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Where to burn to move LAN?


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The "longitude of the ascending node" specifies the rotation of an inclined orbit around the axis perpendicular to the reference plane (usually the equatorial plane, making it the axis going through the poles). It's a value in degrees offset from a reference direction. You're almost exactly 90 degrees off, meaning if Kerbin was viewed from the top, your polar orbit would cross the pole east to west instead of north to south.

...or, you know, something like that. I don't actually know off the top of my head where Kerbin's reference direction actually is. What's important is that you understand that your orbit must rotate 90 degrees around the axis going through the poles.

To do this, you need to perform a plane change (normal or antinormal burn) above one of the poles.

 

EDIT for a more general answer:

In your case with a 89.4° inclination, you can just burn when your orbit crosses a pole and there will be minimal error. However, as the devitation from 90° inclination increases, this maneuver introduces more and more error in other values of your orbit.

In such a situation you will have no other choice than to first plane change at AN/DN into a full polar orbit of around 90° inclination, then turn the orbit around the pole, and then plane change back into your desired inclination.

Alternatively you can also flatten your orbit into a perfectly equatorial one of around 0° inclination, and then plane change at the correct moment in time to "create" your ascending node at the correct longitude and enter a correctly rotated target orbit.

And what if you currently happen to be in an orbit of around 45° inclination? Well, I'm sorry, but then you're royally screwed and need to pay dV out the nose no matter which strategy you pick. :P LAN errors are among the most costly errors to fix in mid-inclination orbits, so if you're going into one of those, be very sure you get it right the first time!

Edited by Streetwind
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16 hours ago, Delusional Poet said:

I have a contract that wants me in a polar orbit with a specific per and apo and LAN.

When you're doing a contract, can't you select the contract orbit as target to make it show the relative ascending node between your orbit and the target orbit? If so, that would be the best place to burn.

In general, it is more efficient to reach the correct orbital plane (inclination, LAN) first, then worry about the rest of the orbit.

 

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I've just attempted something like this and the above is correct (plane change at a pole).  But very expensive.

I believe there is a way  to achieve this by launching at a time of day based on the time of year.  (You could then fine-tune the LAN cheaply at a pole as described.)

I will test this out for your LAN:308 target and report back within 24 hours.

I am using the following references:

  1. https://forum.kerbalspaceprogram.com/index.php?/topic/196048-when-is-the-annual-moho-transfer-again/&do=findComment&comment=3830929
  2. https://wiki.kerbalspaceprogram.com/wiki/The_annual_and_reliable_Moho_Transfer_Opportunity

Stay tuned.

 

Edited by Hotel26
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6 hours ago, Delusional Poet said:

Yikes!  So it sounds like it’s going to take a heck of a lot more than the 6s of fuel I have left can manage.  So much for my idea to launch straight north from the space center for my polar orbit.  Lucky it’s unmanned.

Your idea was entirely valid!

Unfortunately, in contrast to equatorial ones, inclined orbits are sensitive to the exact time of launch. This is because Kerbin itself rotates around its polar axis. Yes, the same axis around which the LAN is defined.

Think about trying to launch directly into a low Kerbin orbit matching Minmus' inclination so you have an easier time successfully getting a Minmus transfer. This is an excellent plan, but it only works twice per Kerbin day - once every three ingame hours. Because only then, the launch site is in a location that actually allows you to match the target plane.

The same is true for your polar orbit with a specific LAN.

 

Edited by Streetwind
Typos, typos everywhere!
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17 hours ago, Hotel26 said:

report back within 24 hours

So if you were starting again from the ground (which it sounds like you will have to), then:

  • If you launch at noon (sun directly above) at (or very close to the equator, e.g. KSC) into a distinctly inclined orbit (e.g. polar)
  • then your resultant LAN will be day-of-year/426*360

I tried this on day 254 (in my Orbit world) and got 254/426*360 ~= a LAN of 213.  I then retried in my Lab world on day 80 and got 80/426*360 being very close to LAN 67.

Having that reference, if you were in my Orbit world on day 254 and wanted to launch into your target orbit with LAN 308, you would need to wait (308-213)/360*6 hours, which is equal to 1h 35m later.  Kerbin turns a full revolution in 6 hours so  the time for KSC to move from LAN 213 to LAN 308 (95 degrees) is just over 1.5 hours.

Noon at KSC appears to be around 5:42 in the day, but that's irrelevant to the objective above.  You can point your camera up and judge when the rocket nose is pointing at the sun above, or you can watch the shadows on the ground disappear under their casters.

Try it: you'll like it!

 

@Streetwindthe gist of this, by the way, is that the absolute reference for LAN is the inertial frame of reference which, in KSP, is the solar ecliptic.   Which is why matching local launch time on LAN involves time of year.  One delightful benefit, of course, is this means that you can match your ejection orbit to an external target Node.  For example, launching at Moho's AN with the intention to rendez-vous at Moho's DN (which is coincidentally close to its PE).

 

Edited by Hotel26
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