1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Buckingham-Pi for "algorithmic" non-dimensionalization

  1. Feb 19, 2015 #1
    I would like to use the Buckingham-Pi theorem in order to "algorithmify" non-dimensionalization of existing equations. I can get things to work for very simple problems, but am running into issues with a harder example. I posted my question on physics.stackexchange.com the day before yesterday, but it has received little attention.

    Could someone here comment, either here or there?

    Alternatively, do you know of existing ways to "algorithmify" non-dimensionalization?
  2. jcsd
  3. Feb 19, 2015 #2
    The way I learned to do this kind of thing is not by use of the Buckingham Pi Theorem. The method I learned always works great. For each of the independent variables (in this case only the time t), define a dimensionless parameter such as τ = t / t0, where t0 is a characteristic time to be defined in terms of the other parameters in the problem. If there are spatial variables, such as x, define X = x/x0.

    In your problem, for example, substitute t = t0τ into your differential equation, and reduce all the terms to dimensionless form (by multiplying or dividing the entire equation by whatever parameters necessary to do this). You will end up with groups involving t0. Select one of the groups involving t0, and choose t0 such that that group is equal to unity. Substitute this expression for t0 into all the other groups involving t0. This will automatically generate the other dimensionless groups involved.

    The choice of t0 or x0 can also come out of the boundary conditions of the problem, if that choice makes better sense.

    In your problem, the dependent variable is already dimensionless. However, in most problems that is not the case. If the dependent variable is not dimensionless, you can do the same trick with that variable.

  4. Feb 20, 2015 #3
    This gives:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook