lon answering to it should be 2683 cubical inches ; 2 gallons Part III. make a peck, and 32 pecks, or 8 bullels, a quarter..

The Scots wheat-firlot contains 21 pints, or 2199 cubical inches, and the barley-firlot 31 pints, or 3208 cubical inches ; 4 firlots make a boll, and 16 bolls a chalder ; a peck is the fourth part of a firlot, and a lippie che fourth part of a peck.

A Paris pint is nearly equal to 2 English pints, and contains 48 cubical Paris inches.

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If it be a rectangular parallelopiped, of which the altitude is one inch, it is evident, that there will be as many cubical inches in its content, as there are square inches in its base ; and if its altitude be any number of inches, the content will be as many times the number of cubical inches ; therefore find the area of the base, and multiply it by the height, and the product is the solid content: and every prism is equal a to a rectangular paral- a 2. Cor. lelopiped, of the fame base and altitude with it. Therefore, if 32.11. E. the base found by the 4th or 5th Problem of the 2d Part be 96 inches, and the height 11 inches, the content is 1056 solid inches.



To find the solid content of a pyramid.

Find the area of the base by the 4th or sth Problem of the 2d Part, and multiply it by s of the height, the product is the content, by 4th Prop. 12th Book E.

Cor. If the solid content of a frustum of a pyramid be required, find the content of the entire pyramid, and of the part that is wanting, and their difference is the content of the fruftum: or find the areas of the bases, and multiply them by corresponding sides of these bases; then, as the difference of these sides to the height of the fruftum, so à is the difference a 4. 6. and of the products to the content.

19. 5. E

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Part III.

Find the area of the base, by Prob. 6. cr 7. of the 2d Part, mand multiply it by the height, the product is the folid content a 5. 12. E. Thus, if the diameter of the base be 14, and the height 12 and 1. of inches, the area of the base is 153-9384, which, multiplied by tuis,

12, gives 1844.2603 folid inches for the content.
Pl. III. Cor. If the bung-diameter EF of a cak ABCD be not much
is. 13. greater than the head-viameter AB; find the areas of the circles

at the head and bung, and take half the sum for a mean base,
which, multiplied by the length, gives the content; only the
dimensions are to be taken within the Itaves.

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12. E.

Find the area of the base, by Prop, 6. ad Part, and multiply å i.Cor.7. it by of the height, to get the folid contenta. For example,

if the diameter of the base be 8 inches, and the height 9 inches,
the area of the base is 50.2656, which, multiplied by 9, gives

452.39 folid inches for the content. Pl. III,

COK. 1. If the content of the frustum of a cone be required, lig. 14. let it be ABCD; draw "AG, parallel to DE; then GC : CD:: bj. 1. E. AC: CE :: AH: EF , the altitude of the complete cone ; and 6 4. 6. E. if the content of the complete cone ECD, and of the part EAB,

be found, their difference is ABCD. Which may be done, by
fubtracting the product of the diameters CD and AB from the
square of their lum, the remainder, multiplied by :7854, and

then by of the height AH, gives the content of the fruftum. P.III.

Cor. 2. Casks, of which the staves are very much bended toFig. 15. wards the middle, and straight towards the ends, as ABEF,


be taken for two frultums of a cone.

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To find the folid content of a sphere.

Because the sphere is of its circumscribed cylinder, find a %. 12. E.

the area of a great circle of the sphere, and multiply it by of
the axis, the product is the content: or, which is the same
thing, multiply the cube of the axis by Ź of .7854, that is,
:5236. For example, let the axis be 9 inches, the cube of it is
729, which, multiplied by.5236, gives 381,7044 for the solid


Cox. I.


Cor. 1. It may be proved, as was done in the 9th Proposition Part III. of the 12th Book, that the folid described by the revolution of the part BGMC of the circle, together with the cone described by the triangle BGP, is equal to the cylinder described by the rectangle BH; and therefore the I


content of the zone described by BGMC
is got by subtracting § of the square of

GP or GB the height, from the square E
of BC", and multiplying the remainder


b z. 12. L. by 3.1416, and then by the height BG.

Cor. 2. Hence the content of a seg-
ment of the sphere may be found, by subtracting the remaining
zone from the hemisphere.

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a 4. 6. L.

To find the solid content of a spheroid.

A spheroid is a solid described by the revolution of a semiellipse about one of its axis.

If ACR be a semiellipse, and the pre-
paration be made as in the gth Proposition A

of the 12th Book, it may be proved, that
the spheroid described by the semiellipse I NA

ACR, about the axis AR, is of the cir- G P/
cumscribing cylinder: For, by the nature
of the ellipse, the square of EL is to the
rectangle AE, ER, as the square of DA
is to that of AB, that is, as a the square


of EO to that of EB; therefore the squares
of EL, EO are to the rectangle AE, ER,
and the square of EB, as the square of DA
to that of AB b; but the rectangle AE,

b 12. S.E. ER, with the square of EB, is equal to

C 5. 2. E.

R the square of AB; therefore the squares of LE, EO are equal d to that of AD or EF;

d 14. s. E and the cylinders described by the revolution of the rectangles BL, BO, are therefore equal to the cylinder described by Cor. From which it may be proved, that the cone described by BAD,

6. 12. f. together with the hemispheroid described by AMC, is equal to the cylinder described by AC, as was done in the gth Prop. of the 12th Book. And the cone is a third part of the cylinderf; f 1. Cor. 7. therefore the hemispheroid is of it. Therefore, find the area of the circle described by the revolving axis, and multiply ic by of the fixed asis.

Cor. I.

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PART III. Cor. I. In the fame manner, it may be proved, that the part

of the spheroid described by the revolution of BGMC, together
with the cone described by BGP, is equal to the cylinder de.
scribed by BGHC. Now, the circle described by GP is the
excess of the circle described by GH or BC, above that descri-
bed by GM; therefore, from three times the area of the greater
base, subtract the excess of this area above the area of the less
base ; or, which is the same thing, to twice the area of the

greater base add the area of the less base, and the sum is the f 1. Cor.7. base of a cone, equal' in height and content to the zone of the

PI. III. Cor. 2. When the staves of a cask are much bended, it is
Fig. 16.

supposed to be the middle zone of a spheroid: Let the length of
such a cask be 40 inches, and its bung-diameter 32 inches, and
its head diameter 26 inches; to 2048, twice the square of 32,
add 676, the square of 16, and multiply the sum 2724 by 40,
the length, and then by .2618, j of 7854, and the product
28525.728 is the solid content in inches, which, divided by
282, gives for gallons I pint of ale ; or if it be divided by
231, the quotient is 123 gallons i pint of wine.

12. E.

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To find the folid content of any regular body.

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The folidity of the tetrahedron, which is a pyramid, is found by the 2d Problem of this Part. The hexahedron, or cube, is found, Prob. 1, of this. The octahedron, being two pyramids of equal heights, upon the same square for a base, is measured by Prob. 2.

The dodecahedron consists of 12 equal pyramids, upon regular pentagons for their bases, and may be measured by Prob. 2. of this. The icosahedron consists of twenty equal pyramids, upon triangular bases, and may be measured by the fame problem. The bases' and heights of these pyramids may be either measured, or else found by Trigonometry:

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To find the folidity of any body, however irregular.

Let the body be immersed in water, in a veffel of the figure of a prism, and take notice how much the water is raised by the immersion of the body: for it is plain, that the space which


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the water fills, after immerfing the body, exceeds the space it Part III,
occupied before, by a space equal to the folid content of the
body; and this excefs may be found, by Prob. 1. by multiplying
the area of the mouth of the vessel, by the difference between the
elevations of the water before and after immersion. In the
fame manner, the solidity of a part of a body may be found, by
immersing that part only.

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O find how much is contained in a cask, that is Pl. III.

in part empty, of which the axis is parallel to Fig. 17.& the horizon.

Let AGBH be the circle made by cutting the cask at the bung, by a plane perpendicular to the axis, AB the bung-diameter, and GBH the segment of it, filled with liquor, of which the depth EB may be found by the guaging rod; and the diameter AB, being also known, the area of the segment GBH may be found, by the ad Cor. to Prob. 6. Part 2. or it may be found, from a table of segments, such as that at the end of guaging.

Find the mean base of the cask, which, if the cask be not of the kind mentioned in Prob. 3. of this part, may always be found, by dividing the content of the cask by its length; let this be the circle CKDL, and let the segment of it, fimilar to GBH, be KDL; this segment may be found, for the circle AGBH is to CKDL, as the segment GBH to the segment KDL; and this segment, multiplied by the length of the cask, gives the quan tity of liquor in it.



to of any country; and this may be done, by means of the preceding problems, by first finding the solid content in inches, and then dividing it by the number of inches in the measure requi. red: but the work may be shortened, by finding multipliers, which will give the content in that measure, instead of inches. For example, if the vessel be contained by planes, instead of dividing by 231, multiply by the decimal .004329, and the product will be the content in wine.gallons; and instead of dividing by 282, multiply by .003546, to get the content expressed in gallons of ale : But those multipliers are feldom used, because they do not abridge the operation; they are found by annexing


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