25+ 2.5 = 27.5 diameter at 5 inches deep 25+ 5·0 = 300 ...........at 10 inches deep 25+ 7.5 = 32.5 ............at 15 inches deep 25+10·0 = 350 ............at 20 inches deep 25+12.5 = 37.5 ..........at 25 inches deep

It is evident that, had the wider end of the frustum been uppermost, the numbers found by the factor would have been decrease of diameter, and must have been severally subtracted from 40 the greatest dia

meter.

The Rule for the frustum of a square pyramid, in respect of intermediate breadths, is manifestly the same as for the cone.

PROBLEM XIV.

To determine the Content of the Frustum of a Pyramid.

RULE.

By the Pen.

Find the area of each end, and, to the sum of the two areas, add the geometrical mean between them; then multiply by one-third the altitude, and the product will be the content in cubic inches; which being divided by the number of cubic inches in any proposed integer, will give the content in the denomination of that integer.

By the Sliding Rule.

Find the area of each of the two ends separately, and set either of them on C, to the same point on D; then against the other on C, will be the geometrical mean between them on D.

Next, set the gage-point, for the proposed integer, on A, to one-third the altitude on B, and against the sum of the three areas on A, will be the content on B.

EXAMPLE.

Let EFGHIKL be a vessel in the form of the frustum of a square-based pyramid, the depth being 48 inches, and the breadth of the ends, 21 inches, and 39 inches, respectively; the content is required in wine gallons, ale gallons, and malt bushels.

39 side of the greater square

Multiply by 39 side

Product 1521 area of the base. Multiply by 441 area of the less end

1521

6084

6084

Product 670761 (819 geometrical mean.

64

161) 307

161

1629) 14661

14661

Lastly, for the sum of the three areas multiplied by one-third the depth, it will be,

441 area of the less end

819 mean area

1521 area of the base

2781 sum.

Multiply by 16 one-third the depth

16686 2781

W.G. divisor 231) 44496 (192.62 wine gallons.

A. G. divisor 282) 44496 (157.79 ale gallons.

M.B. divisor 2150) 44496 (20-69 malt bushels.

BY THE SLIDING RULE.

We set 1 on D, to 1 on C, and opposite to 21 on D, we find 441 on C, for the area of the less end, and opposite to 39 on D, we find 1521 on C, for the area of the greater end.

We now set 1521 on C, to 1521 on D, and looking against 441 on C, we discover 819 as the geometrical mean proportional between the areas of the ends.

Next it is,

441 less end

819 mean proportional 1521 base or greater end

2781 SUM.

110 16

G. P. 282 to 16 against 2781 are 157.79 A. G.

2150

20.69 M.B.J

on B. 192.62 W.G.)

PROBLEM XV.

To determine the Content of a Solid with Parallel (but Unequal and Dissimilar) Ellipses for Ends, and Sides every way straight.

RULE.

To the transverse diameter of the greater end, add half the transverse diameter of the less end, and multiply the sum by the conjugate axis of the greater end; reserving the product.

Next, to the transverse diameter of the less end, add half the transverse diameter of the greater end, and multiply the sum by the conjugate axis of the less end.

Then to this product add the reserved product, and multiply the sum by one-third part the altitude. The result will be the square of the diameter of a circle, which, being an inch deep, shall be of equal content with the given figure. This last product, divided by the circular divisor of the proposed integer, will consequently be the content.