Tax your brain, right here in this thread....
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This is how I worked it out back when I first looked at it:
If you lay bearings in a row front to back of the box, then the first row will contain 400 bearings.
The next row, if you aligned them the same, would also have 400, but...if you align the bearings into the gaps between the first row, you lose one bearing, but the PAIR OF ROWS occupies less space. I calculated using 2p pieces that the reduction in width is 11.539%.
So the first pair of rows contains 799 bearings with a width of 4.42305mm.
We already know that at least 400 bearings will go across, which is 200 rows. 200 X 4.42305 = 884.61mm
The remaining 115.39 if you divide by 4.42305 you get 26.08. So we have a further 26 pairs of rows.
So the first layer in the box will contain 226 pairs of rows (452 rows), each pair containing 799 bearings One long, one short to a pair.
First Layer: 226 X 799 = 180574
The next layer, if stacked the opposite way (short row over long row and vice versa), can contain the same amount of bearings.
So we can have 400 X 180574 = 72,229,600 as the total, if horizontal rows are staggered and vertical columns are stacked.
Now, this is where it gets complicated...thinking in 3D! It had occurred to me that the next layer, could be sat "Cannon Ball Style" over the first.
For this to happen, the next layer would have to move one row to the left, so losing 399 bearings (last row would have been a short row) and one row to the rear, again seeming to lose another row of 400, which would mean that the next layer would contain 799 less bearings at 179775.
So, for the first two layers we have 180574 + 179775 = 360349 bearings in the first two layers.
We know that the reduced height will be the same as the reduced width, so now we can have 226 PAIRS OF LAYERS rather than 200.
226 x 360349 = 81,438,874 bearings in the box.
Total volume of 665932.432 Cubic Litres. 66% density
Total Surface Area 1598.2379m^2
That is how I did it anyway, using basic maths.
If you lay bearings in a row front to back of the box, then the first row will contain 400 bearings.
The next row, if you aligned them the same, would also have 400, but...if you align the bearings into the gaps between the first row, you lose one bearing, but the PAIR OF ROWS occupies less space. I calculated using 2p pieces that the reduction in width is 11.539%.
So the first pair of rows contains 799 bearings with a width of 4.42305mm.
We already know that at least 400 bearings will go across, which is 200 rows. 200 X 4.42305 = 884.61mm
The remaining 115.39 if you divide by 4.42305 you get 26.08. So we have a further 26 pairs of rows.
So the first layer in the box will contain 226 pairs of rows (452 rows), each pair containing 799 bearings One long, one short to a pair.
First Layer: 226 X 799 = 180574
The next layer, if stacked the opposite way (short row over long row and vice versa), can contain the same amount of bearings.
So we can have 400 X 180574 = 72,229,600 as the total, if horizontal rows are staggered and vertical columns are stacked.
Now, this is where it gets complicated...thinking in 3D! It had occurred to me that the next layer, could be sat "Cannon Ball Style" over the first.
For this to happen, the next layer would have to move one row to the left, so losing 399 bearings (last row would have been a short row) and one row to the rear, again seeming to lose another row of 400, which would mean that the next layer would contain 799 less bearings at 179775.
So, for the first two layers we have 180574 + 179775 = 360349 bearings in the first two layers.
We know that the reduced height will be the same as the reduced width, so now we can have 226 PAIRS OF LAYERS rather than 200.
226 x 360349 = 81,438,874 bearings in the box.
Total volume of 665932.432 Cubic Litres. 66% density
Total Surface Area 1598.2379m^2
That is how I did it anyway, using basic maths.
My educated guessing is getting better.
You might get more if you started the first row on the internal diagonal of the cube and worked out like a pyramid in diferent directions to the corners, then again you might not, but I seem to recall it can help in some instances in reducing the voids.
Running computer simulations similar to your 2p method and you can get as low as 60% up to as high as 74% depending on where you start and what order you add the spheres. In real world situations you are highly unlikely to start from the middle and work out in different directions but this might be require to get the maximum numbers in.
Craig
Maxum
Well-known member
I've squashed my box. :-((
bonzodogdoodaband
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64.000.000
12566370.6144sqm
12566370.6144sqm
Les Graham
Well-known member
Anyone fancy a pint..............................well it's boring init
I've still got room in my box for 30 odd crates of beer, if you like!Anyone fancy a pint..............................well it's boring init
It's only boring, if it isn't your cup of tea...find us something you like to peruse...a riddle or whatever, something taxing!
Yeah, I seem to remember when I looked at the close Packing theories, that there were a number of projected lattices that could be used.My educated guessing is getting better.
You might get more if you started the first row on the internal diagonal of the cube and worked out like a pyramid in diferent directions to the corners, then again you might not, but I seem to recall it can help in some instances in reducing the voids.
Running computer simulations similar to your 2p method and you can get as low as 60% up to as high as 74% depending on where you start and what order you add the spheres. In real world situations you are highly unlikely to start from the middle and work out in different directions but this might be require to get the maximum numbers in.
Craig
As I say, my aim when I attempted to answer the question a couple of years ago on another forum, was to see if a mere mortal with basic Maths, could get near the answer...it seems I did OK, which I was happy about! ;cool:
Your starting in the middle bit; is that 2D middle of the bottom of the box, or 3D middle of the centre of the box...in mid air?
Cheers for your input, has been interesting.
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Both the middle of 2D or 3D depending on the ball box size ratio, the best way is for every three spheres touching then a fourth sphere can be placed on top in the hollow between the bottom three spheres. If you can achieve this everywhere then you get the maximum density possible of 74% (ish) . The trick is to get as many doing this as possible in the space you have, but with the size of the box and size of the bearing an ordered approach may not be the best way.
I work in a family owned materials testing laboratory and we get journals sent to us and I remember reading one that had an article from some Professor of Chemistry at Princeton University, he did some simulations on random placement. The idea is to look for any combination of spheres that are so tightly packed that none of the spheres can move - a jammed state. It can look like random placement but actually its a very clever lattice that you would never work out unless you had a fancy computer program for the amount of ball bearings we have here. I think using his program you could get it into the 70 % density range.
Getting 66% density is a good bit of work without using a computer.
I work in a family owned materials testing laboratory and we get journals sent to us and I remember reading one that had an article from some Professor of Chemistry at Princeton University, he did some simulations on random placement. The idea is to look for any combination of spheres that are so tightly packed that none of the spheres can move - a jammed state. It can look like random placement but actually its a very clever lattice that you would never work out unless you had a fancy computer program for the amount of ball bearings we have here. I think using his program you could get it into the 70 % density range.
Getting 66% density is a good bit of work without using a computer.
Well...I must be using 11% of my brain then!Both the middle of 2D or 3D depending on the ball box size ratio, the best way is for every three spheres touching then a fourth sphere can be placed on top in the hollow between the bottom three spheres. If you can achieve this everywhere then you get the maximum density possible of 74% (ish) . The trick is to get as many doing this as possible in the space you have, but with the size of the box and size of the bearing an ordered approach may not be the best way.
I work in a family owned materials testing laboratory and we get journals sent to us and I remember reading one that had an article from some Professor of Chemistry at Princeton University, he did some simulations on random placement. The idea is to look for any combination of spheres that are so tightly packed that none of the spheres can move - a jammed state. It can look like random placement but actually its a very clever lattice that you would never work out unless you had a fancy computer program for the amount of ball bearings we have here. I think using his program you could get it into the 70 % density range.
Getting 66% density is a good bit of work without using a computer.
I appreciate your recognition, thank you.Yes, now I remember...the ideal shape is a hexagon of spheres one central, 6 around around the outside, three and one, top and bottom.
So you would layer the bottom with cones (missing the bottom sphere), fill in the gaps at the edges and corners, then keep adding diamonds till you get to the top, then fill in with more cones and the edges.
Simples...might give that a bash later! :huh:
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Maxum
Well-known member
I'm watching James Bond and even he can't work it out )
)Yeah but he does get the girls, the nice car, the snappy suits...can carry a gun (even if it is a PPK), can shoot people....and gets to go parachuting with The Queen...I'm watching James Bond and even he can't work it out
Name a maths geek he gets to do half of that..
