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Math calculation question

LashL

Goddess of Legaltainment™
LashL
Joined
Aug 12, 2006
Messages
36,711
I am hoping that someone in this sub-forum can help me out with a math conversion/calculation question.

If I want to estimate how many round objects are in a 2 cubic foot square container, what is the best way to do this? In this example, let's say the objects are four inches in diameter.

Initially, I wanted to calculate the area of the round object and then divide the 2 cubic feet by the result. I think that the area of the round object is pi x radius squared (3.14 x 16 square inches) (50.24 square inches), but got stuck when I tried to divide cubic feet by square inches.

I think I'm messing up by mixing area and volume, and that I should be calculating the volume of the round object and then dividing the 2 cubic feet by the result - is that correct? If so, how do I calculate the volume of the sphere?

Is it (pi x diameter cubed)/6? (3.14 x 64)/6 (36.27 cubic inches)?

Then, I could divide 2 cubic feet by 36.27 cubic inches to come up with an estimate of the number of spheres in the box?

Thank you in advance for your assistance. :)
 
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Interesting question. If you aren't going to try to pack the spheres efficiently, then dividing the volume of the box by the volume of a sphere will do the trick. The formula for the volume of a sphere is:

[latex]$V = \frac{4}{3}\pi r^3$[/latex]

However, it's a different game if you want to pack the spheres efficiently. In fact, it's a 400-year-old game:

http://www.maa.org/devlin/devlin_9_98.html

~~ Paul
 
LashL said:
I am hoping that someone in this sub-forum can help me out with a math conversion/calculation question.

If I want to estimate how many round objects are in a 2 cubic foot square container, what is the best way to do this? In this example, let's say the objects are four inches in diameter.

Initially, I wanted to calculate the area of the round object and then divide the 2 cubic feet by the result. I think that the area of the round object is pi x radius squared (3.14 x 16 square inches) (50.24 square inches), but got stuck when I tried to divide cubic feet by square inches.

I think I'm messing up by mixing area and volume, and that I should be calculating the volume of the round object and then dividing the 2 cubic feet by the result - is that correct? If so, how do I calculate the volume of the sphere?

Is it (pi x diameter cubed)/6? (3.14 x 64)/6 (36.27 cubic inches)?

Then, I could divide 2 cubic feet by 36.27 cubic inches to come up with an estimate of the number of spheres in the box?

Thank you in advance for your assistance. :)
Click to expand...

The volume of a sphere is:

[latex]$$\frac{4}{3}\pi r^3$$[/latex]
 
Thank you. Does that formula (I'm afraid I don't know how to make the symbols on the screen) come to a different result than (pi x diameter cubed)/6?

And thank you for the cool link, too!

Edit: I just worked it out and the answer is 33.4 cubic inches rather than 36.27 so I see that I was wrong about pi x diameter cubed. Thanks to both of you :)
 
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wollery said:
No, pi*d3/6 is identical to 4pi*r3/3
Click to expand...

Oh! I must have just messed something up with the calculator then. Much appreciated!

Edit: yes, on recalculating it, I see that they both result in an answer of 33.49 cubic inches. I see that I had the alternative formula right initially but the calculation wrong (I'm in a poorly lit room. Alternatively, I blame Lisa.) Thanks again :)
 
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Disclaimer - the following is not meant as a dig at LashL, but at modern reliance and trust on computational machinery in general.

Rules for using calculators, as I used to teach all my maths students;

1. Never trust the result you get, you may well have pressed a wrong key accidentally.

2. To get round rule 1 make every calculation at least twice.

3. If the answers are the same then you've probably got it right.

4. If they aren't the same do it at least 2 more times.

Of course, the first thing I always did was to make them put their calculators away and relearn how to do maths on paper and in their heads.

Calculators are only as good as the person using them and the care that person takes. They are not, as many people seem to think, magic correct answer machines.

Actual exchange from a lesson I once gave to some (supposedly intelligent) 13 year olds
(Paraphrased, since I can't remember the exact numbers, but otherwise it's true.) -

Student - 22 sir.
Me - 17 plus 25 equals 22? Are you sure?
Student - Yeah.
Me - Really?
Student - Yeah, that's what the calculator says.
Me - How can you add two positive numbers, and get a result that's smaller than one of those numbers?
Student - Ummmm......
Me - I suggest that when typing the numbers in you hit the 2 in 25 too lightly and it didn't register, because 22 is the sum of 17 and 5.
Student - Oh. Is it?
Me - *Sighs*
 
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Dividing the volume of the container by the diameter of the sphere cubed will give you a minimum (essentially assuming the spheres take up as much room as a circumscribed cube), so the answer would be somewhere between that and the answer you get from Paul's method.

Or you can fill the container (with the spheres in it) with water, pour the water out and measure it, and subtract the volume of the water from the volume of the container and divide what's left by the volume of the spheres.
 
The maximum packing ratio for spheres is about 75%. That means that only about 75% of the containing volume will be sphere, and the rest will be air. In practice, the ratio will be somewhat lower for a couple of reasons:

  1. The spheres won't efficiently drop into that perfect matrix.
  2. There are edge effects where you can't pack in a half a sphere.
The second item will be more of a factor as the containing volume approaches the volume of the sphere.

There's a website that talks about packing ratios, but I'm a newbie so I can't name it directly. Go to google and search: sphere packing ratio. It's the top item.

Rob Cockerham (cockeyed.com) also has an article on his site where he competed in a contest to guess the number of ping pong balls that would fit into the interior of an SUV. His experiences are an interesting read, and mostly related to what you're asking about.
 
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From Wolfram: "Random close packing of spheres in three dimensions gives a packing density of only about 0.64 (Jaeger and Nagel 1992), significantly smaller than the optimal packing density for cubic or hexagonal close packing of 0.74048."
 
Of course, the Sphere Packing problem may be undecidable, so we may never know for sure what the best configuration is. That Gödel chap has a lot to answer for.
 
Big Al said:
Of course, the Sphere Packing problem may be undecidable, so we may never know for sure what the best configuration is. That Gödel chap has a lot to answer for.
Click to expand...
I'm not sure I understand. Are you suggesting that you don't believe the proof of the Kepler conjecture, which states that the optimal packing has a density of ~0.74048 (achieved by either cubic or hexagonal close packing)? Or did you mean something else?
 
Thank you all for your helpful answers. Much appreciated.

So, to come up with a reasonable estimate of the number of spheres in the container, I should take the result of the calculation and then adjust it to somewhere between 64-74% of that figure, is that correct?
 
That sounds about right. As long as you're computing a range, you might want to go even a bit lower. About 52% is the scenario where the balls are stacked in a perfect cubic structure:

R = radius of sphere

Cubic Ratio = V(sphere) / V(cube about that sphere)
= ((4/3) pi r^3) / (2 * r)^3
= pi / 6
= 0.5236

So a range of 52% to 74%, with a practical value of 64%.
 
If you win the jar of sweets, i assume you'll be sharing them with everyone that helped out?

:-P
 
Jarom said:
I'm not sure I understand. Are you suggesting that you don't believe the proof of the Kepler conjecture, which states that the optimal packing has a density of ~0.74048 (achieved by either cubic or hexagonal close packing)? Or did you mean something else?
Click to expand...

Well, I understood there was still some controversy about Hales' proof relying on computer breakdown of a finite number of solutions (like Haaken and Appel's proof of the Mordell (Four-Coulour) Conjecture). As far as I know, the peer review process is still going on, nine years later, and it's still a Conjecture, not a Theorem.

The last I heard, the best accepted evidence was a mathematical proof that the packing efficiency could never be better than 74.5% or something, converging in on the face-centred cubic arrangement's efficiency.

Also, how does Hales' proof "stack up" to the abovementioned arrangement? AFAIK, KC concerns itself with the most efficient way of stacking a given pile of spheres (basically, a face-centred pyramid). Not filling a cylinder with spheres of an arbitrary size.

However, I'm more than happy to be enlightened! Maths can move pretty quickly at times.
 
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