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UnHohmann transfers


I'm looking for the max delta-v required to go from orbit to orbit, assuming a Hohmann style transfer. So one burn at each end, and maybe one in the middle for plane change.

The math is looking pretty dense for me to take a run at it myself. And a lot of what I'm seeming to find just calculates the Hohmann Δv. Or I'm not reading it very well... also probably true.

Anybody got some pointers? Absolute preference is a nice precalculated table. But, I'm kind of prepared to do the lifting needed to build it... I just can't seem to find a decent place to start on the calcs.

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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.



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)


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


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


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


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



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