by Conquest » Sat Dec 18, 2004 2:29 am
In order to assist in this process, I'm going to make a few assumptions: 1. the basketball in question is based on a circumference of 29.5" (the NBA standard is 29.5 - 30"); 2. you are wanting to know the inside volume of the basketball as opposed to the outside volume (since we obviously can't fill the ball material itself with air/sand/whatever...); and 3) the ball's outer material is is 1/4" (this is a pure guess as I can find no specification for this on the internet - you could cut open a basketball and get the exact thickness, but this would work best on a new basketball since the wear & tear reduces this measurement and this could prove to be too expensive).
Having made the previous assumptions, we now arrive at a fillable circumference of 28.5" (29.5" - (1/4" * 2). The formula for a circle's cirmcumference is: circumference = Pi * diameter or 28.5" = 3.1415926535 * diameter or 28.5 / 3.1415926535 = diameter or diameter = 9.0718"
The radius is 1/2 of the diameter; therefore, the radius of this circle is: radius = 9.0718" / 2 = 4.5359"
We can now input the known radius into the formula for the volume of a sphere which is: (4/3) * Pi * Radius cubed = (1.333) * 3.1415926535 * 4.5359" cubed = 4.1887 * 93.3233 cubic inches = 390.9033 cubic inches
I understand that I made a few assumptions on the front-end of this; the first two are extremely valid and I can see no one really disputing. The only one that might need to be modified is the third one which is the measurement of the thickness of the ball material. Depending upon the importance of this answer being exactly correct, you may want to contact either a local sporting goods company to see if they have a new ball that's burst which you may measure or contact one of the major sporting goods manufacturers who may give you the measurement. I hope this helps!