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I have perfect circles in lon/lat on a OpenLayers map.

Near the Equator I locate each point on the circle using bearing-calculation for each point from circle-center. Meaning there are 360 bearings as there are 360 lon/lat points:

1.deg on circle = 1deg bearing/2.deg on circle = 2deg bearing....etc).

When I move a circle north of the Equator I'm not able to do this anymore due to the following explanation: https://en.wikipedia.org/wiki/Bearing_(navigation):

enter image description here

How can I "locate" each point on the circle by calculation from the circle center when circles are "off equator". Bearing calculation is not a requirement, I just need to locate every point per degree in a circle?

Background: First I thought this was a projection problem, however I now have discovered that it is a calculation problem as several of the points on the circle gets the same bearing:

Update 11.sept: Attached is an example, i try to locate each point and draw a line between them. For each bearing (0-359) i use "Destination point given distance and bearing from start point" from this site: https://www.movable-type.co.uk/scripts/latlong.html where the center-point is the start point and distance is constant.

enter image description here

This does not work, obviously, as the bearing calculation is off away from Equator. How can I calculate correct bearing away from Equator?

Or any suggestions how to solve this by other means? Trigonometric, math or whatever? ( The map is not imporatant, but the final shape is... )

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    Is this "perfect" circle perfect in Cartesian space or geodetic space? If you know the vertex order you can intuit the angle, but computing angles in angular units is tricky (and counterintuitive). – Vince Aug 28 '19 at 11:36
  • As it is in lon/lat i suppose it is in geodetic space? I'm not a gis-person so terminology is not my strength.... See added picture... – otk Sep 11 '19 at 6:00
  • If the figure appears to be a perfect circle when plotted as Plate Carree, it isn't a true circle. If it appears to be egg-shaped, with the longer part toward the pole, then it is a perfect circle on the ground. Solving bearing and distance on a spheroid is a partial differential eqation, solvable only by iterative means. You are better off finding a library than a formula. – Vince Sep 11 '19 at 10:34
  • If the circle is perfect when plotted on a Web Mercator map, the the units are not degrees, and it isn't a circle, due to the infinite distortion of Mercator at the poles. Your picture doesn't clarify the problem. – Vince Sep 11 '19 at 10:40

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