Feynman exercise - container with steel balls

[hubot]

Homework Statement

You received many steel balls of the same diameter d and container of known volume V. All dimensions of the container are much greater than balls diameter. How many balls can be placed in a container?

Homework Equations

$$(\frac{4}{3}\pi)(\frac{d^3}{8})\approx0,52d^3$$

The Attempt at a Solution

To place balls in a container we should arrange this container with in a certain order. Let's assume that this balls will be organised such that centres every eight of the bullets will be in vertices of cube. Of course, the length of side of the cube will be equal in diameter of bullet d. For each cube one bullet. Calculating using this formula $$(\frac{4}{3}\pi)(\frac{d^3}{8})\approx0,52d^3$$ we conclude that bullets occupy 52% of the available area.

 

[BvU]

Area ?
And sphere packing doesn't go in cubes.

 

[haruspex]

Hubot, you have not made it clear what you are asking of the forum. Nor is it clear what information was provided to you, what exactly was the question asked of you, and what parts are your attempt at solution.
Please clarify.

 

[CalebB-M]

I am working on this problem to. Its from the feynman lectures problem set book. The problem statement is literally find an equation for giving the number of steel balls when you only know the relationship to the diameter of the balls to the dimensions of the box i.e much greater.
My solution was V/d^3 always drop the decimal. This idea came to me similar to what you thought the spheres stack like a cube.

 

[scottdave]

You can gain some insight from this Standupmaths video on the subject. They don't stack on top of one another like a cube.

 

[bitphi]

The volume $V_b$ of an individual ball is given by:

$$ V_b = \frac{4}{3}\pi r^3 $$

The total volume $V_N$ that can be occupied by $N$ balls is given by Gauss's sphere-packing rule:

$$ V_N = \frac{\pi V}{3\sqrt 2} $$

So:

$$ N = \frac{V_N}{V_b} = \frac{3 \pi V}{12 \pi (d/2)^3 \sqrt 2 } = \frac{2V}{d^3 \sqrt 2 } $$

 

[haruspex]

The volume $V_b$ of an individual ball is given by:

$$ V_b = \frac{4}{3}\pi r^3 $$

The total volume $V_N$ that can be occupied by $N$ balls is given by Gauss's sphere-packing rule:

$$ V_N = \frac{\pi V}{3\sqrt 2} $$

So:

$$ N = \frac{V_N}{V_b} = \frac{3 \pi V}{12 \pi (d/2)^3 \sqrt 2 } = \frac{2V}{d^3 \sqrt 2 } $$

Your algebra is fine, but the way you define the variables doesn’t quite work. If $V_N$ is "The total volume … that can be occupied by N balls" then N should appear on the RHS, not V. How about "the volume of ball that can be packed into much larger sphere of volume V"?
Also, you could simplify the answer to $\frac{V\sqrt 2}{d^3}$.

 

[phinds]

The total volume that can be occupied by balls is given by Gauss's sphere-packing rule:

The OP not specify a sphere. It was a vague question.

 

[vela]

The OP not specify a sphere. It was a vague question.

Ball with diameter $d$. To me, it's pretty clear that's a sphere. What other shape did you have in mind?

 

[phinds]

Cylinder and cube come to mind. I do agree that sphere is the most likely intended shape, BUT ... I don't go by what is intended, I go by what is SAID.

 

[haruspex]

Cylinder and cube come to mind.

Described as "balls"? With only one dimension specified, and that described as a diameter?
I think you just invented cubic snooker.

 

[phinds]

Described as "balls"? With only one dimension specified, and that described as a diameter?
I think you just invented cubic snooker.

My comment was about the CONTAINER, not the balls themselves.

 

[haruspex]

My comment was about the CONTAINER, not the balls themselves.

Then I do not understand your post #8. In "Gauss's sphere-packing rule", "sphere" refers to the items being packed, not the container. Nowhere did post #6 assume a spherical container.

 

[hutchphd]

The spirit of the question is clear. In the limit of "large container" the exact shape provides only a surface effect which vanishes in the limit. So the limit is the packing ratio. Much Ado About Nothing

 

[Chestermiller]

A simple cubic array is not the densest packing of spheres. Google face-centered cubic array or close-packed cubic array.

 

[Chestermiller]

I am working on this problem to. Its from the feynman lectures problem set book. The problem statement is literally find an equation for giving the number of steel balls when you only know the relationship to the diameter of the balls to the dimensions of the box i.e much greater.
My solution was V/d^3 always drop the decimal. This idea came to me similar to what you thought the spheres stack like a cube.

This is correct for simple cubic packing.

 

[haruspex]

This is correct for simple cubic packing.

You are replying to a five year old post.

 

[hutchphd]

You are replying to a five year old post.

I am amazed that the question of dense packing of spheres (apparently called the Kepler Conjecture) hexagonal and cubic close packing both give $$\pi/(3\sqrt 2$$ and that this is not proven maximal ?. Is it really not settled?

 

[haruspex]

hexagonal and cubic close packing both give

They don’t. What makes you think that?

Is it really not settled?

"In 1998, Thomas Callister Hales, following the approach suggested by László Fejes Tóth in 1953, announced a proof of the Kepler conjecture. Hales' proof is a proof by exhaustion involving checking of many individual cases using complex computer calculations. Referees said that they were "99% certain" of the correctness of Hales' proof. On 10 August 2014, Hales announced the completion of a formal proof using automated proof checking, removing any doubt.[4]"
- https://en.wikipedia.org/wiki/Sphere_packing

 

[hutchphd]

They don’t. What makes you think that?

This is not simple cubic. I think it is face centered cubic but the coordination number is the same as HCP. My brain does not stretch that far! For instance:

https://www.differencebetween.com/d...nal-close-packing-and-vs-cubic-close-packing/

 

[haruspex]

This is not simple cubic. I think it is face centered cubic but the coordination number is the same as HCP. My brain does not stretch that far! For instance:

https://www.differencebetween.com/d...nal-close-packing-and-vs-cubic-close-packing/

Ok, thanks for the explanation.