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I Take v(t) and h(t) to make v(h)

  1. Feb 10, 2017 #1
    So I have some very complicated functions that do not have trivial inverses, but I'd really like to make a new function from them. What I have is:
    v(t): velocity as a function of time
    h(t): height as a function of time (never mind that is says hb)
    upload_2017-2-10_10-52-29.png

    What I want:
    v(h): velocity as a function of height

    The height function goes as an exponential function, so there are no local minima to worry about:
    upload_2017-2-10_10-51-4.png

    Other than fitting a polynomial to the h(t) function (which I'm not sure would be invertible anyway), does anyone know of a good way to combine these functions into v(h)?
     
  2. jcsd
  3. Feb 10, 2017 #2

    FactChecker

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    I don't know if there is a closed-form analytical solution, but since you are sure that h(t) is monotonically increasing, you could make a table of t as a function of h and fit a function through it.
    From the looks of it, I would first try taking a logarithm of height and see if a regression line gives you the accuracy you want.
    If that doesn't work, you can fit a spline function through it and make the resolution as good as you need. (A spline would only be practical if you make computer code to implement it.)
     
    Last edited: Feb 10, 2017
  4. Feb 10, 2017 #3

    mfb

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    How/where do you need v(h)? It is easy to calculate individual points, and you can also calculate the derivative v'(h) analytically for every point where you know t.
    If (b+k)/k is a small integer, it is possible to write down t(h) in closed form, otherwise it is probably impossible.

    Introducing new variables for t-independent stuff would make the equations much more readable.
     
  5. Feb 10, 2017 #4
    The reason I want this v(h) form is so that I can integrate a function which relies, in part, on v(t).

    So... I have another function:
    [tex]F(t,h)=c_1v(t)+c_2g(h)[/tex]
    where [itex]c_1[/itex] and [itex]c_2[/itex] are constants and [itex]g(h)[/itex] is another well-understood function of height.
    What I want to find is:
    [tex]\int_{h=0}^{h_f}F dh[/tex]
    to find the work done by this force over some height. And since velocity and height are well-defined functions, I figured it would make my life easier to integrate over a single variable, [itex]dh[/itex], instead of trying to figure out how to get work from a function of both time and height.

    So, if I could get [itex]t(h)[/itex] and plug that into [itex]v(t(h))[/itex], my life would be easier...

    Unfortunately, the quantity [itex](b+k)/k[/itex] will be, at best, [itex]1[/itex] so it can't be ignored...
     
  6. Feb 10, 2017 #5
    Ummm... maybe I could do this (please comment your thoughts!):

    [tex]\int_{h=0}^{h_f} F dh = \int_{h=0}^{h_f} F dh \frac{dt}{dt} = \int_{t=0}^{t_f} F v(t) dt[/tex]

    Does that work, or is there a different conversion for the limits of integration or differential element that I'm forgetting?
     
  7. Feb 10, 2017 #6

    FactChecker

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    Do you need it to be closed form or can you use numerical methods?
     
  8. Feb 10, 2017 #7
    Numerical results are perfectly fine!
     
  9. Feb 10, 2017 #8

    mfb

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    If it is 1, h(t) is a polynomial of degree 2, and easy to invert.

    Numerically, you can just integrate over t and don't need anything special.
     
  10. Feb 11, 2017 #9
    Yeah the problem is, it'll never be 1 in reality. It'll be something like (0.05+k)/k which is why it's so hard to invert.
     
  11. Feb 11, 2017 #10

    mfb

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    All that doesn't matter if you just need a numerical value for the integral.
     
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