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I am trying to get the 16 equidistant point on ellipse (equal arc length along the ellipse).

Using some research, I am able to get the angle from which I can get those point by drawing a straight line and getting the intersection point but unable to find the length of line to be drawn from the center. I have also explored https://math.stackexchange.com/questions/172766/calculating-equidistant-points-around-an-ellipse-arc but getting confused in formulaenter image description here

What is the value of φ here?

Can anyone please help me there on getting the points. Thanks

Christoph Rackwitz
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ArenMayank
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    see [compute ellipse parametric angle for specific real angle](https://stackoverflow.com/a/71219436/2521214) look for `ellipse_angle` in **[edit2]** also see [evenly space circles along ellipse circumference](https://stackoverflow.com/a/19560243/2521214) – Spektre May 06 '22 at 08:29
  • This is called the *eccentric angle*, I believe. – Stéphane Laurent May 06 '22 at 08:53

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We can define (axis-aligned) ellipse parametrization as

x = a * cos(φ)
y = b * sin(φ)

where parameter φ has range 0..2*Pi.

Resulting point (x,y) is situated at angle θ relative to the ellipse center. Your linked post shows formula for θ/φ transformation.

Note - θ is real angle, φ is not!, it is just parameter.

You perhaps don't need θ here. To solve the problem, you have to find ellipse circumference length L using elliptic integral for φ = 2*Pi (numerically).

Then find φ values corresponding to arc length L/16, 2*L/16...15*L/16 - numerically again, and calculate corresponding point coordinates from the parametrization equations.

MBo
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    no need to do this numerically ... I did it like that for years too but recently I came up with simple `O(1)` analytic solution see [float ellipse_angle(float rx,float ry,float ang)](https://stackoverflow.com/a/71219436/2521214) at the end of my answer (edit2) there is also slow numeric version before that ... – Spektre May 06 '22 at 08:34
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    @Spektre Emm.. Seems your approach is intended for `t(or fi)-theta` transformation, but I wrote about numerical methods for equal arc length problem – MBo May 06 '22 at 12:33
  • ahhh now I see the `(equal arc length along the ellipse)` I probably stopped reading at `16 equidistant point` ... – Spektre May 06 '22 at 15:33