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Maths Quiz.

gaunt

Member
May 14, 2009
7
0
11
Oops, can't read.

Nealy 117 million? Thats a tad high :p And I doubt 9 million meters of surface area aswell mate.
I don't think so, as you fit 400 ball bearing in just a one meter row, and that means that to put the first layer in the box is 160,000. Then the next layer doesn't sit directly on top for the first layer it sits in the gaps between the lower ball bearings. Which means that you can fit 399 in a row, therefore can fit 159,201.

Now this is where I guessed a little, that the combined hight of the two layers were 1.5 diameter. which means that you can fit 266 dual layers into the box, plus another 160,000 layer with the remaining space.

But yes I did make a mistake with the total surface area, as I used the diameter and not the radius. The total surface area when using the radius is 2,293,911 meters.

Btw I know it sounds odd, however I am using crystal structures and solid state physics.
 

xDansomee

Speedball in the woods!
May 30, 2009
979
35
53
30
Manea, Cambs.
www.iBotModz.net
I don't think so, as you fit 400 ball bearing in just a one meter row, and that means that to put the first layer in the box is 160,000. Then the next layer doesn't sit directly on top for the first layer it sits in the gaps between the lower ball bearings. Which means that you can fit 399 in a row, therefore can fit 159,201.

Now this is where I guessed a little, that the combined hight of the two layers were 1.5 diameter. which means that you can fit 266 dual layers into the box, plus another 160,000 layer with the remaining space.

But yes I did make a mistake with the total surface area, as I used the diameter and not the radius. The total surface area when using the radius is 2,293,911 meters.

Btw I know it sounds odd, however I am using crystal structures and solid state physics.
Well, I have just realised my numbers are totally wrong, again. Even though I did work out 160,000 on the bottom row before hand... It's all written down :p

Surely in the gaps you can only fit 398 in a row?

So the layers would be; 400 x 400 > 398 x 398 > 400 x 400 > 398 x 398 etc, all the way to the top.

If four ball bearings were put into a cube, then one ball bearing would fit into the middle of them, so a 400 x 400 > 398 x 398 > 400 x 400 would only be 5mm tall right? And thats 478404 in the first 5mm in hight?
 

gaunt

Member
May 14, 2009
7
0
11
Well, I have just realised my numbers are totally wrong, again. Even though I did work out 160,000 on the bottom row before hand... It's all written down :p

Surely in the gaps you can only fit 398 in a row?

So the layers would be; 400 x 400 > 398 x 398 > 400 x 400 > 398 x 398 etc, all the way to the top.

If four ball bearings were put into a cube, then one ball bearing would fit into the middle of them, so a 400 x 400 > 398 x 398 > 400 x 400 would only be 5mm tall right? And thats 478404 in the first 5mm in hight?
Almost, however using the hexagonal close packing structure means that it is possible to fit 399 in the row of the second layer. Due the second layer ball bearing sitting in the gaps the bottoms are lower so the over all hight is less. This is where I guessed that the hight of the two layer is 3.5mm but that was just a guess.
 

xDansomee

Speedball in the woods!
May 30, 2009
979
35
53
30
Manea, Cambs.
www.iBotModz.net
Almost, however using the hexagonal close packing structure means that it is possible to fit 399 in the row of the second layer. Due the second layer ball bearing sitting in the gaps the bottoms are lower so the over all hight is less. This is where I guessed that the hight of the two layer is 3.5mm but that was just a guess.
Ahh I see what you mean,

Although, I wasn't really taking that into account, and 'assumed' that the combined height of layer one and two would be 2.5mm + 1.25mm, therefore 3.75mm, not 3.5. Therefore, when the third layer was stacked onto the top, it would add an extra 1.25mm making the overall height of that pattern 5mm.
 

gaunt

Member
May 14, 2009
7
0
11
Ahh I see what you mean,

Although, I wasn't really taking that into account, and 'assumed' that the combined height of layer one and two would be 2.5mm + 1.25mm, therefore 3.75mm, not 3.5. Therefore, when the third layer was stacked onto the top, it would add an extra 1.25mm making the overall height of that pattern 5mm.
Yes your right with the hight. I did use the hight with my working, i think.

Well I hope I was on the right lines, as from the two science questions that I saw. This was the one that I could do in a sleepy state.

I just want to know if I was right or not now.
 

xDansomee

Speedball in the woods!
May 30, 2009
979
35
53
30
Manea, Cambs.
www.iBotModz.net
Yes your right with the hight. I did use the hight with my working, i think.

Well I hope I was on the right lines, as from the two science questions that I saw. This was the one that I could do in a sleepy state.

I just want to know if I was right or not now.
Judging by my total fail in previous calculations I think you are right.

If not, you are pretty close to the answer, as of mis calculations in numbers per layer, and height etc. :p