Tax your brain, right here in this thread....

[Skeet UK]

As one might do in a Cocktail Lounge, I have a suggestion.

I think we should have a thread, that people can post up entertaining stuff, that taxes the brain; that those who are inclined, can while away the lonely hours with.

With that in mind, who likes a bit of Maths?

I am no Maths wizard by any means, but the following question came up somewhere a while back and I very much enjoyed the process that followed...employing only GCSE maths and anything I have learned since then, So...to the conundrum.

I have a box, with internal dimensions of 1M x 1M x 1M (1 cubic meter).

How many 2.5mm ball bearings can I fit inside this box and what is the total surface area of the bearings?

The box must be able to close, it has no internal hinge to impede closure or capacity and the dimensions of the box do not change when the bearings are added (ie: a theoretical box, not a cardboard box that will bend etc).

Have fun!

 

[ExSCA]

Nice idea (brain explodes)

 

[Maxum]

64000

 

[Dick Arbon]

64 million ?

 

[Maxum]

53.3333333333333 of a m3

 

[Maxum]

64 million ?

Need a chuffing big box )

 

[Skeet UK]

Yep, starting to get there with 64 Million 400x400x400....but it is more than that....get out your 2p pieces and a ruler...I have just revisited this question and managed to find even more than I did the first time!

Nope,no chuffing big box required...64 Million of them will easily fit into a 1m cubed box.

 

[ExSCA]

Right - I think its 64,000,00 - with a surface area of 1,256,000 meters squared.

But that assumes alignment like this:

0 0 0

0 0 0

When I think bearings would align

0 0 0 0

0 0 0

0 0 0 0

 

[Maxum]

Doh saw 25 mm not 2.5 hahaha. I've got a chuffing little box )

 

[Maxum]

8,533,333333333333333333333333

 

[Sue Claridge]

zero

 

[Skeet UK]

Right - I think its 64,000,00 - with a surface area of 1,256,000 meters squared.

But that assumes alignment like this:

0 0 0

0 0 0

When I think bearings would align

0 0 0 0

0 0 0

0 0 0 0

64 Million, if stacked as you could cubes.

But as you said they don't stack like cubes.

The second diagram is closer, but with one crucial difference

_0 0 0 0

0 0 0 0

_0 0 0 0

Underscores are so the balls show correctly

So if they stack in that way...what happens to that layer? And by how much?

 

[timps]

Volume of box = 1m3

Diameter of ball is 2.5 mm so radius is 1.25 mm = 0.00125 m

Volume of ball is ((4/3)*3.14159* (0.00125^3))

Volume of ball = 0.00000000818122395833333 m3

So volume of box divided by volume of ball bearing will give you how many ball bearings will fit in with no gaps (if you squashed them into one solid lump).

So 1 / 0.00000000818122395833333 = 122231099.538769.

Round it down as you only want whole ball bearings

I have to do bulk density calculations at work and the theoretical maximum density achievable for spheres is accepted at 74%, you unlikely to achieve this in practice in a box as the walls can interfere with the settlement needed but theoretically you can so...

122231099 * 0.74 = 90451013

90,451,013 whole ball bearings.

0.74 m3 total volume area

I could be wrong but it makes sense to me .

 

[Skeet UK]

Volume of box = 1m3

Diameter of ball is 2.5 mm so radius is 1.25 mm = 0.00125 m

Volume of ball is ((4/3)*3.14159* (0.00125^3))

Volume of ball = 0.00000000818122395833333 m3

So volume of box divided by volume of ball bearing will give you how many ball bearings will fit in with no gaps (if you squashed them into one solid lump).

So 1 / 0.00000000818122395833333 = 122231099.538769.

Round it down as you only want whole ball bearings

I have to do bulk density calculations at work and the theoretical maximum density achievable for spheres is accepted at 74%, you unlikely to achieve this in practice in a box as the walls can interfere with the settlement needed but theoretically you can so...

122231099 * 0.74 = 90451013

90,451,013 whole ball bearings.

0.74 m3 total volume area

I could be wrong but it makes sense to me .

Yeah, when it came up before, many of the guys used Close Packing of Spheres theory. There are 3 accepted theories, one being Mude, all run around the 74% mark.

If you ignore the theories for general estimates and actually get physical with it (3 or more 2p pieces, ruler) you can calculate the first layer almost perfectly.

You cannot fit a fraction of a bearing in, only whole bearings, in rows. If a bearing would bring it outside of the dimensions of the box, it cant be there.

However, I found that you could calculate this more accurately. The question ask how many can I fit in a 1M cubed box. So because of the size of the bearings and the convenient dimensions of the box, you can get down pretty close. It require you place the bearings in a specific lattice.

I had no problems with the first layer and decided on a value for the second layer, then went from there...when I came back to it yesterday, I have come up with another idea for the second layer, which changes the answer...hence why I thought I would bring it on here.

 

[timps]

Granted 74% is just a theoretical but proven maximum, the fraction of the bearing not fitting is not as important as it seems as we are dealing with percentage of voids and this is proportional to the volume occupied, it might cost you a few ball bearings but not thousands upon thousands.

.

At work we test void contents with the ASTM and BSI approved tests, using water to fill the voids, then as we know the molecular weight of water we can measure the voids. Achieving 60 % density is an everyday achievable occurrence with the type of single sized particle scenario you describe even within the constraints of the box. This is just using irregular, random packing, shaking and settlement methods (maximum for this type of packing is considered to be 64% density).

On 64,000,000 bearings you would have :-

Number of balls x volume of ball

64,000,000 * 0.00000000818122395833333 = 0.52 m3 so that gives you 52% ball bearings and 48% voids in 1m3 box, even in the constrains of the box you will be able to achieve 60% density or more.

So you need to get it up to about 73,338,659 ball bearings at least.

At an educated guess 65% density should be achievable.

Craig

 

[Chard]

Granted 74% is just a theoretical but proven maximum, the fraction of the bearing not fitting is not as important as it seems as we are dealing with percentage of voids and this is proportional to the volume occupied, it might cost you a few ball bearings but not thousands upon thousands.

.

At work we test void contents with the ASTM and BSI approved tests, using water to fill the voids, then as we know the molecular weight of water we can measure the voids. Achieving 60 % density is an everyday achievable occurrence with the type of single sized particle scenario you describe even within the constraints of the box. This is just using irregular, random packing, shaking and settlement methods (maximum for this type of packing is considered to be 64% density).

On 64,000,000 bearings you would have :-

Number of balls x volume of ball

64,000,000 * 0.00000000818122395833333 = 0.52 m3 so that gives you 52% ball bearings and 48% voids in 1m3 box, even in the constrains of the box you will be able to achieve 60% density or more.

So you need to get it up to about 73,338,659 ball bearings at least.

At an educated guess 65% density should be achievable.

Craig

Piss off

 

[Skeet UK]

So you need to get it up to about 73,338,659 ball bearings at least.

At an educated guess 65% density should be achievable.

Craig

The formulae are great,because they allow you to quickly estimate, to a high degree of accuracy, what you need to know.

My purpose was, as a Layman who does not know about these formulae (at the time)...how close could I get with basic maths and some practical tests.

I can get over 73 Million ...out of interest, in this scenario, the voids equate to 360L, around 32 crates of lager!

Anyone else got any ideas...Mr Rutherford...?

 

[ExSCA]

I stopped thinking about it a little because I was working on something else last night. But that's the way to do it... the surface area is a red herring - the key calculation is the mass of the bearings. Then divide the volume available by the mass. I'll be back in a bit.

 

[Skeet UK]

I stopped thinking about it a little because I was working on something else last night. But that's the way to do it... the surface area is a red herring - the key calculation is the mass of the bearings. Then divide the volume available by the mass. I'll be back in a bit.

Surface area is just another part of the question that you calculate once you have the quantity.

Mass of bearings is not important unless you want to start factoring compressive strength of steel and how much space that might yeild.

So think along the lines of what you said before, what I said about it and then get your 2p bits out!