Inspired by this question, I want to ask a question about centrifugation.

Suppose a protocol says : 10 min at 2500 rpm . Can we instead centifuge for 5 min at 5000 rpm or 20 min at 1250 rpm ?


No, it isn't that simple, because centrifugal force varies with the square of the rotor speed. Have a look at the Wikipedia page for clearing factor.

Those equations are usually used to determine what you need to do if you are going to use a different rotor, but they can also be used to solve your question.

Because you are looking at a situation where the same particles are being centrifuged, the sedimentation coefficient s is the same in both centrifuge runs. Also all of the factors relating to the rotor properties are the same (rmax etc.). So, if you have a set of conditions that work, t1 and RPM1, and you want to choose new conditions t2 and RPM2, you can use the equation:

RPM2 = SQRT(t1/t2) x RPM1

For the example you mention t1 = 10 min and RPM1 = 2500 rpm. If you change the time to 5 min:

RPM2 = SQRT(10/5) x 2500 = 3536 rpm

Because the centrifugal force varies with the square of the rotor speed you don't need to double the speed if you halve the time.

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