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Dimensional Analysis formula

  1. Nov 21, 2014 #1
    Under the standard form of dimensional analysis, I know that we relate a dependent variable to a function of the independent variable(s). However, what if there is some additive variable needed in the equation? How does this method, which expresses all of the independent variables as a product of the variables (with the exponents being any real number) times a constant, account for the needed addition of operations such as subtraction and addition? How can we derive correct equations if there is a missing sum or difference needed in the formula?
     
  2. jcsd
  3. Nov 21, 2014 #2

    mfb

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    Staff: Mentor

    If you can construct a specific unit with multiple independent expressions, then dimensional analysis for this unit does not work.
    A trivial example is a setup where you have two velocities, or two masses or something similar. There is no way to figure out which velocity/mass/... to use just by dimensional analysis.
     
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