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For an ellipse (centered at 0,0), let a = 5.088 and b = 3.006.

I used an equation to determine the circumference of ellipse. The approximation = 25.850.

Now, I assume it is safe to say that each quadrant has a length = 6.462 because of symmetry. My question involves 1/2 of the arc of the first quadrant. Given the assymetric nature of an ellipse, can I divide the quadrant length (6.462) by 2 to get a 1/2 arc of 3.231? If so, I get an angle of 37 degrees with corresponding point on the ellipse at x = 3.142 y = 2.364.

Are my calculations correct? I've researched this website and I'm beginning to confuse myself with regards to the "t" parameter.

Thanks!

Joe
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    The area of an ellipse is a relatively easy calculation. The length of an elliptical arc is not so easy. Simply chopping the value in half does not guarantee that you have the correct angle. You might consider using a quadratic as an approximation if you want a number that is "good enough". Also, what equation for the circumference did you use? – abiessu Oct 15 '13 at 18:47
  • Got this from Wiki for the circumference: http://en.wikipedia.org/wiki/Ellipse Circ=PI()(3(a+b)-SQRT(10ab+3*(a^2+b^2))) – Joe Oct 15 '13 at 19:33
  • I'm not sure how you would get an angle using that approximation though. What method did you use? – abiessu Oct 15 '13 at 19:42
  • Here is more info: I found a question that was similar to this topic. http://math.stackexchange.com/questions/433094/how-to-determine-the-arc-length-of-ellipse Knowing the arc length, I used numerical integration method provided in that topic. The previous question had an angle and wanted to find the arc length. I had the arc length (or assumed 1/2 of the quadrant length), but wanted to find the corresponding X,Y coordinates after integrating from t=0 to t=t1. Working backwards (and in Excel, by the way), I got t1=0.9051 and used x=acos(t1) and y=bsin(t1) – Joe Oct 15 '13 at 20:11

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