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

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

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

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

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

ADDITION OF DECIMALS.

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.

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

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

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

SUBTRACTION OF DECIMALS.

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.

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

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.

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

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

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

DIVISION OF DECIMALS.

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.

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.

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.

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

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.

OTHER EXAMPLES.

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

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.

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.

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 〉

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.

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.

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.

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.

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 〉

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]

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.

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 3^½, the root of 50 is 50^½, and the third root of it is 50⅓, also the third root of 47—15 is 〈 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 req^d.

〈 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.

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

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

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.

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.

DEFINITIONS.

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

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

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

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.

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

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.

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

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

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

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

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

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

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

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.

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.

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

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.

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

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

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

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

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

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

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.

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

35. A rhomboid is an oblique-angled parallelogram.

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

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

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

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

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.

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.

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

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

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

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

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

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

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.

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.

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.

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.

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

PROBLEMS.

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.

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.

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 〉

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

〈 math 〉

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.

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 〉

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 〉

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 〉

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 〉

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

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 〉

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

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

EXAMPLES.

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

〈 math 〉

〈 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.

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 〉

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.

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 〉

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 〉

〈 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 〉

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

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 〉

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 〉

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.

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.

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.

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.

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

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

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.

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.

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.

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.

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.

Page  125Note also, that, in cubic measure,

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 〉

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 〉

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 〉

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 〉

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

Ans. 90.

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

Ans. 39 13/16.

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

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  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 〉

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

Ans. 464.205.

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

Ans. 19½.

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

Ans. 527.7888.

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

Ans. 24.681722.

PROBLEM IX. To find the Solidity of a Wedge.

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 〉

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 〉

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 〉

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 〉

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 〉

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 〉

Page  141Then, by our rule, 〈 math 〉

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.

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.

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.

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 〉

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

• Ans. superficies 24
• Ans. solidity 8

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

• Ans. superficies 13.85640
• Ans. solidity 3.77120

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

• Ans. superficies 82.58292
• Ans. solidity 61.30496

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

• Ans. superficies 34.64100
• Ans. solidity 17.45352

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 〉

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.

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.

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.

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. 20^f 5′ 8″.

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

Ans. 3s 3¾d.

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

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

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

By the Sliding Rule.

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.

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.

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

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.

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.

Ex. 2. Required the content of a wall 62 feet 6 inches long, and 14 feet 8 inches high, and 2½ bricks thick.

Ans. 169.753 yards.

Ex. 3. A triangular gable is raised 17 1/• feet high, on an end wall whose length is 24 feet 9 inches, the thick∣ness being 2 bricks: required the reduced content.

Ans. 32.08⅓ yds.

Ex. 4. The end wall of a house is 28 feet 10 inches long, and 55 feet 8 inches high to the eaves; 20 feet high is 2½ bricks thick, other 20 feet high is 2 bricks thick, and the remaining 15 feet 8 inches is 1½ brick thick; above which is a triangular gable, which rises 42 courses of bricks, of which every 4 courses make a foot. What is the whole content in standard measure.

Ans. 253.626 yards.

MASONS WORK.

TO masonry belongs all sorts of stone work; and the measure made use of is a foot, either superfi∣cial or solid.

Walls, columns, blocks of stone or marble, &c. are measured by the cubic foot; and pavements, slabs, chimney-pieces, &c. by the superficial or square foot.

Cubic or solid measure is used for the materials, and square measure for the workmanship.

In the solid measure, the true length, breadth, and thickness, are taken, and multiplied continually toge∣ther. In the superficial, there must be taken the length and breadth of every part of the projection, which is seen without the general upright face of the building.

EXAMPLES.

1. Required the solid content of a wall, 53 feet 6 inches long, 12 feet 3 inches high, and 2 feet thick.

〈 math 〉

Page  165By the Sliding Rule.

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

Ans. 21f 10in.

CARPENTERS AND JOINERS WORK.

TO this branch belongs all the wood-work of a house, such as flooring, partitioning, roofing, &c.

Note. Large and plain articles are usually measured by the square foot or yard, &c. but inriched mouldings, and some other articles, are often estimated by running or lineal measure, and some things are rated by the piece.

In measuring of joists, it is to be observed, that only one of their dimensions is the same with that of the floor; and the other will exceed the length of the room by the thickness of the wall, and ⅓ of the same, because each end is let into the wall about ⅔ of its thickness.

No deductions are made for hearths, on account of the additional trouble and waste of materials.

Partitions are measured from wall to wall for one di∣mension, and from floor to floor, as far as they extend, for the other.

No deduction is made for door-ways, on account of the trouble of framing them.

In measuring of joiners work, the string is made to ply close to every part of the work over which it passes.

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.

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•.

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;

Required the expence of glazing at 14d per foot.

Ans. £ 13 11 10^r.

Ex. 4. Required the expence of glazing the windows of a house at 13d a foot; there being three stories, and three windows in each story;

Page  174

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.

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.

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.

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.

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 〉

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 〉

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.

CHAPTER 1. Description and Use of the Instruments.

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.

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

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.

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

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

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

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.

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.

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.

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.

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.

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 ell•pse.

PROBLEM II. In an Ellipse, to find the Transverse, or Conjugate, or Ordinate, or Absciss; having the other three given.

CASE I. To find the Ordinate.
• As the transverse∶
• Is to the conjugate∷
• So is the mean proportional between the two abscisses∶
• To the ordinate.

EXAMPLES.

1. In the ellipse ADBC, the transverse AB is 70, the conjugate CD is 50, and the abscisses AP 14, and PB 56; what is the ordinate PQ?

Page  233First 〈 math 〉

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 〉

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.

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.

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.

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 〉

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.

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 〉

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 〉

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.

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 〉

〈 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 〉

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 〉

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 〉

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 〉

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 〉

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 〉

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 〉

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.

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.

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

Or the particular rule will be for each as in the fol∣lowing problems.

PROBLEM XVIII. To find the Ullage by the Sliding Rule.

By one of the preceding problems find the whole content of the cask. Then set the length on N to 100 on SS for a segment standing, or set the bung diameter on N to 100 on SL for a segment lying; then against the wet inches on N is a number on SS or SL, to be re∣served.

Next, set 100 on B to the reserved number on A; then against the whole content on B will be found the ullage on A.

EXAMPLES.

1. Required the ullage answering to 10 wet inches of a standing cask, the whole content of which is 92 gal∣lons, and length 40 inches.

Having set 40 on N to 100 on SS, then against 10 on N is 23 on SS, the reserved number.

Then set 100 on B to 23 on A, and against 92 on B is 21.2 on A, the ullage required.

Ex. 2. What is the ullage of a standing cask whose whole length is 20 inches, and content 11 •/• gallons; the wet inches being 5?

Ans. 2.65 gallons.

Ex. 3. The content of a cask being 92 gallons, and the bung diameter 32, required the ullage of the seg∣ment lying when the wet inches are 8.

Ans. 16.4 gallons.

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

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 〉

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.