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EXAMPLE.

A solid, 27 inches in depth, whose ends are ellipses (the transverse and conjugate diameters of the one being, respectively, 40 inches, and 30 inches; and of the other, 30 inches, and 20 inches) is given to be determined in wine gallons, ale gallons, and malt bushels.

SOLUTION.

40 transverse of the greater end Add 15 half the less transverse

55 sum

Multiply by 30 greater conjugate
1650 reserved product.

30 transverse of the less end Add 20 half the greater transverse

50 sum

Multiply by 20 the less conjugate

1000 product

Add 1650 reserved product

2650 total

Multiply by 9 one-third the altitude

W.G. divisor 294) 23850 (81.12 wine gallons.

A. G. divisor 359) 23850 (66-43 ale gallons.

M.B. divisor 2738) 23850 (8.71 malt bushels.

COROLLARY.-Either end of the solid may be a circle without impairing the truth of the result.

PROBLEM XVI.

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

RULE.

To the length of the greater end, add half the length of the less, and multiply the sum by the breadth of the greater end; reserving the product.

Next, to the length of the less end, add half the length of the greater, and multiply the sum by the breadth of the less end.

Then to this product add the reserved product, and multiply the sum by one-third the altitude. The result will be the area of a square, which, being an inch deep, shall equal in content, the given solid. This area, divided by the number of cubic inches in the proposed integer, will consequently give the content required.

EXAMPLE.

A solid 15 inches deep, whereof one of the ends is a square, having 60 inches for its side; and the other end, a rectangle, 40 inches by 24; is given to be determined in wine gallons, ale gallons, and malt bushels.

SOLUTION.

60 greater length

Add 20 half the less length

80 sum

Multiply by 60 greater breadth

4800 reserved product.

40 the less length

Add 30 half the greater length

70 sum

Multiply by 24 the less breadth

1680 product

Add 4800 reserved product

6480 total

Multiply by 5 one-third the depth

Product 32400 cubic inches.

Wherefore it will be,

W.G. divisor 231) 32400 (140.26 wine gallons.

A. G. divisor 282) 32400 (114.89 ale gallons.

M. B. divisor 2150) 32400 ( 15.07 malt bushels.

PROBLEM XVII.

To determine the Content of the UNGULA of the Frustum of a Cone.

RULE.

FOR THE GREATER HOOF.

Find a mean proportional between the greatest and least diameters of the frustum, and multiply it by the least diameter: then subtract the product from the square of the greatest diameter, and multiply the remainder by the greatest diameter.

The last product multiplied by one-third the altitude, and divided, first by the difference of the diameters, and afterwards by the circular divisor corresponding to any proposed integer, will be the content in the denomination of that integer.

FOR THE LESS HOOF.

Find a mean proportional between the least and greatest diameters, as for the greater hoof, and multiply this proportional by the greatest diameter; then from the product subtract the square of the least diameter, and multiply the remainder by the least diameter. The last product multiplied by one-third part the frustum's height, and divided, first by the difference of the diameters, and afterwards by the circular divisor

for any proposed integer, will give the content in the denomination of that integer.

EXAMPLE.

Let ABC be the greater hoof of the frustum of a cone, and BCD, the less hoof, the altitude Cc being 39 inches; the diameter AB, where the frustum is the widest, 35 inches; and the least diameter CD, 31 inches: the content of each hoof is required in wine gallons, ale gallons, and malt bushels.

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