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# 5 tube balancing argument

Joe Chou jchou at cgl.ucsf.edu
Wed Oct 9 17:31:41 EST 1996

```jpcd0 at mole.bio.cam.ac.uk (John Dixon) writes:

>Some one here was doing 5 minipreps (equal volume) and was loading a 12
>hole microcentrifuge, when he found that the balance tubes had been
>cleared away. He was about to fill another sixth tube to balance 3
>opposite 3, when I said that you can balance five anyway. This started a
>massive debate, which has now come down to three positions.

>I reckoned that if you put three tubes in, in a triangle ie in holes 1,5,9
>these are balanced. Then you can balance the two others opposite each
>other in holes 2 and 8.

>One guy maintains that this is not balanced at all, on the grounds that
>you can split the fuge into two halves where one half has two tubes and
>the other has three, therefore it must be unbalanced. I am not convinced.

>Then another guy said that he balances five by putting them in as near a
>pentagonal position as he can ie 1, 3, 5, 8, 10. I dont think this is OK
>because you cannot remove a balanced pair to leave a balanced three,
>although I doubt it does much harm.

You are correct.

Physically, tubes in a centrifuge are balanced if their center of mass
is on the axis of rotation. Of course, the forces on the ROTOR are NOT
balanced (e.g., even when you have just 2 tubes opposite one another,
they exert more force on the rotor at those 2 poles, but the TUBES are
balanced in that their center of mass coincides with the axis of rotation.)

Putting in 5 tubes as you described: 3 in an equilateral triangle and
the other 2 opposite one another anywhere else, gives you a center of
mass over the axis of rotation.

You can disprove the guy who says that you can split the fuge into two
unequal halves by pointing out that a fuge balanced with 3 tubes will
be split into 2 unequal halves in all but 3 radially symmetric ways --
all other ways will be asymmetric... but obviously, the fuge is still
balanced.

Another way to see if a set of tubes are balanced would be to draw
a vector for each tube in the balance, with each vector of equal length
but with a direction equal to the direction from the center of the rotor
to the tube. Vector add (head-to-tail) all of the vectors, and show that
the last vector ends up at the point where you began -> it is balanced.

You could probably figure out how to vary the lengths of the vector
based on different masses of the tubes (lets see, centripetal force
equals, umm, (mv^2)/r?, so length should just be proportional to mass),
with the direction figured out as before, to balance your rotor. Then
you could figure out how to put in a bunch of tubes with different mass
and balance it with just one or two appropriately filled and placed tubes.

I *really* have to get back to work... <sigh>

Joe
--
- Joe Chou  (jchou at cgl.ucsf.edu)
http://devbio-mac1.ucsf.edu/
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