LatLon at UT with rotation query

[cyberKerb]

Is it possible to get the Lat & Lon of an object (in orbit) as at a UT time that take into account planetary rotation?

I've tried the following:
Vector3d VESS_Position = ActiveVessel.orbit.getPositionAtUT(FutureUT);
Vector2d FutureVES_LatLon = LatLon.GetLatitudeAndLongitude(ActiveVessel.mainBody.BodyFrame, ActiveVessel.mainBody.position, VESS_Position);

However, this doesn't seem to factor in planet rotation.

[MOARdV]

You know the rotation period of the planet, so you can determine how much you need to adjust the longitude based on that.  I think it's a case of determine the change in time (FutureUT - now), divide that by the Body.rotationPeriod (as long as the period isn't 0), take the fraction from that result, multiply by 360, and add or subtract (don't remember which) to the longitude.  And then normalize the result to [-180, +180].  That should get you the right answer, or at least get you pointed towards it - it's been a while since I messed with that type of problem.

[cyberKerb]

2 hours ago, MOARdV said:

You know the rotation period of the planet, so you can determine how much you need to adjust the longitude based on that.  I think it's a case of determine the change in time (FutureUT - now), divide that by the Body.rotationPeriod (as long as the period isn't 0), take the fraction from that result, multiply by 360, and add or subtract (don't remember which) to the longitude.  And then normalize the result to [-180, +180].  That should get you the right answer, or at least get you pointed towards it - it's been a while since I messed with that type of problem.

Thanks for the tip, I managed to find the code you were trying to remember

https://github.com/Mihara/RasterPropMonitor/blob/b32f65c07168dfe7c6c8f786ecc00e358cc894c3/SCANsatRPM/JSISCANsatRPM.cs#L277

One question - does this method work when at high latitudes?

Edited July 29, 2020 by cyberKerb

[MOARdV]

 

7 hours ago, cyberKerb said:

One question - does this method work when at high latitudes?

As long as latitude +/- 90 define the axis of rotation, yeah.  It doesn't matter if the latitude is 0 or 89.99999 - the period of rotation is constant.  A given line of longitude will sweep across 360 degrees in a single day.