EXAMPLE. The area of the base, de f, Fig. 92, of a cylindrical ungula is 15.5 ins., and its height 20; what are its contents? 15.5×20=310=product of area and height=result required.

Ex. 2. The area of the base of a cylindrical ungula is 168.25 inches, and the height of it 22; what are its contents in cubic feet? Ans. 2.148 cubic feet.

2. When the section passes obliquely through the opposite sides of the cylinder, Fig. 93.

RULE.-Multiply the area of the base of the cylinder by half the sum of the greatest and least lengths of the ungula, and the product will be the contents required.

EXAMPLE. The area of the base de ƒ of a cylindrical ungula is 25 inches, and the greater and less heights of it, a d, be, are 15 and 17 inches; what are its contents?

15+17

2

2x5- =400=product of half the sum of the heights and the area of the base = result required.

Ex. 2. The area of the base of a cylindrical ungula is 75.8 inches, and the greater and less heights of it are 4.25 and 5.65 feet; what are its contents in cubic feet?

Ans. 2.6056 cubic feet.

3. When the section passes through the base of the cylinder and one of its sides, and the versed sine does not exceed the sine, Fig. 94.

RULE. From two thirds of the cube of the sine, a d, of the arc, d f, of the base, subtract the product of the area of the base and the cosine,* a e, of the half arc.

Multiply the difference thus found by the quotient arising from the height divided by the versed sine, and the product will give the contents required.

h

-axcx =S, v s representing the versed sine.

Fig. 94.

d

c

EXAMPLE.-The sine a d of half the arc of the base of an ungula, Fig. 94, is 5, the diameter of the cylinder is 20, and the height of the ungula 10; what are the contents of it? * When the cosine is 0, the product is 0.

of 53=83.333=two thirds of the cube of the sine.

As the versed sine and radius of the base are equal, the cosine is 0. Hence, area of base X cosine=0.

83.333-0×10=166.666=difference of of cube of the sine and the product of area of base and the cosine, the height÷the versed sine the contents required.

Ex. 2. The sine of half the arc of the base of an ungula is 12 inches, the diameter of the cylinder is 25, and the height of the ungula 18; what are its contents?

Ans. 1190.34375 inches.

For rules to ascertain the area of base, see pp. 85-87.

4. When the section passes through the base of the cylinder, and the versed sine exceeds the sine, Fig. 95.

RULE. To two thirds of the cube of the sine of half the arc of the base, add the product of the area of the base and the excess of the versed sine over the sine of the base.

Multiply the sum thus found by the quotient arising from the height divided by the versed sine, and the product will be the contents required.

EXAMPLE. The sine ad of half the arc of an ungula, Fig. 95, is 12 inches, the versed sine a g is 16, the height c g 20, and the diameter of the cylinder 25 inches; what are the contents?

of 123=1152=two thirds of cube of sine of the arc of the base. Area of base (see Rules, p. 84-134)=331.78.

1152+(331.78×16-12.5)=2313.23=sum of 3 of the cube of the sine of the base, and product of area of base, and difference between the versed sine and sine of the base.

2313.23 × 20÷÷16=2891.5375=product of above sum and the height, divided by the versed sine=result required.

5. When the section passes obliquely through both ends of the cylinder, Fig. 96.

RULE.-Conceive the section to be continued till it meets the side of the cylinder produced; then, as the difference of the versed sines of the arcs of the two ends of the ungula is to the versed sine of the arc of the less end, so is the height of the cylinder to the part of the side produced.

Find the contents of each of the ungulas by rules 3 and 4, and their difference will be the contents required.

vxh

Or, =h', v and v' representing the versed sines of the

V

arcs of the two ends, h the height of the cylinder, and h' the height of the part produced.

EXAMPLE.-The versed sines, a e, do, and sines, i k, g r, of the arcs of the two ends of an ungula, Fig. 96, are assumed to be respectively 8.5 and 25, and 11.5 and 0 inches, the length of the ungula within the cylinder, cut from one having 25 inches diameter is 20 inches; what is the height of the un

gula produced beyond the cylinder, and what the contents of the ungula?

258.5:8.5::20: 10.303=height of ungula produced beyond the cylinder. Lower ungula, the sine, g r, being 0, the versed sine the diameter. Base of ungula being a circle of 25 inches diameter, area=490.874. The versed sine and diameter of the base being equal (25), the sine is 0. 490.874×25 6135.925=product of area of base and excess of versed sine over the sine of the base.

30.303-25

1.2121=quotient of height÷versed sine.

Then, 6135.925 × 1.2121=7437.3547=product of above product and quotient the result required.

SPHERE.

Definition. A solid, the surface of which is at a uniform distance from the centre.

To ascertain the Contents of a Sphere, Fig. 97.

RULE.-Multiply the cube of the diameter by .5236, and the product will be the contents required.

Or, d3×.5236 S, d representing the diameter.

EXAMPLE. What are the contents of a sphere, Fig. 97, its diameter, a b, being 10 inches?

103=1000, and 1000 × .5236=523.6 cubic inches.

Ex. 2. The diameter of a sphere is 17 inches; what are its contents? Ans. 1.4887 cubic feet. Ex. 3. What are the contents of a globe 10.5 feet in diameter? Ans. 606.132 cubic feet. Centre of Gravity. Is in its geometrical centre.

NOTE.-.5236 of 3.1416.