Filling a glass with balls?
How many 1.5″ diameter balls can we put in a glass that has the following dimensions.
Height: 72″
Diameter of base: 18″
Diameter of rim (top) 30″
Kindly show the working.
How many 1.5″ diameter balls can we put in a glass that has the following dimensions.
Height: 72″
Diameter of base: 18″
Diameter of rim (top) 30″
Kindly show the working.
StoneHenge
on November 30th, -0001
I do not think there is an analytic solution to this problem. The general problem of “how many spheres (or ellipsoids) can one pack into a volume of arbitrary shape” is a topic of ongoing research in mathematics.
You can bound the number by assuming that the balls are arranged in cubic or hexagonal closest packing. These are the closest possible packing densities for spheres, and in these arrangements, the spheres occupy pi/(3*sqrt(2)) (~= 74.05%) of the total possible volume.
Your “glass” has the shape of a frustrum (a truncated cone). The volume of a frustrum is given by
V = (R^2 + r*R + r^2)*pi*h/3
where R and r are the radii of the top and bottom of the frustrum, and h is the height. In your case,
V = (15^2 + 15*9 + 9^2)*pi*72/3 in^3
V = 10584*pi in^3
The volume occupied by balls if they could be packed in closest packing would be:
10584*pi^2/(3*sqrt(2)) in^3 = 2494.673*pi^2 in^3
Each 1.5″-diameter ball has a volume of:
v = (4*pi*0.75^3)/3 = 0.5625 * pi in^3
The number of these balls that occupy the closest packed filled volume is given by V/v:
V/v = (1494.673 * pi^2)/(0.5625 * pi) ~= 13932.88
but we can’t have a fraction of a ball, so at *most*, you could put 13,932 balls in the glass.
In fact, the dimensions of the glass are such that you can’t fill it with spheres in cubic or hexagonal close packing, so the packing density will be somewhat less than this. For random packing of spheres, the packing density falls to ~64%, so following the logic outlined above, the glass could contain about 12,042 randomly packed balls.
Cat
on November 30th, -0001
ALOT. DO YOUR OWN MATH PROBLEMS. I DID MINE.