# Feynman exercise - container with steel balls

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1. Feb 23, 2016

### hubot

1. The problem statement, all variables and given/known data
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?

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

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.

2. Feb 23, 2016

### BvU

3. Feb 23, 2016

### 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.

4. Sep 26, 2017

### 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.

5. Sep 26, 2017

### 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.