Number of Marbles Needed for Prom

kinterb

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Mar 17, 2016
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Ok, so in my Leadership class, we are planning prom. We are filling our centerpieces with marbles and I was asked to figure out approximately how many marbles we needed, as we have a very limited budget for them. The marbles that we are using are the types with flat bottoms. We need to fill 3 glass vases of size 4x4x8(C1), 4x4x10(C2), and 4x4x12(C3) for each table. There are 27 tables and we want to fill 1/3 of each vase with marbles. The marbles have a diameter of 5/8" and a height of 3/16". I figured the way I could most accurately approximate how many marbles we needed were by figuring out the volume of each size of vase(Vc), finding the volume of one marble(Vm) and dividing them together. (Vm*x=Vc, where x= amount of marbles). I started out by finding the volume of 1/3 of each vase.
VC1=pi(2^2)(8/3)=(32pi)/3
VC2=pi(2^2)(10/3)=(40pi)/3
VC3=pi(2^2)(12/3)=16pi
Next, I found the volume of one marble. I did this by using integration to figure out a solid of rotation. I know that I could have just used half of a sphere volume for an approximation, but this way is more accurate, considering I know the dimensions of the marble. It will give me a closer approximation. I assumed the marble was semi-ellipsoidal in shape.

The formula for the ellipse would be [(x^2)/{(5/16)^2}]+[(y^2)/{(3/16)^2}]=1 Solving for y gives y= + or - sqrt[(9/256)-(9x^2)/25]

So I chose to revolve the positive side of this equation around the y-axis from 0 to 5/16. This would generate a shape of a marble with a flat bottom. I used the shell method.

So I did 2pi times the integral from 0 to 5/16 of x*sqrt[(9/256)-(9x^2)/25]. This has the volume of the marble being 0.038351713. I then plugged the values of Vm and VC1-3 into the equation I generated earlier. This gives me needing around 874 marbles for the small jar, around 1093 marbles for the middle sized jar, and around 1310 marbles for the large jar. This seems completely nonsensical. Why would a jar that has a volume of around 33 in^3 need 874 marbles to fill? Not to mention this would mean that for all the tables, we would need 88,479 marbles total. This would cost us nearly $5,000 dollars just for marbles... In other words, I am doing something wrong. Could someone please help me catch my mistake? Or find a better way to solve this issue? Thanks :)
 
We are filling our centerpieces with marbles...The marbles that we are using are the types with flat bottoms. We need to fill 3 glass vases of size 4x4x8(C1), 4x4x10(C2), and 4x4x12(C3) for each table. There are 27 tables and we want to fill 1/3 of each vase with marbles. The marbles have a diameter of 5/8" and a height of 3/16". I figured the way I could most accurately approximate how many marbles we needed were by figuring out the volume of each size of vase(Vc), finding the volume of one marble(Vm) and dividing them together. (Vm*x=Vc, where x= amount of marbles).
This will work only if the "marbles" are actually right rectangular solids ("brick-shaped"), so that they fill their volume entirely and exactly. Is this the case?
 
This will work only if the "marbles" are actually right rectangular solids ("brick-shaped"), so that they fill their volume entirely and exactly. Is this the case?

It's not, but for the approximation that we need, wouldn't it be close enough? If not is there another way to do it to account for the cracks, maybe only taking like 85% of the final amount or something like that.
 
The mathematics of packing is part of the field of optimization techniques (here). The outputs are very sensitive to inputs and assumptions.

In "real life", I'd recommend getting some marbles and running some tests. Fill one of the boxes to the desired height with marbles. Pour out the marbles. Count them. Do it again. Repeat until you're comfortable with a number as the expected average. Then do the same with the other sizes of boxes. ;)
 
It's not, but for the approximation that we need, wouldn't it be close enough? If not is there another way to do it to account for the cracks, maybe only taking like 85% of the final amount or something like that.

Yes, you absolutely can. The mathematics of packing spherical objects into a closed volume is a subject that has been extensively studied for many many years. Depending on how you pack the spheres, you can utilize as little as 5% or as much as 74% of the available volume. A 1992 paper showed that random packing averages a 64% fill rate, so you could use that as a basis for your approximation, and round up a bit as needed. I'd say that it's better to have too many marbles than not enough.

For more information, you can visit this page about sphere packing on Wolfram MathWorld.
 
And if you are left with some extra marbles

give those to Denis

he had lost some - he plays hockey!!
 
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