Well, here's goes GCSE maths
I also noted that each sphere would likely interlock and not space itself out evenly.
Drawing two circles, orientated thus : in Paint, then placing another circle as close in between them as possible, it appeared, to allow about 1 12th of it's width, between the other two.
Got some 2p pieces to check that and with two next to each other, you can get about 3mm of a third 2p, into the gap. a 2p is about 26mm across, so that is 11.539% of its width.
Calculating the actual numbers, obviously, your first row would contain 400 bearings.
Although I am no maths wizard and I know the question relates to what I now know as "Sphere packing", it does specify ball bearings.
Ball bearings are made of steel which has a density (just Googled it) of 7.85g cm^3.
So the mass of each bearing
By aligning the second row into the gaps of the first, it would occupy less width, but would contain one less bearing.
So if my guess at 11.539% reduction in width is correct, each two rows in the bottom layer, would contain 799 bearings, but would only occupy 4.42305mm.
We know 400 rows at least will fit, so 200 x 4.42305 = 884.61mm
This then leaves a remainder of 115.39mm, which if we divide by 4.42305 we get 26 and a bit.
As those 26 are two rows with 799 bearings in each, plus the 200 pairs of rows we already know about, 226 pairs of rows containing 799 bearings, will occupy the first layer.
First layer contains 180,574 bearings.
So now we need to know what the second layer will hold.
This started to get complicated, thinking in 3 dimensions. Mrs Skeet, brought home some spare wedding cupcakes. I ate one. I noticed that I had had to remove it from a small square box.
I found some old paint, plopped them in and noticed that:
Each bearing will sit on top of the next below it, so each layer, should be identical in cross section, assuming you stack the balls. I had occurred to me, that they would naturally want to fit, "pile o cannonballs" style, but that does not mean that they have to, and would also mean less balls in each second layer, by quite some way.
So, hopefully if we stack them as such, then we can get 400 vertical rows and still be within the confines of the box.
So that would be 400 x 180,574 which is
72,229,600 bearings.
Gonna have to Google the surface area of a sphere.
4piR2 apparently, so that would be 19.625mm (pi @ 3.14)
So total surface area of the bearings would be
72,229,600 x 19.625 =
1,417,505,900 mm squared. (1,417.505900 square meters)
Am I close? If I am, can I have a free Platinum membership please Robbo?
!
SkeetEDIT: Right then. Having slept on this, I have decided to further stretch my GCSE maths. Naturally, I do not expect to compete with those who know about "Close Packing Theory" and what not, but it can be fun to do what one can, with what one has.
Sooo...
If each layer stacked up as you would expect and not in columns, then (thinking in complete layers, and as I type), then each layer will move one half row in two directions, which would mean it would lose one row of bearings from two sides. I cant tell if that will be a long or short row, so I will have to accept a loss of one or 2 bearings per row from my (probably waste of time) calculations.
So the second layer, might contain 399 rows or 199.5 pairs of rows containing 796 bearings.
Vertical reduction of the layer should be the same as the horizontal reduction on the bottom so;
The second layer would contain 158802 bearings.
So the first two layers, would contain 158802 + 180574 bearings = 339376 bearings per two layers at a height of 4.42305mm
Again, we can fit 226 pairs of layers so, drum roll please:
Using GCSE maths, I estimate that the box could contain 339376 x 226 bearings
=
76,698,976 bearings
with a surface area of 76,698,976 x 19.625mm^2
=
1,505,217,404mm^2 (1505.217404m^2)
I then wondered...after all this hard work, would I have room for a beer or two?
Volume of each bearing is (pi @ 3.14) 8.177083 mm^3
So the volume of the box they would effectively fill is
627173.892767008 cm^3
Leaving spare volume in the box of
372826.107232992 cm^3 ( 372.826 L ish)
or about 35 crates of beer.
Which doesn't seem possible
Although I am no maths wizard, I now know that the question relates to "Sphere Packing" which I know nothing of, prior to thsi and as such, did not use it in my calculations. However, the question does state, Ball Bearings.
These are made of steel and as such, are heavy. Google tells me that steel has a "standardised"
density of 7.85g cm^3 which for my calculations, equates
6.2712 Metric tons in the box.
So, who has considered the compressive effects of this mass on the layers, and
how much room would this compression yield?
Bon? How are my calculations?
