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Efficient Hohmann Transfer Altitudes

Mr Shifty

By Mr Shifty
in KSP1 Tutorials

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Mr Shifty

Mr Shifty

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For every Hohmann transfer to or from another celestial body's orbit, there is an optimal altitude to start from. For instance if you start a Hohmann burn from Kerbin to Eve at 12,908km, you'll save about 500m/s relative to starting from a 100km orbit. I thought this information might be useful for planning refueling station altitudes.

Simplifications:
- no inclination changes
- eccentric orbits are assumed circular with radius equal to the semi-major axis

Notes:
- All values in km
- A value of 0 means that the optimum is as close as possible to the planet surface (optimal altitude is negative); a value of infinity means that the optimum is as far away as possible from the planet surface (optimal altitude is outside the planet's SOI)
- These can be used for ejection altitudes or injection altitudes, i.e. transiting from Kerbin to Eve has an optimal ejection altitude from Kerbin orbit at 12,908km, and transiting from Eve to Kerbin has an optimal injection altitude into Kerbin orbit at 12,908km.
- It is typically less efficient overall to transit to a higher altitude in order to begin a transfer because the fuel expended getting to a higher altitude outweighs the gain you get from launching your transfer at the optimal altitude. Nevertheless, this might be useful information for where to put your refueling station.

Optimal Hohmann Transfer Altitudes
From/To Moho Eve Kerbin Mun Minmas Duna Dres Jool Laythe Vall Tylo Bop Pol Eeloo
Moho XXX 0 0 XXX XXX 0 0 0 XXX XXX XXX XXX XXX 0
Eve 4655 XXX 30608 XXX XXX 4674 1229 636 XXX XXX XXX XXX XXX 482
Kerbin 700 12908 XXX XXX XXX 8701 1052 371 XXX XXX XXX XXX XXX 217
Mun XXX XXX XXX XXX 0 XXX XXX XXX XXX XXX XXX XXX XXX XXX
Minmas XXX XXX XXX 366 XXX XXX XXX XXX XXX XXX XXX XXX XXX XXX
Duna 0 0 580 XXX XXX XXX 147 0 XXX XXX XXX XXX XXX 0
Dres 0 0 0 XXX XXX 0 XXX 0 XXX XXX XXX XXX XXX 0
Jool 82501 134265 191969 XXX XXX 368355 Infinity XXX XXX XXX XXX XXX XXX 0
Laythe XXX XXX XXX XXX XXX XXX XXX XXX XXX 0 0 0 0 XXX
Vall XXX XXX XXX XXX XXX XXX XXX XXX 0 XXX 0 2378 1060 XXX
Tylo XXX XXX XXX XXX XXX XXX XXX XXX 0 0 XXX 0 0 XXX
Bop XXX XXX XXX XXX XXX XXX XXX XXX 0 0 24 XXX 481 XXX
Pol XXX XXX XXX XXX XXX XXX XXX XXX 0 0 0 93 XXX XXX
Eeloo 0 0 0 XXX XXX 0 49 2200 XXX XXX XXX XXX XXX XXX

TODO: Add how much m/s you actually save vs. a 100km orbit.

Edited by Mr Shifty
Reformatted the table for the new forum that hurts my eyes.
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Mr Shifty

Mr Shifty

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Don't forget about the DV you use to get to that high orbit. That counts. It will be interesting to see if you actually save anything at all. I suspect you will not save much.

There's a note mentioning that no, you don't save anything. BUT, if you have a refueler stationed at the optimal altitude...

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Conarr

Conarr

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So if I'm at my fuel station around Kerbin and I'm heading to Duna, the optimum orbit is 8701 according to your chart. So would my interplanetary ship have to sweep down to Kerbin 75km orbit and back out to take advantage of the Oberth effect or just burn straight away from the fuel depot?

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Mr Shifty

Mr Shifty

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So if I'm at my fuel station around Kerbin and I'm heading to Duna, the optimum orbit is 8701 according to your chart. So would my interplanetary ship have to sweep down to Kerbin 75km orbit and back out to take advantage of the Oberth effect or just burn straight away from the fuel depot?

Burn straight away from the fuel depot. Oberth effect means you get more change in specific orbital energy per unit of fuel burned at high velocities, but it's balanced by having to climb out of the gravity well, i.e. you need a higher change in specific orbital energy to get to escape velocity from a lower orbit. Essentially what Oberth tells you is that you should always burn at the periapsis of your current orbit for greatest effect, but burning to lower your orbit is usually self-defeating. Example: at 100km around Kerbin, you have to lose about 20x more orbital energy to escape than at 13,000km, but your velocity (and hence available Oberth energy boost) would only be about 6x higher at the lower altitude if you lowered your periapsis to it. Plus you'd have to burn fuel to lower it.

Edited by Mr Shifty
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Mr Shifty

Mr Shifty

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So... I'm planning to send hive ships full of probes to Eve. If I understood correctly - I should assemble my fleet at just above Mun orbit(12 000 000m)?

Correct, but only if have the fuel to spare to refuel the fleet there. As tavert has pointed out to me, it takes more fuel to get to 12908 than you save for the transfer.

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Nao

Nao

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This is quite interesting! Can you share some articles/links that deal with Hohmann efficiency? I kind of understand how that works, but can't think of actual mathematic formulas. Basic googling left me scratching my head still.:P

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British_Rover

British_Rover

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Correct, but only if have the fuel to spare to refuel the fleet there. As tavert has pointed out to me, it takes more fuel to get to 12908 than you save for the transfer.

Well if you were refueling with Kethane at the Mun then yes this would make a lot of sense.

So for example if you were setting up a long term base on Eve/Gilly then the optimal thing to do would be have a Mun base that refines and/or ships Kethane to a facility in Munar orbit. Your large fuel depot orbits Kerbin just above the Mun and when the window is right you simply send a tanker to that orbit to refuel your outbound Eve ships.

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soulsource

soulsource

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Would someone be so nice as to explain how those numbers are being calculated? I tried to reproduce the table and got different results... My calculation yields an optimal orbital radius of 2*GM/dvH2, where GM is the standard gravitational parameter of the origin body, and dvH is the delta-v of the first burn of a Hohmann transfer (without taking into account Oberth effect) between the two bodies.

With this formula I get for instance following optimal ejection altitude (meaning: radius of Kerbin already subtracted) for a trip starting from Kerbin (Unit: km):

From/To Moho Eve Duna Dres Jool Eelo
Kerbin 681 11043 7777 1021 360 209
Edited by soulsource
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DStaal

DStaal

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Useful chart, but it's slightly confusing: Is the planet I'm going to on the horizontal or the vertical axis?  (From discussion, it's the 'leave from vertical while going to the horizontal', but it's hard be sure from the chart layout.)  Perhaps a heading line above ('To') and to the left ('From') would help?

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Mr Shifty

Mr Shifty

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On 4/9/2016 at 10:44 PM, GoSlash27 said:

soulsource,

Your numbers agree with mine. I thought I had invented something new, but apparently you beat me to it by about 11 hours...

:blush:

Best,
-Slashy

Your equations look good from what I can remember. I made this chart 3 years ago, derived the results by hand so could have easily made errors. Our results aren't that different though.

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Nicias

Nicias

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Ok, I use this post all the time, and did some more math on it and found a few things out:

First, as @GoSlash27 points out in the other thread, this works out to r= -2a (where a is the SMA of your departing orbit), It doesn't quite work out as nicely as @Red Iron Crown suggests since we have finite SOI's. What you get instead is:  r = 2 / (v^2/ u -  2/ SOI).

Second, when you use these gate orbits, the maneuver dV is equal to orbital velocity. Surprising but not unusual in these kind of optimization problems.

Third, if you are using a rescale, as long as distances and radii all scale, these orbits scale the same as well.

Finally, for injections, these aren't the cheapest option. It is always better to do a two-burn injection. Burning low to enter an elliptical orbit, then circularizing at apoapsis is always better. Furthermore, it is always better to burn as low as possible and then as high as possible in this situation. So the cheapest injection is to burn just above the surface (or atmosphere) to capture and then circularize just inside the SOI. I haven't done the math on a three burn injection yet.

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