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Finding best fit for a parabolic segment

  1. Jun 8, 2007 #1
    I have a set of points

    x y
    0 0
    1 1
    2 3
    3 4
    4 6
    5 8
    6 9
    7 11
    8 14
    9 16
    10 18
    11 21
    12 24
    13 27
    14 30

    That seem to define, fairly closely, a parabolic segment. How do I find the scale and x/y origin of the parabola that most closely fits these points?
    Last edited: Jun 8, 2007
  2. jcsd
  3. Jun 9, 2007 #2


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    Must the set of points be a parabola? What does the scatter plot of the points look like?

    You might try picking a few points to use for a set of equations based on the general form for conic sections; solve the set of equations for the coefficients and make your best judgement. the relationship of the coefficient values determine what kind of conic section the chosen set of points represent. You might want to try the points which you believe are the most important for your system. Some kind of matrix or linear algebra/curve fitting software would make the task efficient.

    Your general form would be like:
    [tex] \[
    Ax^2 + By^2 + Cxy + Dx + Ey + F = 0
    \] [/tex]
  4. Jun 9, 2007 #3

    Gib Z

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    [tex]y = 1.1\cdot10^{-8}x^7 + 2.5\cdot10^{-7}x^6 + 3.2\cdot10^{-6}x^5 + 3.1\cdot10^{-5}x^4 + 0.00063x^3 + 0.036x^2 + 1.34x - 0.0981[/tex] although y=1.8x isnt bad either.
    Last edited: Jun 9, 2007
  5. Jun 9, 2007 #4
    After working with a larger data set, I don't think it is a parabola. It looks very similar to one when I manually overlay a true parabola in Adobe Illustrator but it can't be made to fit exactly. It's doesn't appear to be an asymptote either.

    When calculated a different way, the points seem to follow

    y = x^(1/m)

    Where m is in the neighborhood of pi/2
  6. Jun 9, 2007 #5

    The results show a concavity towards the y axis. A function such as [tex]x^{1/m} [/tex] with m > 1 gives a concavity towards the x axis.

    PS: Could you PM me the datasheet?
  7. Jun 9, 2007 #6
    Thanks Gib.

    Did you derive that with Mathematica? I only posted a few data points because there are an infinite number :)

    When I calculate them a different way, it appears to be a simple power function with an exponent of 1/m. I tried i/phi which diverges too quickly and 1/(pi/2) which is better but still diverges a little too quickly.
  8. Jun 9, 2007 #7

    Gib Z

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    I used Graphmatica. How do you have an infinite number of points without some sort of generating function...however you do it, send it over.
  9. Jun 9, 2007 #8
    The original data set was a bit contrived to get it to oriented it that way.

    Will do. Thanks for taking a look.
  10. Jun 9, 2007 #9
    Hey ktoz, I'll take a look another time, it's time for me to go to bed... it's 6:19 AM here. :zzz:
  11. Jun 9, 2007 #10

    Gib Z

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    Holy whack...Worst i've ever done was 5:30 :)
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