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Prob. 20. To find the solidity of a sphere or globe.

Multiply the cube of the diameter by -5236, and the product is the solidity. Ex. What is the solidity of a globe, whose diameter is 3 feet?

3X3 X3=27
And •5236

27

36652 10472

14 1372 the solidity. Prob. 21. To find the solidity of a segment of a globe.

To three times the square of half the diameter of the base, add the square of he height ; multiply the sum by the height, then the product multiplied by :5236, will give the solidity.

Er. What is the solidity of a spherical segment, the diarneter of the base being 4 feet, and the height of the segment 3 feet ?

;=2 balf the diameter.
2
X2

3x3=9 the square of the height.

X 3

12 three timos the square of half the diameter. +9

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Prob. 22. To find the solidity of a spherical zone, the radii of the two parallel circles being given, and the distance between them.

To the squares of the two radiuses, add one-third of the square of the height; multiply the sum by the height, and the product by 1-5708, will give the solidity.

Er. What is the solid content of a spherical zone, whose greater radius is 12 inches, and the lesser 10 inches ; and the height or diso tance of the ends 4 inches ? 12+10° +4x4x1.5708=1566:6112 the solidity required.

Prob. 23. To find the solidity of a wedge.

Multiply the area of the base by the perpendicular height, and half the product will give the solidity.

Er. Required the solidity of a wedge, the dimensions of the base being 1f. 3in, and 2f. 6in. and the height 4f.

2.5 X 1.25 X4 Then

=625 the solidity.

2 Note.-The solidity of any prismatic ungnla will be fonnd in the same inanner; that is, half the product of the area of the base multiplied into the height, will give the solidity.

Prob. 23. To find the solidity of a boof, or ungula, from the frustrum of a square pyramid.

To the square of the side of the base, or that end which is complete, add one half of the product of the sides of the two ends ; this being multiplied by otre-third of the height, gives the solidity.

And if the hoofs are any other than that of a square pyranid, find the square root of the area of each end, which will give the side of a square equal in area ; then proceed as above.

Ex. Required the solidity of an ungula, from the frustrum of a square pyramid, the side of the greater end, which is complete, being 1f. 6in. that of the lesser end 1f. 3in. and the height 5f. ?

15 x

5

x=5'3125 which is the solidity required.

2 Prob. 24. To find the solidity of a spheroid ; the fixt axis and the revolving axis being given.

Multiply the fixt axis by the square of the revolving axis, and the product by 5236, will give the solidity.

Er. What is the solidity of a prolate spheroid, the major axis of which is 100 feet, and the minor 60 feet ?

100 x 60° x 5236=188496 the solidity required. Er. What is the solidity of an oblate spheroid, whose transverse axis is 100 feet, and the shortest axis 60 feet ?

60 x 100% * *5236=314160 the solidity required. Prob. 25. To find the solidity of a parabolic conoid, the diameter of the base being given, and the perpendicular height.

Multiply the square of the diameter of the base by •3927, and the product by the height will give the solidity.

Ex. What is the solidity of a par bolic conoid, the height being 50 feet, and the diameter of the base 30 feet?

30% X 3927 x 50=17671.5 the solidity required. Prob. 26. To find the solidity of the frustrum of a parabolic conoid the greater diameter, the lesser, and the perpendicular height being given.

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To the square of the diameter of the greater end, add the square of the diameter of the lesser end ; multiply the sum by '3927, and the product by the height, will give the solidity required.

Er. What is the solidity of a parabolic frustrum, the diameter of the greater end being 60 feet, the lesser end 48 feet, and the distance of the ends 18 feet?

60° +48=5904 the sum of the squares of the ends. Then 5904 x 3927 x 18=41733 0144 the solidity required.

If a circle is carried round a right line as an axis, and in the same plane with the circle, either touching the axis, or at a given distance from it, it will generate a solid, called an annulus, or cylindric ring.

The diameter of a generating circle is called the thickness of the ring.

Ťwice the distance of the generating circle, from the axis of rotation, is called the inner diameter.

Prob. 27. To find the solidity of an annulus, or ring, whose thick. ness and inner diameter are known.

Method 1. To the thickness of the annulus, add the inner diameter; multiply the sum by the square of the thickness, and the product by 2:4674, will give the solidity sought.

Method 2. Multiply the circunference round the middle of the annulus, or that circle generated by the centre of the generating circle, by the area of the generating circle, and the product will give the solidity.

Note. This last method will give the solidity of any part of an annulus, or ring, comprehended between any two planes passing through the fixed axis.

Er. What is the solidity of an annulus, whose inner diameter is 8 inches, and the thickness of the annulus 3 inches ?

Then 8+3 x 3° X 2'4674=244-2726 the solidity required. Prob. 23. To find the solidity of a hollow cylivder, the exterior and interior diameters being given, and the perpendicular height.

Multiply the sum of the diameters by their difference, and the product by :7854 ; then by the altitude of the cylinder, and you will have the solidity required. Er. Required the solidity of a hollow cylinder, the exterior diame

' ter being 14 feet, the interior diameter 12 feet, and the perpendicular height 10 feet.

(14+12)(14-12) 7854 x 10=408 408 the solidity required. Prob. 24. To find the solidity of a frustrum of a hollow cone, of an equal thickness; the exterior and interior diameters at each end being given, and the perpendicular altitude.

Multiply together the sum of the two interior diameters, and the difference of the diameters at either end, by •7854, and by the per. pendicular altitude; then by the said difference, and the continued product will give the solidity required.

Es. Required the solidity of a frustrum of a hollow cone, the boto

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pm diameters being respectively 35 and 32 feet ; w.., diameters respectively 29 and 26 feet; and the perpendicular altitude 25 feet.

Nov 35-32=3 and (32+26+3) = 61.7854 x 61 x 3 x 25 = 8593 205 the solidity required.

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CONVEX SIDE OF SOLIDS. Prob. 25. To find the convex surface of a right cylinder, the cir. cumference and length of the cylinder being given.

Multiply the circumference by the length of the cylinder, and the product will be the area.

Er. What is the convex superficies of a right cylinder, whose circumference is 2ft. 6in, and the length 5f. 3in.

Then (5f. 3i.) x (2f. 6.)=13f. 1i. Gii. the answer. Note.-If the diameter is given, find the circumference, and proceed as before,

Prob. 26. To find the convex superficies of a right cone, the cir• cumference and slant side being given.

Multiply half the circumference by the slant side of the cone, and the product will be the area.

Ex. What is the convex superficies of a right cone, the circumference of the base being 4.5ft. and the slant side 6'25ft. ?

4.5

x 6°25=14:0625. Note.-If the diameter is given, find the circumference, and proceed as before.

Prob. 27. To find the convex surface of the frustrum of a right cone, the circumferences of both ends being given, and the slant side of the cone.

Multiply half the sum of the circumferences by the slaut side of the cone, and the product will be the area.

Er. What is the convex superficies of a frustrum of a right cone, the circumference of the base being 4.6ft. the circumference of the top being 8.25ft. and the altitude 5.75 feet?

4:6+3.25
Here x 5.75=22.56875 the answer.

2 Prob. 28. To find the superficies of a sphere or globe, the greatest circumference being given.

Multiply the square of the circumference by '3183, and the product will be the superficies.

Er. What is the superficies of a globe, the greatest circumference being 106f. 2

Here (106)* x3183=33704188

Prob. 29. To find the convex superficies of the segment of a sphere ór globe, the diameter of the base of the segment, and its height, being given.

To the square of the diameter of the base, add the square of twice the height; and the sum multiplied by 7854, will give the superficies.

Er. What is the convex surface of the segment of a globe, the diameter of the base bcing 17-25 feet, and the height 4:5 feet?

9=81 square of twice the height. (17.25)=297:5625 square of the diameter of the base. Then (297.5625 +81) * •7854=297.3229875 the superficies required.

Prob. 30. To find the convex surfaces of a spherical zone, the diameters of the ends and their distance being given.

Find the diameter of the sphere by Problems 13 and 14; multiply the diameter of the sphere, and the distance of the parallel ends of the zone together, and the product by 3.1416, will give the superficies required.

Er. In a spherical zone, the distance of the parallel ends being 4in. the diameter of the greater end 24in. and that of the lesser end 20in. what is the convex superficies, when the centre of the sphere is without the zone?

The distance of the greater chord from the centre, will be found to be 3.5 inches, by Prob. 13.

The radius will be found to be 25 inches, by Prob. 14, or the diameter 50 inches.

Therefore, 50 x 4x3.1416=628·32 the answer. Note.-If the diameter is given, find the circumference, and proceed as before.

Prob. 31. To find the superficies of an annulus, or ring, whose thickness and inner diameters are known.

Method 1. To the thickness of the ring, add the inner diameter ; multiply the sum by the thickness, and the product by 9:869, will give the superficies required.

Method 2. Multiply the circumference of the generating circle by the circumference round the middle of the ring, or that line generated by the centre of the generating circle, and the product will be the

area.

Note.-This last method will give the convex superficies of any part of an aupulus, or ring, comprehended between two planes passing through the first axis.

Er. What is the convex superficies of an annulus, or ring, whose inner diameter is sin, and the thickness 3in. ?

Then (3+8) X3 x 9 869=325.677 the superficies required

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