A Question of Volume

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
If you take three billiard balls or similar and place them together in such a way that each is in contact with the other two, now all three are in contact with each other.

What is the volume of the space between the balls expressed as a percentage of the volume of just one of the balls?

Ron.
 
Assuming you are talking about the volume between the three balls and a horizontal plane across the top and bottom surface points, and also the vertical planes between the three centre points (you need to have complete boundarys to calculate a volume). This area is 32.7% of the volume of one ball.
 
I have a horrible long formula for it...

sq root 3 * cube root (3v/4PI)* cube root (3v/4PI) - 4/6PI *sq rt (3v/4PI)

Might take a liitle longer to get that factored down though!
 
keynsham1 wrote,..
Assuming you are talking about the volume between the three balls and a horizontal plane across the top and bottom surface points, and also the vertical planes between the three centre points (you need to have complete boundarys to calculate a volume). This area is 32.7% of the volume of one ball.
Click to expand...

Hello Keynsham1,

Yes indeed, you have interpreted the question in the only way in which it can be answered, and the volume of the space (although you said area, I know you meant volume) does indeed represent 32.7% of the volume of just one ball.... :D

Hello Rich,

You can simplify your expression to 3^0.5 * (3V/4PI)^2/3 - PI/3(3V/PI)^0.5, but what have you equated it to?

Ron.
 
SydneyRoverP6B said:
keynsham1 wrote,..
Assuming you are talking about the volume between the three balls and a horizontal plane across the top and bottom surface points, and also the vertical planes between the three centre points (you need to have complete boundarys to calculate a volume). This area is 32.7% of the volume of one ball.
Click to expand...

Hello Keynsham1,

Yes indeed, you have interpreted the question in the only way in which it can be answered, and the volume of the space (although you said area, I know you meant volume) does indeed represent 32.7% of the volume of just one ball.... :D

Hello Rich,

You can simplify your expression to 3^0.5 * (3V/4PI)^2/3 - PI/3(3V/PI)^0.5, but what have you equated it to?

Ron.
Click to expand...


Ok, how do you get 32.7%? I have a theory, but I get it ~30.8%
 
darth sidious wrote,...
Ok, how do you get 32.7%? I have a theory, but I get it ~30.8%
Click to expand...

Hello Darth,

An equilateral triangle of side length 2r (where r is the radius of a sphere) is formed by joining the centres. This represents the cross sectional area of a triangular prism, the volume of which includes the required space. Each sphere occupies 2/9PIr^3 cubic units within the aforementioned prism, totalling 2/3PIr^3 cubic units. Thus the volume of free space within the prism is V = r^3(2(3)^0.5 - 2PI/3) where r is independent of the required ratio. Solving for V and dividing by the volume of one sphere then expressing your answer as a percentage gives 32.699....32.7% to 1dp.

Ron.
 
that's the volume of the space left in the triangle if you take out the 1/6 of each spherewhich is inside the triangle i think....

Rich
 
rockdemon wrote,..
that's the volume of the space left in the triangle if you take out the 1/6 of each spherewhich is inside the triangle i think....
Click to expand...

Hello Rich,

Then what does V in your expression represent? I am sorry Rich,..it's not correct... :(

The volume of free space is V = r^3(2(3)^0.5 - 2PI/3).

Ron.
 
SydneyRoverP6B said:
Hello Darth,

An equilateral triangle of side length 2r (where r is the radius of a sphere) is formed by joining the centres. This represents the cross sectional area of a triangular prism, the volume of which includes the required space. Each sphere occupies 2/9PIr^3 cubic units within the aforementioned prism, totalling 2/3PIr^3 cubic units. Thus the volume of free space within the prism is V = r^3(2(3)^0.5 - 2PI/3) where r is independent of the required ratio. Solving for V and dividing by the volume of one sphere then expressing your answer as a percentage gives 32.699....32.7% to 1dp.

Ron.
Click to expand...

Ok, I started the same way as you, an equilateral triangle, and worked out the volume of that prism. I think my method of attack is the same as yours, just somehow messed up with the formula!

Will revisit this soon... quite heart-warming to me that I started the right way!
 
Ok, I think I see where I messed up. I took the triangle to the outer surface of the balls, not to the centres as you suggested.

That would mean the lines joining the centre points must be an equilateral triangle, each length being 2r.

From Heron's theorem/formula A^2 = p(p-a)(p-b)(p-c), p = Perimeter/2 , a,b,c are the side lengths, which are all 2r in this cae

p = 3 x 2r/2=3 x r, p-a = p-b = p-c = r, so A^2 = 3 x r^4, so A = 3^0.5 x r^2

V = A x 2 x r =2 x 3^0.5 x r^3

Volume of a complete sphere is (4/3) x pi x r^3. As we are using an equilateral triangle, whose interior angles are all 60deg. (which is 1/6th a complete revolution) Volume of one portion of a sphere in our prism = (4/3) x pi x r^3 x (1/6) = (2/9) x pi x r^3

With three spheres, total volume in the prism = 3 x (2/9) x pi x r^3 = (2/3) x pi x r^3.

Volume of space = 2 x r^3 x (3^0.5 - pi/3) = (2/3) x r^3 (3^1.5 - pi)

Volume of one sphere = (4/3) x pi x r^3

Volume of space / Volume of one sphere = [(2/3) x r^3 (3^1.5 - pi)] / [(4/3) x pi x r^3] = (3^1.5 - pi) / (2 x pi) = 0.32699334313268807426698974746945 = 32.7%

Good problem!
 
Hello Darth,

Glad you found your initial error and that it all worked out. I am pleased that you liked the problem and that you enjoy doing them too.. :D

Keep your eyes peeled for some more. :D

Ron.
 
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