UnHohmann transfers

[HebaruSan]

49 minutes ago, Nicias said:

Thanks for giving it a look :). That change is correct, the difficulty is in finding theta. Theta is the angle between the vectors. It is only the angle between the orbits if the velocity is purely tangential (so at apoapsis or periapsis). Take the limit of two orbits that 90degrees inclined to each other, but highly elliptical. If they have 1000 m/s radial and 1m/s tangential, then the angle between the two velocities is only about 0.08 degrees, rather than the 90 degrees.  So you need to find that angle, it isn't just the relative inclination. Or you have to find the tangential velocity, which is easier, because it is just given by  h/r (where h is the angular momentum of the orbit) but then you need to find the radius when you do the burn. Which means you need to find out where on the transfer ellipse the AN/DN is, which means you need to find the AoP of the transfer ellipse, which leads to my question about AoP.

MechJeb's plane change calculator does a cross product, then some other vector operations I don't quite understand, then some dot products to get the burn into ship-relative coordinates (here pasted into Astrogator):

https://github.com/HebaruSan/Astrogator/blob/master/src/PhysicsTools.cs#L657-L684

It skips trying to calculate the angle, though, so if you really need that, that code won't help.

[Nicias]

Ok, so I mathed the heck out of this.  I used Mathematica for all of my calculations. I did everything as if you are going from Eeloo to Kerbin. I could re-run it to go the other way if I just change the AoP of Eeloo to negative its current value.

I first ignored inclination. I determined both the dV for transfer with minimum eccentricity at a given departure angle. I also determined the minimum dV for a departure at the given departure angle with periapsis at Kerbin's orbit. The results range from 3141 m/s to 3740 m/s for both methods, departing at apoapsis for worst case and periapsis for best.  Note that periapsis is at 0 or 360 degrees and apoapsis is at 180 degrees.

https://imgur.com/rzZXYtZ

https://imgur.com/usf927a

As other have mentioned, the dV is dominated by the Eeloo burn. 

Now adding in inclination I first did it by correcting the inclination at the AN or DN that is on the way down to Kerbin. Again I used minimum eccentricity and optimized for dV transfers. For the least eccentric I got a range of 3782 to 5013 m/s. Departing at 226 degrees and 291 degrees. The range in inclination change is the largest of the factors. Breaking down the dv budget as Eeloo departure, inclination change, Kerbin capture gives: 3782 = 1221 + 410 + 2151 and 5013 = 1732 + 1289 + 1993.   As you see the inclination burn is the biggest factor.  (410 versus 1289)

https://imgur.com/4LYot79

The same calculations minimizing dv gives: 3584 @ 100 degrees to 4882 @ 309 degrees. With breakdowns:

https://imgur.com/SXO9sB4

3584 = 1289 + 203 + 2092 and 4882 = 1664 + 1255 + 1964 

The minimum transfer one saves dv on the inclination by going a little farther out. than the mimimum eccentricity one.  If you look at the graphs for these two you'll notice a sawtooth pattern to the inclination changes and hence the total. As the departure location of the transfer orbit approaches the line of nodes, the transfer gets cheaper as you are departing closer and closer to the node. As the departure location passes the node (100 and 280 degrees) then you have to correct at the lower node. So that 1255 in the above example is correcting right outside Kerbins SOI.  This explains the jumps in the graphs. In practice you could roll that correction into the Kerbin capture and then have Pythagorean and Oberth savings.

Finally I allowed for correcting the inclination at either node. That gave: 3619 @ 146 and 4512 @ 10 for least eccentric with breakdowns:

3619 = 1101 + 345 + 2172 and 4158 = 1834 + 417 + 1906

https://imgur.com/iT3COSl

dv optimized were: 3531 @ 125 and 4495 @ 8 degrees, with breakdowns of: 

https://imgur.com/FjXfSgk

3531= 1122 + 265 + 2144 and 4495 = 1835 + 755 + 1906.