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Math Help - Perimeter of an ellipse from radii, and vice-versa

  1. #1
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    Perimeter of an ellipse from radii, and vice-versa

    Hi there,

    I'm new here. Greetings.

    I need to figure out how to calculate the perimeter, or circumference, of an ellipse using the radii, and vice-versa: how to figure out the radii given the perimeter and the ratio of the two radii to each other.

    I found an equation for the perimeter, or circumference here: Circumference/Perimeter of an Ellipse: Formula(s) - Numericana)
    given the radii x and y, where x>y:

    P=3.1415*(((2*((x^2)+(y^2)))-(((x-y)^2)/2))^.5)

    The result will be an approximation but it will be close enough.

    Any chance you could help me solve this equation for x in terms of C and y, and then also for y in terms of C and x?
    Iím pretty rusty on my equations!

    Many Thanks,

    rgesh
    San Francisco, CA
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Perimeter of an ellipse from radii, and vice-versa

    Hello Rgesh. Given values for perimter 'P' and semi-axis 'y' you can rearrange the equation to get a quadratic equation in the other semi-axis 'x':

    P = \pi \sqrt{2(x^2+y2)-\frac 1 2 (x-y)^2}

    Bring the pi factor to the left hand side and square both sides, and expand the (x-y)^2 term:

    \frac {P^2} {\pi^2} = 2(x^2+y^2)-\frac 1 2 (x^2-2xy+y^2)

    Now gather terms to make a quadratic in x:

    \frac 3 2 x^2 +yx +(\frac 3 2 y^2- \frac {P^2}{\pi ^2}) = 0

    Now you can apply the quadratic formula to solve for x. The equivalent equation for y, given values for x and P, is:

    \frac 3 2 y^2 +xy +(\frac 3 2 x^2- \frac {P^2}{\pi ^2}) = 0
    Last edited by ebaines; November 22nd 2013 at 11:02 AM.
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    Re: Perimeter of an ellipse from radii, and vice-versa

    Hi ebaines,

    Thank you for your help, however, I'm a bit lost.
    I need to use these equations to plug into an Excel chart, so I need them in a format such as x=..... and y=......
    I'm not sure how to use the equations you gave, where everything=0.

    Also, I realize that I need the equations to include another variable, R, that would represent the ratio of x to y. Is that possible?

    Thanks.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Perimeter of an ellipse from radii, and vice-versa

    Are you familar with the quadratic equation, which I'm sure you learned in algebra class? Given:

     ax^2+bx+c = 0

    The solution for x is:

     x = \frac {-b \pm \sqrt{b^2-4ac}}{2a}

    Here a = 3/2,\ b = y,\ c = \frac 3 2 y^2-\frac {P^2}{\pi^2}. Plug these values into the quadratic formula and you'll get the value for x. Actually you will get two values (due to the "plus-or-minus" factor), but one value is positive and the other is negative, so you can throw out the negative value.
    Last edited by ebaines; November 22nd 2013 at 10:54 AM.
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    Re: Perimeter of an ellipse from radii, and vice-versa

    I definitely learned that in algebra class. That was in 1972. I'm out of practice.

    Many thanks for your help. Much appreciated.
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: Perimeter of an ellipse from radii, and vice-versa

    As for the ratio R: if you have values for P and R (defined as R=x/y), then starting with the first solution I gave earlier:

    \frac 3 2 x^2 +yx +(\frac 3 2 y^2- \frac {P^2}{\pi ^2}) = 0

    divide through by y^2:

    \frac 3 2 \frac {x^2}{y^2} +\frac {x}{y} +(\frac 3 2 - \frac {P^2}{\pi ^2y^2}) = 0

    Replace x/y with R:

    \frac 3 2 R^2 + R +(\frac 3 2 - \frac {P^2}{\pi ^2y^2}) = 0


    Rearrange to get y^2 by itself:

    y^2 = \frac {P^2}{\pi^2 (\frac 3 2 R^2 +R + \frac 3 2 )}

    Take the square root of both sides:

     y = \frac {P}{\pi \sqrt{\frac 3 2 R^2 +R + \frac 3 2 }}

    This gives you the y semi-axis. You can get the other from x = Ry

    Hope this helps.
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  7. #7
    MHF Contributor ebaines's Avatar
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    Re: Perimeter of an ellipse from radii, and vice-versa

    Quote Originally Posted by rgesh View Post
    I definitely learned that in algebra class. That was in 1972. I'm out of practice.

    Many thanks for your help. Much appreciated.
    Algebra class for me was 1970, but I'm an engineer and so I guess I've managed to stay in practice. Glad to be able to help!
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