The compendious measurer: being a brief, yet comprehensive, treatise on mensuration and practical geometry. ... Adapted to the use of schools ... By Charles Hutton, ...
Hutton, Charles, 1737-1823.
Page  [unnumbered]

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.

    Page  2
  • 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.

Page  3Ciphers 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.

Page  4

ADDITION OF DECIMALS.

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.

Page  5

SUBTRACTION OF DECIMALS.

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.

Page  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 Page  7intend 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.

Page  8

DIVISION OF DECIMALS.

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

    Page  9
  • 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 Page  10will 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.

Page  114. 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 =

Page  12So 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.

Page  13II. 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.

Page  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

Page  15IV. 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:

Page  16Write 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.

Page  17

CIRCULATING DECIMALS.

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.

Page  18If 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 〈 math 〉
  • 2. And 〈 math 〉
  • 3. And 〈 math 〉
  • 4. And 〈 math 〉
Page  19

ADDITION OF REPETENDS.

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

Page  20

SUBTRACTION OF REPETENDS.

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 Page  21lines 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  

Page  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

Page  23〈 math 〉

〈 math 〉

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 〈 math 〉

Or rather thus: Having found 〈 math 〉, then 〈 math 〉

Page  24

INVOLUTION; OR RAISING OF POWERS.

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.

Page  [unnumbered]

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

Page  26The 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, Page  27then 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.

〈 math 〉

Page  28〈 math 〉

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.

Page  29

EVOLUTION; OR EXTRACTION OF ROOTS.

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 〈 math 〉. and sometimes roots are designed like powers, with the reciprocal or Page  30the index of the root above the given number. So the root of 3 is 3½, the root of 50 is 50½, and the third root of it is 50⅓, also the third root of 47—15 is 〈 math 〉. 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 Page  31the 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 Page  32operation 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 Page  3382 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. 〈 math 〉

Ex. 3. For the Root of .000729. 〈 math 〉

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.

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

EXAMPLE 4.

〈 math 〉

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 Page  35pointing 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.

〈 math 〉

Ex. 6. For the root of 5.

〈 math 〉

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 Page  36required. 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 〈 math 〉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. 〈 math 〉is ⅚

Ex. 2. 〈 math 〉or 〈 math 〉is 3/7;

Ex. 3. For the root of 9/12 〈 math 〉

Page  37Ex. 4. For the root of 5/12 〈 math 〉

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.

Page  38Ex. 2. To find a Mean between 2 and 5.

Here 2 x 5=10 (3.162278 the mean reqd.

〈 math 〉

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.

Page  39Thus for the 4th root of 97.41.

〈 math 〉

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.
Page  40

TO EXTRACT THE CUBE ROOT.

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.

Page  416. 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?

〈 math 〉

Page  42If 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.

Page  43

TO FIND TWO MEAN PROPORTIONALS.

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.

Page  44If 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.

Page  45EXAMPLE. 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 〈 math 〉

Page  46Again, for a second operation, the cube of this root is 21035.318645155823, and the process by the latter method will be thus: 〈 math 〉

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, 〈 math 〉.

Or, 〈 math 〉.

Page  47And 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 〈 math 〉.

Page  48OTHER 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.

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

This is 〈 math 〉.

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.

Page  50This 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.
    Page  51
  • 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 〈 math 〉

Page  52
RULE II.

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 〈 math 〉

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.

Page  53Ex. 3 Multiply 〈 math 〉

OTHER EXAMPLES.

〈 math 〉

Page  [unnumbered]

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.

Page  [unnumbered]

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.

[illustration] [diagram]

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

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

Page  576. A right line, or straight line, lies all in the same direction, be∣tween its extremities; and is the shortest distance between two points.

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [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.

[illustration] [diagram]

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

Page  5817. An acute angle is less than a right angle.

[illustration] [diagram]

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

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

Page  5927. A right-angled triangle, is that which has one right angle.

[illustration] [diagram]

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

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

[illustration] [diagram]

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

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

35. A rhomboid is an oblique-angled parallelogram.

[illustration] [diagram]

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

[illustration] [diagram]

Page  6037. A trapezium is a quadrilate∣ral which hath not both its pairs of opposite sides parallel.

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [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.

[illustration] [diagram]

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

Page  6145. The radius of a circle, is a right line drawn from the center to the circumference.

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

Page  6251. A sector is any part of a circle, bounded by an arc, and two radii drawn to its extremities.

[illustration] [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.

[illustration] [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.

[illustration] [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.

[illustration] [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.

Page  6357. 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.

[illustration] [diagram]

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

Page  64

PROBLEMS.

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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  65
PROBLEM IV. To draw a Line Parallel to a given Line AB.

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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  66CASE 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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

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

Page  67
PROBLEM VI. From a given Point A, out of a given Line BC, to let fall a Perpendicular.

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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  68N. 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.

[illustration] [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.

[illustration] [diagram]

Page  69
PROBLEM IX. At a given Point A, in a given Line AB, to make an Angle equal to a given Angle C.

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.

[illustration] [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.

[illustration] [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.

Page  70
PROBLEM XI. To measure a given Angle A. (See the last figure.)

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.

[illustration] [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.

[illustration] [diagram]

Page  71Note. 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.

[illustration] [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:

[illustration] [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.

[illustration] [diagram]

Page  72
PROBLEM XVI. To find a Fourth Proportional to three given Lines, AB, AC, AD.

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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  75
PROBLEM XIX. To make a Triangle with three given Lines AB, AC, BC.

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.

[illustration] [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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  74Note. 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.

[illustration] [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.

[illustration] [diagram]

Page  75A 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.

[illustration] [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.

[illustration] [diagram]

Page  76
PROBLEM XXIV. To make an Octagon on a given Line AB.

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.

[illustration] [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.

[illustration] [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 Page  77sides, 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.

Page  78Another 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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  79
PROBLEM XXVIII. To find the Center of a given Polygon, or the Center of its inscribed or circumscribed Circle.

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.

[illustration] [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.

[illustration] [diagram]

Page  80
PROBLEM XXX. About any given Triangle to circumscribe a Circle.

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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  81Note. 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.

[illustration] [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.

Page  82
PROBLEM XXXIV. In a given Circle to inscribe a Pentagon, or a Decagon.

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.

[illustration] [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.

[illustration] [diagram]

Page  83
PROBLEM XXXVI. To draw a right Line equal to the Circumference of a given Circle.

[illustration] [diagram]

Take III 1 equal to 3 times the diameter and 〈 math 〉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.

[illustration] [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.

[illustration] [diagram]

Page  84
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.

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.

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

Page  85
PROBLEM XL. To make a Figure similar to any other given Figure ABCDE.

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.

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

Page  86

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  87Note. 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.

[illustration] [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.

[illustration] [diagram]

Page  88
PROBLEM XLVI. To make a Square equal to a given Rectangle ABCD.

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.

[illustration] [diagram]

{inverted ⁂} 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.

[illustration] [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.

Page  89
PROBLEM XLVIII. To make a Square equal to the Difference between two given Squares PR. (See the last figure).

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.

[illustration] [diagram]

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

Page  90
PROBLEM L. To make plane Diagonal Scales.

[illustration] [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 Page  91extend 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.

[illustration]
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.

Page  92These, 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.

[illustration] [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.

Page  [unnumbered]

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

Page  94

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

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.

〈 math 〉

[illustration] [diagram]

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

〈 math 〉

[illustration] [diagram]

Page  953. To find the area of a rhombus, whose length is 6.20 chains, and perpendicular height 5.45.

〈 math 〉 Ans. 3 acres, 1 rood, 20 perches.

[illustration] [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.

〈 math 〉

[illustration] [diagram]

Page  965. 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.

Page  97〈 math 〉

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.

〈 math 〉

[illustration] [diagram]

Ans. 91/3 square yards

Page  983. 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.

Page  99〈 math 〉

[illustration] [diagram]

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

〈 math 〉

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.

Page  1006. 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.

〈 math 〉

[illustration] [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.

Page  1013. 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

[illustration] [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.

Page  102EXAMPLES.

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

〈 math 〉

[illustration] [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.

〈 math 〉

    Page  103
  • 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.

〈 math 〉

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.

Page  1046. 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

[illustration] [diagram]

Page  105〈 math 〉

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.

Page  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.

〈 math 〉

[illustration] [diagram]

〈 math 〉

Page  1072. 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.

[illustration] [diagram]

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

Again, when the radius, and height PD are given. Then CD−DP=CP; and 〈 math 〉.

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.

Page  108Then 〈 math 〉; and CD−CP=10−6=4=PD; and 〈 math 〉.

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

Then CD−DP=10−4=6=CP; and 〈 math 〉.

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.

〈 math 〉

[illustration] [diagram]

Page  109〈 math 〉

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.

〈 math 〉ans.

By Rule 2.

〈 math 〉

By Rule 3.

〈 math 〉

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.

Page  110

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

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.

〈 math 〉

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.

〈 math 〉

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

Ans. 2 1234.

Page  111Ex. 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.

〈 math 〉

Page  112〈 math 〉

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.

Page  113RULE 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.

〈 math 〉

[illustration] [diagram]

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

〈 math 〉

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.

Page  115〈 math 〉

[illustration] [diagram]

〈 math 〉

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.

Page  116Ex. 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

〈 math 〉

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.

〈 math 〉

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.

Page  117

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

[illustration] [diagram]

[illustration] [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 〈 math 〉

and 〈 math 〉

〈 math 〉

Page  118Ex. 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,

〈 math 〉

TP=TQ=½DS=13

OP=OT−TP=27−13=14

OQ=OT + TQ=27 + 13=40

〈 math 〉

EQ=OE−OQ=50−40=10

EP=OE−OP=50−14=36

〈 math 〉

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

〈 math 〉

OP=TP−OT=½PQ−OT=15

〈 math 〉

〈 math 〉

GR=OG − OR=25 − 17.677669=7.322331

Page  119〈 math 〉

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.

〈 math 〉

[illustration] [diagram]

Page  120Ex. 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.

〈 math 〉

[illustration] [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.

Page  [unnumbered]

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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

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

[illustration] [diagram]

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

[illustration] [diagram]

Page  1226. 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.

[illustration] [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.

[illustration] [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.

[illustration] [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.

Page  12311. 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.

[illustration] [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, Page  124which 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.

[illustration] [diagram]

[illustration] [diagram]

[illustration] [diagram]

[illustration] [diagram]

[illustration] [diagram]

Page  125Note 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?

〈 math 〉

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, Page  126all 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.

〈 math 〉

[illustration] [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?

Page  127Here 5/2 x 5/2=25/4=6¼

Then .433013 tabular n°.

〈 math 〉

[illustration] [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?

Page  128〈 math 〉

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.

〈 math 〉

[illustration] [diagram]

Page  129Ex. 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.

〈 math 〉

[illustration] [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.

Page  130

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

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

〈 math 〉

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?

〈 math 〉

See Fig. to Def. 6.

Page  131Ex. 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.

Page  130〈1 page duplicate〉Page  131〈1 page duplicate〉

Page  132Or, 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?

〈 math 〉

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.

Page  133〈 math 〉

[illustration] [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.

Page  134

PROBLEM IX. To find the Solidity of a Wedge.

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?

〈 math 〉

[illustration] [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.

Page  135RULE.

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?

〈 math 〉

[illustration] [diagram]

Page  136Ex. 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.

〈 math 〉

[illustration] [diagram]

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

Ans. 151.

Page  137Ex. 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?

〈 math 〉

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

Ans. 9.9288 feet.

Page  138Ex. 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.

〈 math 〉

[illustration] [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.

Page  139EXAMPLES.

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

〈 math 〉

[illustration] [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 Page  140the 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 〈 math 〉, and 〈 math 〉, hence OE=OC−CE=20⅚−15=5⅚.

Also, by problem x, rule 2, 〈 math 〉

[illustration] [diagram]

Page  141Then, by our rule, 〈 math 〉

[illustration] [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.]

Page  142By the work of the last problem, 〈 math 〉

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 Page  143half 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.

[illustration] [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 〈 math 〉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, 〈 math 〉

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

〈 math 〉

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.

Page  145

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

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 Surfacs 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, 〈 math 〉

[illustration] [diagram]

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

  • Ans. superficies 24
  • Ans. solidity 8

[illustration] [diagram]

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

  • Ans. superficies 13.85640
  • Ans. solidity 3.77120

[illustration] [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

[illustration] [diagram]

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

  • Ans. superficies 34.64100
  • Ans. solidity 17.45352

[illustration] [diagram]

Page  147

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

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.

〈 math 〉

[illustration] [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.

Page  148EXAMPLES.

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

〈 math 〉

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

Ans. 789.568.

Page  [unnumbered]

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 Page  150lines, 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 Page  15124 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.
Page  152

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

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.
Page  [unnumbered]

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?

〈 math 〉

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.

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

Ans. 20f 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 Page  155half 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.

〈 math 〉

Page  156By 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 Page  157is 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?

〈 math 〉

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.

Page  158Ex. 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?

〈 math 〉

Page  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.

Page  [unnumbered]

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.

Page  [unnumbered]

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.

Page  162All 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?

〈 math 〉

Page  163By 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.

Page  [unnumbered]

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.

〈 math 〉

Page  165By 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.

Page  [unnumbered]

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.

Page  167The 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 Page  168for 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.

〈 math 〉

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.

Page  169Ex. 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.

Page  170〈 math 〉

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.

Page  171EXAMPLES.

1. How many yards contains the ceiling, which is 43 feet 3 inches long, and 25 feet 6 inches broad?

〈 math 〉

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. 53vd 5t 3i of rendering
  18 5 6 of ceiling
    39 0 1/1 ½ of cornice.

Page  172

PAINTERS WORK.

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?

〈 math 〉

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.

Page  173

GLAZIERS WORK.

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?

〈 math 〉

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 10r.

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;

Page  174

the height of the lower tier is 7f 9
of the middle 6 6
of the upper 5
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?

〈 math 〉

Page  175Ex. 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?

Page  176〈 math 〉

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.

Page  177Domes 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.

〈 math 〉

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.

Page  178

PROBLEM II. To find the Content of the Concavity of a Vaulted Roof.

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.

〈 math 〉

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.

Page  179

PROBLEM III. To find the Superficies of a Dome.

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?

〈 math 〉

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.

Page  180

PROBLEM IV. To find the Solid Content of a Dome.

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.

〈 math 〉

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.

〈 math 〉

Page  181

PROBLEM VI. To find the Solid Content of a Saloon.

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.

〈 math 〉

[illustration] [diagram]

Page  182Again 〈 math 〉

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?

Page  183〈 math 〉

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.

〈 math 〉

Page  1842. 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.

Page  [unnumbered]

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

Page  186The 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.

〈 math 〉

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,

Page  1871. 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 Page  188as 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 Page  189the 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 Page  190at 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 Page  191each 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 Page  192certainly 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. Page  193It 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, &c; 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 Page  194an 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  
  69 16 a spring
foot path 16 810  
cross hedge 18 973 20 a pond

Page  195But 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 Page  196the 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.

Page  197
PROBLEM II. To take Angles and Bearings.

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.

[illustration] [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 pn, 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.

Page  1984. 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

[illustration] [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, Page  199find 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

[illustration] [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, Page  200the 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  

[illustration] [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  

[illustration] [diagram]

Page  201Measure 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.

Page  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  

[illustration] [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.

Page  203Thus, 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

[illustration] [diagram]

PROBLEM VII. To Survey any Field with the Plain Table.

1. From one Station.

Plant the table at any angle, as C, from whence all the other angles, or marks set up, can be seen; and turn the table about till the needle point to the flower-de-luce; and there screw it fast. Make a point for C on the paper on the table, and lay the edge of the index to C, turning it about C till through the sights you see the mark D; and by the edge of the index draw a dry or obscure line: then measure the distance CD, and lay that distance down on the line CD. Then turn the in∣dex about the point C, till the mark E be seen through the sights, by which draw a line, and measure the distance to E, laying it on the line from C to E. In like manner determine the positions of CA and CB, by turn∣ing

[illustration] [diagram]
Page  204the sights successively to A and B; and lay the lengths of those lines down. Then connect the points with the boundaries of the field, by drawing the black lines CD, DE, EA, AB, BC.

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.

[illustration] [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 Page  205do 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 Page  206OA, 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.

[illustration] [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 bePage  207equal 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.

[illustration] [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.

Page  208And 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.

[illustration] [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.

Page  209

[illustration] [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.

Page  210When 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.

Page  2113. 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 Page  212survey 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 Page  213draught, 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 Page  214angles, 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. Page  215And 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 Page  216pole 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.

Page  217

[illustration] [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.

Page  218

CHAPTER III. Of Planning, Casting-up, and Dividing.

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 Page  219therefore 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 Page  220there 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

〈 math 〉

Page  221Or the first example to prob. v, page 200, thus: 〈 math 〉

Or the 2d example to the same prob. v, thus: 〈 math 〉

Page  2223. 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 〈 math 〉

Page  2234. 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: 〈 math 〉

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 Page  224separately, 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 Page  225may 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.

[illustration] [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.

Page  226Then 〈 math 〉

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. Page  227This 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.

Page  228
FIFTH METHOD.

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, bshes, 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.

Page  [unnumbered]

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.

[illustration] [diagram]

3. If the cone be cut parallel to the base, the section will be a cir∣cle.

[illustration] [diagram]

Page  2304. The section is called an ellipsis, when the cone is cut ob∣liquely through both sides.

[illustration] [diagram]

5. The section is a parabola, when the cone is cut parallel to one of its sides.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

[illustration] [diagram]

Page  2319. 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.

[illustration] [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.

[illustration] [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.

[illustration] [diagram]

Page  232Assume 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 ellpse.

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?

Page  233First 〈 math 〉

[illustration] [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.

Page  234First 〈 math 〉

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.

Page  235〈 math 〉

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?

Page  236〈 math 〉

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.

〈 math 〉

Page  237Ex. 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.

〈 math 〉

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.

Page  238EXAMPLES.

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?

〈 math 〉

[illustration] [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.

[illustration] [diagram]

Page  239Then 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.

[illustration] [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.

Page  240EXAMPLES.

1. The absciss VB being 2, and the ordinate AB 6, required the length of the curve AVC.

〈 math 〉

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.

Page  241EXAMPLES.

1. Required the area of the parabola AVCA, the ab∣sciss VB being 2, and the ordinate AB 6.

〈 math 〉

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.

〈 math 〉

Page  242Ex. 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.

[illustration] [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.

Page  243Note. 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?

〈 math 〉

[illustration] [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.

Page  244EXAMPLES.

1. The transverse being 24, and the conjugate 21; required the two abscisses to the ordinate AB 14.

〈 math 〉

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.

Page  245EXAMPLES.

1. The transverse being 24, the less absciss VB 8, and 〈◊〉, ordinate AB 14; what is the conjugate?

〈 math 〉

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.

Page  246〈 math 〉

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.

Page  247〈 math 〉

Then 2358720∶3078720∷48∶62.6520 the whole curve 〈 math 〉

Page  248Ex. 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.

[illustration] [diagram]

Page  249〈 math 〉

Page  250〈 math 〉

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.

〈 math 〉

[illustration] [diagram]

Page  251Ex. 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.

〈 math 〉

[illustration] [diagram]

Page  252〈 math 〉

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.

Page  253〈 math 〉

[illustration] [diagram]

〈 math 〉

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.

Page  254EXAMPLES.

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.

〈 math 〉

[illustration] [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.

Page  255CASE 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.

〈 math 〉

[illustration] [diagram]

Page  256Ex. 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.

Page  2571. For the Central Distance, and Axes of the Ellipse.

〈 math 〉

[illustration] [diagram]

Then as 12∶20 (or AG)∷30 (or CH)∶50=IK the transverse.

2. For the Generating Elliptic Segment.

〈 math 〉

Page  2583. For the Solidity of the Spindle.

〈 math 〉

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.

Page  259Note. 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?

〈 math 〉

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 Page  260formed 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.

〈 math 〉

[illustration] [diagram]

Page  261Ex. 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.

〈 math 〉

Page  262Ex. 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.

〈 math 〉

[illustration] [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.

Page  263EXAMPLES.

1. Required the content of the parabolic spindle ACBD, whose length AB is 40, and the greatest diameter CD 16.

〈 math 〉

[illustration] [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.

Page  264EXAMPLES.

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.

〈 math 〉

[illustration] [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.

Page  [unnumbered]

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 Page  266middle 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,Page  267for 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.

Page  268
PROBLEM IV. To Extract the Square Root, as of 225.

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.

Page  269
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.

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 Page  270to 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 Page  271in 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.

[illustration] [diagram]

Page  272〈 math 〉

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.
Page  273
PROBLEM XIII. To find the Content of a Cask of the Second Form.

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.

[illustration] [diagram]

〈 math 〉

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.

Page  274And 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.

[illustration] [diagram]

〈 math 〉

Page  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.

[illustration] [diagram]

Page  276〈 math 〉

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
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.
Page  277
PROBLEM XVI. To find the Content of a Cask by Four Dimensions.

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.

〈 math 〉

Page  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.

Page  279〈 math 〉

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.

Page  280

OF THE ULLAGE OF CASKS.

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.

Page  281

PROBLEM XIX. To Ullage a Standing Cask by the Pen.

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.

〈 math 〉

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 Page  282corresponding 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.

〈 math 〉

Page  [unnumbered]

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

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, Page  284not 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.

〈 math 〉

Page  285Ex. 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.

〈 math 〉

Page  286Ex. 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.

〈 math 〉

CASE 2. When the body is lighter than water, so that it will not quite sink; assix to it a piece of another Page  287body 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 Page  288the 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.

〈 math 〉

Answer there is 100lb of copper and conseq. 12lb of tin in the composition.

Page  [unnumbered]

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. Page  290Or 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.

〈 math 〉

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.

Page  291EXAMPLES.

1. Required the weight of a leaden ball of 6.6 inches diameter.

〈 math 〉

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.

Page  292〈 math 〉

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.

〈 math 〉

Ex. 2. What is the diameter of a 24lb iron ball?

Ans. 5.54 inches.

Page  293

PROBLEM IV. To find the Diameter of a Leaden Ball.

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.

〈 math 〉

The cube root of which is almost 7, whose cube is 343.

〈 math 〉

Ex. 2. What is the diameter of an 8lb leaden ball?

Ans. 3.303 inches.

Page  294

PROBLEM V. To find the Weight of an Iron Shell.

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.

〈 math 〉

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.

Page  295

PROBLEM VI. To find how much Powder will fill a Shell.

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?

〈 math 〉

Ex. 2. How much powder will fill the shell whose in∣ternal diameter is 7 inches?

Ans. 6lb.

Page  296

PROBLEM VII. To find how much Powder will fill a Rectangular Box.

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.

〈 math 〉

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?

Page  297〈 math 〉

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?

〈 math 〉

Page  298The cube root of this is nearly 9, whose cube is 729.

〈 math 〉

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.

Page  299〈 math 〉

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?

Page  300〈 math 〉

Ex. 2. What length of a cylinder of 8 inches diameter may be filled with 20lb of powder?

Ans. 11 15/16.

Page  [unnumbered]

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 Page  302the 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.

〈 math 〉

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?

Page  303〈 math 〉

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.

Page  304〈 math 〉

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.

Page  305〈 math 〉

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.

Page  [unnumbered]

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.

Page  307EXAMPLES.

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.

〈 math 〉

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?

〈 math 〉

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.

Page  [unnumbered]

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 Page  309same 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 Page  310from 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 Page  311the 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

Page  31231. 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 Page  313one 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.

Page  31448. 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.

Page  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.

Page  31661. 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.

Page  [unnumbered]

A TABLE OF THE AREAS of the SEGMENTS of a CIRCLE, Whose Diameter is Unity, and supposed to be divided into 1000 equal Parts.

Page  318Page  319Page  320Page  321
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    

Page  [unnumbered]

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.

Page  323But, 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.