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How to find terminal velocity from experimental data?

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data

    So I have a set of data, velocity and time, of a falling object. So the question is: how can I get the value for the terminal speed?

    2. Relevant equations
    Since we're talking about terminal velocity an ecuation that descrives a Velocity vs time graph should be of the form:
    [itex]V =Vt\left(1-e^\left(\frac{-bt}{m}\right)\right)[/itex]

    where [itex]Vt[/itex] is ther terminal velocity
    [itex]m[/itex] is the mass
    [itex]b[/itex] is a constant which depends of the caracteristics of the body
    and [itex]t[/itex] is the time


    3. The attempt at a solution
    So I need to fit my data with an ecuation of this style,my first thought was that I had to do a logistic regression but that didn't look right (the ecuation is similiar but is not the same),
    Then I thought doing a little of algebra
    [itex]Vt-V=Vt\left(e^\left(\frac{-bt}{m}\right)\right)[/itex]
    And then All I needed to do was an exponential regression but again that doesn't look like the best way because I'm already giving [itex]Vt[/itex] a value
     
  2. jcsd
  3. May 27, 2012 #2

    ehild

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    If you have got measurement data at time t1 and 2t1 you can determine Vt by solving the system of equations
    Vt-V1=Vte-Bt1
    Vt-V2=Vte-2Bt1.
    (B=b/m)

    Square the first equation and divide by the second. The exponentials cancels, and you get an equation for Vt in terms of V1 and V2, easy to solve.

    Find as many pairs as you can, and take the average of Vt-s.

    ehild
     
  4. May 28, 2012 #3
    How many data points do you have?
    Do you have enough to plot a graph of v against t?
     
  5. May 28, 2012 #4
    So you're saying that I should do the following?
    (Vt-V1)^2=Vt^2*e^-2bt
    and then divided by
    (VT-V2)=Vt*e^-2bt
    and then I should get:
    (Vt-V1)^2/(Vt-V2)=Vt
    but that ecuation is not so easy to solve unless I've misunderstood the procedure
     
  6. May 28, 2012 #5
    yes, I have 56 data points, the problem lies in the graph because the points don't seem to have a tendency to a certain value, I thought about taking the maximum as VT but I don't like that solution.
    Any suggestion?
     
  7. May 28, 2012 #6

    ehild

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    That equation is equivalent to (Vt-V1)^2=Vt(Vt-V2).
    Expand both sides of the equation. : Vt^2-2VtV1+V1^2=Vt^2-VtV2.

    Vt^2 cancels. -2VtV1+V1^2=-VtV2. Isolate Vt.
    Vt=V1^2/(2V1-V2)


    ehild
     
  8. May 28, 2012 #7
    thank you!
    seems obvious now, that would do it
     
  9. May 29, 2012 #8

    ehild

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    I guess you have got some data-processing program, Excel perhaps? As you have enough data points, it is possible to get the approximate derivative of the V(t) function. V' = VtBe-Bt. The plot ln(V') vs time has to be straight line, with slope -VtB and crossing the vertical axis at VtB.

    ehild
     
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