USE OF THE TABLE.

To determine the Content of a Parabolic Spindle by

the Pen.

RULE.

Multiply the square of the middle diameter by the length of the spindle, and divide the product by the parabolic-spindle divisor.

To determine the Content of a Parabolic Spindle by the Sliding Rule.

RULE.

Set the parabolic-spindle gage-point, for the proposed integer, on D, to the transverse or longest diameter, on C, and against the middle diameter on D,` will be the content on C.

EXAMPLE.

The content of a parabolic spindle, 60 inches in length, and 20 inches in diameter at the middle, is required in wine gallons, ale gallons, and malt bushels.

SOLUTION BY THE PEN.

Middle diameter 20

Middle diameter 20

Square of the conjugate 400

Length of the spindle 60

Product 24000

And,

W.G. par. sp. divisor 551-47) 24000 (43.52 W.G.

A. G. par. sp. divisor 673-22) 24000 (35.65 A. G.

M. B. par. sp. divisor 5133.75) 24000 ( 4·67 M.B.

on D.
୮

on C. [43.52 W.G.

P.sp. G.P. 25.9 to 60 against 20 are 35-65 A. G. 4.67 M.B.J

71.6

PROBLEM XII.

To determine the Content of the Frustum of a Cone.

RULE.

By the Pen.

To the square of the difference of the diameters of the two ends, add three times the product of the greater diameter multiplied by the less. Then multiply the sum by one-third the altitude, and divide the result by the circular divisor of the proposed integer, for the

content.

By the Sliding Rule.

Find a mean proportional between the diameters of the ends, and set the circular gage-point, for the proposed integer, on D, to one-third the depth on C, then against each of the diameters, and also the mean

proportional between them, on D, take the content on C. The sum of the three contents will be the content of the frustum.

EXAMPLE.

Let ABDCEFG be the frustum of a cone; BC, the diameter at the top, being 24 inches; the altitude, 75 inches; and EG, the diameter of the base, 64 inches: the content is required in wine gallons, ale gallons, and malt bushels.

SOLUTION BY THE PEN.

64 greater diameter

Subtract 24 less diameter

Remainder 40 difference of the diameters.

40

1600 square of the difference

64 × 24 x 3 = 4608 triple product

Sum 6208

Multiply by 25 one-third the altitude

31040

12416

Product 155200

Hence it will be,

W.G. circ. divisor 294)155200 (527.89 wine gallons.

A. G. circ. divisor 359)155200 (432.31 ale gallons.

M. B. circ. divisor 2738)155200 ( 56.68 malt bushels.

BY THE SLIDING RULE.

1. For a mean proportional between the diameters.

Set 64 on C, to 64 on D, and against 24 on C, are 39.2 on D, for the mean proportional.

2. For the solution by help of the mean proportional.

PROBLEM XIII.

To determine Intermediate Diameters at any assigned Depth of the Frustum of a Cone, having the top and bottom Diameters given.

RULE.

From the greater diameter subtract the less, and divide the remainder by the depth of the frustum : the quotient will be the differential factor for any assigned depth.

To employ this factor, multiply the proposed depth by it, and the product will be the defect of the less diameter to equal the diameter, of the frustum at that depth, when the depth is counted from the smaller end; but the excess of the greater diameter above the length of the diameter of the frustum at that depth, when the depth is counted from the larger end.

EXAMPLE.

Let ABCD be a perpendicular section of the frustum of a cone by a plane co-inciding with the axis: and let the less diameter AB, be 25 inches; the greater diameter, CD, 40 inches; and the depth 30 inches: intermediate diameters at every 5 inches are required.