After seeing this question asked before and not seeing a satisfactory answer I tried to come up with my own answer. First, we need to establish the reference direction that the longitude of ascending node (LAN) is measured against. See for example this post for a definition, the reference direction points from Kerbin to Kerbol on day 1 (lets ignore the tiny difference of 0.00159rad because the mean anomaly of Kerbin is 3.14 and not Pi in this assumption.). The next needed part is that a solar (Kerbol) day is a little longer than the sidereal rotation of Kerbin. Looking at the picture below one can piece together the global longitudinal angle in the Kerbol (different from the coordinates on Kerbin).

At 0 degrees longitude on day 1 year 1 the global longitude is 90 degrees.

At KSC the lat/long coordinates are about 0,-75, so that on day 1 year 1 the global longitude is 90-75 = 15 degrees

Due to the rotation of Kerbin the global longitude increases by 360 degrees over 1 day, i.e. 1 degree per minute, plus ...

The difference between sidereal day and solar day is adding 360 degrees per complete sidereal orbit, or 21600/9203545*360 degrees per full day.

Now putting this all together the time x (in minutes) after 0:00h in UT to have the KSC at global longitude LAN on day d can be calculated with this formula (not showing the algebra):

x = LAN - (90 + <launchlongitude>) - (d-1)*21600/9203545*360

As an example, to launch at global longitude 90 degree, on day 60 the calculation yields x = 25.15m. So, if the Kerbals launch at 0:25h at KSC on day 60 with an inclination their resulting orbit will have a LAN of about 90 degrees.

Hope this helps. I tried it, but please let me know if you find errors.

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## VMQ 3

After seeing this question asked before and not seeing a satisfactory answer I tried to come up with my own answer. First, we need to establish the reference direction that the longitude of ascending node (LAN) is measured against. See for example this post for a definition, the reference direction points from Kerbin to Kerbol on day 1 (lets ignore the tiny difference of 0.00159rad because the mean anomaly of Kerbin is 3.14 and not Pi in this assumption.). The next needed part is that a solar (Kerbol) day is a little longer than the sidereal rotation of Kerbin. Looking at the picture below one can piece together the global longitudinal angle in the Kerbol (different from the coordinates on Kerbin).

Now putting this all together the time

x(in minutes) after 0:00h in UT to have the KSC at global longitudeLANon daydcan be calculated with this formula (not showing the algebra):As an example, to launch at global longitude 90 degree, on day 60 the calculation yields x = 25.15m. So, if the Kerbals launch at 0:25h at KSC on day 60 with an inclination their resulting orbit will have a LAN of about 90 degrees.

Hope this helps. I tried it, but please let me know if you find errors.

Edited by VMQSome clean up.

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