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Thread: Distance along the surface of an oval between two points

  1. #1
    Mar 2009

    Distance along the surface of an oval between two points

    Hi All...

    I am looking for a formula to calculate the distance between 2 points on the outside of an oval.

    I made a small example image.
    I have the X,Y value of both A and B.. so in this example they are:
    A.x = -1.9
    A.y = -2.8
    B.x = 2.0
    B.y = 2.5

    I also have the radius of the oval... we can call R and equals 9.42 (should note I made that up, but I know that the radius of a 6x6 circle)

    I am trying to find a way to calculate the distance from A to B clock wise along the surface of the oval, and not necessarily the shorted distance.

    I found a formula for calculating the distance for two lat and log and tried to adapt that, but it didn't work... here is what I had:

    acos(sin(A.x) * sin(B.x) + cos(A.x) * cos(B.x) * cos(B.y-A.y)) * R

    Honestly.. this type of math is above my head, but I really need the answer so I am giving it all I have... been reading up on cos and sin and know its like the x and y of my A and B... but everything I find talks about how to calculate out the angle from the center to get the x and y.. which I already have ....

    Hope someone can help.. Please
    Last edited by skidx; Mar 14th 2009 at 04:32 PM.
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  2. #2
    Dec 2008
    Auckland, New Zealand
    It's more complicated than you might think...

    The formula for the length of part (L) of a curve is $\displaystyle L = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$

    The equation of an oval, more correctly called an ellipse, is $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

    Rearranging gives $\displaystyle y = \pm \sqrt{b^2 - \frac{b^2x^2}{a^2}}$

    This becomes difficult to work with as an ellipse is not the graph of a function. If one point is below the x-axis and the other above then you have to use both square roots separately, because an ellipse is the joint graph of those two square roots above...
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