Machinery's Handbook, 31st Edition
92
VOLUMES OF SOLIDS
Spherical Zone:
c 1
V 0.5236 h = A 2 π rh 6.2832 rh area of spherical surface = = = r c 2 2 4 --- c 2 2 c 1 2 – 4 h 2 – 8 h ---------------- 2 = + 3 c 2 2 4 ---- h 2 + + 3 c 1 2 4 ----
h
c 2
r
Example: In a spherical zone, let c 1 = 3; c 2 = 4; and h = 1.5 in. Find the volume. V 0.5236 1.5 × 3 3 2 × 4 ------- 3 4 2 × 4 ------- 1.5 2 + + × 0.5236 1.5 × 27 4 --- 48 4 + --- +2.25 × = =
16.493 in 3
=
Spherical Wedge:
V = volume
A = area of spherical surface
α = center angle in degrees
α 360 -----
4 π r 3 3 ------ ×
0.0116 α r 3
=
=
V
α 360 ----- 4 π r 2 ×
0.0349 α r 2
=
=
A
Example: Find the area of the spherical surface and the volume of a wedge of a sphere. The diameter of the sphere is 100 mm, and the center angle a is 45 degrees. V 0.0116 45 × 50 3 × 0.0116 45 × 125,000 × 65 250 mm 3 , 65.25 cm 3 = = = = A 0.0349 45 × 50 2 × 3926.25 mm 2 39.26 cm 2 = = = Hollow Sphere:
V = volume of material used to make hollow sphere
r
4 π 3 --- R 3 r 3 – ( )
4.1888 R 3 r 3 – ( )
=
=
V
d
D
R
π 6 -- D 3 d 3 – ( )
0.5236 D 3 d 3 – ( ) =
=
Example: Find the volume of a hollow sphere, 8 in. in outside diameter, with a thickness of mate rial of 1.5 in. Here R = 4; r = 4 - 1.5 = 2.5. V 4.1888 4 3 2.5 3 – ( ) 4.1888 64 15.625 – ( ) 4.1888 48.375 × 202.63in 3 = = = = Paraboloid:
1 ⁄
2 π r
2 h
0.3927 d 2 h
Volume
V = =
=
r
3
2 π 3 p ---
2
d
4 --- p 2 +
p 3 –
A = =
Area
d
h
where p d 2 8 h = --- Example: Find the volume of a paraboloid in which h = 300 millimeters and d = 125 millimeters. V 0.3927 d 2 h 0.3927 125 2 × 300 × 1,840,781 mm 3 1,840.8 cm 3 = = = =
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