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 Centrifugal Balance II (Posted on 2005-10-18)
A circular centrifuge has 30 slots spaced evenly around its circumference. Thirty samples need to be spun in the centrifuge, their masses being 1g, 2g, 3g, . . . 29g, 30g. How can all the samples be placed in the centrifuge at once while keeping it balanced properly?

For what other values of N is it possible to balance an N slot centrifuge with samples weighing 1g, 2g, 3g, . . . (N-1)g, Ng?

 No Solution Yet Submitted by Brian Smith Rating: 4.2857 (7 votes)

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 re(7): Sums of irrationals | Comment 22 of 30 |
(In reply to re(6): Sums of irrationals by goFish)

Yes, of course you can add and subtract the same irrational number to get a rational number.  But that's not what we're talking about here.

If your axis through that circle is at an odd angle, then your moments won't match like that.  One weight will contribute moment of sin(x) times the weight, another weight will contribute moment of sin(x+60) times a different weight, and so on.

The problem is really much easier to see if you just try to crunch the actual numbers.  Take a six-point wheel with weights A, B, C, D, E and F, arranged in that order around the wheel clockwise.  Draw an axis that runs between A and F, and C and D.  But it doesn't split them evenly.  The axis is 5 degrees from A and D, and 31 degrees from C and F.

The moment on the ABC side of the circle is:
A*sin(5) + B*sin(65) + C*sin(125)
The moment on the opposite side is: D*sin(5) + E*sin(65) + F*sin(125).

The only way to make the two sides balance is if A=D, B=E, and C=F.  No other set of integers will do it.  Probably no other set of rational numbers.  To see why, try to pick six different rational numbers A-F that will make this equation balance:

A(0.87155742)+B(0.906307787)+C(0.819152044)= D(0.87155742)+E(0.906307787)+F(0.819152044)

And if somehow you can make that equation balance, then let's tilt the axis a degree and do it all over again.

Or if you prefer your irrationals as square roots, then pick some integers A and B to make this equation balance:
A*SQRT(2)=B*SQRT(3)

The lesson is that you can't make different irrational numbers equal each other using integer coefficients, and probably not even ratinal coefficients.

The circle cannot balance along every axis.  It can't be done.  The only axes where the circle can balance are straight through the weights, or through the midpoints between the weights.

 Posted by Leonidas on 2005-11-01 17:43:11

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