An Introduction to Mensuration and Practical Geometry

Here (72×3+42) × 4× .5236=(49 × 3+4o)× 4×.5236 =(147+42)× 4× .5236=(147+16) × 4× .5236 = 163 × 4 × .5236=652× .5236=341.3872 solid inches, the answer. 2. What is the solidity of the segment of a sphere, the diameter of whose base is 20, and its height 9?

Ans. 1795.4244.

3. What is the content of a spherical segment, whose height is 4 inches, and the radius of its base 8?

Ans. 435.6352. 4. What is the solidity of a spherical segment, the diameter of its base being 17.23368, and its height 4.5? Ans. 572.5566. 5. The diameter of a sphere being six inches, required the solidity of the segment whose altitude is two inches. Ans. 29.3216 cubic inches.

6. Required the solidity of a spherical segment, the height of which is 15, the diameter of the sphere being 18. Ans. 2827.44.

PROBLEM XV.

To find the solidity of a frustrum or zone of a sphere.

RULE.*

To the sum of the squares of the radii of the two ends, add one third of the square of their distance, or of the

of the segment, which is the same as the rule.

Or, if d=diameter of the sphere, and h=height of the segment; then will .5236h2× (3d—2h) solidity.

* Demon. The difference between two segments of a sphere whose heights are H and h, and the radii of whose bases are R and r, will

- by the last problem=6 × (3R2H+H3—3r2h—h3)—zone whose height is H-k. And therefore by putting a for the altitude of the frustrum,

breadth of the zone, and this sum multiplied by the said breadth, and the product again by 1.5708, will give the solidity.

EXAMPLES.

1. What is the solid content of the zone ABCD, whose greater diameter AB is 20 inches, the less diameter CD 15 inches, and the distance nm of the two ends 10 inches?

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33.33) × 10 × 1.5708 189.58 x 10 x 1.5708 = 1895.8 X 1.5708-2977.92264 solid inches, the answer.

2. What is the solid content of a zone, whose greater diameter is 24 inches, the less diameter 20 inches, and the distance of the ends 4 inches? Ans. 1566.6112 inches.

3. Required the solidity of the middle zone of a sphere, whose top and bottom diameters are each 3 feet, and the breadth of the zone 4 feet? Ans. 61.7848 feet.

and exterminating H and h by the means of the two equations

X 2'

which is the

rule.

If it be the middle zone of the sphere, the solidity will be=d2 - 3h2) x.7854h; where d=diameter of each end, and h=its height.

PROBLEM XVI.

To find the solidity of a spheroid.

RULE.*

Multiply the square of the revolving axe by the fixed axe, and this product again by .5236, and it will give the solidity required.

Where note that .5236 is of 3.1416.

EXAMPLES.

1. In the prolate spheroid ABCD, the transverse, or fixed axe AC is 90, and the conjugate, or revolving axe DB is 70: what is the solidity?

*Demon. Let Ac=a, DB=b, ar=x, rn=y, and p= 3.14159, &c.

Then a2: b2 :: xx (a—x) : 2× (ax—x2)=y2 by the property of the ellipsis.

And therefore the fluxion of the solid (=py';)=b2×

(axx-xx); and its fluent x! × (}ax2x3)=segment

=content of the whole spheroid. Q. E. D.

If ƒ be put fixed axe, r-revolving axe, q=(f2 r2)÷ f', and p=3.1416, &c.

Then will prf√1+}q=surface of the oblate spheroid, and prf√1—1q=that of the prolate spheroid.

Here DB2x AC× .5236=702 × 90 × .5236=4900 × 90 × 5236=441000×.5236=230907.6=solidity required. 2. What is the solidity of a prolate spheroid, whose fixed axe is 100, and its revolving axe 60? Ans. 188496. 3. What is the solidity of an oblate spheroid, whose fixed axe is 60, and its revolving axe is 100?

PROBLEM XVII.

Ans. 314160.

To find the content of the middle frustrum of a spheroid, its length, the middle diameter, and that of either of the ends, being given.

CASE I.

When the ends are circular, or parallel to the revolving

axis.

RULE.*

To twice the square of the middle diameter add the square of the diameter of either of the ends, and this sum

*Demon. Let Aо=a, Do=b, En=h, no=c, ro=x, re= and p=3 3.14159, &c.

·× (a2 — x2)= b2

multiplied by the length of the frustrum, and the product again by .2618, will give the solidity. Where note that .2618 of 3.1416.

EXAMPLES.

1. In the middle frustrum of a spheroid EFGH, the middle diameter DB is 50 inches, and that of either of the ends EF or GH is 40 inches, and its length nm 18 inches: what is its solidity?

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Here (50×2+40°) 18x.2618 (2500x2+1600) × 18.2618 = (5000 + 1600) × 18x.2618 6600 × 18× .2618118800 x .261331101.84 cubic inches, the

answer.

2. What is the solidity of the middle frustrum of a prolate spheroid, the middle diameter being 60, that of either of the two ends 36, and the distance of the ends 80?

Ans. 177940.224.

Whence, by substituting this value of a2 in the former

equation, we shall have y= b2 — ·

b4x2-b2h2x3

b2c2