h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

I have a box 1m wide, 1m long and of variable(?) height.
I also have 1,000, bricks, of 100mm cube.
Of course, if stacked neatly, those bricks would need the box to be 1m high to fill it, level.

Now, if I was to:
1. Tumble the bricks in, randomly
or
2. Place them carefully to take up maximum space

Then,
. Assuming 1 & 2 are not the same (are they?), how high would the box need to be to be filled (to average) level in each case?
. If they're to be placed carefully, what is/are the attitude/s of the bricks?

Again,
. Using 8,000 bricks of 50mm - in the same box
. How high?

And again,
. same box, 64,000 bricks of 25mm


And so on.

The relationship of box heights varies like ... how?

(but not too much complicated maths, eh?)

Whoops,

Please
--- and ---
thank You

rod!!!

enough with the excuses!!!

just put the lego away and tidy your bedroom!!!

Blood yell, Taff - how did you know?

It is my raised beds room, digging out for one then tumbling the clayish sods back in, chopping 'em down, digging out for the next, and, having failed to become one with my shovel, I got intrigued

I know this problem is a lot more complicated than it looks.

Yes, Gnomon. It took me a while (not really surprising, ) to realise that, soon after I determined that maximum space for 1 brick is (presumably?) standing upright on one corner, and a row of 'em (if you could) would collapse as soon as you let go (& tapped the box), 'cos they wouldn't fit across 1m.

Does nobody want to come out to play?


Maximum space - on their corners (with a dab of superglue?) the 100mm cubes are equivalent to cubes of 173mm (ish) edge so 1,000 will take up 1.73 cubic m

That's the same for all - big, medium, small, ie 1.73 cubic m each time.

But none fill length & width of the box, so would tumble to some extent. The smaller the brick, presumably the smaller the tumbled volume.

I've reached my limit at this point - to what extent would they tumble?


My soft sods have bigger voids and, when chopped down*, seem to take up about half the original volume - but there's some compaction happening, too.
* like that old gardening book said - rake until the grains are as ears of wheat.

I think the main problem, Rod, is your "not too much complicated maths" bit. I don't know enough maths to answer the question, but I do know enough to know that it is probably pretty complicated. Might be worth asking toybox for his opinion.

Thanks hd. Yeah, p'raps so. However, I don't want to tout for business from those not volunteering

OK everybody - maths as complicated as you like ( & I'll advertise for a translator).

I have a near enough idea of what happens to my sods (in that sense my Q is triviaI) - just thought it might tickle a few people's curiosity as it did mine.

I'd guess that the maximum space would require just less than 4 metres high.

Arrange your cubes so that they are aligned with the sides of box, and with a spacing of just less than one cube width between them. You'll be able to cover the bottom layer with 25 cubes rather than the 100 that you would need to pack it solid. Now arrange the next layer so that each cube sits on the four corners of a cube on the lower layer. Once again you'll be able to cover the space in 25 cubes. Keep going with 40 layers for a height of 4 metres. I can't imagine how you could arrange the cubes to take any more space than that.

So the random jumble would be somewhere between 1 and 4 metres high. Again at a guess, I'd say it would be about 2.7m.

Hi Gnomon.

Yes, agreed on your maximum - nicely reasoned (I was tempted to say "good job I thought of that", but won't).

Your random jumble is well within my intuitive feel (oh how I wish I'd noted estimates). How come you've guessed at \about 2.7m/?
Ah, pi-ish or 2/3-ish (but not 8/13?)


So my observation of smaller volume after another pass of sodding chopping will be from smaller bits falling into larger voids (as well as compaction).
I'll buy that ... for the moment

Thanks

Rod, meagremath

I'm not sure I understand the problem. I reckon you can find a way to take up very large amounts of space by essentially using those bricks to build walls and having large empty spaces inside. You might need to make some support columns or something, but my guess would be that as the size of the bricks gets smaller and smaller (keeping the total volume the same), the maximum height of the structure you could build with it would get larger and larger without limit.

Dogster, I asked the question initially because...

Dig in clayish soil, set aside the sods in a pile and later, chop that pile of sods down, with your spade, to smaller pieces. Then to smaller pieces, etc

Why does the pile get smaller after each pass you make with the spade?

At first it seemed simple then after a little think, not so simple.


Hence the question.

I see - unfortunately I think I'm gonna also have to say that it looks like a hard problem. These packing problems tend to be very tricky, only some of them have only recently been solved using advanced modern mathematical techniques.

Aye, except that at present (imho), the packing aspect doesn't seem to answer the Q, as packing would appear to apply to all the sizes (if regular) similarly.

At the mo I'm leaning back towards the simple - voids being progressively filled, along with compaction (and settling only coming in later)

N.B. I'm weak in maths anyway.

Hi Rod

Maths and me are almost strangers after an O Level 33 years ago but I am with you on the simple approach.

I would add though that the clay sod have additional properties that need to be accounted for - they will be wet and sticky and not rigid (Oh er this is getting bad now!).

When you put a wet clay sod on a pile it and all the other will stick together to some extent and deform as well. As you chop them with a spade they will get smaller and pack more closely but they will also break up in a random way as well (at least that is my experience) with small bit filling small voids. There will also be some contraction due to drying (assuming it isn't hissing it down where you are).

I'm certain the maths is fiendish

t.

Hi U2~/turvy

I reckon you've about got it for the sods - and the simple approach is always most suitable for me, too!

The more I think about the bricks, though, the less complicated it seems - on the surface anyway. The question as stated was, really, asking how different the packing would be for different sized bricks and it's looking simpler now.

In each \batch/ they're all the same. Between batches their proportions are the same. The box is the same in each case. Tidy packing fills the box level to that same height.

Tumbling them in gets less obvious because I don't know whether or not to expect the bricks to find the same relative attitudes between bricks/batches (because of their size relative to the box).

One other aspect is: when tumbled in, chances are they'll not end up level - there'll be high parts & low parts so a sort of average level might be appropriate.

So, if the tumbled levels actually are different, how do they vary? In a regular (predictable) manner?
That's the complex bit, methinks.

Or not - and that, said he, is the problem.

I'm certainly no expert, but surely the way to approach this is to work out the maximum space an individual brick can take up, and then multiply?

A single brick will take up a space equal to its own diagonal, squared (imagine its standing balanced on one corner, and measure the distance from the corner its standing on, to the corner vertically above). Each subsequent brick, to take up maximum space will have to stand on its own corner, directly on top of the top corner of the one below. This will leave voids between the bricks (which if my visualisations are right) the same size as a brick-so the maximum volume is actually double the volume of the bricks?

Of course ^that doesn't explain the relationship between the differing sizes of bricks, nor does it take into account random positionings of the bricks by tumbling them, but the maximum possible 'fill size' would seem to be the same for each size brick if the close packed size is equal.

Ah, Baggins, have you missed out the prequel posts...? (scroll up to post 1)


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