How does sphere diameter affect bulk density in a given volume? 5

[pskvorc]

I am an EE, not an ME, so please accept my apology, (and point me to where it should be), if this is not the correct place to ask this question. I am a retired engineer looking to solve a 'personal' problem.

I have looked extensively for a solution to the below defined problem, and have found none. Let me save "you" time: I am well aware of the "64 to 74%" rules of thumb regarding packing density of spheres. Neither of those 'solutions' address my problem.

My problem defined is:
1) I have a cylinder whose dimensions (and therefore, volume), cannot be modified.
2) I need to add as much weight to that cylinder as I practically can using spheres of pure lead. (Rho_Pb = 11.34g/cc) (Price renders tungsten and gold "impractical".)
3) It is obvious that the smaller the sphere used, the higher the BULK density (weight) will be. However, there are practical reasons for not using spheres smaller than a given size. (At a point, spheres become "dust".)
4) The small-sphere handling issue does not have a threshold. Meaning that regardless of the starting sphere diameter, as sphere diameter gets smaller, the handling problems increase.
5) I want to maximize the weight of the cylinder, but I don't want to have to deal with the handling problems associated with "small" spheres. In other words; there is a point of diminishing returns where the gain realized in increased weight is offset by handling issues.

I want to be able to calculate BULK density in a CYLINDER as a function of sphere DIAMETER, when sphere density is known (and constant). The cylinder diameter and height are >> than the largest sphere diameter.

My problem restated in "simple" language is:
How does sphere diameter affect bulk density in a cylinder?

I have been to half a dozen "math" sites, and got NO help. Every time I have to talk to a mathematician I am reminded of why I am an Engineer.

Thanks!
Paul

 

[bcd]

An approximate estimate can be calculated by determining the size of square box a sphere will fit in. Then divide the volume of the cylinder to approximate how many "boxes" will fit in. Each box is one sphere of a the related diameter Not exact, but will get you very close and enables you to determine an approximate weight based on sphere diameter.

 

[JohnRBaker]

Yes, but spheres tend to 'nest' so any approximation using the square box analogy will be off by a significant amount.

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[drawoh]

bcd,

Spheres pack hexagonally.

--
JHG

 

[rb1957]

as you've done your research I won't repeat what I've found (I suspect nothing new).

as the density does not depend on the diameter I suspect the counter-intuitive result that diameter doesn't affect this.

however, I'd've thought there was an impact with how many spheres you can place inside your cylinder's circular cross-section. It seems intuitive that smaller spheres leave less empty space around them, but they would also leave more spaces.

I suspect that using the smallest diameter you allow will give you the maximum weight. I suspect the problem is very difficult to solve exactly ... finding the maximum number of spheres that fill a cross-section is almost easy to do but not necessarily the maximum result; does the next layer of spheres sit on three or four or two (and the side of the cylinder) spheres ? does giving up a few spheres in the cross-section allow the next row to sit lower and so fill the space better ??

another day in paradise, or is paradise one day closer ?

 

[SnTMan]

Well, I'm kind of going out an a limb, but it seems the maximum density possible is about 74%.

Apparently independent of sphere radius, unless I'm missing something (highly possible).

It seems your task then is to, using the theoretical max density, determine if the weight of your fixed volume cylinder is adequate, and to identify the smallest sphere diameter acceptable.

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[rb1957]

I suspect that that estimate is high. It doesn't seem to account for the finite limits of the problem ... fitting into the cylinder.

I reckon if you used that to calculate a number of balls, then melted those balls and poured that into the cylinder, that there would be space left over ... ie you can fit more lead atoms into the cylinder than you can lead spheres.

another day in paradise, or is paradise one day closer ?

 

[IRstuff]

These might help.

https://arxiv.org/pdf/1203.3373.pdf

http://www.scielo.org.mx/pdf/rmfe/v61n1/v61n1a5.pdf

http://demonstrations.wolfram.com/PackingSpheresIntoAThinCylinder/

The 74% case is very ideal, and requires spheres that have no friction, so you can shake and roll them about. The first paper basically puts the packing density on the order of 50% for large spheres, relative to the cylinder diameter.



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[ornerynorsk]

As lead is very malleable, squash said spheres into a more optimal shape that will pack and nest more efficiently, and/or pour iodine, one of the heaviest liquids, into the cylinder until all voids are filled.

It is better to have enough ideas for some of them to be wrong, than to be always right by having no ideas at all.

 

[rb1957]

the wolfram link looks good. what does "compression ratio" mean ? that the balls fill 0.17 of the volume of the cylinder ?? (doesn't look right) or that the balls fill 0.83 of the cylinder (and leave 0.17 of the volume empty) ?

another day in paradise, or is paradise one day closer ?

 

[zeusfaber]

Since you've already specified that the cylinder is much larger than the individual spheres, you should have got yourself away from the region where the bulk density goes up and down significantly depending on how close the principle dimensions of the cylinder are to an integer number of row depths.

In practice, the quality of your packing is likely to have a greater impact on the final density than your choice of sphere size. If you're not already familiar with it, suggest you Google "Snowstorm Filling".

A.

 

[SnTMan]

Yes, the OP did say "The cylinder diameter and height are >> than the largest sphere diameter."

As it appears there is no exact solution, I'm sticking with 74%.

The OP will doubtless be kind enough to let us know how it comes out

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[handleman]

As has been mentioned, the perfect sphere packing density for uniform sphere size is simply a function of geometry ratio and is not dependent on sphere size. However, you can only achieve that density either (a)with infinite space or (b)allowing partial spheres within the volume. If you could magically have infinite perfectly packed spheres and then your cylinder instantly appeared and enclosed those partial spheres, it wouldn't matter what size the spheres were or the cylinder. Well, as long as the cylinder wasn't completely contained by one sphere or something.
So, completely worst case would be a ball size such that only one ball fits in the cylinder, quite loosely, but such that one more complete ball cannot fit. Best case is infinitesimal spheres such that all of the spheres are inside the cylinder. At that point you'd have the best packing, assuming perfect arrangement. Since this requires infinitely small spheres but approaches a max volume, you can assume that the approach is asmymptotic, such that as your sphere/cylinder diameter ratio gets <10 or so, additional gains are pretty dang minimal.

So, we know that the smaller the better, and there's an upper limit. To some degree the question becomes, "What size spheres do you want to mess with", and "how important is getting maximum weight". My intuition says that if your cylinder is a Coke can and your spheres are the size of BBs, you are not going to gain more that a couple percent by decreasing to dust size.

Now... All this goes out the window if you allow for variations in diameter. Take, for example, marble-size spheres. Fill your can with those, then add the dust-size spheres. You'll get better fill. It's even better fill that you would get with 100% dust-size spheres. That doesn't seem to make sense. We just said that the bulk density is independent of size, but it's obvious that you can fill the spaces between big balls with little ones? So, just replace the big balls with little ones and you have higher density, right? (but you don't.)

The difference comes because the packing efficiency of the center of the larger spheres. It's 100%. If you replace a 100% dense solid sphere with the equivalent volume of tiny spheres, the density of that section drops to the packed density.

 

[3DDave]

What is the precise problem? There will be no general solution because the form of packing will have significantly different results for non-zero size spheres. This seems like good puzzle to occupy you in your retirement.

The 74% is not a rule of thumb. It's a limit for perfect spheres in an arbitrary volume as the diameter of the spheres approach zero.

 

[TheTick]

There’s certainly a theoretical maximum (74% was bandied about). There is also a point of diminishing returns.

Beyond that, it’s up to you to decide what’s worth the trouble. Probably when you see no significant increases with smaller sizes.

 

[JStephen]

In real life, the cheapest spheres are probably going to be lead shot, which comes in a finite number of sizes, and isn't that expensive. So go buy a few bags and experiment.

You specify one size, but obviously, mixing several sizes would increase the density. I assume there is SOME variation in the lead shot size in a given nominal size as well. And you'd get a certain amount of flattening at the interfaces.

 

[pskvorc]

Thank you all for your responses.

SnTMan - I have a fundamental problem with the "74% solution". I'll explain my 'fundamental problem' this way:

1) Assuming a cylinder of diameter of 1 (d_c=1) and length 10 (L_c=10), and spheres of diameter 1 (d_s=1), I will be able to fit 10 spheres in the cylinder.
The total volume of the spheres is 10 * 4/3 * pi * (d_s/2)^3 or 5.236. The volume of the cylinder is 10 * pi * (d_s/2)^2 or 7.85. The ratio of the volume of the spheres to the volume of the cylinder is 5.236:7.85 or 66.6%. This is within the range provided by the "64 to 74%" rule of thumb. Continue to reduce the size of the spheres, and you approach the 74% figure of "ideal packing". However...

2) Assume the maximum "packing" of a liquid and obviously the volume of the liquid equals the volume of the cylinder. Now, stepping back from a liquid ever so slightly, to REALLY SMALL spheres, and I am unconvinced (via theoretical mathematical gymnastics) that the volume occupied by the spheres is ONLY 74% of the volume of the cylinder. Assuming that the d_s is "microscopic", the volume occupied by the spheres is going to approach equality with the volume of the cylinder.

Moving back even a bit more, let me use the real values I am faced with.

d_c=1.746 cm.
L_c=21.272 cm.
I have lead spheres of the following diameters: (all dimensions in cm)
1.746, 1.575, 1.092, 0.952, 0.782, 0.305, 0.279, 0.241, 0.229, 0.216, 0.203, and 0.127.

As you can see, there is more than an order of magnitude between the smallest (0.127 cm) and the largest (1.746cm).

The ratio of d_c to d_s0.127 is 1.746:0.127 = 13.748 or more than an order of magnitude.

Am I to "believe" that spheres with diameters equal to the diameter of the cylinder are only going to have a "bulk density" that is 10 PERCENT less (64% vs 74%) that spheres that have diameters that are almost fourteen times smaller than the diameter of the cylinder? That seems counter-intuitive. I was hoping for a formula that related bulk density to the ratio of the volume of a cylinder and the diameter of the spheres filling it. Maybe there is no such thing.

In my search for an answer to this, I have seen no reference to the 64-74% ratio being constrained to sphere diameter. The only 'constraint' I have seen on that rule of thumb was the "density" of the "packing". In other words, if the spheres were allow to fill the cylinder "randomly", the ratio was ~64%. If the spheres were "densely packed" (ordered placement, i.e. lattice, not random), then the ratio (bulk density) was 74%. That simply doesn't make sense to me taken to the "limits" (Cylinder diameter and sphere diameter equal at the "large" limit, and sphere diameter "microscopic" at the "small" limit.)

I hope "y'all" realize that I am not trying to be argumentative, rather I am simply looking for the 'right' solution. I could of course use trial and error, but that's not very satisfying in an "engineering" sense, and doesn't help me with the next time I am faced with a similar situation.

Thanks,
Paul

PS - Since the 64-74% rule is 'gaining support', I will "conduct the experiment", and report results here.

Paul

 

[SnTMan]

pskvorc, we calculate what we can, we test the rest

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[3DDave]

Try playing with circles within a square; the geometry is similar and simpler.

If the diameter = square side length, the ratio of area of circle to area of square is pi/4. If you have 4 circles, they each occupy a quarter and have the same ratio, pi/4. This pattern repeats with any decrease in size for square packing. The ratio is the same.

You can get a higher density for very small circles with hexagonal packing, but that pattern also repeats.

To get the highest density you can use nesting circles to fill in the empty spaces, but they have to be in the right places.

 

[GregLocock]

There is no analytical solution that I know of for packing relatively large balls into cylinder, except that for your largest size of ball you actually have the bounding case. You can fit 12 of those in.

The 74% number is an approximation to pi/(3*sqrt(2)) and was proved by Gauss. Don't argue with Gauss.

However, you can simulate the effect of filling a container with billiard balls using a particle physics modeller.

https://www.youtube.com/watch?v=0e3p5iu3zj8

While they are hilarious fun your time would be better spent with a few bags of balls and a set of weighing scales and your cylinder.



Cheers

Greg Locock


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