Calculating Hohmann Transfer dV among Joolian Moons

[Duxwing]

I want to know how much a Hohmann transfer between any two Joolian moons costs.

I am planning a Joolian tour wherein I will send three ships to Jool, one carrying Kerbals, one carrying my landers, and one my wheeled science labs. All three ships will aerobrake at Laythe, where my crew will enter my Laythian science lab and descend. After gathering data, my Laythian lander will descend to my Laythian science lab and lift my Kerbals to low Laythian orbit, wherefrom all three ships will Hohmann transfer to Vall. At Vall, like at all further airless planets, my lab will descend on a tug that I will bring from planet to planet, and I will otherwise follow the Laythian procedure.

How can I calculate how much dV I will need for each transfer?

-Duxwing

[Geschosskopf]

Usually it costs rather more to capture at the airless moons that it does to transfer to them. The whole tour, if you do it brute force and ignorance like me, can add up to quite a lot, but I'm sure the navigation gurus can tell you better ways. IOW, the numbers below are certainly suboptimal.

There are 2 ways to hit all the moons with the same ship: inside-out like you've said or outside-in. There are pros can cons to each method. If you go inside-out, you only have to do Bop's plane change once because you save it for last after Pol. But then you leave for home from way out there. If you start with Pol and work in, you can save at Laythe by aerocapturing there when you might be getting a bit low on fuel, and then you're closer to Jool to start the trip home.

Anyway, from an inside-out perspective, IIRC it's SORTA like this:

Laythe -> Vall:

Transfer: ~250m/s

Capture: ~800m/s

Vall -> Tylo:

Transfer ~300m/s

Capture ~1000m/s

At this point, it's usually better to skip Bop and hit Pol next, so you only have to do the big plane change to Bop once.

Tylo -> Pol

Transfer: ~600-800m/s depending on positions

Capture: ~800-1200m/s depending on angle of intercept

Pol -> Bop

Transfer (including plane change): ~800-1000m/s depending on positions

Capture: ~800-1000m/s depending on angle of intercept

Something to keep in mind is that navigation is really funky between Laythe, Vall, and Tylo. You really can't do smooth, efficient transfers where your departure path is more or less parallel with your starting moon's orbital path. Such transfers look great to start with but as soon as you leave the moon's SOI and swing just slightly wider, the moon you just left will catch up from behind and suck you back in to a lithobrake. To avoid this, you have to use a sub-optimal transfer burn where you give yourself a significant amount of outwards velocity at the expense of forward velocity, to get yourself out beyond the sweep of the moon's SOI as it speeds by inside you. And this means you arrive at the target moon at a more pronounced angle to its orbit, which makes the capture burn bigger than it should be. It's all very annoying, but that's what you get when you cram such big SOIs as these moons have so close together that they almost overlap. But fortunately, once you get beyond Tylo, you no longer have this problem.

[Duxwing]

Thanks for the answer!

A few comments on going inward:

-Going outward may require less fuel. The difference in Oberth effects between any two orbits equals the difference between their kinetic energies. Aerobraking halves the amount of fuel needed to descend from Pol to Jool. If rocket-stopping at Pol rather than aerobraking at Laythe uses more fuel than this final Joolian aerobraking saves, then going outward requires less fuel than going inward.

-Going from Bop to Pol saves fuel because equatorial inclination around Pol is not necessary and burning down to Bop again wastes fuel.

-Perhaps polar parking orbits would allow optimal transfers: your origin moon would zip under (or over) and away instead of catching up to you. Establishing them for an inside-out tour would require only a tiny push while approaching Laythe.

-Duxwing

[Geschosskopf]

-Going outward may require less fuel. The difference in Oberth effects between any two orbits equals the difference between their kinetic energies. Aerobraking halves the amount of fuel needed to descend from Pol to Jool. If rocket-stopping at Pol rather than aerobraking at Laythe uses more fuel than this final Joolian aerobraking saves, then going outward requires less fuel than going inward.

Could well be. But OTOH, you can leave direct from a moon, too, without having to go back to Jool itself. This might change the overall calculation some, too.

-Perhaps polar parking orbits would allow optimal transfers: your origin moon would zip under (or over) and away instead of catching up to you. Establishing them for an inside-out tour would require only a tiny push while approaching Laythe.

Strangely enough, just a couple days ago I managed move a SCANsat probe from Vall to Tylo starting from a polar orbit. This was only possible because, by sheer coincidence, my orbital plane was essentially tangent to Vall's orbital path at the moment the transfer window to Tylo came up. so I could leave Vall more or less parallel to its path. I did not plan that, it just worked out that way. In every other case (and I've mapped the Joolian moons a number of times), the polar orbit was more or less perpendicular to the starting moon's orbit when the transfer window came up, so burning from there would have sent me straight at or away from Jool, meaning I had to make a big plane change before departing.

I really have no clue how you'd get into the initial polar orbit and so arrange it that its plane would be pointing the right way when the transfer window came up. I suppose that if you waited long enough, it would eventually happen but it might be months or years later.

In any case, the starting moon's SOI is just as wide vertically as horizontally, so you have to get far from it quickly vertically when leaving from a polar orbit, just like you have to get away horizontally when leaving from an equatorial orbit. In this case, I left over Vall's north pole heading somewhat upwards. I intercepted Tylo under its south pole almost exactly a full orbit of Jool later, adjacent to where I'd left Vall.

[Duxwing]

Could well be. But OTOH, you can leave direct from a moon, too, without having to go back to Jool itself. This might change the overall calculation some, too.

Leaving from Laythe would make inside-out better because the greater kinetic energy at Pol exceeds the Oberth effect from Jool's gravity at Laythe, where aerobraking is free.

Strangely enough, just a couple days ago I managed move a SCANsat probe from Vall to Tylo starting from a polar orbit. This was only possible because, by sheer coincidence, my orbital plane was essentially tangent to Vall's orbital path at the moment the transfer window to Tylo came up. so I could leave Vall more or less parallel to its path. I did not plan that, it just worked out that way. In every other case (and I've mapped the Joolian moons a number of times), the polar orbit was more or less perpendicular to the starting moon's orbit when the transfer window came up, so burning from there would have sent me straight at or away from Jool, meaning I had to make a big plane change before departing.

I have returned to Kerbin from a sub-optimal Minmusian polar orbit by burning where it touched Minmus' orbital plane; perhaps this method works for Joolian transfers.

I really have no clue how you'd get into the initial polar orbit and so arrange it that its plane would be pointing the right way when the transfer window came up. I suppose that if you waited long enough, it would eventually happen but it might be months or years later.

Put your periapsis over a Laythian pole, lower your periapsis into Laythe's atmosphere, and hold on tight.

In any case, the starting moon's SOI is just as wide vertically as horizontally, so you have to get far from it quickly vertically when leaving from a polar orbit, just like you have to get away horizontally when leaving from an equatorial orbit. In this case, I left over Vall's north pole heading somewhat upwards. I intercepted Tylo under its south pole almost exactly a full orbit of Jool later, adjacent to where I'd left Vall.

The moon's SOI would move from rather than to you.

-Duxwing

Edited March 7, 2014 by Duxwing

[vidboi]

From\To	Laythe	        Vall	        Tylo	        Bop	        Pol
Laythe	5.55E+02	5.80E+02	6.37E+02	7.25E+02	7.65E+02
Vall	2.83E+02	2.37E+02	2.74E+02	3.88E+02	4.51E+02
Tylo	8.17E+02	7.84E+02	7.74E+02	7.88E+02	8.04E+02
Bop	4.81E+02	3.21E+02	1.68E+02	6.00E+01	8.93E+01
Pol	5.25E+02	3.92E+02	2.49E+02	7.51E+01	3.75E+01

Here's a table (sorry for the lack of formatting) of optimal hohmann transfer delta-v costs in the joolian system. Due to the reversibility of orbits, the delta-v cost to decelerate into orbit at the destination is just the same as the transfer burn going the other way, although of course with aerobreaking this can be greatly reduced. Hope it helps!

[Rusty6899]

I made this delta-V map of the system. Start at one moon and follow the line to another, the first number is the ejection burn from moon A, the second is the circularisation burn at B. So from Laythe to Vall it takes 590 m/s to encounter Vall and then 300 to circularise. From Vall to Laythe takes 300 for the ejection burn and 590 to circularise.

The numbers in the middle regard reaching the moons from Jool orbit, I'll go into it if prompted.

I'd also probably trust vidbol's numbers ahead of mine as it was my first attempt at calculating transfers of any sort (although happily they are fairly similar...ish).

[Geschosskopf]

Here's a table (sorry for the lack of formatting) of optimal hohmann transfer delta-v costs in the joolian system.

Is that optimum per the numbers or in actual practice? As I mentioned, due to the size of the SOIs compared to your ship's speed, a by-the-book optimum transfer isn't workable between Laythe and Vall, or Vall and Tylo, due to the sheer size of their SOIs compared to the speed you're moving away from them. It has to do with KSP's 2-body physics and SOIs.

The by-the-book optimum has you leaving the starting moon nearly parallel to its orbit, so it takes a while to get any horizontal separation. While you're still in its SOI, this is no problem because your motion is figured relative to that moon. But once you leave the moon's SOI, now both you and the moon are moving relative to Jool. Because your orbit is slightly wider at this point, the game treats you as moving slower than the moon relative to Jool, even though you were just moving faster than the moon while in its SOI. Thus, the moon catches up with you despite you just having achieved escape velocity from it. And because you haven't yet moved very far outwards or inwards towards the target moon, you'll be swept up by the big SOI and back where you started, only on a lithobraking trajectory heading in the opposite direction you just left from. Wonderful stuff, this KSP physics .

I've only found 2 ways to keep this from happening, both of which involve non-optimal transfers. The 1st method is wait a bit after the optimal transfer window, then crank your ejection angle backwards to give you more inwards or outwards velocity from the get-go so you can sidestep the SOI coming up from behind, while still hitting the target. This involves both a bigger transfer burn and a bigger capture burn. The other is to do the transfer burn from just barely inside the moon's SOI, so even a little bit of inwards or outwards velocity puts you out of reach as the moon goes by behind you. In this case, you use the normal transfer window and ejection angle so the capture burn is also normal. However, the transfer burn is bigger because you're sacrificing Oberth by burning way out there, and if you circularized out there to wait on the transfer window instead of being in a highly eccentric orbit, you waste even more fuel.

[Duxwing]

I may have solved the problem: a Hohmann transfer between any two bodies orbiting a third costs the absolute value of the difference between the costs of Hohmann transferring (->) from the origin (O) to the parent (P) and from the parent to the target (T):

O->T = |O->P - P->T|

This formula considers all Oberth Effects, the first occurring regardless of transfer method and the latter being cancelled by the dV cost of descending to the parent. The formula seems intuitive: the change in velocity necessary to move between two orbits should equal the difference in their velocities.

@vidboi Your dV values seem too large; one is over 8,000m/s.

-Duxwing

[Rusty6899]

@vidboi Your dV values seem too large; one is over 8,000m/s.

-Duxwing

The largest is 817m/s, isn't it?

[Duxwing]

The largest is 817m/s, isn't it?

Oops! I misread because scientific notation is only for values exceeding one thousand or exceeded by one.

-Duxwing