[bonanova]

Four identical cubical boxes, with interior dimensions of a cm on a side, and

weighing b grams each, were packed with wooden spheres of density of d gram/cm³

Box 1 contained 1 single sphere; the largest one that would fit inside the box.

Box 2 contained 8 identical spheres; the largest that would fit inside the box.

Box 3 contained 27 identical spheres; again the largest that would fit.

Box 4 contained 64 identical spheres; again the largest possible.

Which loaded box [box plus contents] is heaviest, and what is its weight?

[Guest]

I guess I have it

All should weigh same and the weight should be b + (pi/6)(a^3)(d)

[Guest]

all should weight equal, as the boxes' weight don;t make a diff

all spheres' density is same and the only thing that would make a diff is their volumes

total volumes for each come out to be 0.16666 (dim of a^3)

Edited February 20, 2009 by tarunark

[Guest]

Box 4 is the heaviest! In fact, the more spheres in the box, the more the box is occupied (in volume) and thus if we imagine a box filled with infinite number of microscopic spheres, then the box is fully loaded and its weight tends to (b+d.a^3). As for the weight of box 4, I'm too lazy to do the calculations...

Edited February 20, 2009 by mojail

[Guest]

Box 4 is the heaviest! In fact, the more spheres in the box, the more the box is occupied (in volume) and thus if we imagine a box filled with infinite number of microscopic spheres, then the box is fully loaded and its weight tends to (b+d.a^3). As for the weight of box 4, I'm too lazy to do the calculations...

Assuming that the boxes are the same size,

box one would be the heaviest because in the other boxes there would be more empty space between the spheres as they can't interlock

[Guest]

Box 4 is the heaviest! In fact, the more spheres in the box, the more the box is occupied (in volume) and thus if we imagine a box filled with infinite number of microscopic spheres, then the box is fully loaded and its weight tends to (b+d.a^3). As for the weight of box 4, I'm too lazy to do the calculations...

I think;

Even with microscopic spheres, the weight doesn't changes, if the spheres are all at same dimension. In practice, there usually exist various size of spheres (i.e sand), so that small sands fit in the space among other greater sands, and will weigh more.

[Guest]

I agree with mojail. The smaller spheres don't weigh the same as the bigger spheres. As the volume decreases only the density remains constant, not the mass. Therefore each sphere is proportionaly as massive as the next. The only variable here is the volume of sphere contained in each box and my guess is that the box with the large number of small spheres wins out.

Edited February 20, 2009 by mpapineau

[Guest]

Box 1: d.((4/3pi).(a^3))

Box 2: d.((4/3pi).(1/2a^3)).8

Box 3: d.((4/3pi).(1/3a^3)).27

Box 4: d.((4/3pi).(1/4a^3)).64

The result is all four boxes weight the same

Replace a with a number. 12 for an example

12^3=1728

(6^3)*8=1728

(4^3)*27=1728

(3^3)*64=1728

Edited February 20, 2009 by Snowman63

[Guest]

Box 1: d.((4/3pi).(a^3))

Box 2: d.((4/3pi).(1/2a^3)).8

Box 3: d.((4/3pi).(1/3a^3)).27

Box 4: d.((4/3pi).(1/4a^3)).64

The result is all four boxes weight the same

Replace a with a number. 12 for an example

12^3=1728

(6^3)*8=1728

(4^3)*27=1728

(3^3)*64=1728

while it would not surprise me if the answer came out as all are equal, I wonder is it possible to have a closer packing than a simple 4x4x4 grid. If that is the case then we might be able to fit slightly larger spheres than diameter a/4 in the box.

Maybe, maybe not.

[Guest]

I work in the oil industry and this problem is one of the first things we get taught because of it's application to rock properties. The porosity is the percentage of open space in the rock. The porosity is actually independent of the grain size. Instead, it is dependent on the standard deviation of the grain size. The greater the diversity in grain size, the lower the porosity. The greatest porosity is an idealized, all-equal-size-spheres system as described above.

[Guest]

yeah.. I was about to change my answer

I have to agree now that the space between each sphere is going to create an exponential (I think) loss of mass. I really wouldn't bet anything on my math right now (especially considering the company), but I figured:

Let a=12cm (side of cube)

Let d=1g/cm^3 (density of sphere)

Box 1 weighs 904.77g

Box 4 weighs 14.08g

[Guest]

i think this is the problem

"The greatest challenge to any thinker is stating the problem in a way that will allow a solution." - Bertrand Russell

i dont think there can be a solution because there is not enough information in the problem. the heaviest might be the first box because it has the biggest sphere, but because all the space that it is left in the box u might think that smaller spheres are the answer because they will fill that space, but in fact because of the shape of a sphere, the spheres will create space between them, so i dont think there can be an answer to this unless more information is provided..... they might all just weigh the same...

[Guest]

I am going to assume that the fact that no information is given regarding the density or weight of the spaces between spheres that the answer must be that they all weigh the same. If the the box was to be held under water then the spaces would be more dense and therefore the lightest box out of water would be the heaviest under water. Seems to fit although when you visualise the problem you would think you could pack in more space using more spheres.

Really keen to find out the right answer to this.

[Guest]

I work in the oil industry and this problem is one of the first things we get taught because of it's application to rock properties. The porosity is the percentage of open space in the rock. The porosity is actually independent of the grain size. Instead, it is dependent on the standard deviation of the grain size. The greater the diversity in grain size, the lower the porosity. The greatest porosity is an idealized, all-equal-size-spheres system as described above.

I love to see when brain teasers have practical applications!

[Guest]

Yeah I thought that was a great post too, kudos voltage. He even gave it a name!

I think I read somewhere that a sphere will take up about 50% of the volume in a cube. If we consider that the best way to pack spheres into a cube would be side by side pressed up against each other evenly stacked, you could exchange these spheres for cubes with sides equal to the sphere's diameters and put the same number into the box. This means that for every sphere you add to the box, you also take up it's volume in empty space.

[Guest]

Four identical cubical boxes, with interior dimensions of a cm on a side, and

weighing b grams each, were packed with wooden spheres of density of d gram/cm³

Box 1 contained 1 single sphere; the largest one that would fit inside the box.

Box 2 contained 8 identical spheres; the largest that would fit inside the box.

Box 3 contained 27 identical spheres; again the largest that would fit.

Box 4 contained 64 identical spheres; again the largest possible.

Which loaded box [box plus contents] is heaviest, and what is its weight?

Agree with the

The 4th box is heaviest.

[bonanova]

If we consider that the best way to pack spheres into a cube would be side by side pressed up against each other evenly stacked, you could exchange these spheres for cubes with sides equal to the sphere's diameters and put the same number into the box. This means that for every sphere you add to the box, you also take up it's volume in empty space.

Excellent.

With that assumption, which certainly holds for the examples in the OP, no math is needed.

[HoustonHokie]

while it would not surprise me if the answer came out as all are equal, I wonder is it possible to have a closer packing than a simple 4x4x4 grid. If that is the case then we might be able to fit slightly larger spheres than diameter a/4 in the box.

Maybe, maybe not.

I think that mpapineau has the elegant answer for a cubical arrangement of spheres. Because one sphere in a cube requires a certain proportion of its volume (approx 50%), 8 spheres in 8 cubes (or 27 s in 27 c, etc) require the same proportion of the overall cube containing them. The only way to change this is to change the spacing, as armcie has suggested. To complete the proof, we need to show that a cubical arrangement is the best.

If, for the 64-sphere arrangement, your spheres have diameter > a/4, the best arrangement you can get for the bottom layer is 3 x 3. The next layer would also be 3 x 3, offset from the bottom layer so the centers of the spheres lined up above the centers of the gaps in the bottom row. So, in order to get the requisite 64 spheres, you'd have to have roundup(64/9) = 8 layers of spheres (7 layers of 9 + 1 layer of 1). For a cube with side a, you can make the following assertions about the spacing:

a = 3.5d + 2.5g_h

a = 4.5d + 3.5g_v

(2d)² = (d + g_h)² + (d + g_v)²

where:

d = sphere diameter

g_h = horizontal gap between spheres in the same layer

g_v = vertical gap between spheres in layers n and n + 2

The diameter of the sphere comes out to be .197a, which is smaller than a/4. Because the spheres are smaller then they could be with the cubical arrangement, they don't satisfy the requirements of the OP.

[Guest]

Excellent.

With that assumption, which certainly holds for the examples in the OP, no math is needed.

Not proven, that is too pat an answer.

The assumption is that a simple cubic packing is the best way to get 64 spheres into the box, but it hasn't been shown mathmatically (sic).

http://www.tiem.utk.edu/~gross/bioed/webmo...herepacking.htm

It is clear that for very small particles they will assume a hexagonal packing, and will occupy about 74% of the space. The post about porosity is interesting, but natural materials would naturally tend towards the most efficient packing - hexagonal.

So the question is, given that we are constrained by the number of spheres (they are relatively large compared to the box) does the space wasted by offsetting the cubes outweigh the increase in packing density?

HH had an interesting approach, but I think the calculation for the first layer is d = a/3.5 rather than a/4. Even if you show that you can't quite fit those spheres, you can always shrink them down and still be bigger than a/4.

For example, what if you did 6 rows of packing that way, and then jumbled the remaining 10 spheres in on top?

I will say that box 1 and 2 are the lightest, that box 3 likely weighs the same, but that box 4 likely weights more. Unless the density of the cubes is less than that of the surrounding meduim (which wasn't specified!).

[HoustonHokie]

Not proven, that is too pat an answer.

The assumption is that a simple cubic packing is the best way to get 64 spheres into the box, but it hasn't been shown mathmatically (sic).

http://www.tiem.utk.edu/~gross/bioed/webmo...herepacking.htm

It is clear that for very small particles they will assume a hexagonal packing, and will occupy about 74% of the space. The post about porosity is interesting, but natural materials would naturally tend towards the most efficient packing - hexagonal.

So the question is, given that we are constrained by the number of spheres (they are relatively large compared to the box) does the space wasted by offsetting the cubes outweigh the increase in packing density?

HH had an interesting approach, but I think the calculation for the first layer is d = a/3.5 rather than a/4. Even if you show that you can't quite fit those spheres, you can always shrink them down and still be bigger than a/4.

For example, what if you did 6 rows of packing that way, and then jumbled the remaining 10 spheres in on top?

I will say that box 1 and 2 are the lightest, that box 3 likely weighs the same, but that box 4 likely weights more. Unless the density of the cubes is less than that of the surrounding meduim (which wasn't specified!).

how tall does 6 layers of spheres get? I would say that the 6 layers are at least 7*radius tall. If d = a/3.5, then r = a/7, and at 6 layers, the spheres have already reached the top of the box. You won't be able to stuff any more in. In fact, if d = a/4, then after 6 layers, the high of the packed spheres would be 7/8 a, and you still have more than a full layer of 9 spheres to dump in the top of the box. Hexagonal packing is efficient in an infinite sense (or if you happen to have a hexagonal or spherical container). In fact, if the OP had specified 63 spheres instead of 64, the answer might be completely different, because then you could use an offset pattern that is 7 layers tall. But then again, it might not. The problem is the large outside gaps that bound two sides of each layer, which greatly reduce the efficiency of the packing.

[Izzy]

Okay, so here's what I'm thinking. 10x10x10cm box. Let's say wood weighs 1 kg per cm^3.

Box one: 4/3(pi)10^3 = 4186.66 kg + b

Since I couldn't figure out how to pack spheres into the cube if there spheres weren't in multiples of perfect squares, I'm gonna skip to box four. We'll be stacking 8 spheres by 8 spheres, making each sphere's diameter 1.25.

(4/3(pi)1.25^3)64 = 523.33.. kg + b

...That doesn't exactly sound right to be, but from what I get, the less spaces, the heavier, so box one is heaviest? *shrugs*

[Guest]

how tall does 6 layers of spheres get? I would say that the 6 layers are at least 7*radius tall. If d = a/3.5, then r = a/7, and at 6 layers, the spheres have already reached the top of the box. You won't be able to stuff any more in. In fact, if d = a/4, then after 6 layers, the high of the packed spheres would be 7/8 a, and you still have more than a full layer of 9 spheres to dump in the top of the box. Hexagonal packing is efficient in an infinite sense (or if you happen to have a hexagonal or spherical container). In fact, if the OP had specified 63 spheres instead of 64, the answer might be completely different, because then you could use an offset pattern that is 7 layers tall. But then again, it might not. The problem is the large outside gaps that bound two sides of each layer, which greatly reduce the efficiency of the packing.

post earlier, but the site was on the fritz.

You are right.

I agree that 6 layers will be too tall, I think it is actually a shade over 9 * radius [ (sqrt(50) + 2) * radius ]. So in this case the extra space added due to the offset is greater than the space removed by more efficient packing in the center, and the simple cubic packing is the best. However, I couldn't let the contention that cubic packing remains the most efficient with arbitrarily small spheres stand without challenge. There is almost certainly some ratio between sphere diameter and cube height where the hex packing becomes more efficient overall.

[Magic_luver101]

Box 1: d.((4/3pi)x(a^3))

Box 2: d.((4/3pi)x(1/2a^3))x8

Box 3: d.((4/3pi)x(1/3a^3))x27

Box 4: d.((4/3pi)x(1/4a^3))x64

24^3=13824

(12^3)x8=13824

(8^3)x27=13824

(6^3)x64=13824

all are the same

[Izzy]

Woah, I was in the shower when I realized my mistake. Thank you shower-tiles. Man the "eureka!" moment feels good.

So, I was think way too two-dimensionally before. Still a 10x10x10 cube, with 1 cm^3 of wood being 1 kg of weight.

Box one: 4/3(pi)10^3 = 4186.66 kg + b

Okay, so for box two we have four spheres, arranged like this:

o o

o o

with another four directly behind them. We can make them big enough to touch (or at least I think. I drew in the fog in the shower ), making each of their diameters 5 cm.

(4/3(pi)5^3)8 = 4186.66 kg + b

For the third box, do the same thing as the second, but with:

o o o

o o o

o o o (three times of course)

(4/3(pi)3.333..^3)27 = 4186.66 kg + b.

Do it again.

(4/3(pi)2.5^3)64 = 4186.66 kg + b.

Too bad that's been posted already. I agree with the others!

[bonanova]

There is almost certainly some ratio between sphere diameter and cube height where the hex packing becomes more efficient overall.

When I posted, I was convinced that 4 would not benefit from hex packing, but I hadn't verified that 5 would.

Else I would have included only two sizes: either 1 and 5, or 4 and 5.

Note that seven layers would have to fit:

25+16+25+16+25+16 = 123 compared with 5 cubic packed layers - 125.

[Guest]

Long time reader but I think this is my first post. I noticed that most people are not doing the math, some at the end though.

Most of you that did it mathematically are correct that all the boxes are the same, but the math is slightly off. "a" is the length of the sides of the cube, which would equal the diameter of the sphere in box 1, this makes r (which is the radius of the sphere) in the equation

4/3pi(r^3) equal to 1/2a.

Thus turning the equation into:

4/3pi(1/2a)^3 or 1/6pi(a)^3 (by simplifying the fractions).

The other boxes get slightly more confusing because we need to remember that "a" is split up more times and to multiply by the number of spheres,

so for box 2:

r=(1/4a)

because we need the radius of one sphere, and there are two spheres on each side and 8 spheres total.

The equation ends up being:

4/3pi(1/4a)^3(8) or 1/6pi(a)^3.

This goes on for 3 and 4 with r equaling 1/6a and 1/8a respectively, and multiplied by 27 and 64 respectively.

But by simplifying the fractions in each one we end up with:

1/6pi(a)^3(# spheres).

And this will work for any perfect cube.