, the breadth at the greater end 1 foot 3 inches, and at the leſs end 11 inches?

Anſ. 13 13/24 feet.

[101]3. Required the area of a trapezoid, the parallel ſides being 21′ feet 3 inches and 18 feet 6 inches, and the diſtance between them 8 feet 5 inches.

Anſ. 167 feet 3 inches 4″ 6‴.

4. In meaſuring along one ſide AB of a quadrangu⯑lar field, that ſide and the two perpendiculars upon it from the oppoſite corners, meaſured as below: required the content.

Anſ. 4 ac 3 r 17.92 p.

chains
AP=1.10
AQ=7.45
AB=11.10
PC=3.62
QD=5.95

[diagram]

PROBLEM V. To find the Area of a Trapezium.

CASE I. For any Trapezium.

Divide it into two triangles by a diagonal; then find the areas of theſe triangles, and add them together.

Note. If two perpendiculars be let fall on the diago⯑nal, from the other two oppoſite angles, the ſum of theſe perpendiculars being multiplied by the diagonal, half the product will be the area of the trapezium.

CASE 2. When the Trapezium can be inſcribed in a Circle.

Add all the four ſides together, and take half the ſum, and ſubſtract each ſide ſeparately from the half ſum; then multiply the four remainders continually together, and the ſquare root of the laſt product will be the area of the trapezium.

[102]EXAMPLES.

1. To find the area of the trapezium ABCD, the diagonal AC being 42, the perpendicular BF 18, and the perpendicular DE 16.

[...]

[diagram]

2. In the trapezium ABCD, the ſide AB is 15, BC 13, CD 14, AD 12, and the diagonal AC is 16; re⯑quired the area.

[...]

The triangle ABC 91.1921
The triangle ADC 81.3326
The trapezium ABCD 172.5247 the anſwer.

3. If a trapezium can be inſcribed in a circle, and have its four ſides 24, 26, 28, 30; required its area.

[...]

4. How many ſquare yards of paving are in the tra⯑pezium, whoſe diagonal is 65 feet, and the two per⯑pendiculars let fall upon it 28 and 33.5 feet?

Anſ. 222 1/12 yards.

5. What is the area of a trapezium, whoſe ſouth ſide is 27.40 chains, caſt ſide 35.75 chains, north ſide 37.55 chains, weſt ſide 41.05 chains, and the diagonal from ſouth-weſt to north-eaſt 48.35 chains?

Anſ. 123 ac or 11.8672 per.

[104]6. What is the area of a trapezium, whoſe diagonal is 108½ feet, and the perpendiculars 56¼ and 60¾ feet?

Anſ. 6347¼ feet.

7. What is the area of a trapezium inſcribed in a circle, the four ſides being 12, 13, 14, 15?

Anſ. 180.9972372.

8. In the four-ſided field ABCD, on account of ob⯑ſtructions in the two ſides AB, CD, and in the perpen⯑diculars BF, DE, the following meaſures only could be taken, namely the two ſides BC 265 and AD 220 yards, the diagonal AC 378 yards, and the two diſtances of the perpendiculars from the ends of the diagonal, namely AE 100, and CF 70 yards: required the area in acres, when 4840 ſquare yards make an acre.

Anſ. 17 ac 2 ro 21 per.

PROBLEM VI. To find the Area of an Irregular Polygon.

Draw diagonals dividing the figure into trapeziums and triangles. Then find the areas of all theſe ſepa⯑rately, and their ſum will be the content of the whole irregular figure.

EXAMPLES.

1. To find the content of the irregular figure ABCDEFGA, in which are given the following dia⯑gonals and perpendiculars: namely

AC 5.5

FD 5.2

GC 4.4

Gm 1.3

Bn 1.8

Go 1.2

Ep 0.8

Dq 2.3

[diagram]

[105] [...]

PROBLEM VII. To find the Area of a Regular Polygon.

RULE 1.

Multiply the perimeter of the figure, or ſum of its ſides, by the perpendicular falling from its center upon one of its ſides, and half the product will be the area.

RULE 2.

Square the ſide of the polygon; multiply that ſquare by the multiplier ſet oppoſite to its name in the follow⯑ing table, and the product will be the area.

[106]

No. of ſides	Names	Multipliers
3	Trigon or equal. ▵	0.4330127
4	Tetragon or ſquare	1.0000000
5	Pentagon	1.7204774
6	Hexagon	2.5980762
7	Heptagon	3.6339124
8	Octagon	4.8284271
9	Nonagon	6.1818242
10	Decagon	7.6942088
11	Undecagon	9.3656399
12	Dodecagon	11.1961524

EXAMPLES.

1. Required the regular pentagon, whoſe ſide AB is 25 feet, and perpendicular CP 17.2047737.

[...]

[diagram]

[...]

[107]2. To find the area of the hexagon, whoſe ſide is 20.

Anſ. 1039.2304.

3. To find the area of the trigon, or equilateral triangle, whoſe ſide is 20.

Anſ. 173.2052.

4. Required the area of an octagon, whoſe ſide is 20.

Anſ. 1931.37084.

5. What is the area of a decagon, whoſe ſide is 20?

Anſ. 3077.68352.

PROBLEM VIII. In a Circular Arc, having any two of theſe given, to find the reſt; namely, the Chord AB of the Arc, the Height or Verſed Sine DP, the Chord AD of Half the Arc, and the Diameter, or the Radius AC or CD.

Here will always be given two ſides of the right-angled triangles, APC, APD; and therefore the other parts will eaſily be found from the pro⯑perty in Problem 111, namely, that the ſquare of the longeſt ſide, is equal to the ſquares of both the two ſhorter added to⯑gether.

[diagram]

Thus, if there be given the radius, and the chord AB, or its half AP. Then [...]; and CD−CP=PD; and [...].

Again, when the radius, and height PD are given. Then CD−DP=CP; and [...].

And when AP and PD are given. Then as DP; PA [...]; PA∶CD+CP=PA²/PD, and 2CD=PA²/PD+PD;

EXAMPLES.

1. Suppoſe the radius AC or CD to be 10, and the half chord AP 8.

[108]Then [...]; and CD−CP=10−6=4=PD; and [...].

Ex. 2. If the radius be 10, and PD 4.

Then CD−DP=10−4=6=CP; and [...].

Ex. 3. When AP is 8 and DP 4.

Then PA²/PD=64/4=16=CD+CP

And 2CD=16+PD=20, or CD=10.

PROBLEM IX. To find the Diameter and Circumference of a Circle, the one from the other.

RULE 1.

As 7 is to 22, ſo is the diameter to the circumference.

As 22 is to 7, ſo is the circumference to the diameter.

RULE 2.

As 113 is to 355, ſo is the diameter to the circumf.

As 355 is to 113, ſo is the circumf. to the diameter.

RULE 3.

As 1 is to 3.1416, ſo is the diameter to the circumf.

As 3.1416 is to 1, ſo is the circumf. to the diameter.

EXAMPLES.

1. To find the circumference of a circle, whoſe dia⯑meter AB is 10.

[...]

[diagram]

[109] [...]

By Rule 3.

1∶3.1416∷10∶31.416 the circumference nearly, the true circumference being 31.4159265358979 &c.

So that the 2d rule is neareſt the truth.

2. To find the diameter when the circumference is 100.

By Rule 1.

[...]anſ.

By Rule 2.

[...]

By Rule 3.

[...]

3. If the diameter of the earth be 7958 miles, as it is very nearly, what is the circumference, ſuppoſing it to be exactly round?

Anſ. 25000.8528 miles.

4. To find the diameter of the globe of the earth, ſup⯑poſing its circumference to be 25000 miles.

Anſ. 7957 ¾ nearly.

PROBLEM X. To find the Length of any Arc of a Circle.

[110]

RULE 1.

As 180 is to the number of degrees in the arc,

So is 3.1416 times the radius, to its length.

Or as 3 is to the number of degrees in the arc,

So is .05236 times the radius, to its length.

Ex. 1. To find the length of an arc ADB of 30 de⯑grees, the radius being 9 feet.

[...]

Fig. 10 Prob. VIII.

RULE 2.

From 8 times the chord of half the arc ſubſtract the chord of the whole are, and ⅓ of the remainder will be the length of the arc nearly.

Ex. 2. The chord AB of the whole are being 4.65874, and the chord AD of the half are 2.34947; required the length of the arc.

[...]

Ex. 3. Required the length of an arc of 12 degrees 10 minutes, the radius being 10 feet.

Anſ. 2 1234 [...].

[111]Ex. 4. To find the length of an arc whoſe chord is 6, and the chord of its half is 3¹.

Anſ. 7⅓.

Ex. 5. Required the length of the arc, whoſe chord is 8, and height PD 3.

Anſ. 10⅔.

Ex. 6. Required the length of the arc, whoſe chord is 6, the radius being 9.

Anſ. 6.11706.

PROBLEM XI. To find the Area of a Circle.

The area of a circle may be found from the diameter and circumference together, or from either of them alone, by theſe rules following.

Rule 1. Multiply half the circumference by half the diameter. Or, take ¼ of the product of the whole circumference and diameter.
Rule 2. Multiply the ſquare of the diameter by .7854.
Rule 3. Multiply the ſquare of the circumference by .07958.
Rule 4. As 14 is to 11, ſo is the ſquare of the dia⯑meter to the area.
Rule 5. As 88 is to 7, ſo is the ſquare of the cir⯑cumference to the area.

EXAMPLES.

1. To find the area of a circle whoſe diameter is 10, and circumference 31.4159265.

[...]

[112] [...]

Ex. 2. Required the area of the circle, whoſe dia⯑meter is 7, and circumference 22.

Anſ. 38½.

Ex. 3. What is the area of a circle, whoſe diameter is 1, and circumference 3.1416?

Anſ. .7854.

Ex. 4. What is the area of a circle, whoſe diameter is 7?

Anſ. 38.4846.

Ex. 5. How many ſquare yards are in a circle whoſe diameter is 3½ feet?

Anſ. 1.069.

Ex. 6. How many ſquare feet does a circle contain, the circumference being 10.9956 yards?

Anſ. 86.19266.

PROBLEM XII. To find the Area of the Sector of a Circle.

RULE I.

Multiply the radius, or half the diameter, by half the arc of the ſector, for the area. Or, take ¼ of the product of the diameter and are of the ſector.

Note. The are may be found by problem x.

[113]RULE 2.

As 360 is to the degrees in the arc of the ſector, ſo is the whole area of the circle, to the area of the ſector.

Note. For a ſemicircle take one half, for a quadrant one quarter, &c. of the whole circle.

EXAMPLES.

1. What is the area of the ſector CAB, the radius being 10, and the chord AB 16.

[...]

[diagram]

[114]Ex. 2. Required the area of the ſector, whoſe are contains 18 degrees; the diameter being 3 ſeet.

[...]

Then, as 360∶18

Or as 20∶ 1∷7.0686∶ .35343 the anſ.

Ex. 3. What is the area of the ſector, whoſe radius is 10, and arc 20?

Anſ. 100.

Ex. 4. What is the area of the ſector, whoſe radius is 9, and the chord of its arc 6?

Anſ. 27.52678.

Ex. 5. Required the area of a ſector, whoſe radius is 25, its arc containing 147 degrees 29 minutes.

Anſ. 804.4017.

Ex. 6. To find the area of a quadrant and a ſemi⯑circle, to the radius 13.

Anſ. 132.7326 and 265.4652.

PROBLEM XIII. To find the Area of a Segment of a Circle.

RULE I.

Find the area of the ſector having the ſame arc with the ſegment, by the laſt problem.

Find the area of the triangle, formed by the chord of the ſegment and the radii of the ſector.

Then the ſum of theſe two will be the anſwer when the ſegment is greater than a ſemicircle; but the dif⯑ference will be the anſwer when it is leſs than a ſemi⯑circle.

EXAMPLE I.

Required the area of the ſegment ACBA, its chord AB being 12, and the radius EA or CE 10.

[115] [...]

[diagram]

[...]

RULE 2.

To the chord of the whole arc, add 4/3 of the chord of half the arc, or add the latter chord and ⅓ of it more.

Multiply the ſum by the verſed ſine or height of the ſegment, and 4/10 of the product will be the area of the ſegment.

[116]Ex. Take the ſame example, in which the radius is 10, and the chord AB 12.

Then, as before, are found CD 2, and the chord of the half arc AC 6.324555

[...]

RULE 3.

Divide the height of the ſegment by the dia [...], and find the quotient in the column of heights or [...] ſines, at the end of the book.

Take out the correſponding area in the next column on the right hand, and multiply it by the ſquare of the diameter, for the anſwer.

Ex. The example being the ſame as before, we have CD equal to 2, and the diameter 20.

[...]

OTHER EXAMPLES.

Ex. 2. What is the area of the ſegment, whoſe height is 2, and chord 20?

Anſ. 26.878787.

Ex. 3. What is the area of the ſegment, whoſe height is 18, and diameter of the circle 50?

Anſ. 636.375.

Ex. 4. Required the area of the ſegment whoſe chord is 16, the diameter being 20.

Anſ. 44.7292.

PROBLEM XIV. To find the Area of a Circular Zone ADCBA.

[117]

[diagram]

Rule 1. Find the areas of the two ſegments AEB, DEB, and their difference will be the zone ADCB.

Rule 2. To the area of the trapezoid ARDQP add the area of the ſmall ſegment ADR; and double the ſum for the area of the zone ADCB.

EXAMPLES.

1. What is the area of the zone, leſs than a ſemi⯑circle, having the greater chord 16, the leſs chord 6, and the diameter of the circle 20?

Here [...]

and [...]

[...]

[118]Ex. 2. If the greater chord be 96, the leſs 60, and the diſtance between them 26; required the area.

Here if R be the middle of the chord AD, and O the center of the circle. Then, in fig. 1, and by the 1ſt rule,

DS=26

AS=AP − DQ=48 − 30=18

RT=½AP + ½DQ=24 + 15=39

And by Note 6 pa. 36,

[...]

TP=TQ=½DS=13

OP=OT−TP=27−13=14

OQ=OT + TQ=27 + 13=40

[...]

EQ=OE−OQ=50−40=10

EP=OE−OP=50−14=36

[...]

Ex. 3. If the greater chord be 40, the leſs 30, and the diſtance between them 35; required the area of the zone in figure 2, and by the 2d rule.

Here, like as before, we have

TR=½AP + ½DQ=10 + 7½=17½

[...]

OP=TP−OT=½PQ−OT=15

[...]

GR=OG − OR=25 − 17.677669=7.322331

[119] [...]

Ex. 4. If one end be 48, the other 30, and the breadth or diſtance 13, what is the area of the zone.

Anſ. 534.4249.

PROBLEM XV. To find the Area of a Circular Ring, or Space included between two Concentric Circles.

The difference between the two circles, will be the ring. Or, multiply the ſum of the diameters by their difference, and multiply the product by .7854 for the anſwer.

EXAMPLES.

1. The diameters of the two concentric circles being AB 10 and DG 6, required the area of the ring con⯑tained between their circumferences AEBA, and DFGD.

[...]

[diagram]

[120]Ex. 2. The diameters of two concentric circles be⯑ing 20 and 10; required the area of the ring between their circumferences.

Anſ. 235.62.

Ex. 3. What is the area of the ring, the diameters of whoſe bounding circles are 6 and 4?

Anſ. 15.708.

PROBLEM XVI. To meaſure long Irregular Figures.

Take the breadth in ſeveral places at equal diſtances. Add all the breadths together, and divide the ſum by the number of them, for the mean breadth; which multiply by the length for the area.

EXAMPLES.

1. The breadths of an irregular figure, at five equi⯑diſtant places being AD 8.1, mP 7.4, nq 9.2, or 10.1, BC 8.6; and the length AB 39; required the area.

[...]

[diagram]

Ex. 2. The length of an irregular figure being 84, and the breadths at 6 places 17.4, 20.6, 14.2, 16.5, 20.1, 24.3; what is the area?

Anſ. 1583.4.

MENSURATION OF SOLIDS.

[]

DEFINITIONS.

1. SOLIDS, or bodies, are figures having length, breadth, and thickneſs.

2. A priſm is a ſolid, or body, whoſe ends are any plane figures, which are equal and ſimilar; and its ſides are parallelograms.

[diagram]

A priſm is called a triangu⯑lar priſm, when its ends are triangles; a ſquare priſm, when its ends are ſquares; a pentagonal priſm, when its ends are pentagons; and ſo on.

[diagram]

3. A cube is a ſquare priſm, having ſix ſides, which are all ſquares. It is like a die, having its ſides perpendicular to one another.

[diagram]

4. A parallelopipedon is a ſolid having ſix rectangular ſides, every oppoſite pair of which are equal and parallel.

[diagram]

5. A cylinder is a round priſm, having circles for its ends.

[diagram]

[122]6. A pyramid is a ſolid having any plane figure for a baſe, and its ſides are triangles whoſe vertices meet in a point at the top, called the vertex of the pyramid.

[diagram]

The pyramid takes names accord⯑ing to the figure of its baſe, like the priſm; being triangular, or ſquare, or hexagonal, &c.

7. A cone is a round pyramid, having a circular baſe.

[diagram]

8. A ſphere is a ſolid bounded by one continued convex ſurface, every point of which is equally diſtant from a point within, called the center.—The ſphere may be con⯑ceived to be formed by the revolu⯑tion of a ſemicircle about its diame⯑ter, which remains fixed.

[diagram]

9. The axis of a ſolid, is a line drawn from the middle of one end, to the middle of the oppoſite end; as between the oppoſite ends of a priſm. Hence the axis of a pyramid, is the line from the vertex to the middle of the baſe, or the end on which it is ſuppoſed to ſtand. And the axis of a ſphere, is the ſame as a diameter, or a line paſſing through the center, and terminated by the ſurface on both ſides.

10. When the axis is perpendicular to the baſe, it is a right priſm or pyramid; otherwiſe, it is oblique.

[123]11. The height or altitude of a ſolid, is a line drawn from its vertex or top, perpendicular to its baſe.—This is equal to the axis in a right priſm or pyramid; but in an oblique one, the height is the perpendicular ſide of a right-angled triangle, whoſe hypotenuſe is the axis.

12. Alſo a priſm or pyramid is regular or irregular, as its baſe is a regular or an irregular plane figure.

13. The ſegment of a pyramid, ſphere, or any other ſolid, is a part cut off the top by a plane parallel to the baſe of that figure.

14. A fruſtum or trunk, is the part that remains at the bottom, after the ſegment is cut off.

15. A zone of a ſphere, is a part intercepted between two parallel planes; and is the difference between two ſegments. When the ends, or planes, are equally diſtant from the center, on both ſides, the figure is called the middle zone.

16. The ſector of a ſphere, is compoſed of a ſeg⯑ment leſs than a hemiſphere or half ſphere, and of a cone having the ſame baſe with the ſegment, and its vertex in the center of the ſphere.

17. A circular ſpindle, is a ſolid generated by the revolution of a ſeg⯑ment of a circle about its chord, which remains fixed.

[diagram]

18. A regular body, is a ſolid contained under a cer⯑tain number of equal and regular plane figures of the ſame ſort.

19. The faces of the ſolid are the plane figures un⯑der which it is contained. And the linear ſides, or edges of the ſolid, are the ſides of the plane faces.

20. There are only five regular bodies: namely, 1ſt, the tetraedron, which is a regular pyramid, having four triangular faces; 2d, the hexaedron, or cube, [124]which has 6 equal ſquare faces; 3d, the octaedron, which has 8 triangular faces; 4th, the dodecaedron, which has 12 pentagonal faces; 5th, the icoſaedron, which has 20 triangular faces.

Note, If the following figures be exactly drawn on paſteboard, and the lines cut half through, ſo that the parts be turned up and glued together, they will repreſent the five regular bodies: namely, figure 1 the tetraedron, figure 2 the hexaedron, figure 3 the octae⯑dron, figure 4 the dodecaedron, and figure 5 the icoſaedron.

[diagram]

[125] Note alſo, that, in cubic meaſure,

1728 inches make	1 foot
27 feet	1 yard
166⅜ yards	1 pole
64000 poles	1 furlong
512 furlongs	1 mile

PROBLEM I. To find the Solidity of a Cube.

Cube one of its ſides for the content; that is, mul⯑tiply the ſide by itſelf, and that product by the ſide again.

EXAMPLES.

1. If the ſide AB, or AC, or BD, of a cube be 24 inches, what is its ſolidity or content?

[...]

See Fig. at Definition 3, p. 121.

Ex. 2. How many ſolid feet are in the cube whoſe ſide is 22 feet?

Anſ. 10648.

Ex. 3. Required how many cubic feet are in the cube, whoſe ſide is 18 inches.

Anſ. 3⅞.

PROBLEM II. To find the Solidity of a Parallelopipedon.

Multiply the length, breadth, and depth, or altitude, [126]all continually together, for the ſolid content; that is, multiply the length by the breadth, and that product by the depth.

EXAMPLES.

1. Required the content of the parallelopipedon whoſe length AB is 6 feet, its breadth AC 2½ feet, and altitude BD 1¾ feet.

[...]

[diagram]

Ex. 2. Required the content of a parallelopipedon, whoſe length is 10.5, breadth 4.2, and height 3.4.

Anſ. 149.94.

Ex. 3. How many cubic feet in a block of marble, whoſe length is 3 feet 2 inches, breadth 2 feet 8 inches, and depth 2 feet 6 inches?

Anſ. 21 1/9.

PROBLEM III. To find the Solidity of any Priſm.

Multiply the area of the baſe, or end, by the height, and it will give the content.

Which rule will do, whether the priſm be triangular or ſquare, or pentagonal, &c. or round, as a cylinder.

EXAMPLES.

1. What is the content of a triangular priſm, whoſe length AC is 12 feet, and each ſide AB of its equilateral baſe 2½ feet?

[127]Here 5/2 x 5/2=25/4=6¼

Then .433013 tabular n°.

[...]

[diagram]

Ex. 2. Required the ſolidity of a triangular priſm, whoſe length is 10 feet, and the three ſides of its tri⯑angular end or baſe, are 5, 4, 3 feet.

Anſ. 60.

Ex. 3. What is the content of a hexagonal priſm, the length being 8 feet, and each ſide of its end 1 foot 6 inches?

Anſ. 46.765368.

Ex. 4. Required the content of a cylinder, whoſe length is 20 feet, and circumference 5½ feet.

Anſ. 48.1459.

Ex. 5. What is the content of a round pillar, whoſe height is 16 feet, and diameter 2 feet 3 inches?

Anſ. 63.6174.

PROBLEM IV. To find the Convex Surface of a Cylinder.

Multiply the circumference by the height of the cylinder.

Note, The upright ſurface of any priſm is found in the ſame manner. And the ſolidity of a cylinder is ſound as the priſm in the laſt problem.

EXAMPLES.

1. What is the convex ſurface of a cylinder, whoſe length is 16 feet, and its diameter 2 feet 3 inches?

[128] [...]

See Fig. [...] Definition 5, p. 121.

Ex. 2. Required the convex ſurface of the cylinder, whoſe length is 20 feet, and its diameter 2 feet.

Anſ. 125.664.

Ex. 3. What is the convex ſurface of a cylinder, whoſe length is 18 feet 6 inches, and circumference 5 feet 4 inches?

Anſ. 98⅔.

PROBLEM V. To find the Convex Surface of a Right Cone.

Multiply the circumference of the baſe by the ſlant height, or length of the ſide, and half the product will be the ſurface.

EXAMPLES.

1. If the diameter of the baſe be AB 5 feet, and the ſide of the cone AC 18, required the convex ſurface.

[...]

[diagram]

[129]Ex. 2. What is the convex ſurface of a cone, whoſe ſide is 20, and the circumference of its baſe 9?

Anſ. 90.

Ex. 3. Required the convex ſurface of a cone, whoſe ſlant height is 50 feet, and the diameter of its baſe 8 feet 6 inches.

Anſ. 667.59.

PROBLEM VI. To find the Convex Surface of the Fruſtum of a Right Cone.

Multiply the ſum of the perimeters of the two ends, by the ſlant height or ſide of the fruſtum, and half the product will be the ſurface.

EXAMPLES.

1. If the circumferences of the two ends be 12.5 and 10.3, and the ſlant height AD 14, required the convex ſurface of the fruſtum ABCD.

[...]

[diagram]

Ex. 2. What is the convex ſurface of the fruſtum of a cone, the ſlant height of the fruſtum being 12.5, and the circumferences of the two ends 6 and 8.4?

Anſ. 90.

Ex. 3. Required the convex ſurface of the fruſtum of a cone, the ſide of the fruſtum being 10 feet 6 inches, and the circumferences of the two ends 2 feet 3 inches and 5 feet 4 inches.

Anſ. 39 13/16.

PROBLEM VII. To find the Solidity of a Cone, or any Pyramid.

[130]

Multiply the area of the baſe by the height, and ⅓ of the product will be the content.

EXAMPLES.

1. What is the ſolidity of a cone, whoſe height CD is 12½ feet, and the diameter AB of the baſe 2½?

Here 2½ × 2½=5/2 × 5/2=25/4=6¼

[...]

See fig. at Prob. v.

Ex. 2. What is the ſolid content of a pentagonal pyramid, its height being 12 feet, and each ſide of its baſe 2 feet?

[...]

See Fig. to Def. 6.

[131]Ex. 3. What is the content of a cone, its height be⯑ing 10½ feet, and the circumference of its baſe 9 feet?

Anſ. 22.56093.

Ex. 4. Required the content of a triangular pyramid, its height being 14 feet 6 inches, and the three ſides of its baſe 5, 6, 7.

Anſ. 71.0352.

Ex. 5. What is the content of a hexagonal pyramid, whole height is 6.4, and each ſide of its baſe 6 inches?

Anſ. 1.38564 feet.

PROBLEM VIII. To find the Solidity of the Fruſtum of a Cone or any Pyramid.

RULES.

1. Add into one ſum, the areas of the two ends, and the mean proportional between them, or the ſquare root of their product; and ⅓ of that ſum will be a mean area; and which multiplied by the height of the fruſ⯑tum, will give the content.

2. When the ends are regular plane figures; the mean area will be found by multiplying ⅓ of the cor⯑reſponding tabular number belonging to the polygon, either by the ſum ariſing by adding together the ſquare of a ſide of each end and the product of the two ſides, or by the quotient of the difference of their cubes di⯑vided by their difference, or by the ſum ariſing from the ſquare of their half difference added to 3 times the ſquare of their half ſum.

3. And in the fruſtum of a cone, the mean area is found by multiplying .2618, or ⅓ of .7854, either by the ſum ariſing by adding together the ſquares of the two diameters and the product of the two, or by the difference of their cubes divided by their difference, or by the ſquare of half their difference added to 3 times the ſquare of their half ſum.

[130]

[...]

[131]

[...]

[132]Or, if the circumferences be uſed in like manner, inſtead of their diameters, the multiplier will be .02654.

EXAMPLES.

1. What is the content of the fruſtum of a cone, whoſe height is 20 inches, and the diameters of its two ends 28 and 20 inches?

[...]

See Fig. 10 prob. 11.

Ex. 2. Required the content of a pentagonal fruſtum, whoſe height is 5 feet, each ſide of the baſe 1 foot 6 inches, and each ſide of the leſs end 6 inches.

[133] [...]

[diagram]

Ex. 3. What is the ſolidity of the fruſtum of a cone, the altitude being 25, the circumference at the greater end being 20, and at the leſs end 10?

Anſ. 464.205.

Ex. 4. How many ſolid feet are in a piece of timber, whoſe baſes are ſquares, each ſide of the greater end being 15 inches, and each ſide of the leſs end 6 inches; alſo the length, or perpendicular altitude is 24 feet?

Anſ. 19½.

Ex. 5. To find the content of the fruſtum of a cone, the altitude being 18, the greateſt diameter 8, and the leaſt 4.

Anſ. 527.7888.

Ex. 6. What is the ſolidity of a hexagonal fruſtum, the height being 6 feet, the ſide of the greater end 18 inches, and of the leſs 12 inches?

Anſ. 24.681722.

PROBLEM IX. To find the Solidity of a Wedge.

[134]

To the length of the edge add twice the length of the back or baſe, and reſerve the ſum; multiply the height of the wedge by the breadth of the baſe; then multiply this product by the reſerved ſum, and 1/ [...] of the laſt product will be the content.

EXAMPLES.

1. What is the content in feet of a wedge, whoſe al⯑titude AP is 14 inches, its edge AB 21 inches, and the length of its baſe DE 32 inches, and its breadth CD 4½ inches?

[...]

[diagram]

Ex. 2. Required the content of a wedge, the length and breadth of the baſe being 70 and 30 inches, the length of the edge 110 inches, and the height 34.29016.

Anſ. 24.8048.

PROBLEM X. To find the Solidity of a Priſmoid. Definition.

A priſmoid differs only from the fruſtum of a pyra⯑mid, in not having its oppoſite ends ſimilar planes.

[135]RULE.

Add into one ſum, the areas of the two ends and 4 times the middle ſection parallel to them, and ⅙ of that ſum will be a mean area; and being multiplied by the height, will give the content.

Note. The length of the middle ſection is equal to half the ſum of the lengths of the two ends; and its breadth is equal to half the ſum of the breadths of the two ends.

EXAMPLES.

1. How many cubic feet are there in a ſtone, whoſe ends are rectangles, the length and breadth of the one being 14 and 12 inches; and the correſponding ſides of the other 6 and 4 inches; the perpendicular height being 30½ feet?

[...]

[diagram]

[136]Ex. 2. Required the content of a rectangular priſ⯑moid, whoſe greater end meaſures 12 inches by 8, the leſſer end 8 inches by 6, and the perpendicular height 5 feet.

Anſ. 2.453 feet.

Ex. 3. What is the content of a cart or waggon, whoſe inſide dimenſions are as follows: at the top the length and breadth 81½ and 55 inches, at the bottom the length and breadth 41 and 29½ inches, and the height 47¼ inches?

Anſ. 126340.59375 cubic inches.

PROBLEM XI. To find the Convex Surface of a Sphere or Globe.

Multiply its diameter by its circumference.

Note. In like manner the convex ſurface of any zone or ſegment is found, by multiplying its height by the whole circumference of the ſphere.

EXAMPLES.

1. Required the convex ſuperficies of a globe, whoſe diameter or axis is 24 inches.

[...]

[diagram]

Ex. 2. What is the convex ſurface of a ſphere, whoſe diameter is 7, and circumference 22?

Anſ. 151.

[137]Ex. 3. Required the area of the ſurface of the earth, its diameter, or axis, being 7957¾ miles, or its cir⯑cumference 25000 miles?

Anſ. 198943750 ſq. miles.

Ex. 4. The axis of a ſphere being 42 inches, what is the convex ſuperficies of the ſegment whoſe height is 9 inches?

Anſ. 1187.5248 inches.

Ex. 5. Required the convex ſurface of a ſpherical zone, whoſe breadth or height is 2 feet, and cut from a ſphere of 12½ feet diameter.

Anſ. 78.54 feet.

PROBLEM XII. To find the Solidity of a Sphere or Globe.

Multiply the cube of the axis by .5236.

EXAMPLES.

1. What is the ſolidity of the ſphere, whoſe axis is 12?

[...]

Ex. 2. To find the content of the ſphere, whoſe axis is 2 feet 8 inches.

Anſ. 9.9288 feet.

[138]Ex. 3. Required the ſolid content of the earth, ſup⯑poſing its circumference to be 25000 miles.

Anſ. 263858149120 miles.

PROBLEM XIII. To find the Solidity of a Spherical Segment.

To 3 times the ſquare of the radius of its baſe add the ſquare of its height; then multiply the ſum by the height, and the product again by .5236.

EXAMPLES.

1. Required the content of a ſpherical ſegment, its height being 4 inches, and the radius of its baſe 8.

[...]

[diagram]

Ex. 2. What is the ſolidity of the ſegment of a ſphere, whoſe height is 9, and the diameter of its baſe 20?

Anſ. 1799.6132.

Ex. 3. Required the content of the ſpherical ſegment, whoſe height is 2¼, and the diameter of its baſe 8.61684.

Anſ. 71.5695.

PROBLEM XIV. To find the Solidity of a Spherical Zone or Fruſium.

Add together the ſquare of the radius of each end and ⅓ of the ſquare of their diſtance, or the height; then multiply the ſum by the ſaid height, and the product again by 1.5708.

[139]EXAMPLES.

1. What is the ſolid content of a zone, whoſe greater diameter is 12 inches, the leſs 8, and the height 10 inches?

[...]

[diagram]

Ex. 2. Required the content of a zone, whoſe greater diameter is 12, leſs diameter 10, and he ght 2.

Anſ. 195.8264.

Ex. 3. What is the content of a middle zone, whoſe height is 8 feet, and the diameter of each end 6?

Anſ. 494.2784 feet.

PROBLEM XV. To find the Surface of a Circular Spindle.

Multiply the length AB of the ſpindle by the radius OC of the revolving are. Multiply alſo the ſaid arc ACB by the central diſtance OE, or diſtance between [140]the center of the ſpindle and center of the revolving arc. Subſtract the latter product from the former, and multiply double the remainder by 3.1416, or the ſingle remainder by 6.2832, for the ſurface.

Note. The ſame rule will ſerve for any ſegment or zone cut off perpendicular to the chord of the revolv⯑ing arc, only uſing the particular length of the part, and the part of the arc which deſcribes it, inſtead of the whole length and whole arc.

EXAMPLES.

1. Required the ſurface of a circular ſpindle, whoſe length AB is 40, and its thickneſs CD 30 inches.

Here, by the notes at pa. 92.

The chord [...], and [...], hence OE=OC−CE=20⅚−15=5⅚.

Alſo, by problem x, rule 2, [...]

[diagram]

[141]Then, by our rule, [...]

[diagram]

Ex. 2. What is the ſurface of a circular ſpindle, whoſe length is 24, and thickneſs in the middle 18?

Anſ. 1177.4485.

PROBLEM XVI. To find the Solidity of a Circular Spindle.

Multiply the central diſtance OE by half the area of the revolving ſegment ACBEA. Subtract the product from ⅓ of the cube of AE, half the length of the ſpindle. Then multiply the remainder by 12.5664, or 4 times 3.1416, for the whole content.

EXAMPLES.

1. Required the content of the circular ſpindle, whoſe length AB is 40, and middle diameter CD 30.

[See the laſt figure.]

[142]By the work of the laſt problem, [...]

Ex. 2. What is the ſolidity of a circular ſpirdle, whoſe length is 24, and middle diameter 18?

Anſ. 3739.93.

PROBLEM XVII. To find the Solidity of the Middle Fruſtum or Zone of a Circular Spindle.

From the ſquare of hall the length of the whole ſpindle, take 1/ [...] of the ſquare of half the length of the middle fruſtum, and multiply the remainder by the ſaid [143]half length of the fruſtum.—Multiply the central diſtance by the revolving area, which generates the middle fruſtum.—Subtract this latter product from the former; and the remainder multiplied by 6.2832, or 2 times 3.1416, will give the content.

EXAMPLES.

1. Required the ſolidity of the fruſtum, whoſe length mn is 40 inches, alſo its greateſt diameter EF is 32, and leaſt diameter AD or BC 24.

[diagram]

Draw DG parallel to mn, then we have DG=½ mn=20, and EG=½ EF−½ AD=4, chord DE²=DG² + GE²=416, and [...]the diameter of the generating circle, or the radius oE=52, hence 01=52−16=36 the central diſtance, and HI²=OH²−01²=52²−36²=1408, [...]

[144]GE ÷ 20E=4/104=1/26=.03846 a ver. ſine

[...]

Ex. 2. What is the content of the middle fruſtum of a circular ſpindle, whoſe length is 20, greateſt diameter 18, and leaſt diameter 8?

Anſ. 3657.1613.

PROBLEM XVIII. To find the Superficies or Solidity of any Regular Body.

[145]

1. Multiply the proper tabular area (taken from the following table) by the ſquare of the linear edge of the ſolid, for the ſuperficies.

2. Multiply the tabular ſolidity by the cube of the linear edge, for the ſolid content.

Surfaces and Solidities of Regular Bodies.
No. of Sides	Names	Surfac [...]s	Solidities
4	Tetraedron	1.73205	0.11785
6	Hexaedron	6.00000	1.00000
8	Octaedron	3.46410	0.47140
12	Dodecaedron	20.64573	7.66312
20	Icoſaedron	8.66025	2.18169

EXAMPLES.

1. If the linear edge or ſide of a tetraedron be 3, re⯑quired its ſurface and ſolidity.

The ſquare of 3 is 9, and the cube 27. Then, [...]

[diagram]

[146]Ex. 2. What is the ſuperficies and ſolidity of the hexaedron, whoſe linear ſide is 2?

Anſ. ſuperficies 24
Anſ. ſolidity 8

[diagram]

Ex. 3. Required the ſuperficies and ſolidity of the octaedron, whoſe linear ſide is 2?

Anſ. ſuperficies 13.85640
Anſ. ſolidity 3.77120

[diagram]

Ex. 4. What is the ſuperficies and ſolidity of the dodecaedron, whoſe linear ſide is 2?

Anſ. ſuperficies 82.58292
Anſ. ſolidity 61.30496

[diagram]

Ex. 5. Required the ſuperficies and ſolidity of the icoſaedron, whoſe linear ſide is 2?

Anſ. ſuperficies 34.64100
Anſ. ſolidity 17.45352

[diagram]

PROBLEM XIX. To find the Surface of a Cylindrical Ring.

[147]

This figure being only a cylinder bent round into a ring, its ſurface and ſolidity may be found as in the cylinder, namely, by multiplying the axis, or length of the cylinder, by the circumference of the ring, or ſection, for the ſurface; and by the area of a ſection, for the ſolidity. Or uſe the following rules.

For the ſurface.—To the thickneſs of the ring add the inner diameter; multiply this ſum by the thick⯑neſs, and the product again by 9.8696, or the ſquare of 3.1416.

EXAMPLES.

1. Required the ſuperficies of a ring, whoſe thick⯑neſs AB is 2 inches, and inner diameter BC is 12 inches.

[...]

[diagram]

Ex. 2. What is the ſurface of the ring whoſe inner diameter is 16, and thickneſs 4?

Anſ. 789.568.

PROBLEM XX. To find the Solidity of a Cylindrical Ring.

To the thickneſs of the ring add the inner diameter; then multiply the ſum by the ſquare of the thickneſs, and the product again by 2.4674, or ¼ of the ſquare of 3.1416, for the ſolidity.

[148]EXAMPLES.

1. Required the ſolidity of the ring whoſe thickneſs is 2 inches, and its inner diameter 12.

[...]

Ex. 2. What is the ſolidity of a cylindrical ring, whoſe thickneſs is 4, and inner diameter 16?

Anſ. 789.568.

OF THE CARPENTER's RULE.

[]

THIS inſtrument is otherwiſe called the ſliding rule; and it is much uſed in timber meaſuring and artificers works, both for taking the dimenſions, and caſting up the contents.

The inſtrument conſiſts of two equal pieces, each a foot in length, which are connected together by a folding joint.

One ſide or face of the rule, is divided into inches, and half-quarters, or eighths. On the ſame face alſo are ſeveral plane ſcales, divided into twelfth parts by diagonal lines; which are uſed in planning dimenſions that are taken in feet and inches. The edge of the rule is commonly divided decimally, or into tenths; namely each foot into 10 equal parts, and each of theſe into 10 parts again: ſo that by means of this laſt ſcale, dimenſions are taken in feet and tenths and hundredths, and multiplied as common decimal numbers, which is the beſt way.

On the one part of the other face are four lines, marked A, B, C, D, the two middle ones B and C being on a ſlider, which runs in a groove made in the ſtock. The ſame numbers ſerve for both theſe two middle lines, the one being above the numbers, and the other below.

Theſe four lines are logarithmic ones, and the three A, B, C, which are all equal to one another, are double [150]lines, as they proceed twice over from 1 to 10. The other or loweſt line D, is a ſingle one, proceeding from 4 to 40. It is alſo called the girt line, from its uſe in caſting up the contents of trees and timber; and upon it are marked WG at 17.15, and AG at 18.95, the wine and ale gage points, to make this inſtrument ſerve the purpoſe of a gaging rule.

Upon the other part of this face there is a table of the value of a load, or 50 cubic feet, of timber, at all prices, from 6 pence to 2 ſhillings a foot.

When 1 at the beginning of any line is accounted 1, then the 1 in the middle will be 10, and the 10 at the end 100; and when 1 at the beginning is accounted 10, then the 1 in the middle is 100, and the 10 at the end 1000; and ſo on. And all the ſmaller diviſions are altered proportionally.

PROBLEM I. To multiply Numbers together.

Suppoſe the two numbers 24 and 13.—Set 1 o [...] B to 13 on A; then againſt 24 on B ſtands 312 on A, which is the required product of the two given num⯑bers 24 and 13.

Note. In any operations, when a number runs be⯑yond the end of the line, ſeek it on the other radius, or other part of the line, that is, take the 10th part of it, or the 100th part of it, &c. and increaſe the reſult proportionally 10 fold, or 100 fold, &c.

In like manner the product of 35 and 19 is 665,
and the product of 270 and 54 is 14580.

PROBLEM II. To divide by the Sliding Rule.

As ſuppoſe to divide 312 by 24.—Set the diviſor [151]24 on B to the dividend 312 on A; then againſt 1 on B ſtands 13 the quotient on A.

Alſo 396 divided by 27 gives 14.6,
and 741 divided by 42 gives 17.6.

PROBLEM III. To Square any Number.

Suppoſe to ſquare 23.—Set 1 on B to 23 on A; then againſt 23 on B ſtands 529 on A, which is the ſquare of 23.

Or, by the other two lines, ſet 1 or 100 on C to the 10 on D, then againſt every number on D ſtands its ſquare in the line C.

So againſt 23 ſtands 529
againſt 20 ſtands 400
againſt 30 ſtands 900
and ſo on.

If the given number be hundreds, &c. reckon the 1 on D for 100, or 1000, &c. then the correſponding 1 on C is 10000, or 1000000, &c. So the ſquare of 230 is found to be 52900.

PROBLEM IV. To extract the Square Root.

Set 1 or 100, &c. on C to 1 or 10, &c. on D; then againſt every number found on C, ſtands its ſquare root on D.

So, againſt 529 ſtands its root 23
againſt 400 ſtands its root 20
againſt 900 ſtands its root 30
againſt 300 ſtands its root 17.3
and ſo on.

PROBLEM V. To find a Mean Proportional between two Numbers.

[152]

As ſuppoſe between 29 and 430.—Set the one num⯑ber 29 on C to the ſame on D; then againſt the other number 430 on C, ſtands their mean proportional 111 on D.

Alſo, the mean between 29 and 320 is 96.3,
and the mean between 71 and 274 is 139.

PROBLEM VI. To find a Third Proportional to two Numbers.

Suppoſe to 21 and 32.—Set the firſt 21 on B to the ſecond 32 on A; then againſt the ſecond 32 on B, ſtands 48.8 on A, which is the third proportional ſought.

Alſo the 3d proportional to 17 and 29 is 49.4,
and the 3d proportional to 73 and 14 is 2.5.

PROBLEM VII. To find a Fourth Proportional to three Numbers. Or, to perform the Rule-of-Three.

Suppoſe to find a fourth proportional to 12, 28, and 114.—Set the firſt term 12 on B to the 2d term 28 on A; then againſt the third term 114 on B, ſtands 266 on A, which is the 4th proportional ſought.

Alſo the 4th proportional to 6, 14, 29, is 67.6,
and the 4th proportional to 27, 20, 73, is 54.0.

TIMBER MEASURING.

[]

PROBLEM I. To find the Area, or Superficial Content of a Board or Plank.

MULTIPLY the length by the mean breadth.

Note. When the board is tapering, add the breadths at the two ends together, and take half the ſum for the mean breadth.

By the Sliding Rule.

Set 12 on B to the breadth in inches on A; then againſt the length in feet on B, is the content on A, in feet and fractional parts.

EXAMPLES.

1. What is the value of a plank, whoſe length is 12 feet 6 inches, and mean breadth 11 inches?

[...]

By the Sliding Rule.

As 12 B∶11 A∷12½ B∶11½ A.

That is, as 12 on B is to 11 on A, ſo is 12½ on B to 11½ on A.

[154]Ex. 2. Required the content of a board, whoſe length is 11 feet 2 inches, and breadth 1 foot 10 inches.

Anſ. 20^f 5′ 8″.

Ex. 3. What is the value of a plank, which is 12 feet 9 inches long, and 1 foot 3 inches broad, at 2½d a foot?

Anſ. 3s 3¾d.

Ex. 4. Required the value of 5 oaken planks at 3d per foot, each of them being 17½ feet long; and their ſeveral breadths are as follows, namely, two of 13½ inches in the middle, one of 14½ inches in the middle, and the two remaining ones, each 18 inches at the broader end, and 11¼ at the narrower. Anſ. £ 1 5 8¼.

PROBLEM II. To find the Solid Content of Squared or Four-ſided Timber.

Multiply the mean breadth by the mean thickneſs, and the product again by the length, and the laſt pro⯑duct will give the content.

By the Sliding Rule.

C	D	D	C
As length∶12 or 10∷quarter girt∶ſolidity.

That is, as the length in feet on C, is to 12 on D when the quarter girt is in inches, or to 10 on D when it is in tenths of feet; ſo is the quarter girt on D, to the content on C.

Note 1. If the tree taper regularly from the one end to the other, either take the mean breadth and thick⯑neſs in the middle, or take the dimenſions at the two ends, and half their ſum will be the mean dimenſions.

2. If the piece do not taper regularly, but is une⯑qually thick in ſome parts and ſmall in others, take ſe⯑veral different dimenſions, add them all together, and divide their ſum by the number of them, for the mean dimenſions.

3. The quarter girt is a geometrical mean proportional between the mean breadth and thickneſs, that is the ſquare root of their product. Sometimes unſkilful mea⯑ſurers uſe the arithmetical mean inſtead of it, that it [155]half their ſum: but this is always attended with error, and the more ſo as the breadth and depth differ the more from each other.

EXAMPLES.

1. The length of a piece of timber is 18 feet 6 inches, the breadths at the greater and leſs end 1 foot 6 inches and 1 foot 3 inches, and the thickneſs at the greater and leſs end 1 foot 3 inches and 1 foot: required the ſolid content.

[...]

[156] By the Sliding Rule.

B	A	B	A
As 1∶	13½∷	16½∶	223, the mean ſquare.
C	D	C	D
As 1∶	1∷	223∶	14.9, quarter girt.
C	D	D	C
As 18½∶	12∷	14.9∶	28.6, the content.

Ex. 2. What is the content of the piece of timber, whoſe length is 24½ feet, and the mean breadth and thickneſs each 1.04 feet?

Anſ. 26½ feet.

Ex. 3. Required the content of a piece of timber, whoſe length is 20.38 feet, and its ends unequal ſquares, the ſide of the greater being 19⅛, and the ſide of the leſs 9⅞.

Anſ. 29.838 feet.

Ex. 4. Required the content of the piece of timber, whoſe length is 27.36 feet; at the greater end the breadth is 1.78, and thickneſs 1.23; and at the leſs end the breadth is 1.04, and thickneſs 0.91.

Anſ. 41.278 feet.

PROBLEM III. To find the Solidity of Round or Unſquared Timber.

Rule 1, or Common Rule.

Multiply the ſquare of the quarter girt, or of ¼ of the mean circumference, by the length, for the content.

By the Sliding Rule.

As the length upon C∶12 or 10 upon D∷ quarter girt, in 12ths or 10ths, on D∶ content on C.

Note. 1. When the tree is tapering, take the mean dimenſions as in the former problems, either by girting it in the middle, for the mean girt, or at the two ends, and take half the ſum of the two. But when the tree [157]is very irregular, divide it into ſeveral lengths, and find the content of each part ſeparately.

2. This rule, which is commonly uſed, gives the anſwer about ¼ leſs than the true quantity in the tree, or nearly what the quantity would be after the tree is hewed ſquare in the uſual way: ſo that it ſeems intend⯑ed to make an allowance for the ſquaring of the tree. When the true quantity is deſired, uſe the 2d rule, given below.

EXAMPLES.

1. A piece of round timber being 9 feet 6 inches long, and its mean quarter girt 42 inches; what is the content?

[...]

By the Sliding Rule.

C	D	D	C
As 9.5∶	10∷	35∶	116⅓
Or 9.5∶	12∷	42∶	116⅓

Ex. 2. The length of a tree is 24 feet, its girt at the thicker end 14 feet, and at the ſmaller end 2 feet; re⯑quired the content.

Anſ. 96 feet.

[158]Ex. 3. What is the content of a tree, whoſe mean girt is 3.15 feet, and length 14 feet 6 inches?

Anſ. 8.9929 feet.

Ex. 4. Required the content of a tree, whoſe length is 17¼ feet, and which girts in five different places as follows, namely, in the firſt place 9.43 feet, in the ſecond 7.92, in the third 6.15, in the fourth 4.74, and in the fifth 3.16.

Anſ. 42.6075.

RULE 2.

Multiply the ſquare of ⅕ of the mean girt by double the length, and the product will be the content, very near the truth.

By the Sliding Rule.

As the double length on C∶12 or 10 on D∷⅕ of the girt, in 12ths or 10ths, on D∶ content on C.

EXAMPLES.

1. What is the content of a tree, its length being 9 feet 6 inches, and its mean girt 14 feet?

[...]

[159]

By the Sliding Rule.
C	D	D	C
As 19∶	10∷	28∶	149.
Or 19∶	12∷	33 6/10∶	149.

Ex. 2. Required the content of a tree, which is 24 feet long, and mean girt 8 feet.

Anſ. 122.88 feet.

Ex. 3. The length of a tree is 14½ feet, and mean girt 3.15 feet; what is the content?

Anſ. 11.51 feet.

Ex. 4. The length of a tree is 17¼ feet, and its mean girt 6.28; what is the content?

Anſ. 54.4065 feet.

Other curious problems relating to the cutting of timber, ſo as to produce uncommon effects, may be found in my large Treatiſe on Menſuration; from which may be ſeen the abſurd and miſchievous conſequences at⯑tending the conſtant uſe of the firſt or common rule for meaſuring round timber in all caſes.

ARTIFICERS WORK.

[]

ARTIFICERS compute the contents of their works by ſeveral different meaſures.

As glazing and maſonry by the foot.

Painting, plaſtering, paving, &c. by the yard, of 9 ſquare feet.

Flooring, partitioning, roofing, tiling, &c. by the ſquare, of 100 ſquare feet.

And brickwork either by the yard of 9 ſquare feet, or by the perch, or ſquare, rod or pole, containing 272¼ ſquare feet, or 30¼ ſquare yards, being the ſquare of the rod or pole of 16½ feet or 5½ yards long.

As this number 272¼ is a troubleſome number to di⯑vide by, the ¼ is often omitted in practice, and the con⯑tent in feet divided only by the 272. But as this is not exact, it will be both better and eaſier to multiply the feet by 4, and then divide ſucceſſively by 9, 11, and 11. Alſo to divide ſquare yards by 30¼, firſt multiply them by 4, and then divide twice by 11.

All works, whether ſuperficial or ſolid, are computed by the rules proper to the figure of them, whether it be a triangle, or rectangle, a parallelopiped, or any other figure.

BRICKLAYERS WORK.

[]

BRICKWORK is eſtimated at the rate of a brick and a half thick; and if a wall be more or leſs than this ſtandard thickneſs, it muſt be reduced to it, as follows:

Multiply the ſuperficial content of the wall by the number of half bricks in the thickneſs, and divide the product by 3.

The dimenſions of a building are uſually taken by meaſuring half round on the outſide, and half round it on the inſide; the ſum of theſe two gives the compaſs of the wall, to be multiplied by the height for the con⯑tent of the materials.

Chimneys are by ſome meaſured as if they were ſolid, deducting only the vacuity from the hearth to the mantle, on account of the trouble of them.

And by others they are girt or meaſured round for their breadth, and the height of the ſtory is their height, taking the depth of the jambs for their thickneſs. And in this caſe no reduction is made for the vacuity from the floor to the mantle tree, becauſe of the gathering of the breaſt and wings, to make room for the hearth in the next ſtory.

To meaſure the chimney ſhafts, which appear above the building; girt them about with a line for the breadth, to multiply by their height. And account their thick⯑neſs half a brick more than it really is, in conſidera⯑tion of the plaſtering and ſcaffolding.

[162]All windows, doors, &c. are to be deducted out of the contents of the walls in which they are placed. But this deduction is made only with regard to mate⯑rials; for the whole meaſure is taken for workmanſhip, and that all outſide meaſure too, namely, meaſuring quite round the outſide of the building, being in conſi⯑deration of the trouble of the returns or angles. They have alſo ſome other allowances, ſuch as double mea⯑ſure for feathered gable ends, &c.

EXAMPLES.

1. How many yards and rods of ſtandard brickwork are in a wall whoſe length or compaſs is 57 feet 3 inches, and height 24 feet 6 inches; the walls being 2½ bricks or 5 half bricks thick?

[...]

[163] By the Sliding Rule.

B	A	B	A
As 1∶	24½∷	57¼∶	1403.

Ex. 2. Required the content of a wall 62 feet 6 inches long, and 14 feet 8 inches high, and 2½ bricks thick.

Anſ. 169.753 yards.

Ex. 3. A triangular gable is raiſed 17 1/ [...] feet high, on an end wall whoſe length is 24 feet 9 inches, the thick⯑neſs being 2 bricks: required the reduced content.

Anſ. 32.08⅓ yds.

Ex. 4. The end wall of a houſe is 28 feet 10 inches long, and 55 feet 8 inches high to the eaves; 20 feet high is 2½ bricks thick, other 20 feet high is 2 bricks thick, and the remaining 15 feet 8 inches is 1½ brick thick; above which is a triangular gable, which riſes 42 courſes of bricks, of which every 4 courſes make a foot. What is the whole content in ſtandard meaſure.

Anſ. 253.626 yards.

MASONS WORK.

[]

TO maſonry belongs all ſorts of ſtone work; and the meaſure made uſe of is a foot, either ſuperfi⯑cial or ſolid.

Walls, columns, blocks of ſtone or marble, &c. are meaſured by the cubic foot; and pavements, ſlabs, chimney-pieces, &c. by the ſuperficial or ſquare foot.

Cubic or ſolid meaſure is uſed for the materials, and ſquare meaſure for the workmanſhip.

In the ſolid meaſure, the true length, breadth, and thickneſs, are taken, and multiplied continually toge⯑ther. In the ſuperficial, there muſt be taken the length and breadth of every part of the projection, which is ſeen without the general upright face of the building.

EXAMPLES.

1. Required the ſolid content of a wall, 53 feet 6 inches long, 12 feet 3 inches high, and 2 feet thick.

[...]

[165] By the Sliding Rule.

B	A	B	A
1∶	53½∷	12¼∶	655
1∶	655∷	2∶	1310.

Ex. 2. What is the ſolid content of a wall, the length being 12 feet 3 inches, height 10 feet 9 inches, and 2 feet thick?

Anſ. 521.375 feet.

Ex. 3. Required the value of a marble ſlab, at 8s per foot; the length being 5 feet 7 inches, and breadth 1 foot 10 inches.

Anſ. £ 4 1 10½.

Ex. 4. In a chimney-piece, ſuppoſe the length of the mantle and ſlab, each 4f 6in

breadth of both together	3	2
length of each jamb	4	4
breadth of both together	1	9
Required the ſuperficial content

Anſ. 21f 10in.

CARPENTERS AND JOINERS WORK.

[]

TO this branch belongs all the wood-work of a houſe, ſuch as flooring, partitioning, roofing, &c.

Note. Large and plain articles are uſually meaſured by the ſquare foot or yard, &c. but inriched mouldings, and ſome other articles, are often eſtimated by running or lineal meaſure, and ſome things are rated by the piece.

In meaſuring of joiſts, it is to be obſerved, that only one of their dimenſions is the ſame with that of the floor; and the other will exceed the length of the room by the thickneſs of the wall, and ⅓ of the ſame, becauſe each end is let into the wall about ⅔ of its thickneſs.

No deductions are made for hearths, on account of the additional trouble and waſte of materials.

Partitions are meaſured from wall to wall for one di⯑menſion, and from floor to floor, as far as they extend, for the other.

No deduction is made for door-ways, on account of the trouble of framing them.

In meaſuring of joiners work, the ſtring is made to ply cloſe to every part of the work over which it paſſes.

[167] The meaſure of centering for cellars is found by mak⯑ing a ſtring paſs over the ſurface of the arch for the breadth, and taking the length of the cellar for the length; but in groin-centering, it is uſual to allow double meaſure, on account of their extraordinary trouble.

In roofing, the length of the houſe in the inſide, to⯑gether with ⅔ of the thickneſs of one gable, is to be conſidered as the length; and the breadth is equal to double the length of a ſtring which is ſtretched from the ridge down the rafter, and along the eaves-board; till it meets with the top of the wall.

For ſtair-caſes, take the breadth of all the ſteps, by making a line ply cloſe over them, from the top to the bottom, and multiply the length of this line by the length of a ſtep for the whole area.—By the length of a ſtep is meant the length of the front and the returns at the two ends, and by the breadth is to be underſtood the girt of its two upper ſurfaces, or the tread and riſer.

For the baluſtrade, take the whole length of the up⯑per part of the hand-rail, and girt over its end till it meet the top of the newel poſt, for the length; and twice the length of the baluſter upon the landing, with the girt of the hand-rail, for the breadth.

For wainſcotting, take the compaſs of the room for the length; and the height from the floor to the cieling, making the ſtring ply cloſe into all the mouldings, for the breadth.—Out of this muſt be made deductions for windows, doors, and chimneys, &c. but workmanſhip is counted for the whole, on account of the extraordi⯑nary trouble.

For doors, it is uſual to allow for their thickneſs, by adding it into both the dimenſions of length and breadth, and then multiply them together for the area.—If the door be pannelled on both ſides, take double its meaſure for the workmanſhip; but if one ſide only be pannelled, take the area and its half for the workmanſhip.—For the ſurrounding architrave, gird it about the outermoſt part [168]for its length; and meaſure over it, as far as it can be ſeen when the door is open, for the breadth.

Window-ſhutters, baſes, &c. are meaſured in the ſame manner.

In the meaſuring of roofing for workmanſhip alone, all holes for chimney ſhafts and ſky-lights are generally deducted.

But in meaſuring for work and materials, they com⯑monly meaſure in all ſky-lights, luthern-lights, and holes for the chimney ſhafts, on account of their trouble and waſte of materials.

EXAMPLES.

1. Required the content of a floor 48 feet 6 inches long, and 24 feet 3 inches broad.

[...]

Ex. 2. A floor being 36 feet 3 inches long, and 16 feet 6 inches broad, how many ſquares are in it?

Anſ. 5 ſq. 98⅛ feet.

Ex. 3. How many ſquares are there in 173 feet 10 inches in length, and 10 feet 7 inches height, of parti⯑tioning?

Anſ. 18.3972 ſquares.

Ex. 4. What coſt the roofing of a houſe at 10s 6d a ſquare; the length, within the walls, being 52 feet 8 inches, and the breadth 30 feet 6 inches; reckoning the roof 3/2 of the flat?

Anſ. £ 12 12 11¾.

[169]Ex. 5. To how much, at 6s per ſquare yard, amounts the wainſcotting of a room; the height, taking in the cornice and mouldings, being 12 feet 6 inches, and the whole compaſs 83 feet 8 inches; alſo the three window ſhutters are each 7 feet 8 inches by 3 feet 6 inches, and the door 7 feet by 3 feet 6 inches; the door and ſhut⯑ters, being worked on both ſides, are reckoned work and half work?

Anſ. £ 36 12 2½

SLATERS AND TILERS WORK.

IN theſe articles, the content of a roof is found by multiplying the length of the ridge by the girt over from eaves to eaves; making allowance in this girt for the double row of ſlates at the bottom, or for how much one row of ſlates or tiles is laid over another.

In angles formed in a roof, running from the ridge to the eaves, when the angle bends inwards, it is called a valley; but when outwards, it is called a hip. And in tiling and ſlating, it is common to add the length of the valley or hip to the content in feet.

Deductions are ſeldom made for chimney ſhafts or ſmall window holes.

EXAMPLES.

1. Required the content of a ſlated roof, the length being 45 feet 9 inches, and whole girt 34 feet 3 inches.

[170] [...]

Ex. 2. To how much amounts the tiling of a houſe, at 25s 6d per ſquare; the length being 43 feet 10 inches, and the breadth on the ſlat 27 feet 5 inches, alſo the eaves projecting 16 inches on each ſide?

Anſ. £ 24 9 5½.

PLASTERERS WORK.

PLASTERERS work is of two kinds, namely, ceiling, which is plaſtering upon laths; and ren⯑dering, which is plaſtering upon walls: which are meaſured ſeparately.

The contents are eſtimated either by the foot, or yard, or ſquare of 100 feet. Inriched mouldings, &c. are rated by running or lineal meaſure.

Deductions are to be made for chimneys, doors, windows, &c. But the windows are ſeldom deducted, as the plaſtered returns at the top and ſides are allowed to compenſate for the window opening.

[171]EXAMPLES.

1. How many yards contains the ceiling, which is 43 feet 3 inches long, and 25 feet 6 inches broad?

[...]

Ex. 2. To how much amounts the ceiling of a room, at 10d per yard; the length being 21 feet 8 inches, and the breadth 14 feet 10 inches?

Anſ. £ 1 9 8¾.

Ex. 3. The length of a room is 18 feet 6 inches, the breadth 12 feet 3 inches, and height 10 feet 6 inches; to how much amounts the ceiling and rendering, the former at 8d and the latter at 3d per yard; allowing for the door of 7 feet by 3 feet 8, and a fire place of 5 feet ſquare?

Anſ. £ 8 5 6½.

Ex. 4. Required the quantity of plaſtering in a room, the length being 14 feet 5 inches, breadth 13 feet 2 inches, and height 9 feet 3 inches to the under ſide of the cornice, which girts 8½ inches, and projects 5 inches from the wall on the upper part next the ceiling: de⯑ducting only for a door 7 feet by 4.

Anſ.	53^vd	5^t	3ⁱ	of rendering
	18	5	6	of ceiling
		39	0 1/1 ½	of cornice.

PAINTERS WORK.

[172]

PAINTERS work is computed in ſquare yards.

Every part is meaſured where the colour lies; and the meaſuring line is forced into all the mouldings and corners.

Windows are done at ſo much a piece. And it is uſual to allow double meaſure for carved mouldings, &c.

EXAMPLES.

1. How many yards of painting contains the room which is 65 feet 6 inches in compaſs, and 12 feet 4 inches high?

[...]

Ex. 2. The length of a room being 20 feet, its breadth 14 feet 6 inches, and height 10 feet 4 inches; how many yards of painting are in it, deducting a fire-place of 4 feet by 4 feet 4 inches, and two windows each 6 feet by 3 feet 2 inches?

Anſ. 73 2/2 [...] yards.

Ex. 3. What coſt the painting of a room, at 6d per yard; its length being 24 feet 6 inches, its breadth 16 feet 3 inches, and height 12 feet 9 inches; alſo the door is 7 feet by 3 feet 6, and the window ſhutters to two windows each 7 feet 9 by 3 feet 6, but the breaks of the windows themſelves are 8 feet 6 inches high, and 1 foot 3 inches deep: deducting the fire-place of 5 feet by 5 feet 6?

Anſ. £ 3 3 10 [...].

GLAZIERS WORK.

[173]

GLAZIERS take their dimenſions either in feet, inches and parts, or feet, tenths and hundredths. And compute their work in ſquare feet.

In taking the length and breadth of a window, the croſs bars between the ſquares are included. Alſo windows of round or oval forms are meaſured as ſquare, meaſuring them to their greateſt length and breadth, on account of the waſte in cutting the glaſs.

EXAMPLES.

1. How many ſquare feet contains the window which is 4.25 feet long, and 2.75 feet bread?

[...]

2. What will the glazing a triangular ſky-light come to, at 10d per foot; the baſe being 12 feet 6 inches, and the perpendicular height 6 feet 9 inches?

Anſ. £ 4 7 2¼.

3. There is a houſe with three tier of windows, three windows in each tier, their common breadth 3 feet 11 inches;

now the height of the firſt tier is	7^•	10^••
of the ſecond	6	8
of the third	5	4

Required the expence of glazing at 14d per foot.

Anſ. £ 13 11 10^r.

Ex. 4. Required the expence of glazing the windows of a houſe at 13d a foot; there being three ſtories, and three windows in each ſtory;

[174]

the height of the lower tier is	7^f	9^•
of the middle	6	6
of the upper	5	3¼
and of an oval window over the door	1	10½

The common breadth of all the windows being 3 feet 9 inches.

Anſ. £ 12 5 6.

PAVERS WORK.

PAVERS work is done by the ſquare yard. And the content is found by multiplying the length by the breadth.

EXAMPLES.

1. What coſt the paving a foot-path at 3s 4d a yard; the length being 35 feet 4 inches, and breadth 8 feet 3 inches?

[...]

[175]Ex. 2. What coſt the paving a court, at 3s 2d per yard; the length being 27 feet 10 inches, and the breadth 14 feet 9 inches?

Anſ. £ 7 [...]/ [...] 4 4½.

Ex. 3. What will be the expence of paving a rec⯑tangular court yard, whoſe length is 63 feet, and breadth 45 feet; in which there is laid a foot-path of 5 feet 3 inches broad, running the whole length, with broad ſtones, at 3s a yard; the reſt being paved with pebbles at 2s 6d a yard?

Anſ. £ 40 5 10½.

PLUMBERS WORK.

PLUMBERS work is rated at ſo much a pound, or elſe the hundred weight, of 112 pounds.

Sheet lead uſed in roofing, guttering, &c. is from 7 to 12lb to the ſquare foot. And a pipe of an inch bore is commonly 13 or 14lb to the yard in length.

EXAMPLES.

1. How much weighs the lead which is 39 feet 6 inches long, and 3 feet 3 inches broad, at 8½lb to the ſquare foot?

[176] [...]

Ex. 2. What coſt the covering and guttering a roof with lead, at 18s the cwt; the length of the roof being 43 feet, and breadth or girt over it 32 feet; the gut⯑tering 57 feet long, and 2 feet wide; the former 9.831lb, and the latter 7.373lb to the ſquare foot?

Anſ. £ 115 9 1½.

VAULTED AND ARCHED ROOFS.

ARCHED roofs are either vaults, domes, ſaloons, or groins.

Vaulted roofs are formed by arches ſpringing from the oppoſite walls, and meeting in a line at the top.

[177] Domes are made by arches ſpringing from a circular or polygonal baſe, and meeting in a point at the top.

Saloons are formed by arches connecting the ſide walls to a flat roof, or ceiling, in the middle.

Groins are formed by the interſection of vaults with each other.

Vaulted roofs are commonly of the three following ſorts:

1. Circular roofs, or thoſe whoſe arch is ſome part of the circumference of a circle.
2. Elliptical or oval roofs, or thoſe whoſe arch is an oval, or ſome part of the circumference of an ellipſis.
3. Gothic roofs, or thoſe which are formed by two circular arcs, ſtruck from different centers, and meet⯑ing in a point directly over the middle of the breadth, or ſpan of the arch.

PROBLEM I. To find the Surface of a Vaulted Roof.

Multiply the length of the arch by the length of the vault, and the product will be the ſuperficies.

Note. To find the length of the arch, make a line or ſtring ply cloſe to it, quite acroſs from ſide to ſide.

EXAMPLES.

1. Required the ſurface of a vaulted roof, the length of the arch being 31.2 feet, and the length of the vault 120 feet.

[...]

Ex. 2. How many ſquare yards are in the vaulted roof, whoſe arch is 42.4 feet, and the length of the vault 100 feet?

Anſ. 499.37 yds.

PROBLEM II. To find the Content of the Concavity of a Vaulted Roof.

[178]

Multiply the length of the vault by the area of one end, that is, by the area of a vertical tranſverſe ſection, for the content.

Note. When the arch is an oval, multiply the ſpan by the height, and the product by .7854 for the area.

EXAMPLES.

1. Required the content of the concavity of a ſemi⯑circular vaulted roof, the ſpan or diameter being 30 feet, and the length of the vault 150 feet.

[...]

Ex. 2. What is the content of the vacuity of an oval vault, whoſe ſpan is 30 feet, and height 12 feet; the length of the vault being 60 feet?

Anſ. 1694.64.

Ex. 3. Required the content of the vacuity of a go⯑thic vault, whoſe ſpan is 50 feet, the chord of each are 50 feet, and the diſtance of each arch from the middle of theſe chords 15 feet; alſo the length of the vault 20.

Anſ. 42988.8.

PROBLEM III. To find the Superficies of a Dome.

[179]

Find the area of the baſe, and double it; then ſay, as the radius of the baſe, is to the height of the dome, ſo is the double area of the baſe, to the ſuperficies.

Note. For the ſuperficies of a hemiſpherical dome, take the double area of the baſe only.

EXAMPLES.

1. To how much comes the painting of an octagonal ſpherical dome, at 8d per yard; each ſide of the baſe being 20 feet?

[...]

Ex. 2. Required the ſuperficies of a hexagonal ſpherical dome, each ſide of the baſe being 10 feet.

Anſ. 519.6152.

Ex. 3. What is the ſuperficies of a dome with a cir⯑cular baſe, whoſe circumference is 100 feet, and height 20 feet?

Anſ. 2000 feet.

PROBLEM IV. To find the Solid Content of a Dome.

[180]

Multiply the area of the baſe by ⅔ of the height.

EXAMPLES.

1. Required the ſolid content of an octagonal dome, each ſide of the baſe being 20 feet, and the height 21 feet.

[...]

Ex. 2. What is the ſolid content of a ſpherical dome, the diameter of whoſe circular baſe is 30 feet?

Anſ. 7068.6 feet.

PROBLEM V. To find the Superficies of a Saloon.

Find its breadth by applying a ſtring cloſe to it acroſs the ſurface. Find alſo its length by meaſuring along the middle of it, quite round the room.

Then multiply theſe two together for the ſurface.

EXAMPLE.

The girt acroſs the face of a ſaloon being 5 feet, and its mean compaſs about 100 feet, required the area or ſuperficies.

[...]

PROBLEM VI. To find the Solid Content of a Saloon.

[181]

Multiply the area of a tranſverſe ſection by the com⯑paſs taken around the middle part. Subtract this pro⯑duct from the whole vacuity of the room, ſuppoſing the walls to go upright all the height to the flat ceiling. And the difference will be the anſwer.

EXAMPLE.

If the height AB of the ſaloon be 3.2 feet, the chord ADC of its front 4.5, and the diſtance DE of its middle part from the arch be 9 inches; required the ſolidity, ſuppoſing the mean compaſs to be 50 feet.

[...]

[diagram]

[182]Again [...]

Then this taken from the whole upright ſpace, will leave the content of the vacuity contained within the room.

PROBLEM VII. To find the Concave Superficies of a Groin.

To the area of the baſe add 1/7 part of itſelf, for the ſuperficial content.

EXAMPLES.

1. What is the ſuperficial content of the groin arch, raiſed on a ſquare baſe of 15 feet on each ſide?

[183] [...]

Ex. 2. Required the ſuperficies of a groin arch, raiſed on a rectangular baſe, whoſe dimenſions are 20 feet by 16.

Anſ. 365 [...]/7;.

PROBLEM VIII. To find the Solid Content of a Groin Arch.

Multiply the area of the baſe by the height; from the product ſubtract 1/10 of itſelf, and the remainder will be the content of the vacuity.

EXAMPLES.

1. Required the content of the vacuity within a groin arch, ſpringing from the ſides of a ſquare baſe, each ſide of which is 16 feet.

[...]

[184]2. What is the content of the vacuity below an oval groin, the ſide of its ſquare baſe being 24 feet, and its height 8 feet?

Anſ. 4147⅓.

NOTES.

1. To find the ſolid content of the brick or ſtone work, which forms any arch or vault: Multiply the area of the baſe by the height, including the work over the top of the arch; and from the product ſubtract the content of the vacuity found by the foregoing pro⯑blems; and the remainder will be the content of the ſolid materials.

2. In groin arches however, it is uſual to take the whole as ſolid, without deducting the vacuity, on ac⯑count of the trouble and waſte of materials, attending the cutting and fitting them to the arch.

LAND SURVEYING.

[]

CHAPTER 1. Deſcription and Uſe of the Inſtruments.

1. OF THE CHAIN.

LAND is meaſured with a chain, called Gunter's chain, of 4 poles or 22 yards in length, which con⯑ſiſts of 100 equal links, each link being 22/100 of a yard or 66/100 of a foot, or 7.92 inches long, that is, nearly 8 inches or ⅔ of a foot.

An acre of land, is equal to 10 ſquare chains, that is, 10 chains in length and 1 chain in breadth. Or it is 220 x 22 or 4840 ſquare yards. Or it is 40 x 4 or 160 ſquare poles. Or it is 1000 x 100 or 100000 ſquare links. Theſe being all the ſame quantity.

Alſo, an acre is divided into 4 parts called roods, and a rood into 40 parts called perches, which are ſquare poles, or the ſquare of a pole of 5¼ yards long, or the ſquare of ¼ of a chain, or of 25 links, which is 625 ſquare links. So that the diviſions of land meaſure will be thus:

625 ſq. links=1 pole or perch
40 perches=1 rood
4 roods=1 acre

[186]The length of lines, meaſured with a chain, are ſet down in links as integers, every chain in length being 100 links; and not in chains and decimals. Therefore after the content is found, it will be in ſquare links; then cut off five of the figures on the right-hand for decimals, and the reſt will be acres Thoſe decimals are then multiplied by 4 for roods, and the decimals of theſe again by 40 for perches.

EXAMPLES.

Suppoſe the length of a rectangular piece of ground be 792 links, and its breadth 385; to find the area in acres, roods, and perches.

[...]

2. OF THE PLAIN TABLE.

This inſtrument conſiſts of a plain rectangular board of any convenient ſize, the center of which, when uſed, is fixed by means of ſcrews to a three-legged ſtand, hav⯑ing a ball and ſocket, or joint, at the top, by means of which, when the logs are fixed on the ground, the table is inclined in any direction.

To the table belongs,

[187]1. A frame of wood, made to fit round its edges, and to be taken off, for the convenience of putting a ſheet of paper upon the table. The one ſide of this frame is uſually divided into equal parts, for drawing lines acroſs the table, parallel or perpendicular to the ſides; and the other ſide of the frame is divided into 360 degrees from a center which is in the middle of the table; by means of which the table is to be uſed as a theodo⯑lite, &c.

2. A needle and compaſs ſcrewed into the ſide of the table, to point out the directions, and to be a check upon the ſights.

3. An index, which is a braſs two-foot ſcale, with either a ſmall teleſcope, or open ſights erected perpendi⯑cularly upon the ends. Theſe ſights, and one edge of the index are in the ſame plane, and that edge is called the fiducial edge of the index.

Before you uſe this inſtrument, take a ſheet of paper which will cover it, and wet it to make it expand; then ſpread it flat upon the table, preſſing down the frame upon the edges, to ſtretch it and keep it fixed there; and when the paper is become dry, it will, by con⯑tracting again, ſtretch itſelf ſmooth and flat from any cramps or unevenneſs. Upon this paper is to be drawn the plan or form of the thing meaſured.

In uſing this inſtrument, begin at any part of the ground you think the moſt proper, and make a point upon a convenient part of the paper or table, to repre⯑ſent that point of the ground; then ſix in that point one leg of the compaſſes, or a fine ſteel pin, and apply to it the fiducial edge of the index, moving it round till through the ſights you perceive ſome remarkable object, as the corner of a field, &c. and from the ſtation point draw a line with the point of the compaſſes along the [...]iducial edge of the index; then ſet another object or corner, and draw its line; do the ſame by another, and ſo on till as many objects are ſet as may be thought neceſſary. Then meaſure from your ſtation towards [188]as many of the objects as may be neceſſary, and no more, taking the requiſite offsets to corners or crooks in the hedges, &c and lay the meaſures down upon their reſpective lines upon the table. Then, at any conve⯑nient place, meaſured to, fix the table in the ſame po⯑ſition, and ſet the objects which appear from thence, &c. as before; and thus continue till your work is fi⯑niſhed, meaſuring ſuch lines as are neceſſary, and de⯑termining as many as you can by interſecting lines of direction drawn from different ſtations.

And in theſe operations obſerve the following parti⯑cular cautions and directions.

1. Let the lines upon which you make ſtarions be directed towards objects as far diſtant as poſſible; and when you have ſet any ſuch object, go round the table and look through the ſights from the other end of the index, to ſee if any other remarkable object lie directly oppoſite; if there be not ſuch an one, endeavour to find another forward object, ſuch as ſhall have a remark⯑able backward oppoſite one, and make uſe of it rather than the other; becauſe the back object will be of uſe in fixing the table in the original poſition either when you have meaſured too near to the forward object, or when it may be hid from your ſight at any neceſſary ſtation by intervening hedges, &c.

2. Let the ſaid lines, upon which the ſtations are taken, be purſued as far as you conveniently can; for that will be the means of preſerving more accuracy in the work.

3. At each ſtation it will be neceſſary to prove the truth of it; that is, whether the table be ſtraight in the line towards the object, and alſo whether the diſ⯑tance be rightly meaſured and laid down on the paper.— To know if the table be ſet down ſtraight in the line; lay the index upon the table in any manner, and more the table about till through the ſights you perceive either the fore or back object; then, without moving the table, go round it and look through the ſights by [189]the other end of the index, to ſee if the other object can be perceived; if it be, the table is in the line; if not, it muſt be ſhifted to one ſide, according to your judgment, till through the ſights both objects can be ſeen.—The aforeſaid operation only informs you if the ſtation be ſtraight in the line; but to know if it be in the right part of the line, that is, if the diſtance has been rightly laid down; fix the table in the original poſition, by laying the index along the ſtation line and turning the table about till the fore and back objects appear through the ſights, and then alſo will the needle point at the ſame degree as at firſt; then lay the index over the ſtation point and any other point on the paper repreſenting an object which can be ſeen from the ſtation; and if the ſaid object appear ſtraight through the ſights, the ſtation may be depended on as right; if not, the diſtance ſhould be examined and corrected till the object can be ſo ſeen. And for this very uſeful purpoſe, it is adviſeable to have ſome high object or two, which can be ſeen from the greateſt part of the ground, accurately laid down on the paper from the beginning of the ſurvey, to ſerve continually as proof objects.

When, from any ſtation, the fore and back objects cannot both be ſeen, the agreement of the needle with one of them may be depended on for placing the table ſtraight on the line, and for fixing it in the original poſition.

Of ſhifting the Paper on the Plain Table.

When one paper is full, and you have occaſion for more; draw a line in any manner through the fartheſt point of the laſt ſtation line, to which the work can be conveniently laid down; then take the ſheet off the table, and fix another on, drawing a line upon it, in a part the moſt convenient for the reſt of the work; then ſold or cut the old ſheet by the line drawn on it, apply the edge to the line on the new ſheet, and, as they lie in that poſition, continue the laſt ſtation line upon the new paper, placing upon it the reſt of the meaſure, be⯑ginning [190]at where the old ſheet left off. And ſo on from ſheet to ſheet.

When the work is done, and you would faſten all the ſheets together into one piece, or rough plan, the afore⯑ſaid lines are to be accurately joined together, as when the lines were transferred from the old ſheets to the new ones.

But it is to be noted, that if the ſaid joining lines, upon the old and new ſheet, have not the ſame inclina⯑tion to the ſide of the table, the needle will not point to the original degree when the table is rectified; and if the needle be required to reſpect ſtill the ſame degree of the compaſs, the eaſieſt way of drawing the lines in the ſame poſition, is to draw them both parallel to the ſame ſides of the table, by means of the equal diviſions mark⯑ed on the other two ſides.

3. OF THE THEODOLITE.

The theodolite is a brazen circular ring, divided in⯑to 360 degrees, and having an index with ſights, or a teleſcope, placed upon the center, about which the in⯑dex is moveable; alſo a compaſs fixed to the center, to point out courſes and check the ſights; the whole be⯑ing fixed by the center upon a ſtand of a convenient height for uſe.

In uſing this inſtrument, an exact account, or field-book, of all meaſures and things neceſſary to be re⯑marked in the plan, muſt be kept, from which to make out the plan upon your return home from the ground.

Begin at ſuch part of the ground, and meaſure in ſuch directions, as you judge moſt convenient; taking angles or directions to objects, and meaſuring ſuch diſtances as appear neceſſary, under the ſame reſtrictions as in the uſe of the plain table. And it is ſafeſt to ſix the theodolite in the original poſition at every ſtation by means of fore and back objects, and the compaſs, ex⯑actly as in uſing the plain table; regiſtering the num⯑ber of degrees cut off by the index when directed to [191]each object; and, at any ſtation, placing the index at the ſame degree as when the direction towards that ſta⯑tion was taken from the laſt preceding one, to fix the theodolite there in the original poſition, after the ſame manner as the plain table is fixed in the original poſi⯑tion, by laying its index along the line of the laſt di⯑rection.

The beſt method of laying down the aforeſaid lines of direction, is to deſcribe a pretty large circle, quarter it, and lay upon it the ſeveral numbers of degrees cut off by the index in each direction; then, by means of a parallel ruler, draw, from ſtation to ſtation, lines parallel to lines drawn from the center to the reſpective points in the circumference.

4. OF THE CROSS.

The croſs conſiſts of two pair of ſights ſet at right angles to each other, upon a ſtaff having a ſharp point at the bottom to ſtick in the ground.

The croſs is very uſeful to meaſure ſmall and crooked pieces of ground. The method is to meaſure a baſe or chief line, uſually in the longeſt direction of the piece from corner to corner; and while meaſuring it, finding the places where perpendiculars would fall up⯑on this line from the ſeveral corners and bends in the boundary of the piece, with the croſs, by fixing it, by trials, upon ſuch parts of the line as that through one pair of the ſights both ends of the line may appear, and through the other pair you can perceive the correſpond⯑ing bends or corners; and then meaſuring the lengths of the ſaid perpendiculars.

REMARKS.

Beſides the fore-mentioned inſtruments, which are moſt commonly uſed, there are ſome others; as the circumferentor, which reſembles the theodolite in ſhape and uſe; and the ſemi-circle, for taking angles, &c. But of all the inſtruments for meaſuring, the plain table is [192]certainly the beſt; not only becauſe it may be uſed as a theodolite or ſemi-circle, by turning uppermoſt that ſide of the frame which has the 360 degrees upon it; but becauſe it is, in its own proper uſe, by much the eaſieſt, ſafeſt, and moſt accurate for the purpoſe; for by planning every part immediately upon the ſpot, as ſoon as meaſured, there is not only ſaved a great deal of writing in the field-book, but every thing can alſo be planned more eaſily and accurately while it is in view, than it can afterwards from a field-book, in which many little things muſt be either neglected or miſtaken; and beſides, the opportunities which the plain table afford of correcting your work, or proving if it be right, at every ſtation, are ſuch advantages as can never be balanced by any other method. But although the plain table be the moſt generally uſeful inſtrument, it is not always ſo; there being many caſes in which ſometimes one inſtrument is the propereſt, and ſome⯑times another; nor is that ſurveyor maſter of his bu⯑ſineſs who cannot in any caſe diſtinguiſh which is the fitteſt inſtrument or method, and uſe it accordingly: nay ſometimes no inſtrument at all, but barely the chain itſelf, is the beſt method, particularly in regular open fields lying together; and even when you are uſing the plain table, it is often of advantage to meaſure ſuch large open parts with the chain only, and from thoſe meaſures lay them down upon the table.

The perambulator is uſed for meaſuring roads, and other great diſtances on level ground, and by the ſides of rivers. It has a wheel of 8¼ feet, or half a pole, in circumference, upon which the machine turns; and the diſtance meaſured, is pointed out by an index, which is moved round by clock work.

Levels, with teleſcople or other ſights, are uſed to find the level between place and place, or how much one place is higher or lower than another.

An offſet-ſtaff is a very uſeful and neceſſary inſtru⯑ment, for meaſuring the offſets and other ſhort diſtances. [193]It is 10 links in length, being divided and marked at each of the 10 links.

Ten ſmall arrows, or rods of iron or wood, are uſed to mark the end of every chain length, in meaſuring lines. And ſometimes pickets, or ſtaves with flags, are ſet up as marks or objects of direction.

Various ſcales are alſo uſed in protracting and mea⯑ſuring on the plan or paper; ſuch as plane ſcales, line of chords, protractor, compaſſes, reducing ſcale, pa⯑rallel and perpendicular rules, &c. Of plane ſcales, there ſhould be ſeveral ſizes, as a chain in 1 inch, a chain in ¾ of an inch, a chain in ½ an inch, &c. And of theſe, the beſt for uſe, are thoſe that are laid on the very edges of the ivory ſcale, to prick off diſtances by, without compaſſes.

THE FIELD BOOK.

In ſurveying with the plain table, a field-book is not uſed, as every thing is drawn on the table immediately when it is meaſured. But in ſurveying with the theo⯑dolite, or any other inſtrument, ſome ſort of a field-book muſt be uſed, to write down in it a regiſter or ac⯑count of all that is done and occurs relative to the ſurvey in hand.

This book every one contrives and rules as he thinks fitteſt for himſelf. The following is a ſpecimen of a form very generally uſed. It is ruled into 3 columns: the middle, or principal column, is for the ſtations, angles, bearings, diſtances meaſured, and thoſe on the right and left are for the offsets on the right and left, which are ſet againſt their correſponding diſtances in the middle column; as alſo for ſuch remarks as may oc⯑cur, and be proper to note in drawing the plan, &c.

Here ☉ 1 is the firſt ſtation, where the angle or bear⯑ing is 105° 25′. On the left, at 73 links in the diſtance or principal line, is an offset of 92; and at 610 an offset of 24 to a croſs hedge. On the right, at o, or the beginning, an offset 25 to the corner of the field; at 248 Brown's boundary hedge commences; at 610 [194]an offset 35; and at 954, the end of the firſt line, the o denotes its terminating in the hedge. And ſo on for the other ſtations.

Draw a line under the work, at the end of every ſta⯑tion line, to prevent confuſion.

Form of the Field-Book.
Offsets and Remarks on the left.	Stations, Bearings, and Diſtances.	Offsets and Remarks on the right.
	☉ 1
	105° 25′
	00	25 corner
92	73
	248	Brown's hedge
croſs a hedge 24	610	35
	954	00
	☉ 2
	53° 10′
	00	00
houſe corner 51	25	21
	120	20 a tree
34	734	40 a ſtyle
	☉ 3
	67° 20′
	61	35
a brook 30	248
	6 [...]9	16 a ſpring
foot path 16	810
croſs hedge 18	973	20 a pond

[195]But in ſmaller ſurveys and meaſurements, a very good way of ſetting down the work, is, to draw, by the eye on a piece of paper, a figure reſembling that which is to be meaſured; and ſo write the dimenſions, as they are found, againſt the correſponding parts of the figure. And this method may be practiſed to a conſiderable ex⯑tent, even in the larger ſurveys.

CHAPTER II. THE PRACTICE OF SURVEYING.

THIS part contains the ſeveral works proper to be done in the field, or the ways of meaſuring by all the inſtruments, and in all ſituations.

PROBLEM I. To Meaſure a Line or Diſtance.

To meaſure a line on the ground with the chain: Having provided a chain, with 10 ſmall arrows, or rods, to ſtick one into the ground, as a mark, at the end [...] every chain; two perſons take hold of the chain, one at each end of it, and all the 10 arrows are taken by one of them, who is to go foremoſt, and is called the l [...]der; the other being called the follower, for diſtinc⯑tion's ſake.

A picket, or ſtation-ſtaff, being ſet up in the direction of the line to be meaſured, if there do not appear ſome marks naturally in that direction; the follower ſtands at the beginning of the line, holding the ring at the end of the chain in his hand, while the leader drags for⯑ward [196]the chain by the other end of it, till it is ſtretched ſtraight, and laid or held level, and the leader directed, by the follower waving his hand, to the right or left, till the follower ſee him exactly in a line with the mark or direction to be meaſured to; there both of them ſtretch⯑ing the chain ſtraight, and ſtooping and holding it le⯑vel, the leader having the head of one of his arrows in the ſame hand by which he holds the end of the chain, let him there ſtick one of them down with it while he holds the chain ſtretched. This done, he leaves the arrow in the ground, as a mark for the fol⯑lower to come to, and advances another chain forward, being directed in his poſition by the follower, ſtanding at the arrow, as before; as alſo by himſelf now, and at every ſucceeding chain's length, by moving himſelf from ſide to ſide, till he brings the follower and the back mark into a line. Having then ſtretched the chain, and ſtuck down an arrow, as before, the fol⯑lower takes up his arrow, and they advance again in the ſame manner another chain-length. And thus they proceed till all the 10 arrows are employed, and are in the hands of the follower; and the leader, without an arrow, is arrived at the end of the 11th chain length. The follower then ſends or brings the 10 arrows to the leader, who puts one of them down at the end of his chain, and advances with the chain as before. And thus the arrows are changed from the one to the other at every 10 chains length, till the whole line is finiſhed; when the number of changes of the arrows ſhews the number of tens, to which the follower adds the arrows he holds in his hand, and the number of links of an other chain over to the mark or end of the line. So if there have been 3 changes of the arrows, and the fol⯑lower hold 6 arrows, and the end of the line cut off 45 links more, the whole length of the line is ſet down in links thus 3645.

PROBLEM II. To take Angles and Bearings.

[197]

Let B and C be two objects, or two pickets ſet up perpendi⯑cular, and let it be required to take their bearings, or the angle formed between them at any ſtation A.

[diagram]

1. With the Plain Table.

The table being covered with a paper, and fixed on its ſtand; plant it at the ſtation A, and fix a ſine p [...]n, or a point of the compaſſes in a proper point of the paper, to repreſent the point A: Cloſe by the ſide of this pin lay the fiducial edge of the index, and turn it about, ſtill touching the pin, till one object B can be ſeen through the ſights: then by the fiducial edge of the index draw a line. In the very ſame manner draw another line in the direction of the other object C. And it is done.

2. With the Theodolite, &c.

Direct the fixed ſights along one of the lines, as AB, by turning the inſtrument about till you ſee the mark B through theſe ſights; and there ſcrew the inſtrument faſt. Then turn the moveable index about till, through its ſights, you ſee the other mark C. Then the degrees cut by the index, upon the graduated limb or ring of the inſtrument, ſhews the quantity of the angle.

3. With the Magnetic Needle and Compaſs.

Turn the inſtrument, or compaſs, ſo, that the north end of the needle point to the flower-de-luce. Then direct the ſights to one mark as B, and note the degrees cut by the needle. Then direct the ſights to the other mark C, and note again the degrees cut by the needle. Then their ſum or difference, as the caſe is, will give the quantity of the angle BAC.

[198]4. By Meaſurement with the Chain, &c.

Meaſure one chain length, or any other length, along both directions, as to b and c. Then meaſure the diſtance bc, and it is done.—This is eaſily trans⯑ferred to paper, by making a triangle Abc with theſe three lengths, and then meaſuring the angle A as in Practical Geometry, prob. XI.

PROBLEM III. To Meaſure the Offsets.

Ahiklmn being a crooked hedge, or river, &c. From A meaſure in a ſtraight direction along the ſide of it to B. And in meaſuring along this line AB obſerve when you are directly oppoſite any bends or corners of the hedge, as at c, d, e, &c. and from thence meaſure the perpendicular offsets ch, di, &c. with the offset-ſtaff, if they are not very large, otherwiſe with the chain itſelf. And the work is done. And the regiſter, or field-book, may be as follows:

Offs. left.	Baſe line AB.
0	☉ A
ch 62	45 AC
di 84	220 Ad
ek 70	340 Ae
fl 98	510 Af
gm 57	634 Ag
hn 91	785 AB

[diagram]

Note. When the offsets are not very large, their places c, d, e, &c. on the baſe line, can be very well determined by the eye, eſpecially when aſſiſted by lay⯑ing down the offset-ſtaff in a croſs or perpendicular di⯑rection. But when theſe perpendiculars are very large, [199]find their poſitions by the croſs, or by the inſtrument which you happen to be uſing, in this manner: As you meaſure along AB, when you come about C, where you judge a perpendicular will ſtand, plant your inſtrument in the line, and turn the index till the marks A and B can be ſeen through both the ſights, looking both back⯑ward and forward; then look along the croſs ſights, or the croſs line on the index; and if it point directly to the corner or bend h, the place of C is right. Otherwiſe, move the inſtrument backward or forward on the line AB, till the croſs line points ſtraight to h. This being found, ſet down the diſtance meaſured from A to C; then meaſure the offset ch, and ſet it down oppoſite the former, and on the left hand ſide.

Then proceed forward in the line AB, till you arrive oppoſite another corner, and determine the place d of the perpendicular as before. And ſo on throughout the whole length.

PROBLEM IV. To Survey a Triangular Field ABC.

1. By the Chain.

AP	794
AB	1321
PC	826

[diagram]

Having ſet up marks at the corners, which is to be done in all caſes where there are not marks naturally; meaſure with the chain from A to P, where a perpendi⯑cular would fall from the angle C, and ſet up a mark at P, noting down the diſtance AP. Then complete the diſtance AB by meaſuring from P to B. Having ſet down this meaſure, return to P, and meaſure the per⯑pendicular PC. And thus, having the baſe and perpen⯑dicular, [200]the area from them is eaſily found. Or hav⯑ing the place P of the perpendicular, the triangle is eaſily conſtructed.

Or, meaſure all the three ſides with the chain, and note them down. From which the content is eaſily found, or the figure conſtructed.

2. By taking one or more of the Angles.

Meaſure two ſides AB, AC, and the angle A between them. Or meaſure one ſide AB, and the two adjacent angles A and B. From either of theſe ways the figure is eaſily planned; then by meaſuring the perpendicular CP on the plan, and multiplying it by half AB, you have the content.

PROBLEM V. To meaſure a Four-ſided Field.

1. By the Chain.

AE 214	210 DE
AF 362	306 BF
AC 592

[diagram]

Meaſure along either of the diagonals, as AC; and either the two perpendiculars DE, BF, as in the laſt pro⯑blem; or elſe the ſides AB, BC, CD, DA. From either of which the figure may be planned and computed as before directed.

Otherwiſe by the Chain.

AP 110	352 PC
AQ 745	595 QD
AB 1110

[diagram]

[201]Meaſure on the longeſt ſide, the diſtances AP, AQ, AB; and the perpendiculars PC, QD.

2. By taking one or more of the Angles.

Meaſure the diagonal AC (ſee the firſt fig. above), and the angles CAB, CAD, ACB, ACD.—Or meaſure the four ſides, and any one of the angles as BAD.

Thus	Or thus
AC 591	AB 486
CAB 37°20′	BC 394
CAD 41 15	CD 410
ACB 72 25	DA 462
ACD 54 40	BAD 78°35′

PROBLEM VI. To Survey any Field by the Chain only.

Having ſet up marks at the corners, where neceſſary, of the propoſed field ABCDEFG. Walk over the ground, and conſider how it can beſt be divided into triangles and trapeziums; and meaſure them ſeparately as in the laſt two problems. And in this way it will be proper to divide it into as few ſeparate triangles, and as many trapeziums as may be, by drawing diagonals from cor⯑ner to corner; and ſo as that all the perpendiculars may fall within the figure. Thus, the following figure is divided into the two trapeziums ABCG, GDEF, and the triangle GCD. Then, in the firſt, beginning at A, meaſure the diagonal AC, and the two perpendiculars Gm, Bn. Then the baſe GC, and the perpendicular Dq. Laſtly the diagonal DF, and the two perpendicu⯑lars PE, OG. All which meaſures write againſt the correſponding parts of a rough figure drawn to reſemble the figure to be ſurveyed, or ſet them down in any other form you chooſe.

[202]

Am 135	130 mG
An 410	180 nB
AC 550
Cq 152	230 qD
CG 440
FO 206	120 OG
FP 288	80 PE
FD 520

[diagram]

Or thus.

Meaſure all the ſides AB, BC, CD, DE, EF, FG, and GA; and the diagonals AC, CD, GD, DF.

Otherwiſe.

Many pieces of land may be very well ſurveyed, by meaſuring any baſe line, either within or without them, together with the perpendiculars let fall upon it from every corner of them. For they are by thoſe means di⯑vided into ſeveral triangles and trapezoids, all whoſe pa⯑rallel ſides are perpendicular to the baſe line; and the ſum of theſe triangles and trapeziums will be equal to the figure propoſed if the baſe line fall within it; if not, the ſum of the parts which are without being taken from the ſum of the whole which are both within and with⯑out, will leave the area of the figure propoſed.

In pieces that are not very large, it will be ſufficiently exact to find the points, in the baſe line, where the ſe⯑veral perpendiculars will fall, by means of the croſs, and from thence meaſuring to the corners for the lengths of the perpendiculars.—And it will be moſt convenient to draw the line ſo as that all the perpendi⯑culars may fall within the figure.

[203]Thus, in the following figure, beginning at A, and meaſuring along the line AG the diſtances and perpen⯑diculars, on the right and left, are as below.

Ab 315	350 bB
AC 440	70 CC
Ad 585	320 dD
Ae 610	50 eE
Af 990	470 fF
AG 1020	0

[diagram]

PROBLEM VII. To Survey any Field with the Plain Table.

1. From one Station.

[diagram]

2. From a Station within the Field.

When all the other parts cannot be ſeen from one angle, chooſe ſome place O within; or even without, if more conveni⯑ent, from whence the other parts can be ſeen. Plant the table at O, then fix it with the needle north, and mark the point O on it. Apply the index ſucceſſively to O, turning it round with the ſights to each angle A, B, C, D, E, drawing dry lines to them by the edge of the index, then meaſuring the diſtances OA, OE, &c. and laying them down upon thoſe lines. Laſtly, draw the boundaries AB, BC, CD, DE, EA.

[diagram]

3. By going round the Figure.

When the figure is a wood, or water, or from ſome other obſtruction you cannot meaſure lines acroſs it; be⯑gin at any point A, and meaſure round it, either within or without the figure, and draw the directions of all the ſides thus: Plant the table at A, turn it with the needle to the north or flower-de-luce, fix it, and mark the point A. Apply the index to A, turning it till you can ſee the point E, there draw a line; and then the point B, and there draw a line: then meaſure theſe lines, and lay them down from A to E and B. Next, move the table to B, lay the index along the line AB, and turn the table about till you can ſee the mark A, and ſcrew faſt the table; in which poſition alſo the needle will again point to the flower-de-luce, as it will [205]do indeed at every ſtation when the table is in the right poſition. Here turn the index about B till through the ſights you ſee the mark C; there draw a line, meaſure BC, and lay the diſtance upon that line after you have ſet down the table at C. Turn it then again into its proper poſition, and in like manner find the next line CD. And ſo on quite round by E to A again. Then the proof of the work will be the joining at A: for if the work is all right, the laſt direction EA on the ground, will paſs exactly through the point A on the paper; and the meaſured diſtance will alſo reach exactly to A. If theſe do not coincide, or nearly ſo, ſome error has been committed, and the work muſt be examined over again.

PROBLEM VIII. To Survey a Field with the Theodolite, &c.

1. From one Point or Station.

When all the angles can be ſeen from one point, as the angle E (firſt fig. to laſt prob.) place the inſtrument at o, and turn it about till, through the fixed ſights, you ſee the mark B, and there fix it. Then turn the moveable index about till the mark A is ſeen through the ſights, and note the degrees cut on the inſtrument. Next turn the index ſucceſſively to E and D, noting the degrees cut off at each; which gives all the angles BCA, BCE, BCD. Laſtly, meaſure the lines CB, CA, CE, CD; and enter the meaſures in a field-book, or rather againſt the correſponding parts of a rough figure drawn by gueſs to reſemble the field.

2. From a Point within or without.

Plant the inſtrument at o, (laſt fig.) and turn it about till the fixed ſights point to any object, as A; and there ſcrew it faſt. Then turn the moveable index round till the ſights point ſucceſſively to the other points E, D, C, B, noting the degrees cut off at each of them; which gives all the angles round the point O. Laſtly, meaſure the diſ⯑tances [206]OA, OB, OC, OD, OE, noting them down as be⯑fore, and the work is done.

3. By going round the Field.

By meaſuring round, either within or without the field, proceed thus. Having ſet up marks at B, C, &c. near the corners as uſual, plant the inſtru⯑ment at any point A, and turn it till the ſixed index be in the direction AB, and there ſcrew it faſt: then turn the moveable index to the direction AF; and the degrees cut off will be the angle A. Meaſure the line AB, and plant the inſtrument at B, and there in the ſame manner obſerve the angle A. Then mea⯑ſure BC, and obſerve the angle C. Then meaſure the diſtance CD, and take the angle D. Then meaſure DE, and take the angle E. Then meaſure E [...], and take the angle F. And laſtly meaſure the diſtance FA.

[diagram]

To prove the work; add all the inward angles A, B, C, &c. together, and when the work is right, their ſum will be equal to twice as many right angles as the figure has ſides, wanting 4 right angles. And when there is an angle, as F, that bends inwards, and you meaſure the external angle, which is leſs than two right angles, ſubtract it from 4 right angles, or 360 degrees, to give the internal angle greater than a ſemicircle or 180 de⯑grees.

Otherwiſe.

Inſtead of obſerving the internal angles, you may take the external angles, formed without the ſigure by producing the ſides further out. And in this caſe, when the work is right, their ſum altogether will be[207]equal to 360 degrees. But when one of them, as F, runs inwards, ſubtract it from the ſum of the reſt, to leave 360 degrees.

PROBLEM IX. To Survey a Field with Crooked Hedges, &c.

With any of the inſtruments meaſure the lengths and poſitions of imaginary lines running as near the ſides of the field as you can; and in going along them meaſure the offsets in the manner before taught; and you will have the plan on the paper in uſing the plain table, drawing the crooked hedges through the ends of the off⯑ſets; but in ſurveying with the theodolite, or other inſtrument, ſet down the meaſures properly in a field-book, or memorandum-book, and plan them after re⯑turning from the field, by laying down all the lines and angles.

[diagram]

So, in ſurveying the piece ABCDE, ſet up marks a, b, c, d, dividing it into as few ſides as may be. Then begin at any ſtation a, and meaſure the lines ab, bc, cd, da, and take their poſitions, or the angles a, b, c, d; and in going along the lines meaſure all the offsets, as at m, n, o, p, &c. along every ſtation line.

[208]And this is done either within the field, or without, as may be moſt convenient. When there are obſtruc⯑tions within, as wood, water, hills, &c. then meaſure without, as in the figure here below.

[diagram]

PROBLEM X. To Survey a Field or any other Thing, by Two Stations.

This is performed by chooſing two ſtations, from whence all the marks and objects can be ſeen, then meaſuring the diſtance between the ſtations, and at each ſtation taking the angles formed by every object, from the ſtation line or diſtance.

The two ſtations may be taken either within the bounds, or in one of the ſides, or in the direction of two of the objects, or quite at a diſtance and without the bounds of the objects, or part to be ſurveyed.

In this manner, not only grounds may be ſurveyed, without even entering them, but a map may be taken of the principal parts of a country, or the chief places of a town, or any part of a river or coaſt ſurveyed, or any other inacceſſible objects; by taking two ſtations, on two towers, or two hills, or ſuch like.

[diagram]

When the plain table is uſed; plant it at one ſtation m, draw a line mn on it, along which lay the edge of the index, and turn the table about till the ſights point directly to the other ſtation; and there ſcrew it faſt. Then turn the ſights round m ſucceſſively to all the ob⯑jects ABC, &c. drawing a dry line by the edge of the index at each, as mA, mB, mC, &c. Then meaſure the diſtance to the other ſtation, there plant the table, and lay that diſtance down on the ſtation line from m to n. Next lay the index by the line nm, and turn the table about till the ſights point to the other ſtation m, and there ſcrew it faſt. Then direct the ſights ſuc⯑ceſſively to all the objects A, B, C, &c. as before, draw⯑ing lines each time, as nA, nB, nC, &c. and their in⯑terſection with the former lines will give the places of all the objects, or corners, A, B, C, &c.

When the theodolite, or any other inſtrument for taking angles, is uſed; proceed in the ſame way, mea⯑ſuring the ſtation diſtance mn, planting the inſtrument firſt at one ſtation, and then at another; then placing the fixed ſights in the direction m n, and directing the moveable ſights to every object, noting the degrees cut off at each time. Then, theſe obſervations being plan⯑ned, the interſections of the lines will give the objects as before.

[210]When all the objects, to be ſurveyed, cannot be ſeen from two ſtations; then three ſtations may be uſed, or four, or as many as is neceſſary; meaſuring always the diſtance from one ſtation to another; placing the inſtrument in the ſame poſition at every ſtation, by means deſcribed before; and from each ſtation obſerv⯑ing or ſetting every object that can be ſeen from it, by taking its direction or angular poſition, till every object be determined by the interſection of two or more lines of direction, the more the better. And thus may very extenſive ſurveys be taken, as of large commons, rivers, coaſts, countries, hilly grounds, and ſuch like.

PROBLEM XI. To Survey a Large Eſtate.

If the eſtate be very large, and contain a great num⯑ber of fields, it cannot well be done by ſurveying all the fields ſingly, and then putting them together; nor can it be done by taking all the angles and boundaries that incloſe it. For in theſe caſes, any ſmall errors will be ſo multiplied, as to render it very much diſ⯑torted.

1. Walk over the eſtate two or three times, in order to get a perfect idea of it, and till you can carry the map of it tolerably in your head. And to help your me⯑mory, draw an eye draught of it on paper, or at leaſt, of the principal parts of it, to guide you.

2. Chooſe two or more eminent places in the eſtate, for your ſtations, from whence you can ſee all the prin⯑cipal parts of it: and let theſe ſtations be as far diſtant from one another as poſſible; as the fewer ſtations you have to command the whole, the more exact your work will be: and they will be ſitter for your purpoſe, if theſe ſtation lines be in or near the boundaries of the ground, and eſpecially if two lines or more proceed from one ſtation.

[211]3. Take what angles, between the ſtations, you think neceſſary, and meaſure the diſtances from ſtation to ſtation, always in a right line: theſe things muſt be done, till you get as many angles and lines as are ſuffi⯑cient for determining all your points of ſtation. And in meaſuring any of theſe ſtation diſtances, mark accu⯑rately where theſe lines meet with any hedges, ditches, roads, lanes, paths, rivulets, &c. and where any re⯑markable object is placed, by meaſuring its diſtance from the ſtation line; and where a perpendicular from it cuts that line; and always mind, in any of theſe ob⯑ſervations, that you be in a right line, which you will know by taking backſight and foreſight, along your ſta⯑tion line. And thus as you go along any main ſtation line, take offsets to the ends of all hedges, and to any pond, houſe, mill, bridge, &c. omitting nothing that is remarkable. And all theſe things muſt be noted down: for theſe are your data, by which the places of ſuch objects are to be determined upon your plan. And be ſure to ſet marks up at the interſections of all hedges with the ſtation line, that you may know where to meaſure from, when you come to ſurvey theſe particular fields, which muſt immediately be done, as ſoon as you have meaſured that ſtation line, whilſt they are freſh in memory. By this means all your ſtation lines are to be meaſured, and the ſituation of all places adjoining to them determined, which is the firſt grand point to be ob⯑tained. It will be proper for you to lay down your work upon paper every night, when you go home, that you may ſee how you go on.

4. As to the inner parts of the eſtate, they muſt be de⯑termined in like manner, by new ſtation lines: for, after the main ſtations are determined, and every thing ad⯑joining to them, then the eſtate muſt be ſubdivided into two or three parts by new ſtation lines; taking inner ſtations at proper places, where you can have the beſt view. Meaſure theſe ſtation lines as you did the firſt, and all their interſections with hedges, and all offsets to ſuch objects as appear. Then you may proceed to [212]ſurvey the adjoining fields, by taking the angles that the ſides make with the ſtation line, at the interſections, and meaſuring the diſtances to each corner, from the in⯑terſections. For every ſtation line will be a baſis to all the future operations; the ſituation of all parts being entirely dependent upon them; and therefore they ſhould be taken as long as poſſible; and it is beſt for them to run along ſome of the hedges or boundaries of one or more fields, or to paſs through ſome of their angles. All things being determined for theſe ſtations, you muſt take more inner ſtations, and continue to di⯑vide and ſubdivide till at laſt you come to ſingle fields; repeating the ſame work for the inner ſtations, as for the outer ones, till all be done; and cloſe the work as often as you can, and in as few lines as poſſible. And that you may chooſe ſtations the moſt conveniently, ſo as to cauſe the leaſt labour, let the ſtation lines run as far as you can along ſome hedges, and through as many corners of the fields, and other remarkable points, as you can. And take notice how one field lies by ano⯑ther; that you may not miſplace them in the draught.

5. An eſtate may be ſo ſituated, that the whole can⯑not be ſurveyed together; becauſe one part of the eſtate cannot be ſeen from another. In this caſe, you may di⯑vide it into three or four parts, and ſurvey the parts ſe⯑parately, as if they were lands belonging to different perſons; and at laſt join them together.

6. As it is neceſſary to protract or lay down your work as you proceed in it, you muſt have a ſcale of a due length to do it by. To get ſuch a ſcale, you muſt meaſure the whole length of the eſtate in chains; then you muſt conſider how many inches long the map is to be; and from theſe you will know how many chains you muſt have in an inch; then make your ſcale, or chooſe one already made, accordingly.

7. The trees in every hedge row muſt be placed in their proper ſituation, which is ſoon done by the plain table; but may be done by the eye without an inſtru⯑ment; and being thus taken by gueſs, in a rough [213]draught, they will be exact enough, being only to look at; except it be ſuch as are at any remarkable places, as at the ends of hedges, at ſtiles, gates, &c. and theſe muſt be meaſured. But all this need not be done till the draught is finiſhed. And obſerve in all the hedges, what ſide the gutter or ditch is on, and to whom the fences belong.

8. When you have long ſtations, you ought to have a good inſtrument to take angles with, and the plain table may very properly be made uſe of, to take the ſe⯑veral ſmall internal parts, and ſuch as cannot be taken from the main ſtations: as it is a very quick and ready inſtrument.

PROBLEM XII. To Survey a County, or Large Tract of Land.

1. Chooſe two, three, or four eminent places for ſta⯑tions; ſuch as the tops of high hills or mountains, towers, or church ſteeples, which may be ſeen from one another; and from which moſt of the towns, and other places of note, may alſo be ſeen. And let them be as far diſtant from one another as poſſible. Upon theſe places raiſe beacons, or long poles, with flags of different colours flying at them; ſo as to be viſible from all the other ſtations.

2. At all the places, which you would ſet down in the map, plant long poles with flags at them of ſeveral colours, to diſtinguiſh the places from one another; fixing them upon the tops of church ſteeples, or the tops of houſes, or in the centers of leſſer towns.

But you need not have theſe marks at many places at once, as ſuppoſe half a ſcore at a time. For when the angles have been taken, at the two ſtations, to all theſe places, the marks may be moved to new ones; and ſo ſucceſſively to all the places you want. Theſe marks then being ſet up at a convenient number of places, and ſuch as may be ſeen from both ſtations; go to one of theſe ſtations, and with an inſtrument to take [214]angles, ſtanding at that ſtation, take all the angles be⯑tween the other ſtation, and each of theſe marks, ob⯑ſerving which is blue, which red, &c. and which hand they lie on; and ſet all down with their colours. Then go to the other ſtation, and take all the angles between the firſt ſtation, and each of the former marks, and ſet them down with the others, each againſt his fellow with the ſame colour. You may, if you can, alſo take the angles at ſome third ſtation, which may ſerve to prove the work, if the three lines interſect in that point, where any mark ſtands. The marks muſt ſtand till the obſervations are finiſhed at both ſtations; and then they muſt be taken down, and ſet up at freſh places. And the ſame operations muſt be performed, at both ſtations, for theſe freſh places; and the like for others. Your inſtrument for taking angles muſt be an exceeding good one, made on purpoſe with teleſcopic ſights; and of three, four, or five feet radius. A circumferentor is reckoned a good inſtrument for this purpoſe.

3. And though it is not abſolutely neceſſary to mea⯑ſure any diſtance, becauſe any ſtationary line being laid down from any ſcale, all the other lines will be propor⯑tional to it; yet it is better to meaſure ſome of the lines, to aſcertain the diſtances of places in miles; and to know how many geometrical miles there are in any length; and from thence to make a ſcale to meaſure any diſtance in miles. In meaſuring any diſtance, it will not be exact-enough to go along the high roads; by rea⯑ſon of their turnings and windings, and hardly ev [...] lying in a right line between the ſtations, which muſt cauſe infinite reductions, and create endleſs trouble to make it a right line; for which reaſon it can never be exact. But a better way is to meaſure in a right line with a chain, between ſtation and ſtation, over hills and dales or level fields, and all obſtacles. Only in caſe of water, woods, towns, rocks, banks, &c. where one cannot paſs; ſuch parts of the line muſt be meaſured by the methods of inacceſſible diſtances; and beſides, allow⯑ing for aſcents and deſcents, when we meet with them. [215]And a good compaſs that ſhews the bearing of the two ſtations, will always direct you to go ſtraight, when you do not ſee the two ſtations; and in your progreſs, if you can go ſtraight, you may take offsets to any re⯑markable places, likewiſe note the interſection of your ſtationary line with all roads, rivers, &c.

4. And from all your ſtations, and in your whole pro⯑greſs, be very particular in obſerving ſea coaſts, river mouths, towns, caſtles, houſes, churches, windmills, watermills, trees, rocks, ſands, roads, bridges, fords, ferries, woods, hills, mountains, rills, brooks, parks, beacons, ſluices, floodgates, looks, &c. and in general all things that are remarkable.

5. After you have done with your firſt and main ſtation lines, which command the whole county; you muſt then take inner ſtations, at ſome places already determined; which will divide the whole into ſeveral partitions: and from theſe ſtations you muſt determine the places of as many of the remaining towns as you can. And if any remain in that part, you muſt take more ſtations, at ſome places already determined; from which you may determine the reſt. And thus we muſt go through all the parts of the county, taking ſtation after ſtation, till we have determined all we want. And in general the ſtation diſtances muſt always paſs through ſuch remarkable points as have been determined before, by the former ſtations.

6. Laſtly, the poſition of the ſtation line you mea⯑ſure, or the point of the compaſs it lies on, muſt be de⯑termined by aſtronomical obſervation. Hang up a thread and plummet in the ſun, over ſome part of the ſtation line, and obſerve when the ſhadow runs along that line, and at that moment take the ſun's altitude; then having his declination, and the latitude, the azimuth will be found by ſpherical trigonometry. And the azimuth is the angle the ſtation line makes with the meridian; and therefore a meridian may eaſily be drawn through the map. Or a meridian may be drawn through it by hanging up two threads in a line with the [216]pole ſtar, when he is juſt north, which may be known from aſtronomical tables. Or thus; obſerve the ſtar Alioth, or that in the rump of the great bear, be⯑ing that next the ſquare; or elſe Caſſiopeia's hip; I ſay, obſerve by a line and plummet when either of theſe ſtars and the pole ſtar come into a perpendicular; and at that time they are due north. Therefore two per⯑pendicular lines being fixed at that moment, towards theſe two ſtars, will give the poſition of the meridian.

PROBLEM XIII. To Survey a Town or City.

This may be done with any of the inſtruments for taking angles, but beſt of all with the plain table, where every minute part is drawn while in ſight. It is beſt alſo to have a chain of 50 feet long, divided into 50 links, and an offset-ſtaff of 10 feet long.

Begin at the meeting of two or more of the principal ſtreets, through which you can have the longeſt pro⯑ſpects, to get the longeſt ſtation lines. There having fixed the inſtrument, draw lines of direction along thoſe ſtreets, uſing two men as marks, or poles ſet in wooden pedeſtals, or perhaps ſome remarkable places in the houſes at the further ends, as windows, doors, cor⯑ners, &c. Meaſure theſe lines with the chain, taking offsets with the ſtaff, at all corners of ſtreets, bendings, or windings, and to all remarkable things, as churches, markets, halls, colleges, eminent houſes, &c. Then remove the inſtrument to another ſtation along one of theſe lines; and there repeat the ſame proceſs as be⯑fore. And ſo on till the whole is finiſhed.

[diagram]

Thus, fix the inſtrument at A, and draw lines in the direction of all the ſtreets meeting there; and meaſure AB, noting the ſtreet on the left at m. At the ſecond ſtation B, draw the directions of the ſtreets meeting there; meaſure from B to C, noting the places of the ſtreets at n and o as you paſs by them. At the 3d ſtation C take the direction of all the ſtreets meeting there, and meaſure CD. At D do the ſame, and meaſure DE, noting the place of the croſs ſtreets at P. And in this manner go through all the principal ſtreets. This done, proceed to the ſmaller and intermediate ſtreets; and laſtly to the lanes, alleys, courts, yards, and every part that it may be thought proper to repreſent.

CHAPTER III. Of Planning, Caſting-up, and Dividing.

[218]

PROBLEM I. To Lay down the Plan of any Survey.

IF the ſurvey was taken with a plain table, you have a rough plan of it already on the paper which cover⯑ed the table. But if the ſurvey was with any other in⯑ſtrument, a plan of it is to be drawn from the meaſures that were taken in the ſurvey, and firſt of all a rough plan upon paper.

To do this, you muſt have a ſet of proper inſtruments, for laying down both lines and angles, &c. as ſcales of various ſizes, the more of them, and the more accurate, the better; ſcales of chords, protractors, perpendicular and parallel rulers, &c. Diagonal ſcales are beſt for the lines, becauſe they extend to three figures, or chains and links, which are hundredth parts of chains. But in uſing the diagonal ſcale, a pair of compaſſes muſt be employed to take off the lengths of the principal lines very accurately. But a ſcale with a thin edge divided, is much readier for laying down the perpendicular off⯑ſets to crooked hedges, and for marking the places of thoſe offsets upon the ſtation line; which is done at only one application of the edge of the ſcale to that line, and then pricking off all at once the diſtances along it. Angles are to be laid down either with a good ſcale of chords, which is perhaps the moſt accurate way; or with a large protractor, which is much readier when many angles are to be laid down at one point, as they are pricked off all at once round the edge of the pro⯑tractor.

Very particular directions for laying down all ſorts of figures cannot be neceſſary in this place, to any perſon who has learned practical geometry, and the conſtruc⯑tion of figures, and the uſe of his inſtruments. It may [219]therefore be ſufficient to obſerve, that all lines and angles muſt be laid down on the plan in the ſame order in which they were meaſured in the field, and in which they are written in the field-book; laying down firſt the angles for the poſition of lines, then the lengths of the lines, with the places of the offsets, and then the lengths of the offsets themſelves, all with dry or obſcure lines; then a black line drawn through the extremities of all the offsets, will be the hedge or bounding line of the field, &c. After the principal bounds and lines are laid down, and made to fit or cloſe properly, proceed next to the ſmaller objects, till you have entered every thing that ought to appear in the plan, as houſes, brooks, trees, hills, gates, ſtiles, roads, lanes, mills, bridges, wood⯑lands, &c. &c.

The north ſide of a map or plan is commonly placed uppermoſt, and a meridian ſomewhere drawn, with the compaſs or flower-de-luce pointing north. Alſo, in a va⯑cant part, a ſcale of equal parts or chains muſt be drawn, and the title of the map in conſpicuous characters, and embelliſhed with a compartment. All hills muſt be ſhadowed, to diſtinguiſh them in the map. Colour the hedges with different colours; repreſent hilly grounds by broken hills and valleys; draw ſingle dotted lines for foot-paths, and double ones for horſe or carriage roads. Write the name of each field and remarkable place within it, and, if you chooſe, its content in acres, roods, and perches.

In a very large eſtate, or a county, draw vertical and horizontal lines through the map, denoting the ſpaces between them by letters, placed at the top, and bottom, and ſides, for readily finding any field or other object, mentioned in a table.

In mapping counties, and eſtates that have uneven grounds of hills and valleys, reduce all oblique lines, meaſured up hill and down hill, to horizontal ſtraight lines, if that was not done during the ſurvey, before they were entered in the field-book, by making a pro⯑per allowance to ſhorten them. For which purpoſe [220]there is commonly a ſmall table engraven on ſome of the inſtruments for ſurveying.

PROBLEM II. To Caſt up the Contents of Fields.

1. Compute the contents of the figures, whether triangles, or trapeziums, &c. by the proper rules for the ſeveral figures laid down in meaſuring; multiplying the lengths by the breadths, both in links; the product is acres after you have cut off five figures on the right, for decimals; then bring theſe decimals to roods and perches, by multiplying firſt by 4, and then by 40. An example of which is given in the deſcription of the chain, page 187.

2. In ſmall and ſeparate pieces, it is uſual to caſt up their contents from the meaſures of the lines taken in ſurveying them, without making a correct plan of them.

Thus, in the triangle in prob. iv, page 199, where we had AP=794, and AB=1321

[...]

[221]Or the firſt example to prob. v, page 200, thus: [...]

Or the 2d example to the ſame prob. v, thus: [...]

[222]3. In pieces bounded by very crooked and winding hedges, meaſured by offsets, all the parts between the offsets are moſt accurately meaſured ſeparately as ſmall trapezoids. Thus, for the example to prob. 111, p. 198, where

AC 45	62 ch
Ad 220	84 di
Ae 340	70 ek
Af 510	98 fl
Ag 634	57 gm
AB 785	91 Bn

Then [...]

[223]4. Sometimes ſuch pieces as that above, are computed by finding a mean breadth, by dividing the ſum of the offsets by the number of them, accounting that for one of them where the boundary meets the ſtation line, as at A; then multiply the length AB by that mean breadth. Thus: [...]

5. But in larger pieces, and whole eſtates, conſiſting of many fields, it is the common practice to make a rough plan of the whole, and from it compute the con⯑tents quite independent of the meaſures of the lines and angles that were taken in ſurveying. For then new lines are drawn in the fields in the plan, ſo as to divide them into trapeziums and triangles, the baſes and per⯑pendiculars of which are meaſured on the plan by means of the ſeales from which it was drawn, and ſo multiplied together for the contents. In this way the work is very expeditiouſly done, and ſufficiently correct; for ſuch dimenſions are taken, as afford the moſt eaſy method of calculation; and, among a number of parts, thus taken and applied to a ſeale, it is likely that ſome of the parts will be taken a ſmall matter too little, and others too great; ſo that they will, upon the whole, in all probability, very nearly balance one another. After all the fields, and particular parts, are thus com⯑puted [224]ſeparately, and added all together into one ſum, calculate the whole eſtate independent of the fields, by dividing it into large and arbitrary triangles and trape⯑ziums, and add theſe alſo together. Then if this ſum be equal to the former, or nearly ſo, the work is right; but if the ſums have any conſiderable difference, it is wrong, and they muſt be examined, and recomputed, till they nearly agree.

A ſpecimen of dividing into one triangle, or one tra⯑pezium, which will do for moſt ſingle fields, may be ſeen in the examples to the laſt problem; and a ſpeci⯑men of dividing a large tract into ſeveral ſuch trapezi⯑ums and triangles, in prob. VI of chapter II of Sur⯑veying, page 201, where a piece is ſo divided, and its dimenſions taken and ſet down; and again at prob. VI of menſuration of ſurfaces, where the contents of the ſame piece are computed.

6. But the chief ſecret in caſting up, conſiſts in find⯑ing the contents of pieces bounded by curved, or very irregular lines, or in reducing ſuch crooked ſides of fields or boundaries to ſtraight lines, that ſhall incloſe the ſame or equal area with thoſe crooked ſides, and ſo obtain the area of the curved figure by means of the right-lined one, which will commonly be a trapezium. Now this reducing the crooked ſides to ſtraight ones, is very eaſily and accurately performed thus: Apply the ſtraight edge of a thin, clear piece of lanthorn-horn to the crooked line, which is to be reduced, in ſuch a man⯑ner, that the ſmall parts cut off from the crooked figure by it, may be equal to thoſe which are taken in: which equality of the parts included and excluded, you will preſently be able to judge of very nicely by a little practice: then with a pencil draw a line by the ſtraight edge of the horn. Do the ſame by the other ſides of the field or figure. So ſhall you have a ſtraight ſided figure equal to the curved one; the content of which, being computed as before directed, will be the content of the curved figure propoſed.

Or, inſtead of the ſtraight edge of the horn, a horſe-hair [225]may be applied acroſs the crooked ſides in the ſame manner; and the eaſieſt way of uſing the hair, is to ſtring a ſmall ſlender bow with it, either of wire, or cane, or whale-bone, or ſuch like ſlender ſpringy matter; for, the bow keeping it always ſtretched, it can be eaſily and neatly applied with one hand, while the other is at liberty to make two marks by the ſide of it, to draw the ſtraight line by.

EXAMPLE.

Thus, let it be required to find the contents of the ſame figure as in prob. IX of the laſt chapter, page 207, to a ſcale of 4 chains to an inch.

[diagram]

Draw the four dotted ſtraight lines AB, BC, CD, DA, cutting off equal quantities on both ſides of them, which they do as near as the eye can judge: ſo is the crooked figure reduced to an equivalent right-lined one of four ſides ABCD. Then draw the diagonal BD, which, by applying a proper ſcale to it, meaſures 1256. Alſo the perpendicular, or neareſt diſtance, from A to this diagonal, meaſures 456; and the diſtance of C from it, is 428.

[226]Then [...]

And thus the content of the trapezium, and conſe⯑quently of the irregular figure, to which it is equal, is eaſily found to be 5 acres, 2 roods, 8 perches.

PROBLEM III. To Transfer a Plan to another Paper, &c.

After the rough plan is completed, and a fair one is wanted; this may be done, either on paper or vellum, by any of the following methods.

FIRST METHOD.

Lay the rough plan upon the clean paper, and keep them always preſſed flat and cloſe together, by weights laid upon them. Then, with the point of a fine pin or pricker, prick through all the corners of the plan to be copied. Take them aſunder, and connect the pricked points, on the clean paper, with lines; and it is done. [227]This method is only to be practiſed in plans of ſuch figures as are ſmall and tolerably regular, or bounded by right lines.

SECOND METHOD.

Rub the back of the rough plan over with black lead powder; and lay the ſaid black part upon the clean pa⯑per, upon which the plan is to be copied, and in the pro⯑per poſition. Then with the blunt point of ſome hard ſubſtance, as braſs, or ſuch like, trace over the lines of the whole plan; preſſing the tracer ſo much as that the black lead under the lines may be transferred to the clean paper: after which take off the rough plan, and trace over the leaden marks with common ink, or with indian ink, &c.—Or, inſtead of blacking the rough plan, you may keep conſtantly a blacked paper to lay between the plans.

THIRD METHOD.

Another method of copying plans, is by means of ſquares. This is performed by dividing both ends and ſides of the plan, which is to be copied, into any con⯑venient number of equal parts, and connecting the correſponding points of diviſion with lines; which will divide the plan into a number of ſmall ſquares. Then divide the paper, upon which the plan is to be copied, into the ſame number of ſquares, each equal to the for⯑mer when the plan is to be copied of the ſame ſize, but greater or leſs than the others, in the proportion in which the plan is to be increaſed or diminiſhed, when of a different ſize. Laſtly, copy into the clean ſquares, the parts contained in the correſponding ſquares of the old plan; and you will have the copy either of the ſame ſize, or greater or leſs in any proportion.

FOURTH METHOD.

A fourth method is by the inſtrument called a penta⯑graph, which alſo copies the plan in any ſize required.

FIFTH METHOD.

[228]

But the neateſt method of any is this. Procure a copying frame or glaſs, made in this manner; namely, a large ſquare of the beſt window glaſs, ſet in a broad frame of wood, which can be raiſed up to any angle, when the lower ſide of it reſts on a table. Set this frame up to any angle before you, facing a ſtrong light; fix the old plan and clean paper together with ſeveral pins quite around, to keep them together, the clean paper being laid uppermoſt, and upon the face of the plan to be copied. Lay them, with the back of the old plan, upon the glaſs, namely, that part which you intend to begin at to copy firſt; and, by means of the light ſhining through the papers, you will very diſtinctly perceive every line of the plan through the clean pa⯑per. In this ſtate then trace all the lines on the paper with a pencil. Having drawn that part which covers the glaſs, ſlide another part over the glaſs, and copy it in the ſame manner. And then another part. And ſo on till the whole be copied.

Then, take them aſunder, and trace all the pencil⯑lines over with a ſine pen and Indian ink, or with com⯑mon ink.

And thus you may copy the fineſt plan, without in⯑juring it in the leaſt.

When the lines, &c. are copied upon the clean pa⯑per or vellum, the next buſineſs is to write ſuch names, remarks, or explanations as may be judged neceſſary; laying down the ſcale for taking the lengths of any parts, a flower-de-luce to point out the direction, and the proper title ornamented with a compartment; and illuſtrating or colouring every part in ſuch manner as ſhall ſeem moſt natural, ſuch as ſhading rivers or brooks with crooked lines, drawing the repreſentations of trees, b [...]ſhes, hills, woods, hedges, houſes, gates, roads, &c. in their proper places; running a ſingle dotted line for a foot path, and a double one for a car⯑riage road; and either repreſenting the baſes or the elevations of buildings, &c.

CONIC SECTIONS; AND THEIR SOLIDS.

[]

DEFINITIONS.

1. CONIC ſections are the plane figures formed by cutting a cone.

According to the different poſitions of the cutting plane there will ariſe five different figures or ſections.

2. If the cutting plane paſs through the vertex, and any part of the baſe, the ſection will be a triangle.

[diagram]

3. If the cone be cut parallel to the baſe, the ſection will be a cir⯑cle.

[diagram]

[230]4. The ſection is called an ellipſis, when the cone is cut ob⯑liquely through both ſides.

[diagram]

5. The ſection is a parabola, when the cone is cut parallel to one of its ſides.

[diagram]

6. The ſection is an hyperbola, when the cutting plane meets the oppoſite cone continued above the vertex, where it will make another ſection or hyperbola equal to the lower one.

[diagram]

7. The vertices of any ſection, are the points where the cutting plane meets the oppoſite ſides of the cone.

8. The tranſverſe axis is the line between the two vertices. And the middle point of the tranſverſe, is the center of the conic ſection.

[diagram]

[231]9. The conjugate axis, is a line drawn through the center, and perpendicular to the tranſverſe.

10. An ordinate is a line perpendicular to the axis.

11. An abſciſs is a part of the axis between the ordi⯑nate and the vertex.

12. A ſpheroid, or ellipſoid, is a ſolid generated by the revolution of an ellipſe about one of its axes. It is a prolate one, when the revolution is made about the tranſverſe axis; and oblate, when about the conjugate.

[diagram]

13. A conoid is a ſolid formed by the revolution of a parabola, or hyper⯑bola, about the axis. And is accord⯑ingly called parabolic, or hyperbolic.— The parabolic conoid is alſo called a paraboloid; and the hyperbolic conoid, an hyperboloid.

[diagram]

14. A ſpindle is formed by any of the three ſections revolving about a double ordinate, like the circular ſpindle.

15. A ſegment, of any of theſe figures, is a part cut off at the top, by a plane parallel to the baſe.

16. And a fruſtum is the part left next the baſe, af⯑ter the ſegment is cut off.

PROBLEM I. To deſcribe an Ellipſe.

Let TR be the tranſ⯑verſe, CO the conjugate, &c the center. With the radius TC and cen⯑ter C, deſcribe an arc cutting TR in the points F, f; which are called the two foci of the ellipſe.

[diagram]

[232]Aſſume any point P in the tranſverſe; then with the radii PT, PR, and centers F, f, deſcribe two arcs inter⯑ſecting in 1; which will be a point in the curve of the ellipſe.

And thus, by aſſuming a number of points P in the tranverſe, there will be found as many points in the curve as you pleaſe. Then with a ſteady hand draw the curve through all theſe points.

Otherwiſe, with a Thread.

Take a thread of the length of the tranſverſe TR, and faſten its ends with two pins in the foci F, f. Then ſtretch the thread, and it will reach to I in the curve: and by moving a pencil round, within the thread, keep⯑ing it always ſtretched, it will trace out the ell [...]pſe.

PROBLEM II. In an Ellipſe, to find the Tranſverſe, or Conjugate, or Ordinate, or Abſciſs; having the other three given.

CASE I. To find the Ordinate.

As the tranſverſe∶
Is to the conjugate∷
So is the mean proportional between the two abſciſſes∶
To the ordinate.

EXAMPLES.

1. In the ellipſe ADBC, the tranſverſe AB is 70, the conjugate CD is 50, and the abſciſſes AP 14, and PB 56; what is the ordinate PQ?

[233]Firſt [...]

[diagram]

Then 70∶50∷28∶20=PQ the ordinate.

Ex. 2. If the tranſverſe be 80, the conjugate 60, and an abſciſs 16; required the ordinate.

Anſ. 24.

CASE II. To find the Abſciſs.

From the ſquare of half the conjugate take the ſquare of the ordinate, and extract the ſquare root of the remainder. Then

As the conjugate∶
Is to the tranſverſe∷
So is that ſquare root∶
To half the difference of the abſciſſes.

Then add this half difference to half the tranſverſe, for the greater abſciſs; and ſubtract it for the leſs.

EXAMPLES.

1. The tranſverſe AB is 70, the conjugate CD is 50, and the ordinate PQ is 20; required the abſciſſes AP and PB.

[234]Firſt [...]

Ex. 2. What are the two abſciſſes to the ordinate 24, the axes being 80 and 60?

Anſ. 16 and 64.

CASE III. To find the Conjugate.

As the mean proportional between the abſciſſes∶

Is to the ordinate∷

So is the tranſverſe∶

To the conjugate.

Note. In the ſame manner the tranſverſe may be found from the conjugate, uſing here the abſciſſes of the conjugate, and their ordinate perpendicular to the con⯑jugate.

EXAMPLES.

1. The tranſverſe being 180, the ordinate 16, and the greater abſciſs 144; required the conjugate.

[235] [...]

Ex. 2. The tranſverſe being 70, the ordinate 20, and abſciſs 14; what is the conjugate?

Anſ. 50.

CASE IV. To find the Tranſverſe.

From the ſquare of half the conjugate ſubtract the ſquare of the ordinate, and extract the root of the re⯑mainder. Next add this root to the half conjugate, if the leſs abſciſs be given, but ſubtract it when the greater abſciſs is given, reſerving the ſum or difference. Then,

As the ſquare of the ordinate∶

Is to the rectangle of the abſciſs and conjugate∷

So is the reſerved ſum or difference∶

To the tranſverſe.

EXAMPLES.

1. If the conjugate be 50, the ordinate 20, and the leſs abſciſs 14; what is the tranſverſe?

[236] [...]

Ex. 2. The conjugate being 40, the ordinate 16, and the leſs abſciſs 36; required the tranſverſe.

Anſ. 180.

PROBLEM III. To find the Circumference of an Ellipſe.

Add the two axes together, and multiply the ſum by 1.5708, for the circumference, nearly.

EXAMPLES.

1. Required the circumference of the ellipſe whoſe two axes are 70 and 50.

[...]

[237]Ex. 2. What is the periphery of an ellipſe whoſe two axes are 24 and 20?

Anſ. 69.1152.

PROBLEM IV. To find the Area of an Ellipſe.

Multiply the tranſverſe by the conjugate, and that multiplied by .7854, will be the area.

Or multiply .7854 firſt by the one axe, and the product by the other.

EXAMPLES.

1. To find the area of the ellipſe whoſe two axes are 70 and 50.

[...]

Ex. 2. What is the area of the ellipſe whoſe two axes are 24 and 18?

Anſ. 339.2928.

PROBLEM V. To find the Area of an Elliptic Segment.

Divide the height of the ſegment by that axis of the ellipſe of which it is a part; and find in the table of circular ſegments at the end of the book, a circular ſegment having the ſame verſed ſine as this quotient. Then multiply continually together this ſegment and the two axes, for the area required.

[238]EXAMPLES.

1. What is the area of an elliptic ſegment RAQ, whoſe height AP is 20; the tranſverſe AB being 70, and the conjugate CD 50?

[...]

[diagram]

Ex. 2. What is the area of an elliptic ſegment cut off parallel to the ſhorter axis, the height being 10, and the axes 25 and 35?

Anſ. 162.0210.

Ex. 3. What is the area of the elliptic ſegment, cut off parallel to the longer axis, the height being 5, and the axes 25 and 35?

Anſ. 97.8458.

PROBLEM VI. To Deſcribe or Conſtruct a Parabola.

VP being an abſciſs, and PQ its given ordinate; biſect PQ in A, join AV, and draw AP perpendicular to it; and tranſ⯑fer PB to VF and VC in the axis produced. So ſhall r be what is called the focus.

Draw ſeveral double ordi⯑nates SRS, &c. Then with the radii CR, &c. and the cen⯑ter F, deſcribe arcs cutting the correſponding ordinates in the points s, &c.

[diagram]

[239]Then draw the curve through all the points s, &c.

PROBLEM VII. To find any Parabolic Abſciſs or Ordinate.

The abſciſſes are to each other as the ſquares of their ordinates; that is,

As an [...] abſciſs is to the ſquare of its ordinate,
So is any other abſciſs to the ſquare of its ordinate.
Or as the ſquare root of any abſciſs is to its ordinate,
So is the ſquare root of another abſciſs to its ordinate.

EXAMPLES.

1. The abſciſs VB is 9, and its or⯑dinate AB is 6; required the or⯑dinate DE whoſe abſciſs VE is 16.

Here √9 is 3, and √16 is 4.

Then 3∶6∷4∶8=DE required.

Or if the ordinate DE were given=8, to find its abſciſs VE.

Then 6²=36, and 8²=64, Hence 36∶64∷9∶16=VE required.

[diagram]

Ex. 2. If an abſciſs be 8, and its ordinate 10; re⯑quired the ordinate whoſe abſciſs is 18.

Anſ. 15.

Ex. 3. If an abſciſs be 18, and its ordinate 18; what is the abſciſs whoſe ordinate is 10?

Anſ. 8.

PROBLEM VIII. To find the length of a Parabolic Curve.

To the ſquare of the ordinate add 4/3 of the ſquare of the abſciſs, extract the ſquare root of the ſum, and double it for the length of the curve, cut off by the double ordinate, nearly.

[240]EXAMPLES.

1. The abſciſs VB being 2, and the ordinate AB 6, required the length of the curve AVC.

[...]

Ex. 2. What is the length of the parabolic curve whoſe abſciſs is 3, and ordinate 8?

Anſ. 17.435.

PROBLEM IX. To find the Area of a Parabola.

Multiply the baſe by the height, and [...]/ [...] of the pro⯑duct will be the area.

[241]EXAMPLES.

1. Required the area of the parabola AVCA, the ab⯑ſciſs VB being 2, and the ordinate AB 6.

[...]

Ex. 2. What is the area of a parabola whoſe abſciſs is 10, and ordinate 8?

Anſ. 106⅔.

PROBLEM X. To find the Area of a Parabolic Fruſtum.

Multiply the difference of the cubes of the two ends of the fruſtum by double its altitude, and divide the product by triple the difference of their ſquares, for the area.

EXAMPLES.

1. Required the area of the parabolic fruſtum ACFD, AC being 6, DF 10, and the altitude BE 4.

[...]

[242]Ex. 2. What is the area of the parabolic fruſtum, whoſe two ends are 6 and 10, and its altitude 3?

Anſ. 24½.

PROBLEM XI. To Conſtruct or Deſcribe an Hyperbola.

Let D be the cen⯑ter of the hyper⯑bola, or the middle of the tranſverſe AB; and BC perpen⯑dicular to AB, and equal to half the conjugate.

With center D, and radius DC, de⯑ſcribe an arc meet⯑ing AB produced in F and f, which are the two focus points of the hyperbola.

[diagram]

Then, aſſuming ſeveral points E in the tranſverſe produced, with the radii AF, BE, and centers f, r, de⯑ſcribe arcs interſecting in the ſeveral points G; through all which points draw the hyperbolic curve.

PROBLEM XII. In an Hyperbola to find the Tranſverſe, or Conjugate, or Ordinate, or Abſciſs.

CASE I. To find the Ordinate.

As the tranſverſe∶
Is to the conjugate∷
So is the mean propor. between the two abſciſſes∶
To the ordinate.

[243] Note. In the hyperbola, the leſs abſciſs added to the axis, gives the greater abſciſs.

EXAMPLES.

1. If the tranſverſe be 24, the conjugate 21, and the leſs abſciſs VB 8; what is the ordinate AB?

[...]

[diagram]

Then 24∶21∷16∶14=AB required.

Ex. 2. The tranſverſe being 60, the conjugate 36, and the leſs abſciſs 20; required the ordinate. Anſ. 24.

CASE II. To find the Abſciſs.

To the ſquare of half the conjugate add the ſquare of the ordinate, and extract the ſquare root of the ſum. Then

As the conjugate∶
Is to the tranſverſe∷
So is that ſquare root∶
To half the ſum of the abſciſſes.

Then to this half ſum add half the tranſverſe, for the greater abſciſs; and ſubtract it for the leſs.

[244]EXAMPLES.

1. The tranſverſe being 24, and the conjugate 21; required the two abſciſſes to the ordinate AB 14.

[...]

Ex. 2. The tranſverſe being 60, the conjugate 36; required the two abſciſſes to the ordinate 24.

Anſ. 80 and 20.

CASE III. To find the Conjugate.

As the mean propor. between the abſciſſes∶
Is to the ordinate∷
So is the tranſverſe∶
To the conjugate.

[245]EXAMPLES.

1. The tranſverſe being 24, the leſs abſciſs VB 8, and [...], ordinate AB 14; what is the conjugate?

[...]

Ex. 2. What is the conjugate to the hyperbola, whoſe tranſverſe is 60, and ordinate 24, and the leſs ab⯑ſciſs 20?

Anſ. 36.

CASE IV. To find the Tranſverſe.

To the ſquare of half the conjugate add the ſquare of the ordinate, and extract the ſquare root of the ſum.

Next, to this root add the half conjugate when the leſs abſciſs is uſed, but ſubtract it when the greater abſciſs is uſed; reſerving the ſum or difference. Then

As the ſquare of the ordinate∶
Is to the product of the abſciſs and conjugate∷
So is the reſerved ſum or difference∶
To the tranſverſe.

EXAMPLES.

1. The leſs abſciſs VB being 8, and its ordinate AB 14; required the tranſverſe to the conjugate 21.

[246] [...]

Ex. 2. What is the tranſverſe of the hyperbola whoſe conjugate is 36; the leſs abſciſs being 20, and its ordinate 24?

Anſ. 60.

PROBLEM XIII. To find the Length of an Hyperbolic Curve.

1. To 21 times the ſquare of the conjugate, add 9 times the ſquare of the tranſverſe; and to the ſame 21 times the ſquare of the conjugate, add 19 times the ſquare of the tranſverſe; and multiply each ſum by the abſciſs.

2. To each of theſe two products add 15 times the product of the tranſverſe and ſquare of the conjugate.

3. Then as the leſs ſum is to the greater, ſo is the double ordinate to the length of the curve, nearly.

EXAMPLES.

1. Required the length of the curve AVC to the ab⯑ſciſs VB 20 and ordinate AB 24; the two axes being 60 and 36.

[247] [...]

Then 2358720∶3078720∷48∶62.6520 the whole curve [...]

[248]Ex. 2. What is the length of the whole curve to the ordinate 10, the tranſverſe and conjugate axes being 80 and 60?

Anſ. 20.601.

PROBLEM XIV. To find the Area of an Hyperbola.

1. To 5/7 of the abſciſs add the tranſverſe; multiply the ſum by the abſciſs; extract the ſquare root of the product.

2. Multiply the tranſverſe by the abſciſs, and ex⯑tract the root of that product alſo.

3. To 21 times the firſt root add 4 times the ſecond root; multiply the ſum by double the product of the conjugate and abſciſs; then divide by 75 times the tranſverſe, for the area nearly.

EXAMPLES.

1. Required the area of the hyperbola AVCA, whoſe abſciſs VB is 10, the tranſverſe and conjugate being 30 and 18.

[diagram]

[249] [...]

[250] [...]

Ex. 2. What is the area of the hyperbola to the ab⯑ſciſs 25, the two axes being 50 and 30?

Anſ. 805.090868.

PROBLEM XV. To find the Solidity of a Spheroid.

Square the revolving axis, multiply that ſquare by the fixed axis, and multiply the product by .5236 for the content.

EXAMPLES.

1. Required the ſolidity of the prolate ſpheroid [...], whoſe axes arc AB 50 and CD 30.

[...]

[diagram]

[251]Ex. 2. What is the content of an oblate ſpheroid, whoſe axes are 50 and 30?

Anſ. 39270.

Ex. 3. What is the ſolidity of a prolate ſpheroid, whoſe axes are 9 and 7?

Anſ. 230.9076.

PROBLEM XVI. To find the Solidity of the Segment of a Spheroid.

CASE I. When the Baſe is Circular, or Parallel to the Revolving Axis.

Multiply the difference between triple the fixed axe and double the height of the ſegment, by the ſquare of the height, and the product again by .5236.

Then as the ſquare of the fixed axis is to the ſquare of the revolving axe, ſo is the laſt product to the content of the ſegment.

EXAMPLES.

1. Required the content of the ſegment of a prolate ſpheroid, the height AG being 5, the ſixed axe AB 50, and the revolving axe CD 30.

[...]

[diagram]

[252] [...]

Ex. 2. If the axes of a prolate ſpheroid be 10 and 6, required the area of the ſegment whoſe height is 1, and its baſe parallel to the revolving axe.

Anſ. 5.277888.

Ex. 3. The axes of an oblate ſpheroid being 50 and 30, what is the content of the ſegment, the height be⯑ing 6, and its baſe parallel to the revolving axe?

Anſ. 4084.07

CASE II. When the Baſe is Perpendicular to the Revolving Axe.

Multiply the difference between triple the revolving axe and double the height of the ſegment, by the ſquare of the height, and the product again by .5236.

Then as the revolving axe is to the ſixed axe,

So is the laſt product to the content.

EXAMPLES.

1. In the prolate ſpheroid ACBD, the ſixed axe AB is 50, the revolving axe CD 30; required the ſolidity of the ſegment CE [...], its height CG being 6.

[253] [...]

[diagram]

[...]

Ex. 2. In an oblate ſpheroid, whoſe axes are 50 and 30, required the content of the ſegment whoſe height is 5, its baſe being perpendicular to the revolving axe.

Anſ. 1099.56

PROBLEM XVII. To find the Content of the Middle Fruſtum of a Spheroid.

CASE I. When the Ends are Circular, or Parallel to the Revolving Axe.

To double the ſquare of the middle diameter, add the ſquare of the diameter of one end; multiply this ſum by the length of the fruſtum, and the product again by .2618 for the content.

[254]EXAMPLES.

1. Required the ſolidity of the middle fruſtum EGHF of a ſpheroid, the greateſt diameter CD being 30, the diameter of each end EF or GH 18, and the length AB 40.

[...]

[diagram]

Ex. 2. What is the ſolidity of the middle fruſtum of an oblate ſpheroid, having the diameter of each circu⯑lar end 40, the middle 50, and the length 18?

Anſ. 31101.84.

[255]CASE II. When the Ends are Elliptical, or Perpendicular to the Revolving Axe.

To double the product of the tranſverſe and conju⯑gate diameters of the middle ſection, add the product of the tranſverſe and conjugate of one end; multiply the ſum by the length of the fruſtum, and the product again by .2618 for the content.

EXAMPLES.

1. In the middle fruſtum EFHG of an oblate ſpheroid, the diameters of the middle or greateſt elliptic ſection AB are 50 and 30, and of one end EF or GH 40 and 24; required the content, the height IK being 9.

[...]

[diagram]

[256]Ex. 2. In the middle fruſtum of an oblate ſpheroid, the axes of the middle ellipſe are 50 and 30, and thoſe of each end are 30 and 18; required the content, the height being 40.

Anſ. 37070.88.

PROBLEM XVIII. To find the Solidity of an Elliptic Spindle.

RULE I.

1. Take the difference between 3 times the ſquare of the middle or greateſt diameter, and 4 times the ſquare of the diameter at ¼ of the length, or equally diſtant between the middle and one end; as alſo the difference between 3 times the greateſt diameter, and 4 times the ſaid middle diameter. Then the former difference di⯑vided by the latter, will be quadruple the central diſtance, or diſtance between the center of the ſpindle and center of the generating ellipſe.

2. Then find the axes of the ellipſe by problem 11, and the area of the ſegment which generated the ſpindle by problem v.

3. Divide 3 times that area by the length of the ſpindle; from the quotient ſubtract the greateſt diame⯑ter; and multiply the remainder by the quadruple cen⯑tral diſtance, before found.

4. Subtract this product from the ſquare of the greateſt diameter; and multiply the remainder by the length of the ſpindle, and again by .5236, for the ſo⯑lidity.

EXAMPLES.

1. Required the ſolidity of the elliptic ſpindle A [...], the length AB being 40, the greateſt diameter CD [...], and the diameter E [...], at ¼ of the length, 9.49546.

[257]1. For the Central Diſtance, and Axes of the Ellipſe.

[...]

[diagram]

Then as 12∶20 (or AG)∷30 (or CH)∶50=IK the tranſverſe.

2. For the Generating Elliptic Segment.

[...]

[258]3. For the Solidity of the Spindle.

[...]

Ex. 2. Required the ſolidity of the elliptic ſpindle, whoſe length is 20, the greateſt diameter 6, and the diameter at ¼ of the length 4.74773.

Anſ. 322.32.

RULE II.

To the ſquare of the greateſt diameter, add the ſquare of double the diameter at ¼ of the length; mul⯑tiply the ſum by the length, and the product again by .1309 for the ſolidity, very nearly.

[259] Note. This rule will alſo ſerve for any other ſolid formed by the revolution of any conic ſection.

EXAMPLE.

What is the ſolid content of the elliptic ſpindle whoſe length is 20, the greateſt diameter 6, and the diameter at ¼ of the length 4.74773?

[...]

PROBLEM XIX. To find the Solidity of a Fruſtum or Segment of an Elliptic Spindle.

Proceed as in the laſt rule, for this, or any other ſolid [260]formed by the revolution of a conic ſection about an axis, namely,

Add together the ſquares of the greateſt and leaſt diameters, and the ſquare of double the diameter in the middle between the two; multiply the ſum by the length, and the product again by .1309 for the ſolidity.

Note. For all ſuch ſolids, this rule is exact when the body is formed by the conic ſection, or a part of it, revolved about the axis of the ſection. And will always be very near the truth when the figure revolves about another line.

EXAMPLES.

1. Required the content of the middle fruſtum EG [...] of any ſpindle, the length AB being 40, the greateſt [...] middle diameter CD 32, the leaſt or diameter at eith [...] end EF or GH 24, and the diameter 1K, in the mid [...] between EF and CD, 30.157568.

[...]

[diagram]

[261]Ex. 2. What is the content of the ſegment of any ſpindle, the length being 10, the greateſt diameter 8, and the middle diameter 6?

Anſ. 272.272.

Ex. 3. Required the ſolidity of the fruſtum of an hyperbolic conoid, the height being 12, the greateſt diameter 10, the leaſt diameter 6, and the middle dia⯑meter 8½.

Anſ. 667.59.

Ex. 4. What is the content of the middle fruſtum of an hyperbolic ſpindle, the length being 20, the middle or greateſt diameter 16, the diameter at each end 12, and the diameter at ¼ of the length 14½?

Anſ. 3248.938.

PROBLEM XX. To find the Solidity of a Parabolic Conoid.

Multiply the ſquare of the diameter of the baſe by the altitude, and the product again by .3927, for the content.

EXAMPLES.

1. Required the ſolidity of the paraboloid whoſe height BD is 30, and the diameter of its baſe AC is 40.

[...]

[262]Ex. 2. What is the content of the parabolic conoid whoſe altitude is 42, and the diameter of its baſe 24?

Anſ. 9500.1984.

PROBLEM XXI. To find the Solidity of the Fruſtum of a Paraboloid.

Multiply the ſum of the ſquares of the diameters of the two ends by the height, and the product again by .3927, for the content.

EXAMPLES.

1. Required the content of the paraboloidal fruſtum ABCD, the diameter AB being 20, the diameter DC 40, and the height EF 22½.

[...]

[diagram]

Ex. 2. What is the content of the fruſtum of a para⯑boloid, the greateſt diameter being 30, the leaſt 24, and the altitude 9?

Anſ. 5216.6266.

PROBLEM XXII. To find the Solidity of a Parabolic Spindle.

Multiply the ſquare of the middle or greateſt diameter by the length, and the product again by .41888, for the content.

[263]EXAMPLES.

1. Required the content of the parabolic ſpindle ACBD, whoſe length AB is 40, and the greateſt diameter CD 16.

[...]

[diagram]

Ex. 2. What is the ſolidity of a parabolic ſpindle, whoſe length is 18, and its middle diameter 6 feet?

Anſ. 271.4336.

PROBLEM XXIII. To find the Solidity of the Middle Fruſtum of a Parabolic Spindle.

Add all together 8 times the ſquare of the greateſt diameter, 3 times the ſquare of the leaſt diameter, and 4 times the product of the two diameters; multiply the ſum by the length, and the product again by .05236 for the ſolidity.

[264]EXAMPLES.

1. Required the content of the fruſtum of a para⯑bolic ſpindle EGHF, the length AB being 20, the great⯑eſt diameter CD 16, and the leaſt diameter EF 12.

[...]

[diagram]

Ex. 2. What is the content of the fruſtum of a para⯑bolic ſpindle, whoſe length is 18, greateſt diameter 18, and leaſt diameter 10?

Anſ. 3404.23776.

Note. The ſolidities of the hyperboloid and hyper⯑bolic ſpindle, are to be found by rule 2 to prob. XVIII. And thoſe of their fruſtums by prob. XIX; where ſome examples of them are given.

OF GAUGING.

[]

THE buſineſs of caſk-gauging is commonly per⯑formed by two inſtruments, namely, the gauging or ſliding-rule, and the gauging or diagonal rod.

1. OF THE GAUGING RULE.

This inſtrument ſerves to compute the contents of caſks, &c. after the dimenſions have been taken. It is a ſquare rule, having various logarithmic lines on its four ſides or faces; and three ſliding pieces, running in grooves, in three of them.

Upon the firſt face are three lines, namely, two mark⯑ed A, B, for multiplying and dividing; and the third, MD, for malt depth, becauſe it ſerves to gauge malt. The middle one B is upon the ſlider, and is a kind of double line, being marked at both the edges of the ſlider, for applying it to both the lines A and MD. Theſe three lines are all of the ſame radius, or diſtance from 1 to 10, each containing twice the length of the radius. A and B are placed and numbered exactly alike, each beginning at 1, which may be either 1, or 10, or 100, &c. or .1, or .01, or .001, &c. but whatever it is, the [266]middle diviſion 10, will be 10 times as much, and the laſt diviſion 100 times as much. But 1 on the line MD is oppoſite 215, or more exactly 2150.4 on the other lines, which number 2150.4 denotes the cubic inches in a malt buſhel; and its diviſions numbered retrograde to thoſe of A and B. Upon theſe two lines are alſo ſeveral other marks and letters: thus, on the line A are M [...], for malt buſhel, at the number 2150.4; and A for ale, at 282, the cubic inches in an ale gallon; and upon the line B is W, for wine, at 231, the cubic inches in a wine gallon; alſo si, for ſquare inſcribed, at .707, the ſide of a ſquare inſcribed in a circle whoſe diameter is 1; se, for ſquare equal, at .886, the ſide of a ſquare which is equal to the ſame circle; and c, for circum⯑ference, at 3.1416, the circumference of the ſame circle.

Upon the ſecond face, or that oppoſite the firſt, are a ſlider and four lines, marked D, C, D, E, at one end, and root, ſquare, root, cube, at the other; the lines C and E containing reſpectively the ſquares and cubes of the oppoſite numbers on the lines D, D; the radius of D being double to that of A, B, C, and triple to that of E: ſo that whatever the firſt 1 on D denotes, the firſt on C is the ſquare of it, and the firſt on E the cube of it; ſo if D begin with 1, C and E will begin with 1; but if D begin with 10, C will begin with 100, and E with 1000; and ſo on. Upon the line C are marked [...] at .0796, for the area of the circle whoſe circumference is [...]; and 0 d at .7854 for the area of the circle whoſe diameter is 1. Alſo, upon the line D are WG, for wine gauge, at 17.15; and AG for ale gauge, at 18.95; and MR, for malt round, at 52.32; theſe three being the gauge points for round or circular meaſure, and are found by dividing the ſquare roots of 231, 282, and 2150.4 by the ſquare root of .7854: alſo MS, for malt ſquare, are marked at 46.37, the malt gauge point for ſquare mea⯑ſure being the ſquare root of 2150.4.

Upon the third face are three lines, one upon a ſlider marked N; and two on the ſtock, marked SS and SL,[267]for ſegment ſtanding and ſegment lying, which ſerve for ullaging ſtanding and lying caſks.

And upon the fourth, or oppoſite face, are a ſcale of inches, and three other ſcales, marked ſpheroid or 1ſt variety, 2d variety, 3d variety; the ſcale for the 4th, or conic variety, being on the inſide of the ſlider in the third face. The uſe of theſe lines is to find the mean diameters of caſks.

Beſides all thoſe lines, there are two others on the in⯑ſides of the two firſt ſliders, being continued from the one ſlider to the other. The one of theſe is a ſcale of inches, from 12½ to 36; and the other is a ſcale of ale gallons between the correſponding numbers .435 and 3.61; which form a table to ſhew, in ale gallons, the contents of all cylinders whoſe diameters are from 12½ to 36 inches, their common altitude being 1 inch.

The Uſe of the Gauging Rule.

PROBLEM I. To Multiply two Numbers, as 12 and 25.

Set 1 on B to either of the given numbers, as 12, on A; then againſt 25 on B ſtands 300 on A; which is the product.

PROBLEM II. To Divide one Number by another, as 300 by 25.

Set 1 on B to 25 on A; then againſt 300 on A ſtands 12 on B, for the quotient.

PROBLEM III. To find a Fourth Proportional; as to 8, 24, and 96.

Set 8 on B to 24 on A; then againſt 96 on B is 288 on A, the 4th proportional to 8, 24, 96.

PROBLEM IV. To Extract the Square Root, as of 225.

[268]

The firſt 1 on C ſtanding oppoſite the 1 on D on the ſtock, then againſt 225 on C ſtands its ſquare root 15 on D.

PROBLEM V. To Extract the Cube Root, as of 3375.

The line D on the ſlide being ſet ſtraight with E, then oppoſite 3375 on E ſtands its cube root 15 on D.

PROBLEM VI. To find a Mean Proportional, as between 4 and 9.

Set 4 on C to the ſame 4 on D; then againſt 9 on C ſtands the mean proportional 6 on D.

PROBLEM VII. To find Numbers in Duplicate Proportion. As to find a Number which ſhall be to 120, as the Square of 3 to the Square of 2.

Set 2 on D to 120 on C; then againſt 3 on D ſtands 270 on C, for the anſwer.

PROBLEM VIII. To find Numbers in Subduplicate Proportion. As to find a Number which ſhall be to 2 as the Root of 270 to the Root of 120.

Set 2 on D to 120 on C; then againſt 270 on C ſtands 3 on D, for the anſwer.

PROBLEM IX. To find Numbers in Triplicate Proportion. As to find a Number which ſhall be to 100 as the Cube of 36 is to the Cube of 40.

[269]

Set 40 on D to 100 on E; then againſt 36 on D ſtands 72.9 on E, for the anſwer.

PROBLEM X. To find Numbers in Subtriplicate Proportion.

As to find a Number which ſhall be to 40, as the Cube Root of 72.9 is to the Cube Root of 100.

Set 40 on D to 100 on E; then againſt 72.9 on E ſtands 36 on D, for the anſwer.

PROBLEM XI. To Compute Malt Buſhels by the Line MD.

As to find the Malt Buſhels in the Couch, Floor, or Ciſtern, whoſe Length is 230, Breadth 58.2, and Depth 5.4 Inches.

Set 230 on B to 5.4 on MD; then againſt 58.2 on A ſtands 33.6 buſhels on B, for the anſwer.

Note. The uſes of the other marks on the rule, will appear in the examples farther on.

OF THE GAUGING OR DIAGONAL ROD.

The diagonal rod is a ſquare rule, having four faces; being commonly 4 feet long, and folding together by joints. This inſtrument is uſed both for gauging or meaſuring caſks, and computing their contents, and that from one dimenſion only, namely the diagonal of the caſk, or the length from the middle of the bung-hole [270]to the meeting of the head of the caſk with the ſtave oppoſite to the bung; and is the longeſt line that can be drawn within the caſk from the middle of the bung. And, accordingly, on one face of the rule is a ſcale of inches, for meaſuring this diagonal; to which are placed the areas, in ale gallons, of circles to the cor⯑reſponding diameters, in like manner as the lines on the under ſides of the three ſlides in the ſliding rule.

On the oppoſite face are two ſcales of ale and wine gallons, expreſſing the contents of caſks having the cor⯑reſponding diagonals. And theſe are the lines which chiefly form the difference between this inſtrument and the ſliding rule; for all their other lines are the ſame, and to be uſed in the ſame manner.

EXAMPLE.

The rod being applied within the caſk at the bung-hole, the diagonal was found to be 34.4 inches; re⯑quired the content in gallons.

Now to 34.4 inches correſpond, on the rod, 90¼ ale gallons or 111 wine gallons; the content required.

Note. The contents exhibited by the rod anſwer to the moſt common form of caſks, and fall in between the 2d and 3d varieties following.

OF CASKS AS DIVIDED INTO VARIETIES.

It is uſual to divide caſks into four caſes or varieties, which are judged of from the greater or leſs apparent curvature of their ſides, namely,

1. The middle fruſtum of a ſpheroid,
2. The middle fruſtum of a parabolic ſpindle,
3. The two equal fruſtums of a paraboloid,
4. The two equal fruſtums of a cone.

And if the content of any of thoſe be computed in inches, by their proper rules, and this be divided by 282, or 231, or 2150.4, the quotient will be the con⯑tent [271]in ale gallons, or wine gallons, or malt buſhels, reſpectively. Becauſe

282	cubic inches make 1 ale gallon
231	— 1 wine gallon
2150.4	— 1 malt buſhel.

Or the particular rule will be for each as in the fol⯑lowing problems.

PROBLEM XII. To find the Content of a Caſk of the Firſt Form.

To the ſquare of the head diameter add double the ſquare of the bung diameter, and multiply the ſum by the length of the caſk. Then let the product

be multiplied by .0009¼ or divided by 1077 for ale gallons,
and multiplied by .0011⅓ or divided by 882 for wine gallons.

EXAMPLES.

1. Required the content of a ſpheroidal caſk, whoſe length is 40, and bung and head diameters 32 and 24 inches.

[diagram]

[272] [...]

By the Gauging Rule.

Having ſet 40 on C to the ale gauge 32.82 on D, againſt

24 on D ſtands 21.3 on C
32 on D ſtands 38.0 on C
the ſame 38.0
ſum 97.3 ale gallons.

And having ſet 40 on C to the wine gauge 29.7 on D, againſt

24 on D ſtands 26.1 on C
32 on D ſtands 46.5 on C
the ſame 46.5
ſum 119.1 wine gallons.

Ex. 2. Required the content of the ſpheroidal caſk, whoſe length is 20, and diameters 12 and 16 inches.

Anſwer 12.136 ale gallons.
Anſwer 14.869 wine gallons.

PROBLEM XIII. To find the Content of a Caſk of the Second Form.

[273]

To the ſquare of the head diameter add double the ſquare of the bung diameter, and from the ſum take ⅖ or 4/10 of the ſquare of the difference of the diameters; then multiply the remainder by the length, and the product again by .0009¼ for ale gallons, or by .0011⅓ for wine gallons.

EXAMPLES.

1. The length being 40, and diameters 24 and 32, required the content.

[diagram]

[...]

By the Gauging Rule.

Having ſet 40 on C to 32.82 on D, againſt 8 on D ſtands 2.4 on C; the [...]/ [...] of which is 0.96. This taken from the 97.3 in the laſt form, leaves 90.3 ale gallons.

[274]And having ſet 40 on C to 29.7 on D, againſt 8 on D ſtands 2.9 on C; the 4/10 of which is 1.16. This taken from the 119.1 in the laſt form, leaves 117.9 wine gallons.

Ex. 2. Required the content when the length is 20, and the diameters 12 and 16.

Anſwer 12.018 ale gallons,
Anſwer 14.724 wine gallons.

PROBLEM XIV. To find the Content of a Caſk of the Third Form.

To the ſquare of the bung diameter add the ſquare of the head diameter; multiply the ſum by the length, and the product again by .0014 for ale gallons, or by .0017 for wine gallons.

EXAMPLES.

1. Required the content of a caſk of the third form, when the length is 40, and the diameters 24 and 32.

[diagram]

[...]

[275]

By the Gauging Rule.
Set 40 on C to 26.8 on D; then againſt
24 on D ſtands 32.0 on C
32 on D ſtands 57.3 on C
ſum 89.3 ale gallons.

And having ſet 40 on C to 24.25 on D; then againſt

24 on D ſtands 39.1 on C

32 on D ſtands 69.8 on C

ſum 108.9 wine gallons.

Ex. 2. Required the content when the length is 20, and the diameters 12 and 16.

Anſwer 11.2 ale gallons,
Anſwer 13.6 wine gallons.

PROBLEM XV. To find the Content of a Caſk of the Fourth Form.

Add the ſquare of the difference of the diameters to 3 times the ſquare of their ſum; then multiply the ſum by the length, and the product again by .00023⅕ for ale gallons, or by .00028⅓ for wine gallons.

EXAMPLES.

1. Required the content when the length is 40, and the diameters 24 and 32 inches.

[diagram]

[276] [...]

By the Sliding Rule.
Set 40 on C to 65.64 on D; then againſt
8 on D ſtands 0.6 on C
56 on D ſtands 29.1 on C
29.1
29.1
ſum 87.9 ale gallons.

And ſet 40 on C to 59.41 on D; then againſt

8 on D ſtands 0.7

56 on D ſtands 35.6

35.6

ſum 107.5 wine gall.

Ex. 2. What is the content of a conical caſk, the length being 20, and the bung and head diameters 16 and 12 inches?

Anſwer 10.985 ale gallons,
Anſwer 13.416 wine gallons.

PROBLEM XVI. To find the Content of a Caſk by Four Dimenſions.

[277]

Add together the ſquares of the bung and head dia⯑meters, and the ſquare of double the diameter taken in the middle between the bung and head; then multiply the ſum by the length of the caſk, and the product again by .0004⅔ for ale gallons, or by .0005⅔ for wine gallons.

EXAMPLES.

1. Required the content of any caſk whoſe length is 40, the bung diameter being 32, the head diameter 24, and the middle diameter between the bung and head 28¾ inches.

[...]

[278]

By the Sliding Rule.
Set 40 on C to 46.4 on D; then againſt
24 on D ſtands 10.5
32 on D ſtands 19.0
57½ on D ſtands 62.0
ſum 91.5 ale gallons.

Set 40 on C to 42.0 on D; then againſt

24 on D ſtands 13.0

32 on D ſtands 23.2

57½ on D ſtands 75.0

ſum 111.2 wine gallons.

Ex. 2. What is the content of a caſk whoſe length is 20, the bung diameter being 16, the head diameter 12, and the diameter in the middle between them 14⅜?

Anſwer 11.4479 ale gallons,
Anſwer 13.9010 wine gallons.

PROBLEM XVII. To find the Content of any Caſk from Three Dimenſions only.

Add into one ſum 39 times the ſquare of the bung diameter, 25 times the ſquare of the head diameter, and 26 times the product of the two diameters; then multi⯑ply the ſum by the length, and the product again by .00034/9 for ale gallons, or by .00034/11 or .00003 1/11 for wine gallons.

EXAMPLES.

1. Required the content of a caſk whoſe length is 40, and the bung and head diameters 32 and 24.

[279] [...]

Ex. 2. What is the content of a caſk, whoſe length is 20, and bung and head diameters 16 and 12?

Anſwer 11.4833 ale gallons,
Anſwer 14.0352 wine gallons.

Note. This is the moſt exact rule of any, for three dimenſions only; and agrees nearly with the diagonal rod.

OF THE ULLAGE OF CASKS.

[280]

The ullage of a caſk, is what it contains when only partly filled. And it is conſidered in two poſitions, namely, as ſtanding on its end with the axis perpendi⯑cular to the horizon, or as lying on its ſide with the axis parallel to the horizon.

PROBLEM XVIII. To find the Ullage by the Sliding Rule.

By one of the preceding problems find the whole content of the caſk. Then ſet the length on N to 100 on SS for a ſegment ſtanding, or ſet the bung diameter on N to 100 on SL for a ſegment lying; then againſt the wet inches on N is a number on SS or SL, to be re⯑ſerved.

Next, ſet 100 on B to the reſerved number on A; then againſt the whole content on B will be found the ullage on A.

EXAMPLES.

1. Required the ullage anſwering to 10 wet inches of a ſtanding caſk, the whole content of which is 92 gal⯑lons, and length 40 inches.

Having ſet 40 on N to 100 on SS, then againſt 10 on N is 23 on SS, the reſerved number.

Then ſet 100 on B to 23 on A, and againſt 92 on B is 21.2 on A, the ullage required.

Ex. 2. What is the ullage of a ſtanding caſk whoſe whole length is 20 inches, and content 11 [...]/ [...] gallons; the wet inches being 5?

Anſ. 2.65 gallons.

Ex. 3. The content of a caſk being 92 gallons, and the bung diameter 32, required the ullage of the ſeg⯑ment lying when the wet inches are 8.

Anſ. 16.4 gallons.

PROBLEM XIX. To Ullage a Standing Caſk by the Pen.

[281]

Add all together the ſquare of the diameter at the ſurface of the liquor, the ſquare of the diameter of the neareſt end, and the ſquare of double the diameter taken in the middle between the other two; then multi⯑ply the ſum by the length between the ſurface and neareſt end, and the product again by .0004 2/ [...] for ale gallons, or by .0005⅔ for wine gallons, in the leſs part of the caſk, whether empty or filled.

EXAMPLE.

The three diameters being 24, 27, and 29 inches, re⯑quired the ullage for 10 wet inches.

[...]

PROBLEM XX. To Ullage a Lying Caſk by the Pen.

Divide the wet inches by the bung diameter; find the quotient in the column of verſed ſines, in the table of circular ſegments at the end of the book, taking out its [282]correſponding ſegment. Then multiply this ſegment by the whole content of the caſk, and the product again by 1¼ for the ullage required, nearly.

EXAMPLE.

Suppoſing the bung diameter 32, and content 92 ale gallons; to find the ullage for 8 wet inches.

[...]

OF SPECIFIC GRAVITY.

[]

THE ſpecific gravities of bodies, are their relative weights contained under the ſame given magnitude, as a cubic foot, or a cubic inch, &c.

The ſpecific gravities of ſeveral ſorts of matter are expreſſed by the numbers annexed to their names in the following table.

A Table of the Specific Gravities of Bodies.
Fine gold	19640	Brick	2000
Standard gold	18888	Light earth	1984
Quick-ſilver	14000	Solid gun-powder	1745
Lead	11325	Sand	1520
Fine ſilver	11091	Pitch	1150
Standard ſilver	10535	Box-wood	1030
Copper	9000	Sea-water	1030
Gun metal	8784	Common water	1000
Caſt braſs	8000	Oak	925
Steel	7850	Gun-powder, ſhaken	922
Iron	7645	Aſh	800
Caſt iron	7425	Maple	755
Tin	7320	Elm	600
Marble	2700	Fir	550
Common ſtone	2520	Cork	240
Loom	2160	Air	1¼

Note. The ſeveral ſorts of wood are ſuppoſed to be dry. Alſo, as a cubic foot of water weighs juſt 1000 ounces avoirdupois, the numbers in this table expreſs, [284]not only the ſpecific gravities of the ſeveral bodies, but alſo the weight of a cubic foot of each, in avoirdupois ounces; and thence, by proportion, the weight of any other quantity, or the quantity of any other weight, may be known, as in the following problems.

PROBLEM I. To find the Magnitude of any Body, from its Weight.

As the tabular ſpecific gravity of the body,
Is to its weight in avoirdupois ounces,
So is one cubic foot, or 1728 cubic inches,
To its content in feet, or inches, reſpectively.

EXAMPLES.

1. Required the content of an irregular block of com⯑mon ſtone which weighs 1 cwt, or 112lb.

[...]

[285]Ex. 2. How many cubic inches of gun-powder are there in 1lb weight.

Anſ. 30 cubic inches nearly.

Ex. 3. How many cubic feet are there in a ton weight of dry oak?

Anſ. 38¼ ⅜ 8/5 cubic feet.

PROBLEM II. To find the Weight of a Body from its Magnitude.

As one cubic foot, or 1728 cubic inches,
Is to the content of the body,
So is its tabular ſpecific gravity,
To the weight of the body.

EXAMPLES.

1. Required the weight of a block of marble, whoſe length is 63 feet, and breadth and thickneſs each 12 feet; being the dimenſions of one of the ſtones in the walls of Balbeck.

[...]

[286]Ex. 2. What is the weight of 1 pint, ale meaſure, of gun-powder?

Anſ. 19 oz. nearly.

Ex. 3. What is the weight of a block of dry oak, which meaſures 10 feet in length, 3 feet broad, and 2½ feet deep?

Anſ. 4335 15/16lb.

PROBLEM III. To find the Specific Gravity of a Body.

CASE 1. When the body is heavier than water, weigh it both in water and out of water, and take the difference, which will be the weight loſt in water. Then

As the weight loſt in water,

Is to the whole weight,

So is the ſpecific gravity of water,

To the ſpecific gravity of the body.

EXAMPLE.

A piece of ſtone weighed 10lb, but in water only 6¾lb, required its ſpecific gravity.

[...]

CASE 2. When the body is lighter than water, ſo that it will not quite ſink; aſſix to it a piece of another [287]body heavier than water, ſo that the maſs compounded of the two may ſink together. Weigh the heavier body, and the compound maſs, ſeparately, both in water and out of it; then find how much each loſes in water, by ſubtracting its weight in water from its weight in air; and ſubtract the leſs of theſe remainders from the greater. Then

As this laſt remainder,

Is to the weight of the light body in air,

So is the ſpecific gravity of water,

To the ſpecific gravity of the body.

EXAMPLE.

Suppoſe a piece of elm weighs 15lb in air, and that a piece of copper, which weighs 18lb in air and 16lb in water, is affixed to it, and that the compound weighs 8lb in water; required the ſpecific gravity of the elm.

Copper		Compound
18	in air	33
16	in water	6
2	loſs	27
		2
		As 25∷15∶1000∷600 anſ.

PROBLEM IV. To find the Quantities of Two Ingredients in a given Compound.

Take the three differences of every pair of the three ſpecific gravities, namely, the ſpecific gravities of the compound and each ingredient; and multiply the differ⯑ence of every two ſpecific gravities by the third. Then, as the greateſt product is to the whole weight of [288]the compound, ſo is each of the other products to the two weights of the ingredients.

EXAMPLE.

A compoſition of 112lb being made of tin and cop⯑per, whoſe ſpecific gravity is found to be 8784; re⯑quired the quantity of each ingredient, the ſpecific gra⯑vity of tin being 7320, and of copper 9000.

[...]

Anſwer there is 100lb of copper and conſeq. 12lb of tin in the compoſition.

OF THE WEIGHT AND DIMENSIONS OF BALLS AND SHELLS.

[]

THE weight and dimenſions of balls and ſhells might be found from the problems laſt given concerning ſpecific gravity. But they may be found ſtill eaſier by means of the experimented weight of a ball of a given ſize, from the known proportion of ſimilar figures, namely, as the cubes of their diameters.

PROBLEM I. To find the Weight of an Iron Ball, from its Diameter.

An iron ball of 4 inches diameter weighs 9lb, and the weights being as the cubes of the diameters, it will be as 64 (which is the cube of 4) is to 9, ſo is the cube of the diameter of any other ball, to its weight. Or take 9/64 of the cube of the diameter, for the weight. [290]Or take ⅛ of the cube of the diameter, and ⅛ of that again, and add the two together, for the weight.

EXAMPLES.

1. The diameter of an iron ſhot being 6.7, required its weight.

[...]

Ex. 2. What is the weight of an iron ball whoſe diameter is 5.54 inches?

Anſ. 24lb.

PROBLEM II. To find the Weight of a Leaden Ball.

A leaden ball of 4¼ inches diameter weighs 17lb; therefore as the cube of 4¼ to 17, or nearly as 9 to 2, ſo is the cube of the diameter of a leaden ball, to its weight.

Or take 2/ [...] of the cube of the diameter, for the weight, nearly.

[291]EXAMPLES.

1. Required the weight of a leaden ball of 6.6 inches diameter.

[...]

Ex. 2. What is the weight of a leaden ball of 5.24 inches diameter?

Anſ. 32lb nearly.

PROBLEM III. To find the Diameter of an Iron Ball.

Multiply the weight by 7 1/ [...], and the cube root of the product will be the diameter.

EXAMPLES.

1. Required the diameter of a 42lb iron ball.

[292] [...]

The cube root of this is almoſt 7. Suppoſe 7, whoſe cube is 343. Then, by the rule for the cube root at page 121, of my arithmetic.

[...]

Ex. 2. What is the diameter of a 24lb iron ball?

Anſ. 5.54 inches.

PROBLEM IV. To find the Diameter of a Leaden Ball.

[293]

Multiply the weight by 9, and divide the product by 2; then the cube root of the quotient will be the diameter.

EXAMPLES.

1. Required the diameter of a 64lb leaden ball.

[...]

The cube root of which is almoſt 7, whoſe cube is 343.

[...]

Ex. 2. What is the diameter of an 8lb leaden ball?

Anſ. 3.303 inches.

PROBLEM V. To find the Weight of an Iron Shell.

[294]

Take 9/64 of the difference of the cubes of the exter⯑nal and internal diameter, for the weight of the ſhell.

That is, from the cube of the external diameter take the cube of the internal diameter, multiply the re⯑mainder by 9, and divide the product by 64.

EXAMPLES.

1. The outſide diameter of an iron ſhell being 12.8, and the inſide diameter 9.1 inches; required its weight.

[...]

Ex. 2. What is the weight of an iron ſhell, whoſe ex⯑ternal and internal diameters are 9.8 and 7 inches?

Anſ. 84 [...]/ [...] lb.

PROBLEM VI. To find how much Powder will fill a Shell.

[295]

Divide the cube of the internal diameter, in inches, by 57.3, for the lbs of powder.

EXAMPLES.

1. How much powder will fill the ſhell whoſe inter⯑nal diameter is 9.1 inches?

[...]

Ex. 2. How much powder will fill the ſhell whoſe in⯑ternal diameter is 7 inches?

Anſ. 6lb.

PROBLEM VII. To find how much Powder will fill a Rectangular Box.

[296]

Find the content of the box in inches, by multiplying the length, breadth, and depth all together. Then divide by 30 for the pounds of powder.

EXAMPLES.

1. Required the quantity of powder that will ſill a box, the length being 15 inches, the breadth 12, and the depth 10 inches.

[...]

Ex. 2. How much powder will fill a cubical box whoſe ſide is 12 inches?

Anſ. 57⅗lb.

PROBLEM VIII. To find how much Powder will fill a Cylinder.

Multiply the ſquare of the diameter by the length, then divide by 38.2 for the pounds of powder.

EXAMPLES.

1. How much powder will the cylinder hold whoſe diameter is 10 inches, and length 20 inches?

[297] [...]

Ex. 2. How much powder can be contained in the cylinder, whoſe diameter is 4 inches, and length 12 inches?

Anſ. 5 5/191lb.

PROBLEM IX. To find the Size of a Shell to contain a given Weight of Powder.

Multiply the pounds of powder by 57.3, and the cube root of the product will be the diameter in inches.

EXAMPLES.

1. What is the diameter of a ſhell that will hold. 13⅙lb of powder?

[...]

[298]The cube root of this is nearly 9, whoſe cube is 729.

[...]

Ex. 2. What is the diameter of a ſhell to contain 6lb of powder?

Anſ. 7 inches.

PROBLEM X. To find the Size of a Cubical Box to contain a given Weight of Powder.

Multiply the weight in pounds by 30, and the cube root of the product will be the ſide of the box in inches.

EXAMPLES.

1. Required the ſize of a cubical box to hold 50lb of gun-powder.

[299] [...]

Ex. 2. Required the ſize of a cubical box to hold 400lb of gun-powder.

Anſ. 22.89 inches.

PROBLEM XI. To find what Length of a Cylinder will be filled by a given Weight of Gun-powder.

Multiply the weight in pounds by 38.2, and divide the product by the ſquare of the diameter in inches, for the length.

EXAMPLES.

1. What length of a 36 pounder gun, of 6⅔ inches diameter, will be filled with 12lb of powder?

[300] [...]

Ex. 2. What length of a cylinder of 8 inches diameter may be filled with 20lb of powder?

Anſ. 11 15/16.

OF THE PILING OF BALLS AND SHELLS.

[]

IRON balls and ſhells are commonly piled, by hori⯑zontal courſes, either in a pyramidical or wedge-like form; the baſe being either an equilateral triangle, a ſquare, or a rectangle. In the triangle and ſquare, the pile will finiſh in a ſingle ball; but in the rectangle, it will finiſh in a ſingle row of balls, like an edge.

In triangular and ſquare piles, the number of hori⯑zontal rows, or courſes, is always equal to the number of balls in one ſide of the bottom row. And in rectan⯑gular piles, the number of rows is equal to the number of balls in the breadth of the bottom row. Alſo the number in the top row, or edge, is one more than the difference between the length and breadth of the bot⯑tom row.

PROBLEM I. To find the Number of Balls in a Triangular Pile.

Multiply continually together the number in one ſide of the bottom row, that number increaſed by 1, and [302]the ſame number increaſed by 2; and ⅙ of the laſt pro⯑duct will be the anſwer.

EXAMPLES.

1. Required the number of balls in a triangular pile, each ſide of the baſe containing 30 balls.

[...]

Ex. 2. How many balls are in the triangular pile, each ſide of the baſe containing 20?

Anſ. 1540.

PROBLEM II. To find the Number of Balls in a Square Pile.

Multiply continually together the number in one ſide of the bottom courſe, that number increaſed by 1, and double the ſame number increaſed by 1; then ⅙ of the laſt product will be the anſwer.

EXAMPLES.

1. How many balls are in a ſquare pile of 30 rows?

[303] [...]

Ex. 2. How many balls are in a ſquare pile of 20 rows?

Anſ. 2870.

PROBLEM III. To find the Number of Balls in a Rectangular Pile.

From 3 times the number in the length of the baſe row, ſubtract one leſs than the breadth of the ſame, multiply the remainder by the ſaid breadth, and the product by one more than the ſame; and divide by 6 for the anſwer.

EXAMPLES.

1. Required the number of balls in a rectangular pile, the length and breadth of the baſe row being 46 and 15.

[304] [...]

Ex. 2. How many ſhot are in a rectangular complete pile, the length of the bottom courſe being 59, and its breadth 20?

Anſ. 11060.

PROBLEM IV. To find the Number of Balls in an Incomplete Pile.

From the number in the whole pile, conſidered as complete, ſubtract the number in the upper pile which is wanting at the top, both computed by the rule for their proper form; and the remainder will be the num⯑ber in the fruſtum, or incomplete pile.

EXAMPLES.

1. To find the Number of ſhot in the incomplete triangular pile, one ſide of the bottom courſe being 40, and the top courſe 20.

[305] [...]

Ex. 2. How many ſhot are in the incomplete trian⯑gular pile, the ſide of the baſe being 24, and of the top 8?

Anſ. 2516.

Ex. 3. How many balls are in the incomplete ſquare pile, the ſide of the baſe being 24, and of the top 8?

Anſ. 4760.

Ex. 4. How many ſhot are in the incomplete rectan⯑gular pile, of 12 courſes, the length and breadth of the baſe being 40 and 20?

Anſ. 6146.

OF DISTANCES BY THE VELOCITY OF SOUND.

[]

BY various experiments it has been found that ſound flies, through the air, uniformly at the rate of about 1142 feet in 1 ſecond of time, or a mile in 4⅔ ſe⯑conds. And therefore, by proportion, any diſtance may be found correſponding to any given time; namely, multiply the given time in ſeconds, by 1142, for the correſponding diſtance in feet; or take 3/14 of the given time for the diſtance in miles.

Note. The time for the paſſage of ſound in the in⯑terval between ſeeing the flaſh of a gun, or lightning, and hearing the report, may be obſerved by a watch, or a ſmall pendulum. Or it may be obſerved by the beats of the pulſe in the wriſt, counting, on an average, about 70 to a ſecond in perſons in moderate health, or 5½ pulſations to a mile, and more or leſs according to cir⯑cumſtances.

[307]EXAMPLES.

1. After obſerving a flaſh of lightning, it was 12 ſe⯑conds before I heard the thunder; required the diſtance of the cloud from whence it came.

[...]

Ex. 2. How long, after firing the tower guns, may the report be heard at Shooters-Hill, ſuppoſing the diſtance to be 8 miles in a ſtraight line?

[...]

Ex. 3. After obſerving the firing of a large cannon at a diſtance, it was 7 ſeconds before I heard the re⯑port; what was its diſtance?

Anſ. 1½ mile.

Ex. 4. Perceiving a man at a diſtance hewing down a tree with an axe, I remarked that 6 of my pulſations paſſed between ſeeing him ſtrike and hearing the report of the blow; what was the diſtance between us, allow⯑ing 70 pulſes to a minute?

Anſ. 1 mile and 198 yards.

Ex. 5. How far off was the cloud from which thun⯑der iſſued, whoſe report was 5 pulſations after the flaſh of lightning; counting 75 to a minute?

Anſ. 1523 yards.

MISCELLANEOUS QUESTIONS.

[]

QU. 1. WHAT difference is there between a floor 28 feet long by 20 broad, and two others each of half the dimenſions; and what do all three come to at 45s. per 100 ſquare feet?

Anſ. dif. 280 ſq. feet. Amount 18 guineas.

2. An elm plank is 14 feet 3 inches long, and I would have juſt a ſquare yard ſlit off it; at what diſtance from the edge muſt the line be ſtruck?

Anſ. 7 99/171 inches.

3. A ceiling contains 114 yards 6 feet of plaſtering, and the room 28 feet broad; what was the length of it?

Anſ. 36 6/7 feet.

4. A common joiſt is 7 inches deep, and 2½ thick; but I want a ſcantling juſt as big again, that ſhall be 3 inches thick; what will the other dimenſion be?

Anſ. 11⅔ inches.

5. A wooden trough coſt me 3s 6d painting within, at 6d per yard; the length of it was 102 inches, and the depth 21 inches; what was the width?

Anſ. 27¼ inches.

6. If my court yard be 47 feet 9 inches ſquare, and I have laid a foot-path with Purbeck ſtone, of 4 feet wide, along one ſide of it; what will paving the reſt with flints come to at 6d per ſquare yard?

Anſ. £ 5 16 0⅓.

7. A ladder, 40 feet long, may be ſo planted, that it ſhall reach a window 33 feet from the ground on one ſide of the ſtreet; and, by only turning it over, with⯑out moving the foot out of its place, it will do the [309]ſame by a window 21 feet high on the other ſide: what is the breadth of the ſtreet?

Anſ. 56 feet 7¾ inches.

8. The paving of a triangular court, at 18d per foot, came to 100l; the longeſt of the three ſides was 88 feet; required the ſum of the other two equal ſides.

Anſ. 106.85 feet.

9. There are two columns in the ruins of Perſepolis left ſtanding upright; the one is 64 feet above the plain, and the other 50: in a ſtraight line between theſe ſtands an ancient ſmall ſtatue, the head of which is 97 feet from the ſummit of the higher, and 86 feet from the top of the lower column, the baſe of which meaſures juſt 76 feet to the center of the figure's baſe. Required the diſtance between the tops of the two columns.

Anſ. 157 feet nearly.

10. The perambulator, or ſurveying wheel, is ſo contrived, as to turn juſt twice in the length of a pole, or 16½ feet; required the diameter.

Anſ. 2.626 feet.

11. In turning a one-horſe chaiſe within a ring of a certain diameter, it was obſerved that the outer wheel made two turns while the inner made but one: the wheels were both 4 feet high; and, ſuppoſing them fixed at the ſtatutable diſtance of 5 feet aſunder on the axletree, what was the circumference of the track de⯑ſcribed by the outer wheel?

Anſ. 63 feet nearly.

12. What is the ſide of that equilateral triangle whoſe area coſt as much paving at 8d a foot, as the palliſading the three ſides did at a guinea a yard?

Anſ. 72.746 feet.

13. In the trapezium ABCD are given AB=13, BC=31⅓, CD=24, and DA=18, alſo B a right angle; required the area.

Anſ. 410.122.

14. A roof, which is 24 feet 8 inches by 14 feet 6 inches, is to be covered with lead at 8lb to the ſquare foot: what will it come to at 18s per cwt?

Anſ. £ 22 19 10 [...]/ [...];.

15. Having a rectangular marble ſlab, 58 inches by 27, I would have a ſquare foot cut off parallel to the ſhorter edge; I would then have the like quantity di⯑vided [310]from the remainder parallel to the longer ſide; and this alternately repeated, till there ſhall not be the quantity of a foot left: what will be the dimenſions of the remaining piece?

Anſ. 20.7 inches by 6.086.

16. Given two ſides of an obtuſe-angled triangle, which are 20 and 40 poles; required the third ſide that the triangle may contain juſt an acre of land.

Anſ. 58.876 or 23.099.

17. The end wall of a houſe is 24 feet 6 inches in breadth, and 40 feet to the eaves; ⅓ of which is 2 bricks thick, ⅓ more is 1½ brick thick, and the reſt 1 brick thick. Now the triangular gable riſes 38 courſes of bricks, 4 of which uſually make a foot in depth, and this is but 4½ inches, or half a brick thick: what will this piece of work come to at 5l 10s per ſtatute rod?

Anſ. £ 20 11 7½

18. If from a right-angled triangle, whoſe baſe is 12, and perpendicular 16 feet, a line be drawn parallel to the perpendicular cutting off a triangle whoſe area is 24 ſquare feet; required the ſides of this triangle.

Anſ. 6, 8, and 10.

19. The ellipſe in Groſvenor-ſquare meaſures 840 links acroſs the longeſt way, and 612 the ſhorteſt, within the rails: now the walls being 14 inches thick, what ground do they incloſe, and what do they ſtand upon?

Anſ. incloſe 4 ac 0 r 6 p
Anſ. ſtand on 1760½ ſq feet.

20. If a round pillar, 7 inches over, has 4 feet of ſtone in it; of what diameter is the column, of equal length, that contains 10 times as much?

Anſ. 22.136 inches.

21. A circular fiſh-pond is to be made in a garden, that ſhall take up juſt half an acre; what muſt be the length of the cord that ſtrikes the circle?

Anſ. 27 [...]/ [...] yards.

22. When a roof is of a true pitch, the rafters are [...] of the breadth of the building: now ſuppoſing the eaves-boards to project 10 inches on a ſide, what will [311]the new ripping a houſe coſt, that meaſures 32 feet 9 inches long, by 22 feet 9 inches broad on the flat, at 15s per ſquare?

Anſ. £ 8 15 9½

23. A cable which is 3 feet long, and 9 inches in compaſs, weighs 22lb; what will a fathom of that cable weigh, which meaſures a foot about?

Anſ. 78½ lb.

24. My plumber has put 28lb per ſquare foot into a ciſtern 74 inches and twice the thickneſs of the lead long, 26 inches broad, and 40 deep; he has alſo put three ſtays acroſs it within, 16 inches deep, of the ſame ſtrength, and reckons 22s per cwt, for work and materials. I, being a maſon, have paved him a workſhop, 22 feet 10 inches broad, with Purbeck ſtone, at 7d per foot; and upon the balance I find there is 3s 6d due to him. What was the length of the workſhop?

Anſ. 32 f 0¾ inches.

25. The diſtance of the centers of two circles, whoſe diameters are each 50, being given equal to 30; what is the area of the ſpace incloſed by their circumfer⯑ences?

Anſ. 559.119.

26. If 20 feet of iron railing weigh half a ton when the bars are an inch and quarter ſquare, what will 50 feet come to at 3½d per lb, the bars being but ⅞ of an inch ſquare?

Anſ. £ 20 0 2.

27. The area of an equilateral triangle, whoſe baſe falls on the diameter, and its vertex in the middle of the arc of a ſemicircle, is equal to 100: what is the diame⯑ter of the ſemicircle?

Anſ. 26.32148.

28. It is required to find the thickneſs of the lead in a pipe of an inch and quarter bore, which weighs 14 lb per yard in length; the cubic foot of lead weighing 11325 ounces.

Anſ. .20737 inches.

29. Suppoſe the expence of paving a ſemicircular plot, at 2s 4d per foot, come to 10l, what is the diame⯑ter of it?

Anſ. 14.7737

30. What is the length of a chord which cuts off ⅓ of the area from a circle whoſe diameter is 289?

Anſ. 278.6716

[312]31. My plumber has ſet me up a ciſtern, and, his ſhop-book being burnt, he has no means of bringing in the charge, and I do not chooſe to take it down to have it weighed; but by meaſure he finds it contains 64 3/10 ſquare feet, and that it is preciſely ⅛ of an inch in thickneſs. Lead was then wrought at 21l per fother of 19½ cwt. It is required from theſe items to make out the bill, allowing 6 5/9 oz for the weight of a cubic inch of lead.

Anſ. £ 4 11 2

32. What will the diameter of a globe be, when the ſolidity and ſuperficial content are expreſſed by the ſame number?

Anſ. 6.

33. A ſack, that would hold 3 buſhels of corn, is 22½ inches broad when empty; what will that ſack contain which, being of the ſame length, has twice its breadth or circumference?

Anſ. 12 buſhels.

34. A carpenter is to put an oaken curb to a round well, at 8d per foot ſquare: the breadth of the curb is to be 7¼ inches, and the diameter within 13½ feet: what will be the expence?

Anſ. 5s 2¼d.

35. A gentleman has a garden 100 feet long, and 80 feet broad; and a gravel walk is to be made of an equal width half round it: what muſt the breadth of the walk be to take up juſt half the ground?

Anſ. 25.968 feet.

36. A may-pole whoſe top, being broken off by a blaſt of wind, ſtruck the ground at 15 feet diſtance from the foot of the pole; what was the height of the whole may-pole, ſuppoſing the length of the broken piece to be 39 feet?

Anſ. 75 feet.

37. Seven men bought a grinding ſtone of 60 inches diameter, each paying 1/7 part of the expence; what part of the diameter muſt each grind down for his ſhare?

Anſ. the 1ſt 4.4508, 2d 4.8400, 3d 5.3535, 4th 6.0765, 5th 7.2079, 6th 9.3935, 7th 22.6778.

38. A maltſter has a kiln that is 16 feet 6 inches ſquare: but he wants to pull it down, and build a new [313]one that may dry three times as much at once as the old one; what muſt be the length of its ſide?

Anſ. 28 f 7 inc.

39. How many 3 inch cubes may be cut out of a 12 inch cube?

Anſ. 64.

40. How long muſt be the tether of a horſe that will allow to graze, quite around, juſt an acre of ground?

Anſ. 39¼ yards.

41. What will the painting of a conical ſpire come to at 8d per yard; ſuppoſing the height to be 118 feet, and the circumference of the baſe 64 feet?

Anſ. £ 14 0 8¾

42. The diameter of a ſtandard corn buſhel is 18½ inches, and its depth 8 inches; what muſt the diameter of that buſhel be whoſe depth is 7½ inches?

Anſ. 19.1067.

43. Suppoſe the ball on the top of St. Paul's church is 6 feet in diameter; what did the gilding of it coſt at 3½d per ſquare inch?

Anſ. £ 237 19 10½

44. What will a fruſtum of a marble cone come to at 12s per ſolid foot; the diameter of the greater end being 4 feet, that of the leſs end 1½, and the length of the ſlant ſide 8 feet?

Anſ. £ 30 1 10¼

45. To divide a cone into three equal parts by ſec⯑tions parallel to the baſe, and to find the altitudes of the three parts, the height of the whole cone being 20 inches.

Anſ. the upper part 13.867
Anſ. the middle part 3.604
Anſ. the lower part 2.528

46. A gentleman has a bowling-green, 300 feet long, and 200 feet broad, which he would raiſe 1 foot higher, by means of the earth to be dug out of a ditch that goes round it: to what depth muſt the ditch be dug, ſuppoſing its breadth to be every where 8 feet?

Anſ. 7⅜ 3/6 feet.

47. How high above the earth muſt a perſon be raiſed, that he may ſee ⅓ of its ſurface?

Anſ. to the height of the earth's diameter.

[314]48. A cubic foot of braſs is to be drawn into a wire of 1/40 of an inch in diameter; what will the length of the wire be, allowing no loſs in the metal?

Anſ. 97784.797 yards, or 55 miles 984.797 yards.

49. Of what diameter muſt the bore of a cannon be, which is caſt for a ball of 24 lb weight, ſo that the dia⯑meter of the bore may be 1/10 of an inch more than that of the ball?

Anſ. 5.757 inches.

50. Suppoſing the diameter of an iron 9lb ball to be 4 inches, as it is very nearly; it is required to find the diameters of the ſeveral balls weighing 1, 2, 3, 4, 6, 12, 18, 24, 36, and 42 lb, and the caliber of their guns, allowing 1/50 of the caliber, or 1/49 of the ball's diameter, for windage.

Anſwer.
Wt ball	Diameter ball	Caliber gun
1	1.9230	1.9622
2	2.4228	2.4723
3	2.7734	2.8301
4	3.0526	3.1149
6	3.4943	3.5656
9	4.0000	4.0816
12	4.4026	4.4924
18	5.0397	5.1425
24	5.5469	5.6601
36	6.3496	6.4792
42	6.6844	6.8208

51. Suppoſing the windage of all mortars be allowed to be 1/60 of the caliber, and the diameter of the hollow part of the ſhell to be 7/10 of the caliber of the mortar: It is required to determine the diameter and weight of the ſhell, and the quantity or weight of powder requi⯑ſite to fill it, for each of the ſeveral ſorts of mortars, namely, the 13, 10, 8, 5.8, and 4.6 inch mortar.

[315]

Anſwer.
Calib. mort.	Diameter ſhell	Wt ſhell empty	Wt of powder	Wt ſhell filled
4.6	4.523	8.320	0.583	8.903
5.8	5.703	16.677	1.168	17.845
8	7.867	43.764	3.065	46.829
12	9.833	85.476	5.986	91.462
13	12.783	187.791	13.151	200.942

52. How many ſhot are in a complete ſquare pile, each ſide of the baſe containing 29?

Anſ. 8555.

53. How many ſhot are in a complete oblong pile, the length of the baſe containing 49, and the breadth 19?

Anſ. 8170.

54. How many ſhot are in a triangular pile, each ſide of the baſe being 50?

Anſ. 22100.

55. How many ſhot are in an unfiniſhed triangular pile, the ſide of the bottom being 50, and top 20?

Anſ. 20770.

56. How many ſhot are in an unfiniſhed oblong pile, having the corner row 12, and the ſides of the top 40 and 10?

Anſ. 8606.

57. If a heavy ſphere, whoſe diameter is 4 inches, be let fall into a conical glaſs, full of water, whoſe dia⯑meter is 5, and altitude 6 inches; it is required to de⯑termine how much water will run over.

Anſ. 26.272 cubic inches, or near 35/47 parts of a pint.

58. The dimenſions of the ſphere and cone being the ſame as in the laſt queſtion, and the cone only ⅕ full of water; required what part of the axis of the ſphere is immerſed in the water.

Anſ. .546 parts of an inch.

59. The cone being ſtill the ſame, and 1/ [...] full of water; required the diameter of a ſphere which ſhall be juſt all covered by the water.

Anſ. 2.445996.

60. If I ſee the flaſh of a cannon, fired by a ſhip in diſtreſs at ſea, and hear the report 33 ſeconds after, how far is ſhe off?

Anſ. 7 1/14 miles.

[316]61. Being one day ordered to obſerve how far a battery of cannon was from me, I counted by my watch 17 ſeconds between the time of ſeeing the flaſh and hearing the report; how far was the battery from me?

Anſ. 3½ miles.

62. An irregular piece of lead ore weighs in air 12 ounces, but in water only 7; and another fragment of one weighs in air 14½ ounces, but in water only 9; required their comparative denſities.

Anſ. as 145 to 132.

63. Suppoſing the cubic inch of common glaſs weigh 1.36 ounces troy, the ſame of ſalt water .5427, and of brandy .48926; then a ſeaman having a gallon of that liquor in a glaſs bottle, which weighs 3 [...]/ [...] lb troy out of water, and to conceal it from the officers of the cuſ⯑toms, throws it overboard. It is required to determine, if it will ſink, how much force will juſt buoy it up?

Anſ. 12.8968 oz.

64. Suppoſe by meaſurement it be found that a man of war, with its ordnance, rigging, and appointments, draws ſo much water as to diſplace 50000 cubic feet of water; required the weight of the veſſel.

Anſ. 1395 1/10 tons.

A TABLE OF THE AREAS of the SEGMENTS of a CIRCLE, Whoſe Diameter is Unity, and ſuppoſed to be divided into 1000 equal Parts.

[]

Heights	Area Seg.	Height	Area Seg.	Height	Area Seg.
.001	.000042	.027	.005867	.053	.016007
.002	.000119	.028	.006194	.054	.016457
.003	.000219	.029	.006527	.055	.016911
.004	.000337	.030	.006865	.056	.017369
.005	.000470	.031	.007209	.057	.017831
.006	.000618	.032	.007558	.058	.018296
.007	.000779	.033	.007913	.059	.018766
.008	.000951	.034	.008273	.060	.019239
.009	.001135	.035	.008638	.061	.019716
.010	.001329	.036	.009008	.062	.020196
.011	.001533	.037	.009383	.063	.020680
.012	.001746	.038	.009763	.064	.021168
.013	.001968	.039	.010148	.065	.021659
.014	.002199	.040	.010537	.066	.022154
.015	.002438	.041	.010931	.067	.022652
.016	.002685	.042	.011330	.068	.023154
.017	.002940	.043	.011734	.069	.023659
.018	.003202	.044	.012142	.070	.024168
.019	.003471	.045	.012554	.071	.024680
.020	.003748	.046	.012971	.072	.025195
.021	.004031	.047	.013392	.073	.025714
.022	.004322	.048	.013818	.074	.026236
.023	.004618	.049	.014247	.075	.026761
.024	.004921	.050	.014681	.076	.027289
.025	.005230	.051	.015119	.077	.027821
.026	.005546	.052	.015561	.078	.028356
.079	.028894	.114	.049528	.149	.073161
.080	.029435	.115	.050165	.150	.073874
.081	.029979	.116	.050804	.151	.074589
.082	.030526	.117	.051446	.152	.075306
.083	.031076	.118	.052090	.153	.076026
.084	.031629	.119	.052736	.154	.076747
.085	.032186	.120	.053385	.155	.077469
.086	.032745	.121	.054036	.156	.078194
.087	.033307	.122	.054689	.157	.078921
.088	.033872	.123	.055345	.158	.079649
.089	.034441	.124	.056003	.159	.080380
.090	.035011	.125	.056663	.160	.081112
.091	.035585	.126	.057326	.161	.081846
.092	.036162	.127	.057991	.162	.082582
.093	.036741	.128	.058658	.163	.083320
.094	.037323	.129	.059327	.164	.084059
.095	.037909	.130	.059999	.165	.084801
.096	.038496	.131	.060672	.166	.085544
.097	.039087	.132	.061348	.167	.086289
.098	.039680	.133	.062026	.168	.087036
.099	.040276	.134	.062707	.169	.087785
.100	.040875	.135	.063389	.170	.088535
.101	.041476	.136	.064074	.171	.089287
.102	.042080	.137	.064760	.172	.090041
.103	.042687	.138	.065449	.173	.090797
.104	.043296	.139	.066140	.174	.091554
.105	.043908	.140	.066833	.175	.092313
.106	.044522	.141	.067528	.176	.093074
.107	.045139	.142	.068225	.177	.093836
.108	.045759	.143	.068924	.178	.094601
.109	.046381	.144	.069625	.179	.095366
.110	.047005	.245	.070328	.180	.096134
.111	.047632	.146	.071033	.181	.096903
.112	.048262	.147	.071741	.182	.097674
.113	.048894	.148	.072450	.183	.098447
.184	.099221	.219	.127285	.254	.157019
.185	.099997	.220	.128113	.255	.157890
.186	.100774	.221	.128942	.256	.158762
.187	.101553	.222	.129773	.257	.159636
.188	.102334	.223	.130605	.258	.160510
.189	.103116	.224	.131438	.259	.161386
.190	.103900	.225	.132272	.260	.162263
.191	.104685	.226	.133108	.261	.163140
.192	.105472	.227	.133945	.262	.164019
.193	.106261	.228	.134784	.263	.164899
.194	.107051	.229	.135624	.264	.165780
.195	.107842	.230	.136465	.265	.166663
.196	.108636	.231	.137307	.266	.167546
.197	.109430	.232	.138150	.267	.168430
.198	.110226	.233	.138995	.268	.169315
.199	.111024	.234	.139841	.269	.170202
.200	.111823	.235	.140688	.270	.171089
.201	.112624	.236	.141537	.271	.171978
.202	.113426	.237	.142387	.272	.172867
.203	.114230	.238	.143238	.273	.173758
.204	.115035	.239	.144091	.274	.174649
.205	.115842	.240	.144944	.275	.175542
.206	.116650	.241	.145799	.276	.176435
.207	.117460	.242	.146655	.277	.177330
.208	.118271	.243	.147512	.278	.178225
.209	.119083	.244	.148371	.279	.179122
.210	.119897	.245	.149230	.280	.180019
.211	.120712	.246	.150091	.281	.180918
.212	.121529	.247	.150953	.282	.181817
.213	.122347	.248	.151816	.283	.182718
.214	.123167	.249	.152680	.284	.183619
.215	.123988	.250	.153546	.285	.184521
.216	.124810	.251	.154412	.286	.185425
.217	.125634	.252	.155280	.287	.186329
.218	.126459	.253	.156149	.288	.187234
.289	.188140	.324	.220404	.359	.253590
.290	.189047	.325	.221340	.360	.254550
.291	.189955	.326	.222277	.361	.255510
.292	.190864	.327	.223215	.362	.256471
.293	.191775	.328	.224154	.363	.257433
.294	.192684	.329	.225093	.364	.258395
.295	.193596	.330	.226033	.365	.259357
.296	.194509	.331	.226974	.366	.260320
.297	.195422	.332	.227915	.367	.261284
.298	.196337	.333	.228858	.368	.262248
.299	.197252	.334	.229801	.369	.263213
.300	.198168	.335	.230745	.370	.264178
.301	.199085	.336	.231689	.371	.265144
.302	.200003	.337	.232634	.372	.266111
.303	.200922	.338	.233580	.373	.267078
.304	.201841	.339	.234526	.374	.268045
.305	.202761	.340	.235473	.375	.269013
.306	.203683	.341	.236421	.376	.269982
.307	.204605	.342	.237369	.377	.270951
.308	.205527	.343	.238318	.378	.271920
.309	.206451	.344	.239268	.379	.272890
.310	.207376	.345	.240218	.380	.273861
.311	.208301	.346	.241169	.381	.274832
.312	.209227	.347	.242121	.382	.275803
.313	.210154	.348	.243074	.383	.276775
.314	.211082	.349	.244026	.384	.277748
.315	.212011	.350	.244980	.385	.278721
.316	.212940	.351	.245934	.386	.279694
.317	.213871	.352	.246889	.387	.280668
.318	.214802	.353	.247845	.388	.281642
.319	.215733	.354	.248801	.389	.282617
.320	.216666	.355	.249757	.390	.283592
.321	.217599	.356	.250715	.391	.284568
.322	.218533	.357	.251673	.392	.285544
.323	.219468	.358	.252631	.393	.286521
.394	.287498	.430	.322928	.466	.358725
.395	.288476	.431	.323918	.467	.359723
.396	.289453	.432	.324909	.468	.360721
.397	.290432	.433	.325900	.469	.361719
.398	.291411	.434	.326892	.470	.362717
.399	.292390	.435	.327882	.471	.363715
.400	.293369	.436	.328874	.472	.364713
.401	.294349	.437	.329866	.473	.365712
.402	.295330	.438	.330858	.474	.366710
.403	.296311	.439	.331850	.475	.367709
.404	.297292	.440	.332843	.476	.368708
.405	.298273	.441	.333836	.477	.369707
.406	.299255	.442	.334829	.478	.370706
.407	.300238	.443	.335822	.479	.371705
.408	.301220	.444	.336816	.480	.372704
.409	.302203	.445	.337810	.481	.373703
.410	.303187	.446	.338804	.482	.374702
.411	.304171	.447	.339798	.483	.375702
.412	.305155	.448	.340793	.484	.376702
.413	.306140	.449	.341787	.485	.377701
.414	.307125	.450	.342782	.486	.378701
.415	.308110	.451	.343777	.487	.379700
.416	.309095	.452	.344772	.488	.380700
.417	.310081	.453	.345768	.489	.381699
.418	.311068	.454	.346764	.490	.382699
.419	.312054	.455	.347759	.491	.383699
.420	.313041	.456	.348755	.492	.384699
.421	.314029	.457	.349752	.493	.385699
.422	.315016	.458	.350748	.494	.386699
.423	.316004	.459	.351745	.495	.387699
.424	.316992	.460	.352742	.496	.388699
.425	.317981	.461	.353739	.497	.389699
.426	.318970	.462	.354736	.498	.390699
.427	.319959	.463	.355732	.499	.391699
.428	.320948	.464	.356730	.500	.392699
.429	.321938	.465	.357727

THE USE OF THE TABLE.

[]

IN the foregoing table, each number in the column of area ſeg. is the area of the circular ſegment whoſe height, or the verſed fine of its half are, is the number immediately on the left of it, in the column of heights; the diameter of the circle being 1, and its whole area .785398.

The uſe of this table is to find, by it, the area of the ſegment of any other circle, whatever be the diameter. And this is done by firſt dividing the height of any pro⯑poſed ſegment by its own diameter, and the quotient is a decimal to be ſought in the column of heights, and againſt it is the tabular area to be taken out, which is ſimilar to the propoſed ſegment. Then this tabular area being multiplied by the ſquare of the given diameter, will be the area of your ſegment required; becauſe ſimilar areas are to each other as the ſquares of their diameters.

EXAMPLE.

So if it be required to find the area of a ſegment of a circle, whoſe height is 3¼, the diameter being 50.

Here 50) 3.25 (.065 quot. or tab. height, and the tab. ſeg. is .021659 which multiply by 2500 the ſquare of the diam. gives 54.147500 the area required.

[323]But, in dividing the given height by the diameter, if the quotient do not terminate in three places of deci⯑mals without a fractional remainder, then the area for that fractional part ought to be proportioned for, thus: Having found the tabular area anſwering to the firſt three decimals of the quotient, take the difference be⯑tween it and the next following tabular area, which difference multiply by the fractional remaining part of the quotient, and the product will be the correſponding proportional part, to be added to the firſt tabular area.

So if the height of a propoſed ſegment were 3⅓, to the diameter 50.

Here 50) 3⅓	(.066⅔ Then
to .066 anſwers	.022154
the next area is	.022652
their difference is	498
⅔ of which is	232
which added to	.022154
gives the whole tab. area	.022386
And mult. by	2500
gives the area	55.965000 ſought.

FINIS.

Appendix A BOOKS by the ſame AUTHOR, lately publiſhed.

[]

1. MATHEMATICAL TABLES, containing the Common, Logiſtic, and Hyperbolic Logarithms. Alſo Sines, Tangents, Secants, and Verſed Sines, &c. 8vo. Price 14s in Boards.
2. MENSURATION both in Theory and Practice. 4to, 16s in Boards.
3. A SYSTEM OF PRACTICAL ARITHMETIC, AND BOOK-KEEPING, both by Single and Double Entry, for the Uſe of Schools. 2s 6d.
4. A KEY TO THE ARITHMETIC, containing the So⯑lutions, at full Length, of all the Queſtions propoſed in the Arithmetic. For the Uſe of Preceptors and others.
5. THE MATHEMATICAL PARTS OF THE LADIES DIARIES. 3 vols. 15s.
6. THE POETICAL PARTS OF THE LADIES DIARIES. 2 vols. 9s.
7. MATHEMATICAL MISCELLANY. 5s.
8. TRACTS, MATHEMATICAL AND PHILOSOPHICAL. 4to.

Notes

Only 4 parts are here drawn, for want of room.

Distributed by the University of Oxford under a Creative Commons Attribution-ShareAlike 3.0 Unported License

Citation Suggestion for this Object: TextGrid Repository (2020). TEI. 5415 The compendious measurer being a brief yet comprehensive treatise on mensuration and practical geometry Adapted to the use of schools By Charles Hutton. University of Oxford Text Archive. . https://hdl.handle.net/21.T11991/0000-001A-6026-D

PREFACE.

CONTENTS.

INTRODUCTION.

DECIMAL FRACTIONS.

ADDITION OF DECIMALS.

SUBTRACTION OF DECIMALS.

MULTIPLICATION OF DECIMALS.

CONTRACTION.

DIVISION OF DECIMALS.

CONTRACTIONS.

REDUCTION OF DECIMALS.

OTHER EXAMPLES.

CIRCULATING DECIMALS.

REDUCTION OF REPEATING DECIMALS.

ADDITION OF REPETENDS.

SUBTRACTION OF REPETENDS.

MULTIPLICATION OF REPETENDS.

DIVISION OF REPETENDS.

INVOLUTION; OR RAISING OF POWERS.

RULE.

EVOLUTION; OR EXTRACTION OF ROOTS.

TO EXTRACT THE SQUARE ROOT.

TO FIND A MEAN PROPORTIONAL.

TO EXTRACT THE CUBE ROOT.

TO FIND TWO MEAN PROPORTIONALS.

TO EXTRACT ANY ROOT WHATEVER.

GENERAL RULES for extracting any Root out of a Vulgar Fraction or Mixt Number.

DUODECIMALS: OR CROSS MULTIPLICATION.

RULE I.

RULE II.

RULE III.

MENSURATION.

PRACTICAL GEOMETRY.

DEFINITIONS.

PROBLEMS.

PROBLEM I. To divide a given Line AB into two Equal Parts,

PROBLEM II. To divide a given Angle ABC into two Equal Parts.

PROBLEM III. To divide a Right Angle ABC into three Equal Parts.

PROBLEM IV. To draw a Line Parallel to a given Line AB.

PROBLEM V. To erect a Perpendicular from a given Point A in a given Line BC.

PROBLEM VI. From a given Point A, out of a given Line BC, to let fall a Perpendicular.

PROBLEM VII. To divide a given Line AB into any propoſed Number of Equal Parts.

PROBLEM VIII. To divide a given Line AB in the ſame Proportion as another Line CD is divided.

PROBLEM IX. At a given Point A, in a given Line AB, to make an Angle equal to a given Angle C.

PROBLEM X. At a given Point A, in a given Line AB, to make an Angle of any propoſed Number of Degrees.

PROBLEM XI. To meaſure a given Angle A. (See the laſt figure.)

PROBLEM XII. To find the Center of a Circle.

PROBLEM XIII. To deſcribe the Circumference of a Circle through thre [...] given Points, A, B, C.

PROBLEM XIV. Through a given Point A, to draw a Tangent to a given Circle.

PROBLEM XV. To find a Third Proportional to two given Lines AB, AC.

PROBLEM XVI. To find a Fourth Proportional to three given Lines, AB, AC, AD.

PROBLEM XVII. To find a Mean Proportional between two given Lines, AB, BC.

PROBLEM XVIII. To make an Equilateral Triangle on a given Line AB.

PROBLEM XIX. To make a Triangle with three given Lines AB, AC, BC.

PROBLEM XX. To make a Square upon a given Line AB.

PROBLEM XXI. To deſcribe a Rectangle, or a Parallelogram, of a given Length and Breadth.

PROBLEM XXII. To make a regular Pentagon on a given Line AB.

PROBLEM XXIII. To make a Hexagon on a given Line AB.

PROBLEM XXIV. To make an Octagon on a given Line AB.

PROBLEM XXV. To make any regular Polygon on a given Line AB.

PROBLEM XXVI. In a given Circle to inſcribe any regular Polygon; or, to divide the Circumference into any number of equal Parts. (See the laſt figure.)

PROBLEM XXVII. About a given Circle to circumſcribe any Polygon.

PROBLEM XXVIII. To find the Center of a given Polygon, or the Center of its inſcribed or circumſcribed Circle.

PROBLEM XXIX. In any given Triangle to inſcribe a Circle.

PROBLEM XXX. About any given Triangle to circumſcribe a Circle.

PROBLEM XXXI. In, or about, a given Square, to deſcribe a Circle.

PROBLEM XXXII. In, or about, a given Circle, to deſcribe a Square, or an Octagon.

PROBLEM XXXIII. In a given Circle, to inſcribe a Trigon, a Hexagon, or a Dodecagon.

PROBLEM XXXIV. In a given Circle to inſcribe a Pentagon, or a Decagon.

PROBLEM XXXV. To divide the Circumference of a given Circle into 12 equal Parts, each of 30 Degrees.¶ Or to inſcribe a Dodecagon by another Method.

PROBLEM XXXVI. To draw a right Line equal to the Circumference of a given Circle.

PROBLEM XXXVII. To find a Right Line equal to any given Are AB of a Circle.

PROBLEM XXXVIII. To divide a given Circle into any propoſed Number of Parts by equal Lines, ſo that thoſe Parts ſhall be mutu⯑ally equal, both in Area and Perimeter.

PROBLEM XXXIX. To make a Triangle ſimilar to a given Triangle ABC.

PROBLEM XL. To make a Figure ſimilar to any other given Figure ABCDE.

PROBLEM XLI. To reduce a Complex Figure from one Scale to another, mechanically by means of Squares.

PROBLEM XLII. To make a Triangle equal to a given Trapezium ABCD.

PROBLEM XLIII. To make a Triangle equal to the figure ABCDEA.

PROBLEM XLIV. To make a Triangle equal to a given Circle.

PROBLEM XLV. To make a Rectangle, or a Parallelogram equal to a given Triangle ABC.

PROBLEM XXXV. To divide the Circumference of a given Circle into 12 equal Parts, each of 30 Degrees.
¶ Or to inſcribe a Dodecagon by another Method.