The math is something like
(Radius( x) pie squared)times the height, this gives you a cylinder you divide that in half this gives you a cone.
Something like that, but that's not it.
The Smithsonian Institute says this comes out to 210 cubic feet.
No they don't. All sources say the CM's
habitable volume was 210 cubic feet. This is the volume left over after you subtract for all the stuff you said wouldn't fit.
DOES ANYBODY DISAGREE WITH THESE FIGURES???
Yes
I am still coming up ith 210 cubic feet not 490 Maybe you should check your math
No, you need to check your math.
Someone said the CM was 11.4’ high, but this includes the boost protective cover, which is jettisoned during launch. Without the cover, the CM was essentially two attached geometric shapes – a truncated cone and a dished bottom. The truncated cone was 12.8333’ diameter at the base and 6.75’ high (not including the docking probe sticking out the top), with a sidewall slope of 32.5 degrees. The dished bottom also had a diameter of 12.8333’ and its height was 2.0833’.
The volume of a truncated cone is,
V = pi*H/3 * (R
2 + Rr + r
2)
Where H is the height, R is the radius of the bottom, and r is the radius of the top. We have H = 6.75’, R = 12.8333/2 = 6.4167’, and r = 6.4167–6.75*tan(32.5) = 2.1165’. Therefore,
V = pi*6.75/3 * (6.4167
2 + 6.4167*2.1165 + 2.1165
2) = 418.7 ft
3The volume of the dished base is more complicated to calculate, but we can simplify by assuming it is half of an oblate spheroid. The volume of a spheroid is,
V = 4*pi/3 * a * b * c
Where a, b and c are the radii in each of the three axes. In our case, a = b = 12.8333/2 = 6.4167’, and c = 2.0833’. Therefore, a half spheroid with our dimensions has the volume
V = 4*pi/3 * 6.4167
2 * 2.0833 * ½ = 179.7 ft
3Thus, the total volume of the CM is approximately
V = 418.7 + 179.7 = 598 ft
3