(42+3°+4×3) × 2.598076 × 9 865.159306

3

solid feet, the solidity required.

3

=288.386436

3. What is the solidity of the frustrum of a cone, the diameter of the greater end being 4 feet, that of the less end 2 feet, and the altitude 9 feet? Ans. 65.9736.

The following cases contain all the rules for finding the superficies and solidities of conical ungulas.

1. When the section passes through the opposite extremities of the ends of the frustrum.

Let D=AB the diameter of the greater end; d=cD, the diameter of the less end; h=perpendicular height of the frustrum, and n=.7854.

d2-dDd nDh

D-d

Then

solidity of the greater elliptic

II. When the section cuts off part of the base, and makes the angle DrB less than the angle CAB.

E

B
Fi

4. What is the solidity of the frustrum of a cone, the circumference of the greater end being 40, that of the less end 20, and the length or height 50? Ans. 3713.7333.

Let s tabular segment, whose versed sine is Br÷D, 8— tab. seg. whose versed sine is Br—(D—d)÷d, and the other letters as above.

solidity of the elliptic hoof EFbd.

1

And √ 4h2 + (D—d)2 × (seg. FBE /

D- d

III. When the section is parallel to one of the sides of the frustrum.

Let A area of the base FBE, and the other letters as before.

Then

=

AXD

bolic hoof EFBD.

-îd √ (B—d) × d) × }h=solidity of the para

IV. When the section cuts off part of the base, and makes the angle DrB greater than the angle CAB.

5. What is the solidity of the frustrum of a square pyramid, one side of the greater end being 18 inches, that of the less end 15 inches, and the height 60 inches?

Ans. 16380 inches. 6. What is the solidity of the frustrum of an hexagonal pyramid, the side of whose greater end is 3 feet, that of the less end 2 feet, and the length 12 feet?

PROBLEM X.

Ans. 197.453776 feet.

To find the solidity of a cuneus or wedge.

RULE.*

Add twice the length of the base to the length of the edge, and reserve the number.

B

Let the area of the hyperbolic section EDF-A, and the area of the

1

Dd

And ¡× √ 4h2 + (D—d)23× (cir. seg. EBF

Br-(D-d) Br

curve surface of EFBD.

BrD-d Br-d

Note. The transverse diameter of the hpy. seg. =

dx cr

D-d-Br

and the conjugate=d√

Br

from which its

D -d-Br

area may be found by the former rules.

* Demon. When the length of the base is equal to half of the wedge, the wedge is evidently equal to half a prism of the same base and altitude.

Multiply the height of the wedge by the breadth of the base, and this product by the reserved number; of the last product will be the solidity.

EXAMPLES.

1. How many solid feet are there in a wedge, whose base is 5 feet 4 inches long, and 9 inches broad, the length of the edge being 3 feet 6 inches, and the perpendicular height 2 feet 4 inches?

And 7140-1728 4.1319 solid feet, the content required.

And according as the edge is shorter or longer than the base, the wedge is greater or less than half a prism, by a pyramid of the same height and breadth at the base with the wedge, and the length of whose base is equal to the difference of the lengths of the edge and base of the wedge.

Therefore, let the length of the base BC=L; the length of the edge EF=1; the breadth of the base BA=b; and the height of the wedge EP=h; and we shall have by the former rules

blh

+bhx

2

+L+1 blh
2

+bhx

L-l
3

-bhx

31+21-21
6

=bhx

2. The length and breadth of the base of a wedge are 35 and 15 inches, and the length of the edge is 55 inches: what is the solidity, supposing the perpendicular height to be 17.14508 inches? Ans. 3.1006 feet.

'PROBLEM XI.

To find the solidity of a prismoid.

RULE.*

To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends, and this last sum multiplied by of the height will give the solidity.

Note. The length of the middle rectangle is equal to half the sum of the lengths of the rectangles of the two ends, and its breadth equal to half the sum of the breadths of those rectangles.

* Demon. The rectangular prismoid is evidently composed of two wedges, whose heights are equal to the height of the prismoid, and their bases its two ends. Wherefore, by the last problem its solidity will be=(21+1×в+27+L× b)×☎,

L+1,
2

which, by putting м = and m =

h

B+b,
2

h

becomes

BL+bl+4mm× ☎; which is the rule, as was to be shown.

The solidities of the two parts, commonly called the ungulas, or hoofs, into which the frustrum of a rectangular pyramid is divided, may be found by the last two rules, as they are only composed of wedges and prismoids.

A very elegant demonstration of this rule for the prismoid may be seen in Simpson's Fluxions, p. 179, 2d Edition.

If the bases of the prismoid are dissimilar rectangles, of which L, 1 and M, m are corresponding dimensions, and h the height; Then (2+Ll.M+27+L.m)× th=solidity.