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5 tube balancing argument

Dima Klenchin klenchin at facstaff.wisc.edu
Thu Oct 10 10:35:23 EST 1996

In article <jpcd0-0910961641090001 at macr1-5.welc.cam.ac.uk>, jpcd0 at mole.bio.cam.ac.uk (John Dixon) wrote:
#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.

Of course it is balanced in your case and  is not in the other. Adding one
balanced situation on top of existing one does not dusturb anything.  
Forget about 12-position rotor and think about regular 2-way balances:
if you equilibrated it with something, adding equal weights to both sides 
will not disturb equilibrium. 

Strict solution: rotor is balanced if a sum of vectors connecting tubes 
and axis of rotation equals zero (this is for the case of tubes of equal 
weight; for unequal weights, vectors have to be weighted). 

Enjoy your beer. 

- Dima

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