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How many balls can fit into a jar

daigo

Member
Jun 27, 2012
60
The jar has a radius of 6" and a height of 24" and each ball has a radius of 1".

So I found the volume of the jar which is [tex]\pi6^{2}(24) = \approx 2,714.33605[/tex] and the volume of the balls which is [tex]\frac{4}{3}\pi1^{3} = \approx 4.1887902[/tex]

And then I divided how many of the balls can go into the jar by dividing:

[tex]2714.33605 \div 4.1887902 = 648 balls[/tex]

Does that number take into account the spaces between the balls when put into the jar? Like the small gaps when spheres are placed next to each other.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
The jar has a radius of 6" and a height of 24" and each ball has a radius of 1".

So I found the volume of the jar which is [tex]\pi6^{2}(24) = \approx 2,714.33605[/tex] and the volume of the balls which is [tex]\frac{4}{3}\pi1^{3} = \approx 4.1887902[/tex]

And then I divided how many of the balls can go into the jar by dividing:

[tex]2714.33605 \div 4.1887902 = 648 balls[/tex]

Does that number take into account the spaces between the balls when put into the jar? Like the small gaps when spheres are placed next to each other.
No, the maximum packing fraction that can be achieved with regular packing and no boundary effects is ~74%, and the random close packing packing fraction is ~63%.

See: >>here<< and >>here<<

CB
 

soroban

Well-known member
Feb 2, 2012
409
Hello, daigo!

The jar has a radius of 6" and a height of 24" and each ball has a radius of 1".
How many balls can fit into the jar?

So I found the volume of the jar which is: [tex]\pi6^{2}(24) = \approx 2,714.33605[/tex]
and the volume of the balls which is: [tex]\frac{4}{3}\pi1^{3} = \approx 4.1887902[/tex]

And then I divided how many of the balls can go into the jar by dividing:
. . ][tex]2714.33605 \div 4.1887902 = 648\:balls[/tex]

Does that number take into account the spaces between the balls when put into the jar? . . . . no
Like the small gaps when spheres are placed next to each other.

First of all, I'd do your math like this:

Volume of jar: [tex]\pi(6^2)(24) \:=\:864\pi[/tex]

Volume of ball: [tex]\tfrac{4}{3}\pi(1^3)\:=\:\tfrac{4}{3}\pi[/tex]

Therefore: .[tex]864\pi \div \tfrac{4}{3}\pi \:=\:648\text{ balls}[/tex]

You see, I hate long (and incomplete) decimals.
. . I hate writing them down, I hate entering them on my calculator.
I would do anything (even Algebra) to avoid that.


Second,you have melted the 648 balls into a puddle.
. . Then you poured the liquid into the jar.
 

llabnor

New member
Nov 7, 2018
1
The jar has a radius of 6" and a height of 24" and each ball has a radius of 1".

So I found the volume of the jar which is [tex]\pi6^{2}(24) = \approx 2,714.33605[/tex] and the volume of the balls which is [tex]\frac{4}{3}\pi1^{3} = \approx 4.1887902[/tex]

And then I divided how many of the balls can go into the jar by dividing:

[tex]2714.33605 \div 4.1887902 = 648 balls[/tex]

Does that number take into account the spaces between the balls when put into the jar? Like the small gaps when spheres are placed next to each other.
If the ratio of the dimensions of the jar compared to the dimensions of the diameter of the ball were large (it is not in this case), then the answer to the question of how many balls of 1 " radius could you put inside a jar of radius 6 " and a height of 24 " would be 648 times a "density constant" of about 0.5, or, 648 * 0.5 = 324. Let's call it 330 plus/minus 10. The larger the jar (and/or the smaller the marble), the higher the "density constant" would be. If you were figuring how many 5/8" diameter marbles (2 cc each) would go into a 1 gallon jar (3785 cc's), for example, the calculations would be:

3785/2*0.6 = 1324.75

or about 1330 plus/minus 10 marbles. The "density constant" here is 0.6. If you have a VERY LARGE container with VERY SMALL marbles, the density constant would be a maximum of 0.74.