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JasonMewes

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relationship between angle/length of ellipse arc

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For an ellipse of size Lx, Ly,
when ¿start is known,

how do I calculate ¿arc from Larc and vice versa?

Would appreciate an answer in algorithmic/pseudocode form.
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d-glitch
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JasonMewes

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Then I will not feel overly disappointed with myself that I did not manage to find a solution.

I do not need to be very accurate, this is for drawing so accuracy needs only be enough for things to "look good". Approximating the arc length is easy enough, approximating the angle from the length is the part I'd like help with.

I am using cartesian for the most part, polar when necessary for readability/simplicity.
You need to specify the problem before you can solve it.

If you have an ellipse, how do you specify the arc length and position?
If you have one end point and a length, you can crawl along the curve (keeping track of distance) until you find the other end.  The two end points define the angle.
You can set up an Excel spreadsheet to calculate points on the ellipse at 1 degree (or better) increments and calculate the distance along the curve.

Then you have a look-up table for arc length versus angle.
D-glitch,

As specified in the first post, the arc length and position is specified by ¿start and Larc. The graphic illustrates what these parameters describe. Is there anything else you are missing for the "problem to be defined"?

Oh, and "¿" is supposed to be theta if unicode doesn't come through nicely (it didn't last time).

I know I can estimate this by traversing the arc (or using a number of other schemes) and I could certainly implement this in Excel, although this would not help me much as I could simply put the code straight into the target application. Using this approach (calculating a LUT) however would be a (relatively) very expensive way of doing this, as the results of each calculation is only needed once, and only for a small portion of the ellipse.

The reason for asking was not an inability to implement a "good enough" solution to the problem, rather seeking a "perfect" (ie mathematically correct, clean) solution, which of course if overkill in this case - but one always seeks to learn.

Thank you for clarifying.