1. Basic Maths Question

Sure someone like Manan/Pickup will be able to have a decent stab.

If i have a hundred small balls and i want to join them together to make a large ball, how many balls would you put on each layer of the ball to make it actually ball shaped?

All i'm getting with my efforts are pyramids, which aren't strictly balls.

2. hmm, interesting Q

a ball is a sphere, therefore it could hv a ball in the middle and that ball is surrounded by other balls .... the middle layer would hv maximum balls, say n number of balls and then the next layers on its two sides would be n-1, something like the equation below:

...., n-2, n-1, n, n-1, n-2, ....

so when you take, n = 10 ... the equation gets solved

3. My 13 years of maths taught me one thing

The word "Pythagoras"

4. All you need is TWO balls, and you will find a way.

5. mitchell knows how to plan an orgy...

****in rofl

6. Dude,that's anything but a basic maths question... And I am not quite sure I understood the problem correctly,and if I think what I think it means...then I am as useless to you as the problem itself.

7. If you need to just make a sphere you if you can attach each ball with three others to form triangles you eventually make a hollow sphere-like shape.

8. Originally Posted by Josh
My 13 years of maths taught me one thing

The word "Pythagoras"
Haha the countless number of times you open a text book and find someones drew an extra body part on his picture.

9. In the surds chapter of my current book, some bogan's changed all the s's to t's.

10. Originally Posted by Josh
My 13 years of maths taught me one thing

The word "Pythagoras"
Ancient Babylonians ftw.

11. Look, I haven't done maths since 1987 when I left school, though now I'm doing times tables with my 6 yo twins.
But seriously, if you're not into your maths as part of your job or you have a pretty big interest in it, how is the question posed in this thread a "basic" one? Seems damn hard to me, frankly.

12. Originally Posted by Burgey
Look, I haven't done maths since 1987 when I left school, though now I'm doing times tables with my 6 yo twins.
But seriously, if you're not into your maths as part of your job or you have a pretty big interest in it, how is the question posed in this thread a "basic" one? Seems damn hard to me, frankly.
It's not basic if you bloody teach it... why else do you think I haven't answered?

Going to sit down and have a (google-aided) crack now, and see what I get.

13. the diameter of a circle is the longest chord in a circle, so this would be like making two opposite pyramids from a single base, so let's say the diameter has 10 balls there the next chords on either side of the diameter would have 9 balls and so on like shown below

1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1 = 100 balls

from the eq that i gave in my earlier post

...., n-2, n-1, n, n-1, n-2, ....; taking n =10

14. Having researched it, no it isn't basic. It's an application of the Kepler Conjecture.

Basically, good layers to use are:
(hexagonally): 1 [blue], 7 [green], 19 [beige], 37 [pink]
(broadly triangular): 3 [olive], 12 [red], 27 [turquoise]

This is two sequences: one with the formula 3n(n+1)+1, the other more simply 3n².

With these formulae you can extend the sequences thus:
1, 7, 19, 37, 61, 91, 127
3, 12, 27, 48, 75, 108, 147

This provides a fairly exhaustive list of numbers of balls in the layers to construct your sphere, maintaining of course a symmetrical layout.

Therefore, your smallest sphere would have one layer of one sphere.
The second would be a 1 - 3 - 1 = making five spheres.
The next would contain 1 - 3 - 7 - 3 - 1 = making fifteen.

This pattern continues as follows: 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379... and loads of people have found this sequence before, in loads of other things, and it turns out be ½n(n²+1), also known as Sequence A006003.

I actually enjoyed finding all that out. Is that wrong?

15. Originally Posted by ret
the diameter of a circle is the longest chord in a circle, so this would be like making two opposite pyramids from a single base, so let's say the diameter has 10 balls there the next chords on either side of the diameter would have 9 balls and so on like shown below

1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1 = 100 balls

from the eq that i gave in my earlier post

...., n-2, n-1, n, n-1, n-2, ....; taking n =10
No, it wouldn't. That would give you a slice through a sphere.

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