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Cubic Spline Interpolation Tutorial

  1. Jul 31, 2007 #1

    hotvette

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    Attached below are two cubic spline tutorials:

    1. Explanation of the classic tri-diagonal cubic spline formulation. Included are 2 example problems.

    2. Extension to parametric cubic splines. Included are 2 example problems
    .
    :smile:
     

    Attached Files:

    Last edited by a moderator: Mar 3, 2008
  2. jcsd
  3. Oct 19, 2007 #2
    wow,

    After a very long search in google i found this tutorial.

    And it was the first one, which describes parametric cubic splines in a good way.

    Thank you very much :-)

    Now i can lay streets through my landscape :-)
     
  4. Oct 31, 2007 #3
    Sigh....Thank You!

    I was missing something very stupid and now I know what it is!
     
  5. Jan 15, 2008 #4
    Thank you for tutorials. This has helped me very mush.
     
  6. Mar 2, 2008 #5
    Cubic Spline Interpolation of a Circle

    Hey there -
    Thanks for the great tutorials - they really helped me! I'm trying to duplicate your results for cubic interpolation of a circle with 4 points and I got the same solution for the 2nd derivatives in the x and y directions. However, when I solve for the coefficients and plot the cubic polynomials I can't seem to get the same result as you - Heres the code that calculates the coefficients (in MATLAB)

    mx - x''
    my - y''
    Sx - x coefficients
    Sy - y coefficients
    s - arc length [0 0.25 0.5 0.75 1]

    for i = 1:n-1
    Sx(i,1) = (mx(i+1) - mx(i))/6*h;
    Sx(i,2) = mx(i)/2;
    Sx(i,3) = (x(i+1) - x(i))/h - h*(mx(i+1) + 2*mx(i))/6;
    Sx(i,4) = x(i);

    Sy(i,1) = (my(i+1) - my(i))/6*h;
    Sy(i,2) = my(i)/2;
    Sy(i,3) = (y(i+1) - y(i))/h - h*(my(i+1) + 2*my(i))/6;
    Sy(i,4) = y(i);

    end


    %Plotting the function
    ds = 0.01;
    m=1;
    i=1;
    for i = 1:n-1
    for k = s(i):ds:s(i+1)
    genpnx(m) = Sx(i,1)*(k - s(i))^3 + Sx(i,2)*(k-s(i))^2 + Sx(i,3)*(k-s(i)) + Sx(i,4);
    genpny(m) = Sy(i,1)*(k - s(i))^3 + Sy(i,2)*(k-s(i))^2 + Sy(i,3)*(k-s(i)) + Sy(i,4);
    m = m+1;
    end
    i
    end

    plot(genpnx,genpny,'r')

    Do you have any idea what I might be doing wrong? (Ive attached an image of what the spline looks like for 4 points)

    Thanks!
     

    Attached Files:

    Last edited: Mar 2, 2008
  7. Mar 3, 2008 #6

    hotvette

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    Though I'm not a MATLAB person, your expressions for calculating the coefficients look correct. Make sure you are using 0.25 for h. Also, for plotting, t should vary from 0 to 0.25 for each segment.

    Here are the coefficient values I got for the 1st segment:

    ax = 32
    bx = -24
    cx = 0
    dx = 1

    ay = -32
    by = 0
    cy = 6
    dy = 0

    Actually, this isn't the best way to fit a circle with cubic polynomials. It is better to fit a quarter circle with a single parametric cubic. The tutorial has been updated - check it out.
     
    Last edited: Mar 3, 2008
  8. Mar 3, 2008 #7
    Solution

    Thanks for the coefficients - they helped me figure out what the problem was - a syntax problem - the correct expression for the coefficients are

    Sx(i,1) = (mx(i+1) - mx(i))/(6*h)

    while I had

    Sx(i,1) = (mx(i+1) - mx(i))/6*h

    Thanks a ton!
     
  9. Mar 3, 2008 #8

    hotvette

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    Glad to help. Fyi, the problem can be made even simpler by making h = 1, meaning s = [0 1 2 3 4]. Second deriveratives are x" = [-3 0 3 -3], y" = [0 -3 0 3 0] and plotting can be done for t varying from 0 to 1 for each segment.

    I encourage you to review the updated tutorial on circle fitting.
     
  10. Mar 12, 2008 #9

    hotvette

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    Attached is an update to the Cubic Spline Tutorial.

    To Staff,

    Would someone be willing to replace the 1st attachment in this thread with the below? Thanks.
     
    Last edited: Dec 17, 2008
  11. Nov 22, 2008 #10
    thanks a lot buddy.
     
  12. Dec 17, 2008 #11
    Hi hotvette,
    I have been looking for a good tutorial on Cubic Splines, nice work. I have found what I think might be an error in one of your examples, however. Example Problem #1 shows c2 = 0, but when I plug in the values:
    y3 = 1
    y2 = 0.125
    h2 = .5
    y3" = 0
    y2" = 4.5
    I come up with c2 = 1, not 0. Plugging 0.5 into the second polynomial with c2 = 1, I get y2 = 1, thus resulting in a discontinuity with the first polynomial. I'm too lazy to try to work out the root of the problem, but I figured I'd flag the discrepancy.
    Cheers
     
  13. Dec 17, 2008 #12

    hotvette

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    Agree that C2 = 1.0 (typo), but I think the evaluation of the 2nd poly is OK:

    [tex]y = a2(x - x_2)^3 + b2(x - x_2)^2 + c2(x - x_2) + d2[/tex]

    Update with typo fixed is attached.
     

    Attached Files:

  14. Dec 17, 2008 #13
    Yep, my mistake. I used ax^3 + bx^2 + cx + d rather than using (x-x0), etc.
    Thanks again for the excellent tutorial.
     
  15. Dec 29, 2008 #14
    For the cubic spline interpolation if we have
    A=(x2-x)/h; B=(x-x1/h); C=1/6*(A^3-A)*h^2; D=1/6*(B^3-B)*h^2;
    y=A y1 + B y2 + C y1'' + D y2''

    What is the interpolation error of above?


    For example for Hermitian cubic spline it is smaller than 5/384 ||f^{(4)}|| h^4
     
  16. Apr 8, 2009 #15
    Thanks very much my friend! The tutorial for the parametric C Splines was very helpful to me!
    You see, every book I ve read doesn t say how we can relate the derivatives of the actual curve to the derivatives of the parametric equations of x & y. Ok, its easy to proof with the chain rule but what about the values of these derivatives on bounds!! Simply, x'(t)/y'(t) must be constant on bounds, so give x'(t),y'(t) any value you want to keep this ratio constant! HAHAHAH, i feel stupid...
    Very helpful!
    Thanks again!
     
  17. Oct 16, 2009 #16
    Thanks! Just what i was looking for! :)
     
  18. Nov 15, 2009 #17
    Any reading suggestion to smooth out a curve using spline or similar approach.
     
  19. Nov 24, 2009 #18
    great tuto, very precise. Thanks a lot
     
  20. Jul 19, 2010 #19
    Thanks
    That was helpful
     
  21. Jul 21, 2010 #20
    I have trouble implementing the interpolation. not sure what is wrong. appreciate if someone can point out which part I'm wrong. Im pretty sure I have the coefficients correct but somehow im using the spline equations incorrectly (my guess).
    So here are my original x and y
    x y
    0.90 1.30
    1.30 1.50
    1.90 1.85
    2.10 2.10
    3.00 1.95
    3.80 0.40
    4.30 0.25

    Here are my coefficients
    a b c d
    Equation 1 -0.282 0.000 0.545 1.30
    Equation 2 1.046 -0.338 0.410 1.50
    Equation 3 -4.815 1.545 1.134 1.85
    Equation 4 -0.162 -1.344 1.174 2.10
    Equation 5 1.757 -1.780 -1.638 1.95
    Equation 6 -1.625 2.437 -1.112 0.40

    Now If I want to interpolate x=1, my formula would read
    y=-0.282*(1-0.9)^3 + 0.000*(1-0.9)^2 +0.545*(1-0.9) + 1.30
    y=1.354230991

    I think the answer is wrong because I validated it against the utility at http://www.akiti.ca/CubicSpline.html.. Would appreciate if someone can tell me if my coefficients are correct and if i am using the formulas correctly.

    Thanks
     
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