Anyone have the days for AN/DN between planets?

[EdFred]

They should be the same every Kerbal year, but I can't get them quite right. When going to the other non-atmosphere bodies which have inclination changes (I'm looking mostly at you Moho, Dres, and Eeloo) I prefer to launch at the AN or DN from Kerbin, and make my transfer and inclination change at the same time. I don't worry about hitting the planet on the first intercept, I simply reduce my next orbital period to match up the second time I'm in that spot (and I have to make that burn for injection anyway, so it shouldn't be much wasted dV). However, doing it by eye, I'm still off quite a bit when leaving Kerbin, and still end up with AN/DN of 0.5 or so So what I am looking for it what days in the Kerbal year do the planets' nodes match up with Kerbin.

Moho

AN:

DN:

Eve

AN:

DN:

Duna

AN:

DN:

Dres

AN:

DN:

Jool

AN:

DN:

Eeloo

AN:

DN:

Eve, Duna, and Jool aren't that important as aerobraking is more efficient than burning to reduce orbital period.

(I will fill these in for reference when I have the answers)

Edited August 4, 2014 by EdFred

[Mr Shifty]

Since Kerbin's orbital plane is on the ecliptic, the LAN of a planet relative to the sun should be the same longitude as the AN relative to Kerbin. And, since Kerbin's orbit is circular, angle will be proportional to time (where 360° = 1 Kerbin year). Problem is, while we know the LANs of all the planets, I'm not sure what the reference direction is for the LANs, nor am I sure where Kerbin starts on its orbit relative to the reference direction. (Mean anomaly at epoch is approximately pi, but that's not a huge help.) If you can figure out one of these ANs or DNs, we can calculate the rest pretty easily.

[Guest]

I usually just put a simple little probe just outside of Kerbin's soi and then use that to find the an/dn of any particular planet.

[EdFred]

Since Kerbin's orbital plane is on the ecliptic, the LAN of a planet relative to the sun should be the same longitude as the AN relative to Kerbin. And, since Kerbin's orbit is circular, angle will be proportional to time (where 360° = 1 Kerbin year). Problem is, while we know the LANs of all the planets, I'm not sure what the reference direction is for the LANs, nor am I sure where Kerbin starts on its orbit relative to the reference direction. (Mean anomaly at epoch is approximately pi, but that's not a huge help.) If you can figure out one of these ANs or DNs, we can calculate the rest pretty easily.

Yep, and going through the Wiki pages, I found the AN of the inclined planets and was going to calculate it (rather easy to do), but like you said, where does Kerbin start? Does it start at 0? 17? 230?

[Taki117]

Yep, and going through the Wiki pages, I found the AN of the inclined planets and was going to calculate it (rather easy to do), but like you said, where does Kerbin start? Does it start at 0? 17? 230?

For all intents and purposes does it really matter? IF you are basing everything off of Kerbins orbit, then you can set up your circle with Kerbin at 0, regardless of whether or not it is actually at zero. Kerbins orbit is a perfect circle, so it's starting position should be irrelevant, unless my thinking is completely off..

[Mr Shifty]

For all intents and purposes does it really matter? IF you are basing everything off of Kerbins orbit, then you can set up your circle with Kerbin at 0, regardless of whether or not it is actually at zero. Kerbins orbit is a perfect circle, so it's starting position should be irrelevant, unless my thinking is completely off..

Yeah, it matters. Eve, for instance, has a LAN (longitude of the ascending node) of 15°. This means that Eve's orbit crosses the sun's equator going northbound at a longitude of 15°. And since, Kerbin is on the ecliptic (it orbits around the sun's equator) Eve's orbit crosses the sun's equator and Kerbin's orbit at the same angle: 15°. This introduces two questions though:

1) 15° relative to what? In the real solar system, heliocentric LANs are measured relative to the First Point of Aries. This is arbitrary, and there's a similar arbitrary origin of longitude in the KSP solar system, but we don't exactly know what it is.

2) The above wouldn't be important if we knew where Kerbin started: what longitude is Kerbin at at 0s UT? Without knowing this, we can't figure out when it will cross 15° of longitude.

[sonicsst]

I think LAN of Kerbin is 0 and other LANs are measured ccw to Kerbin at epoch = 0. Each planet starts at its apoapsis. I'm pretty sure you can use this and the argument of periapsis to determine the angle to Kerbin at t = 0 but I cant figure it out.

edit: I don't think you need that to determine the dates of the AN/DN. For Moho, argument of periapsis is 15 degrees and the periapsis is 180 degrees ccw, so the AN is 165 degrees ccw at t = 0. Orbital period of 2215754 s multiplied by (165/360) gives us 1015554 s UT for the AN. This is close but wrong, probably because I forgot that angular speed isn't constant in an eliptical orbit. You probably shouldn't trust me too much with this.

another edit: The AN/DNs won't fall on the same day each year. To get subsequent nodes, you'll have to add Moho years in my example, not Kerbin years.

other edit: Oops, I'm doing this whole thing backwards anyway. Nevermind.

Edited August 6, 2014 by sonicsst

[LethalDose]

With one exception (Moho, I'll get to that in a moment), I'm not sure knowing the days when the planets cross the ecliptic (aka Kerbin's orbit) or when Kerbin is aligned with the planets' AN/DN with the ecliptic is useful information for interplanetary transfers (It seems to me that the OP has stated that this is the purpose of his inquiry quite clearly). This information is not useful because the launch windows for the planets will not necessarily coincide with the timing of either Kerbin being aligned to the AN/DN axis or the planet being in that location at arrival.

Also, 2 of the 3 airless bodies orbit Kerbol at higher altitudes than Kerbin (Dres & Eeloo), so vessels in transfer orbits between Kerbin and these targets are very likely to cross their respective AN/DN nodes at velocities much lower than at the vessel's velocity at ejection (Remember: The cost of inclination changes are related to the speed of the vessel where the change is made, it costs more when the vessel is going faster). I acknowledge that it is possible to make the inclination change at launch, but this is very tricky to set up, even with information about when Kerbin is at an AN/DN node.

IMO midcourse corrections are preferable to inclination changes at launch when approaching Dres and Eeloo due to slower orbital speeds and more dV savings due to launching during appropriate windows.

Now Moho. IMO, Moho rotates so fast that there basically aren't transfer windows for it, and its high inclination makes traditional Hohmann transfers a poor choice. Seriously, please don't try Hohmann transfers to Moho... they make Jeb sad. There are two other options: Bi-ecliptic transfers, and braking orbits. Knowing when the Kerbin crosses Moho's AN/DN axis with the ecliptic is useful. For a bi-elliptic transfer, you want to escape kerbin at that time so your apoapsis is also on that that axis so your inclination change is cheaper. For an approach with a braking orbit, you don't even necessarily need to a plane change if your periapsis will intersect Moho along the AN/DN axis, but you made have a higher velocity relative to Moho when you're getting captured.

I think the first time Kerbin crosses Moho's AN/DN node is around Kerbin day 50-60 (~ 12-15 earth days), and then every 1/2 Kerbin year after that. I just launched a Moho mission on Kerbal day 100, and I SWAG (Scientific Wild @$$ Guess) that I had passed it by ~ 30-45 degrees of Kerbin's orbit.

[EdFred]

I am not worried about launch windows. I don't need the planet to be there when I get there. Granted this is mostly for Moho, because there is such a high dV, but I figured might as well have it for all of them. I have tried to eyeball where Moho and Kerbin cross by looking at the ecliptic and rotating the view until Moho's orbit becomes a line, but I haven't gotten close enough for my likes. I'm also convinced that a mid-orbit correction to Moho costs more than when doing it with the ejection burn.

[Mr Shifty]

I think LAN of Kerbin is 0 and other LANs are measured ccw to Kerbin at epoch = 0. Each planet starts at its apoapsis. I'm pretty sure you can use this and the argument of periapsis to determine the angle to Kerbin at t = 0 but I cant figure it out.

edit: I don't think you need that to determine the dates of the AN/DN. For Moho, argument of periapsis is 15 degrees and the periapsis is 180 degrees ccw, so the AN is 165 degrees ccw at t = 0. Orbital period of 2215754 s multiplied by (165/360) gives us 1015554 s UT for the AN. This is close but wrong, probably because I forgot that angular speed isn't constant in an eliptical orbit. You probably shouldn't trust me too much with this.

This might work; you just need to use Kerbin's orbital period rather than Moho's. Kerbin's Mean Anomaly at Epoch is 3.14(=179.909°). For a non-inclined, circular orbit, starting longitude is LAN+AoP+MAaE, so Kerbin's would be 0°+0°+179.909°. Moho's LAN is 70°, which will take Kerbin (360°-179.909°)+70°=250.091° to reach. (The DN will only take 70.091°.) Since Kerbin's period is 9,203,545 seconds, you'd see DN at 1,791,904s UT, which is 20d17h45m05s Kerbin time (assuming UT starts at 0d0h0m0s and days are 24 hours; I can't remember if UT days are 24 or 6 hours) and AN would be at 6,393,677s UT = 74d00h01m17s. If those check out, then you should be able to calculate the rest from there.

[Sirrobert]

I'm pritty sure it's way more efficient to just burn out at the launc window, and make your plane change along the way

[EdFred]

I'm pritty sure it's way more efficient to just burn out at the launc window, and make your plane change along the way

When I've done that, I've gotten plane changes that are VERY substantial.

[Sirrobert]

Still easier than waiting for both the AN AND the transfer window to line up at the same time

[PLAD]

The Wiki gives us the orbital parameters for the planets, including their anomalies on Day 1 at 0:00:00. But as someone pointed out there is nothing we can refer to as '0 degrees'. For instance Kerbin starts with an anomaly of 3.14 radians, or about 180 degrees on day 1. That's great to measure from on day 1, but after that you have to calculate what angle Kerbin is at at a given time before you can use it as a reference. No good. So I do it like this (For reference I'm looking from above the plane of Kerbin's orbit, defined as the direction from which Kerbin appears to be traveling counter-clockwise around Kerbol. I also do all calculations using Earth-time, 24 hour days.):

Moho first passes through the plane of Kerbin's orbit moving up (so ascending node) on day 13.128. Defining "180 degrees" as the position Kerbin is at on day 1:00:00:00, Moho's ascending node is at 70.00 degrees. It passes through descending node on day 25.09, at angle 250.00 degrees. From this we can compute when Kerbin passes through the plane of Moho's orbit,, since we know when Kerbin was at 180 degrees and we know how many degrees it travels in a (24-hour) day. This is what I think you want to know- Kerbin passes through the plane of Moho's orbit on day 21.74 and day 75.00, and every 106.5225 days after those times.

You can't say that Kerbin is always in the plane of Moho's orbit on a given Kerbal-time (6-hour days) day, because a Kerbal year is an integer number of Kerbal days, but Kerbin's real orbit is not an integer number of days, so as the years go by you'll be off by more and more.

I used my 'Lambert E' spreadsheet to figure out this data, get it here. The page 'Calc positions' gives the x, y, and z positions of a planet at a given time as well as the anomaly as measured with the 'Kerbin-starts-at-180-degrees method. Look for the times that the z-position of a planet passes through 0 meters and see what 'approx absolute angle' it's at on that day.

In summary I can give you the 1st UT day that Kerbin passes through the ascending and descending nodes of a planet, you would have to make a little calculator that adds Kerbin years to the 1st pass to determine all later passes.

[EdFred]

Still easier than waiting for both the AN AND the transfer window to line up at the same time

I'm not waiting for the transfer window, just the AN/DN.

[PLAD]

It struck me that it is a bit tricky to edit this spreadsheet from scratch, so here is some detail on how I used it to find when Kerbin crosses the plane of a planet, Eve in this example. I added some colors to it- I use green to show the cells you are supposed to change and orange or yellow for outputs you need to check.

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And to answer the original question, here is when Kerbin crosses the orbital planes of these planets, given in 24-hour UT days:

Moho: Day 21.74 and 75.00.

Eve: Day 5.47 and 58.73.

Dres: Day 30.62 and 83.88.

Eeloo: Day 15.82 and 69.08.

Add Multiples of Kerbin's orbital period (106.5225 days) to these numbers to find more plane-crossings unto infinity.

I didn't do Duna or Jool because their orbits have such low inclinations.