Understanding AN/DN of contract orbits

[NFunky]

Hi guys, I'm doing some sat contracts (in RSS) and need some advice.  I'm talking, of course, about launching into the appropriate inclination.  I know there have been threads about this topic before, but I did pretty extensive searching and couldn't find what I was looking for, so here goes.

Since I'm using RSS, I'm launching from Cape Canaveral AFS in Florida, which is at 28.6 degrees.  The orbit I'm going for is a tundra orbit 71,000x460km, inclined at 110 degrees.  The orbit appears to cross the equator about half way between Ap and Pe.  I've noticed though, that the ascending/descending nodes slide around and change value as I time warp on the pad.  As a full day passes on the pad, the nodes swing back and forth along the target orbit (~45 degrees), and vary between 88 and 145 degrees.

I looked up a launch azimuth calculator and it gave me ~202.9 as a heading, which sounds right to me.  I rotated the rocket in the VAB 115 degrees so I'd only have to do a little yaw correction achieve the proper heading.  Does all this sound right?

I assumed it would make sense to launch when the nodes read 110/-110, but when I tried it, I was off by more than 15 degrees.  I then tried launching when they read 138/-138 (110+28.6), but I was off by 40 degrees.

So does anyone know when I should actually launch?  Am I not taking travel distance during launch into account?  How should I calculate this?

[bewing]

You're obviously looking at it in map mode .... you launch when the target orbit is right over your head.

 

[NFunky]

It's a very eccentric orbit, and I'm visually impaired, so I haven't had much luck with the eyeballing method.

[Capt. Hunt]

the AN/DN are the points where your current orbit passes under the tundra orbit.  The reason they're fluctuating is because your current "orbit" is on the ground and the target orbit is highly  eccentric.  You'll find that once you're in a stable orbit the AN/DN will settle somewhat.  Since this is a very high delta-v insertion though, you're best bet is to launch when the Ascending Node is directly overhead.

Edited March 26, 2017 by Capt. Hunt

[bewing]

The angle of the node at any time is basically independent of its position. So, I do not think it is possible to estimate the time to a node merely by looking at its angle, because that node can occur anywhere in its orbit. The best bet I can see for getting a "time to node" value is to place a ship in equatorial geostationary orbit at the longitude of your launchpad, and set a maneuver node on that orbit to get an approximate time.

[diomedea]

Guess the tundra orbit is inclined 110° to the equator (KSP doesn't really handle axial tilt of bodies, in reality Earth has 23.4° tilt and orbits could be defined in relation to the ecliptic instead of the equator as a reference plane; if this was in relation to the ecliptic the solution would require a different approach tied to spherical geometry).

What is required is to launch, at the correct azimuth (which is already known, e.g. using equation here) when the launch location lies on that orbital plane (or rather, just a bit before, to allow for the amount of time while ascent brings to match inclination with the orbit from the initial inclination = 0°; generally by the time a vessel gets suborbital inclination is matched, the burn to bring to orbital speed shouldn't change inclination).

Knowing where an orbital plane crosses the equatorial plane is easy: the AN/DN nodes of the orbit give that. Should the launch location lie on the equator (kind of like KSC on Kerbin), one needs to wait until the hour angle of KSC matches the LAN (for a launch at positive inclination) or the LAN+180° (for launch at negative inclination). For a launch location at latitude, one needs to compute the difference in hour angle, for where the launch location lies on the orbital plane instead (a spherical trig problem, using 4-part cotangent formula (CT-5) with inclination-90° (angle opposite to the ΔHour_angle side of the triangle), latitude, 90° (angle from meridian where latitude is taken to the parallel where the ΔHour_angle is measured) gives, once simplified: ΔHour_angle = ArcCot(cot(inclination)/sin(latitude). With latitude = 28.6 and inclination = 110°, ΔHour_angle = 9.8834°. Believe one knows that difference is positive when latitude is positive and the node is AN, or latitude negative and the node DN.

So, if one can see where the orbit crosses the equatorial plane, and applies that ΔHour_angle difference, should be launching exactly when the launch position lies on that orbital plane.