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Thread: Calculating the end angle of an elliptic curve with given lenght

  1. #1

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    Calculating the end angle of an elliptic curve with given lenght

    I need a formula to find the end angle of an elliptic curve with given start angle and length.

    i.e. Consider an elliptic arc with major axis 100 and minor axis 50, if the start angle of the curve is 0 and the lenght is 70,5 so is there a way to calculate the end angle.

    I`ve read an article about Jacobi`s studies on inverses of elliptic integrals. Jacobi Amplitude is the inverse of elliptic integral of the first kind, but I couldn`t find anything about the inverse of elliptic integral of 2nd kind, that could be a solution for this problem.

    Would you please share your ideas with me to find a formula for this.
    Thanks in advance.

  2. #2
    Fanatic Member riis's Avatar
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    I've no idea what you're talking about (elliptic integrals and such), but I guess you're right. Look here at formula 60, and on.

    Good luck!

  3. #3

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    Firstly, thank you for your response.

    I want to calculate the end angle of the elliptic arc with given lenght, is what the inverse of the elliptic integral of the second kind does. I can calculate the lenght of the elliptic curve using:

    k = sqrt(1-a^2/b^2)


    and

    Lenght = b * E(end_angle, k) = integral (from 0 to end_angle) sqrt(1-k^2sin(t)^2) dt

    But I want to calculate end angle using the lenght, simply inverse of that operation above, but I couldn`t have found a formula for this.

    There is a formula for the inverse of the elliptic integral of the first kind:

    if

    u = F(phi, k) = integral (from 0 to phi) dt / sqrt(1-k^2sin(t)^2)


    so,

    phi = am(u, k) = integral (from 0 to u) dn(u, k) du


    where dn(u, k) is the Jacobi elliptic function.

    But I couldn`t find the inverse of the elliptic integral of second kind, which could be the solution of my problem:

    l = E(phi, k) = integral (from 0 to phi) sqrt(1-k^2sin(t)^2) dt

    phi = ???
    Last edited by Techno-Logic; May 11th, 2003 at 02:51 PM.

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