Optimal Munar Orbit Insertion Challenge

[closette]

Very nice PakledHostage. I\'ve been wrestling with the orbital equations off and on (hard to find some distraction-free time) but your experiments bear out a reasonable guess, that the optimum angle for an impulse to reduce periapsis is somewhere towards 'backwards and inwards'.

Periapsis distance is given by a(1-e) where semi-major axis a is related directly to orbital energy and eccentricity e is related less simply to both energy and angular momentum. For a hyperbola a is negative and e is greater than 1. The optimum angle you found reduces the magnitude of a and reduces e at the same time.

[Tarmenius]

Back in the Optimal Descent challenge, Majiir suggested that an insertion which planned for a bi-elliptic transfer to descent could yield improved results. I\'m pretty sure I\'ve been doing it wrong, so I wanted to ask you guys for input on the correct way. I doubt it\'ll be more efficient than what we\'ve already accomplished, but I want to try it for my own sense of thoroughness.

[PakledHostage]

I don\'t know much about bi-elliptic transfers either. I did read in the Wikipedia article on bi-elliptic transfers that:

While they require one more engine burn than a Hohmann transfer and generally requires a greater travel time, some bi-elliptic transfers require a lower amount of total delta-v than a Hohmann transfer when the ratio of final to initial semi-major axis is 11.94 or greater, depending on the intermediate semi-major axis chosen.

The reference given in the Wikipedia article for the above quote is: Vallado, David Anthony (2001). Fundamentals of Astrodynamics and Applications. Springer. p. 317. ISBN 0-7923-6903-3.

At 1102 km for our initial semi-major axis and 203 km for our final semi-major axis, the ratio of final to initial semi-major axis is only 5.44:1. Much less than the 11.94:1 quoted above. It is probably worth doing a test to be sure, but maybe we can\'t do much better than the current record?

[Tarmenius]

I read that article a while back and again recently, but even with the quite significant increase of orbital altitude in the example, there wasn\'t much savings to be had (16.18 m/s Delta-v). I just wonder how I\'d set up the Insertion to perform the manuever correctly. I was thinking that I\'d wait until Munar Pe to establish an orbit of 990km x 2,000km (approximately), then from Ap I\'d lower Pe to the target 3km and circularize from there. My gut says that wouldn\'t result in any savings, but if that\'s the correct concept for insertion and bi-elliptic transfer, I\'ll give it a shot.

[PakledHostage]

I just did some number crunching and it looks like the bi-elliptic transfer can be more efficient than a simple Hohmann transfer, but the case I tried is not as good as any of the results on the leader board.

My sample calculation predicts an efficiency that is roughly the same as burning normal to the velocity at 2150 km from the Mun to lower the hyperbolic periapsis to 3 km (my 'Test 2'). I imagine it would be hard to do much better because we\'re restricted to a maximum semi-major axis length by the Mun\'s SOI radius and a minimum semi-major axis length by the Mun\'s surface altitude.

Here\'re the numbers:

Bi-elliptic transfer insertion:

Speed at hyperbolic periapsis at 901 km = 463.6 m/s

Speed at periapsis of 901 km x 2138.5 km orbit = 283.6 m/s (requires -180.0 m/s delta-V)

speed at apoapsis of 901 km x 2138.5 km orbit = 133.5 m/s

speed at apoapsis of 3 km x 2138.5 km orbit = 66.7 m/s (requires - 66.8 m/s delta-V)

Delta-V required to close down orbit to 3 km x 3 km circular orbit at periapsis = -202.0 m/s

Total delta-V = 448.8 m/s

Corresponds to ~183.6 kg fuel remaining

Hohmann transfer insertion:

Speed at hyperbolic periapsis at 901 km = 463.6 m/s

Speed at apoapsis of 3 km x 901 km orbit = 135.7 m/s (requires -327.9 m/s delta-V)

Delta-V required to close down orbit to 3 km x 3 km circular orbit at periapsis = -169.6 m/s

Total delta-V = 497.5 m/s

Corresponds to ~176.7 kg fuel remaining

[Tarmenius]

...not as good as any of the results on the leader board.

Well, those were achieved by lowering Munar Pe right from the scenario\'s start so no insertions that begin later will be quite as efficient. That aside, your results show a greater difference between the bi-elliptic and Hohmann transfers than I would have thought. Excellent work as usual!

[closette]

PakledHostage, your delta-V numbers check out just fine. (I used the KSP Orbit Mechanic but it still took ~ 10 mins to check \'em). Nicely presented too.

The Hohmann vs. bielliptic efficiency quoted is for circular to circular orbits, as we had in the Munar Descent challenge, where I believe the Hohmann method is still the most efficient since a 100 km altitude to ~ 0km altitude transfer has a radius ratio (after adding in the Mun\'s 200km radius) of only 300 km/ 200 km =1.5, clearly favoring Hohmann\'s 2-impulse method.

For hyperbolic -> circular transfers though, PakledHostage has shown that a bi-elliptic 3-impulse maneuver is slightly better. I think this might be because a hyperbolic orbit close to periapsis looks only slightly flatter than a really, really big ellipse. I\'m sure this is a 'known' result but it\'s good to see here, and applicable to every Mun mission.

On my challenge attempts, when I reached the hyperbola\'s periapsis, I have found it difficult to get the apoapsis down to even 40km in a single impulse. Now I can try it in 2 stages. It will also go faster because of time-warp at higher altitudes.

[Tarmenius]

Well, since we\'ve shown that the ideal efficiency really depends on the mission scenario, it\'s time I looked at the bigger picture. I\'ll be taking what we\'ve learned here and running a few different Mun missions starting from a 100km Kerbin orbit in a lander capable of performing its own TMI burn. Hopefully, this will end up illustrating the struggle between mission parameters and efficient flightplans and how to get the most from each type of scenario.

[closette]

@PakledHostage:

I imagine it would be hard to do much better because we\'re restricted to a maximum semi-major axis length by the Mun\'s SOI radius and a minimum semi-major axis length by the Mun\'s surface altitude...

Bi-elliptic transfer insertion:

Speed at hyperbolic periapsis at 901 km = 463.6 m/s

Speed at periapsis of 901 km x 2138.5 km orbit = 283.6 m/s (requires -180.0 m/s delta-V)

speed at apoapsis of 901 km x 2138.5 km orbit = 133.5 m/s

speed at apoapsis of 3 km x 2138.5 km orbit = 66.7 m/s (requires - 66.8 m/s delta-V)

Delta-V required to close down orbit to 3 km x 3 km circular orbit at periapsis = -202.0 m/s

Total delta-V = 448.8 m/s

Corresponds to ~183.6 kg fuel remaining

This result was bothering me for a while, since the large intermediate orbit seemed arbitrary, but just like a bi-elliptic transfer, the larger the better. For a very large intermediate ellipse you can do this transfer with a delta-V of only 354 m/s, except you can\'t because it would go outside the Mun\'s SOI. So for an ellipse which just fits inside the Mun\'s SOI (apoapsis at 2400 km) I calculated that you can get the delta-V down to 441 m/s if you wait long enough. I think that will save you an additional ~ 1 kerbalgram of fuel over PakledHostage\'s chosen transfer orbit.

It may still be more efficient to lower the hyperbola\'s periapsis and periapsis speed from afar, with a back-and-inwards impulse as you carefully plotted in Reply 48.

I also learned that apparently, for a 1-burn capture into a circular orbit from a hyperbola, the most efficient capture occurs when r_pv_inf²/GM equals exactly 2. I\'m waiting for my library to get an orbital mechanics book which I hope has the proof of this.

In the meantime I\'ve been avoiding real work by tracking down papers such as the two attached. Figures 2 through 4 in each paper look very much like what we (sometimes) do on a Munar approach. Having finite thrust means that sometimes it takes a few orbits to circularize. Enjoy.

[PakledHostage]

This result was bothering me for a while, since the large intermediate orbit seemed arbitrary, but just like a bi-elliptic transfer, the larger the better.

You caught me. It was totally arbitrary... I used the last value that was still in my spreadsheet. I was too lazy to change it because I judged that it was sufficiently close to the Mun\'s SOI boundary that it wouldn\'t make much difference anyway.

A further savings of one kerbalgram gets us closer to the easiest to accomplish (given our limited instruments) 'burn towards 90 degrees at 0 degrees elevation on the navball' method, but it still doesn\'t match even that less than optimal result.

Maybe if the Mun was further from Keribn (so the SOI was bigger), we\'d do better with the bi-elliptic method than the 'burn towards 90 degrees at 0 degrees elevation' method. Maybe we should do a sample calculation for a similar approach to Minmus? I wonder if the results would be different there due to Minmus\' larger SOI? Unfortunately I won\'t have time to do it myself though. I\'m away on vacation for 2 weeks. I\'m probably going to miss all the fun of the v0.16 release. Work out the bugs for me, will you guys?

[Tarmenius]

More great resources, closette! It\'s always neat to see how our messing around in this game tends to resemble real-world methods.

I haven\'t had much chance to spend the time required to run the Kerbin-Mun transfers, but the next couple days will be pretty free so I\'m sure I\'ll have some good results coming soon. Also, as PakledHostage mentioned, a comparison to Minmus would be quite interesting. If anyone wants it, I can make a setup for Minmus similar to what I\'ve done for this challenge. It\'s not much trouble, and it\'ll give me another interesting thing to do while I wait for 0.16

[closette]

Well I realize that everyone has moved on to 0.16 which doesn\'t work with the parts provided, but I fired up my precious copy of 0.15 and with a more carefully tuned strategy achieved:

1. 202.92 kg remaining to get into a 3.2x3.8km orbit (first screenshot attached, with the MunArch in the background).

This time, after lowering Pe while far away, I was more careful in changing from hyperbolic to circular orbit. I took two orbits to do it so that I could have engines firing as close to periapsis as possible. This time I also took care to start the burn slightly before periapsis such that the periapsis point would always remain just in front of the spacecraft. If it advanced too far in front, I would stop firing briefly and coast a little.

2. Landed with 115.0 kg remaining (second screenshot attached). This would make a great separate challenge. I did some research on terminal descent onto the Moon from lunar orbit - plenty of papers but very few show the actual throttle profiles. But rather than 'stop, then drop', they seem to employ a 100% throttle braking burn to kill off most but not all of horizontal speed, then a slowly decreasing throttle for a gravity turn, possibly going to zero, followed by a brief throttle-up burst just before the vertical landing. This is what I did, and despite a 'bounce' on landing I got away with an intact spacecraft.

I learned a lot from this challenge - looking forward to similar ones in 0.16 once the LFT fuel bug is fixed.