# Box Packing!

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What is the maximum number of 1-inch diameter spheres that can be packed in a box 10 X 10 X 5 inches?

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cant we make it 1 inch cubes instead?

pleaaaaaseee?

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Think of a honey comb format and take in to account the dimensions.

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Ah an interesting question.

brings the memories flooding back, packing fraction results, fcc, bcc and hcp.

I remember the results for large volumes where that face centered cubic and hexagonal close packing the accepted packing methods

HOWEVER

I also remember that for limited volumes psuedo-randomn packing based on a frustrated icosihedron can be better

for the cuboid you suggest body centered cubic might even be the best.

I cant see any easier way to answer the problem without trying the options out with some marbles and a cardboard box. but I'll think about it first.

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Well if they were cubes then it'd be 500.... but as they're spheres I'd estimate at around 550...

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I agree with gnpatterson to use close-packed hexagonal arrangement. I think the answer is 525.

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It is more I think. I would suggest 607.

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I denigh saying that hcp is the solution!

the problem that I see with hcp is the layer positions

If you think about the the hexagonal layers hcp repeats in the form ABCABCABC while fcc repeats ABABABAB

the best way to show this is is with a diagram, here goes nothing

__A_____A_____A_____A

_____C_____C_____C

__B_____B_____B

_____A_____A_____A

__C_____C_____C_____C

_____B_____B_____B

__A_____A_____A_____A

_____C_____C_____C

now if you print that out and join all the A's up you should have a hexagonal pattern, same with the B's and the C

when you pack hexagonal layers you can pack the layers in one of two equivalent positions. If you draw a close packed layer and look at the equilateral triangles formed by the holes one set points up (or left) and one set points down (or right).

You can then place the next layer either on all the triangles pointing up or the triangles pointing down.

NO BIG deal at this stage

HOWEVER when you place the third layer you have a choice, you can place it either back in the same position as the first layer (ABA) which is in triangles pointing in the opposite direction OR you can go on to place in the triangles pointing in the same direction (ABC).

This is easiest to see if you build a model

This does not effect the packing fraction in the large scale

It does however effect the packing in a cuboid

Firstly choice of which direction to pack in is critical

for the purposes of demonstration I will consider starting with a layer on the 5 by 5 face of a cuboid.

In the first layer we can pack in rows of 5,4,5,4,5 using up a total height of 4.464 = 4* 0.866 +1

The second layer of 4,5,4,5,4 reaches to 4.75277675= 4.464101615 + 1/3 0.866025404

If you choose the third layer to be type C (hcp) it would be 5,4,5,4 and the next row would be absent as if present it would push the height to 5.041451884

HOWEVER the fcc option would give you 5,4,5,4,5 ie a total of 23+22+23 for three layers rather than 23+24+18 (hcp).

This analysis does not calcualate the "thickness" of the layers nor does it say that this is the optimal approach for the cuboid give 10 x 10 x 5. However the appearence of 5.04 which is so frustratingly close to 5 is suggestive that the problem is going to be interesting.

It would in reality be possible to add that fifth row (third hcp layer) if you allow the spheres to rise up above the layer slightly or raised the whole layer up and back a bit.

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OOPs I seem to have my hcp and fcc totally mixed up, it is other way round!?

although the terminology is wrong confused the geometry is probably OK

can anyone with a better notion help (in my defense it was many many years ago that I did this stuff)

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Now looking at the actual problem

here are some results

close packed layer spacing = 0.816496581

spacing of rows = 0.866025404

sphere packing results http://mathworld.wolfram.com/SpherePacking.html

if spheres could be packed without consideration of the walls then a volume of 10*10*5 could hold 628.5393611 spheres = 500 * packing fraction / volume of sphere

HOWEVER considering real packing schemes

1) 12 layers of 10,9,10,9,10 and 9,10,9,10,9 gives 570

2) 5 layers of 10,9,10,9,10,9,10,9,10,9,10 and 9,10,9,10,9,10,9,10,9,10,9 gives 523

3) 12 layers of 5,4,5,4,5,4,5,4,5,4,5 and 4,5,4,5,4,5,4,5,4,5,4 gives 594

considering the amount of apparent wasted space in these "close packed" schemes I have to say that a "random" packing arrangement MUST exist that can do better

BUT I cant say that it could be found easily

I would like to try some different orientations of the layers but don't see this is easy enough to do.

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594 is the correct answer gnpatternson. Well done! How long did it take you?

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I am not yet convinced that 594 _is_ correct.

Obviously I have nothing better to do ALL day, but most of the time was spent typing the conversation about how to think about doing it.

also I did a few su doku puzzles in between.

the 594 solution has a lot going for it in that the 12 layers use up 11* 8.1649 + 1 = 9.98146239 of the 10 inches available. This gives very little room for manouver and would appear to lock this solution in place.

HOWEVER this is not enough to _prove_ that another configuration is not possible.

the mathsworld article references a paper

http://www.combinatorics.org/Volume_11/PDF/v11i1r33.pdf

which shows the actual arrangements for some spheres in cubes but for very low sphere numbers <40 were some computational work has been done to get solutions.

it is obviously not likely to be able to assemble a workable simulation of 600 odd spheres.

how did you come up with the problem?

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The problem was on the site http://www.highiqsociety.com, test for exceptional intelligence. By the way, the 24 square question has never been solved!

I have achieved 12/25 questions correct.

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Where did you get the close pack layer spacing digit from?

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thanks johny321

I visited the site and the "free IQ test" gave the usual flattering results, allowing me to join the site. HOWEVER i agree with grocho marx "i wouldn't want to be in a club that would have someone like me as a member"

BUT what is the 24 square question? it doesnt seem to be in the puzzles section of the site, or arent i looking in the right place?

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Go to the Test for Exceptional intelligence, question 20 something. What were your scores

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Gnpatterson, if you get into the Top scores on the Test for Exceptional intelligence, can you tell me your answers.

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Had a look at the square question,

it is not really fair because I recognised it straight away, it is the disection of a "perfect square"

were it is one of the examples given

I really can't imagine getting many of the questions "right" without taking a large amount of time. Of course some of them I also recognize as mathmatical problems that I can do easily but that is hardly a fair test for example the cannon ball one is just a bit of discreet mathematics. others would seem to just require a significant amount of work the decryption type ones. most just seem to be based on the difficulty of having to find the correct strategy to avoid the hard work.

How long are you supposed to take for the test?

the squares one would be possible to do without "knowing" the answer again beacuse it is a bit of discreet maths that has to set up very carefully for it to get the correct answer. the diagram gives you the template to use.

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oh yes the close pack layer spacing is simple pythagorus to get the height of a unit tetrahedron

The height of the triangle is sqrt(3)/2

then construct the triangle with this as the hypot and 1/3 this as the base, the height is then the number you require.

this would be so much easier with a diagram

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Initially, I let x= the smallest square and let the next smallest square = y. From there, I worked around the center until I came up with a formular for (?), inwhich should be 17x + 7y. But, I didn't know what length unit the smallest square should be. Now, I know y=3x, it shouldn't be a worry.

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looking at your results I think I can spot your mistake. you have accidentally equated the "?" with the second square on the top row.

NB you should not trust the relative sizes of the squares on the diagram

this means while the square on the top right hand corner is shown as being smaller than the one below, it is in fact bigger.

To solve the problem you have to go all the way to equating the total width and total height AND to finding the smallest set of integers for all the squares, this includes the small ones at the bottom.

Care should also be taken to make sure that none of these measures slip into the negative integers if you have rashly canceled them out.

You can only get the value of "?" if you solve the whole thing unless you take a leap of faith.

btw did you get y=3x yourself? If so how did you do it without working out the whole thing, it doesn't seem possible?

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I originally did workout all the lengths, but it took too long. If you want to know how I got x and y on question 21, look at the very smallest square and let it = x and the next smallest one ( to the right) = y. From that, we should get ( ?) as 17x + 7y, I was stuck because I didn't know exactly the value of y relative to x. So, i printed the sheet and measured it out at 200x magnification to get an accurate result. I found y=3x, from there I assumed ( ?)= 38x, removing the variable we should get ( ? )= 38. Which High IQ Society said the answer is.

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you are right, mathworld is wrong

I just did the maths and you are spot on

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Ah spotted the problem the two diagrams are slightly different, with one (sub) rectangle being flipped.

Don't you feel however that the method you used lacks the finesse required by a high IQ society?

I know IF the answer is the only thing that is important then it doesnt matter how you get it.

BUT for me the method is the only thing i will take away from the problem, so to me it is the important thing.

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