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Optimal Munar Orbit Insertion Challenge


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Re-reading your post from earlier this evening, Tarmenius, I realised that my previous result actually isn\'t that interesting. At least not in the way that I thought it was.

I wrote earlier that I thought I could start the 'turn' method prior to entering the Munar SOI, but I was wrong. I did some tests and found that something else entirely works best while still outside the Mun\'s SOI.

I was able to duplicate your fuel burns for the Pe lowering phase by burning retrograde and towards the navball’s nadir immediately after the scenario starts. I also tried burning in a few other directions. I recorded the results and plotted them below:

ZNSop.png

Clearly, burning close to retrograde yields the best results. In some ways this makes sense. Burning retrograde lowers the apoapsis. It also has a slight effect on the periapsis and the angular position of the orbit\'s major axis when, as in this case, it is done at something other than Pe or Ap. The effect isn\'t significant though for the small delta-Vs that we\'re talking about.

If I had to guess what is going on, I would predict that we\'re 'undoing' the excessive TMI burn and lowering the Munar Pe back to 3 km. Had we done the TMI burn more carefully (i.e. targeting our MECO such that the Munar Pe was at 3 km at the end of the TMI burn), it would have taken less fuel than it took to yield the 900 km Munar Pe that this scenario starts with. And had we realised the mistake after MECO and dealt with it then, it would have taken less than the ~30 m/s delta-v to correct than it took at this late stage in the transfer. I would further guess that this is a consequence of the Oberth effect.

But of course in this case, the Munar periapsis was intentionally set at 900 km to yeild a free return with survivable re-entry. In case, say, there was an explosion during the cryo-stir proceedure. Or more likely, because the flight control computer locked up and wouldn\'t reset itself because nobody bothered to configure the watchdog timer...

Also, if you need a spreadsheet program, you could try OpenOffice. I\'ve never tried it but I know people who use it. It is free and seems to have a pretty good reputation.

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202.57 kerbalgrams remaining on board for a 3.37 x 3.43 km orbit, which was a pain to establish and took a long time to circularize at maximum x2 time-warp. But patience paid off.

Same strategy as others - retrograde before entering the Mun\'s SOI, with ~ 10km periapsis, then after entering SOI a short impulse towards Kerbin to reduce periapsis to about 6km, then lots of 'apse-chasing' at low time warp to finalize the orbit.

After all that time I missed a nice shot of the Mun arch as I flew over it, and I messed up an attempted return to Kerbin, almost making it but killing the crew with a 23 m/s impact. Totally feasible for others to succeed from Mun orbit though.

By the way I noticed that for this almost perfect circular orbit, rotating the spacecraft or even toggling ASAS on/off can change the peri- and apo-apsis by ~ 10 meters.

P.S. @PakledHostage - very nice results and must have taken you a while to do so many trials. Your explanation is reasonable too.

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Nice results, Kosmo-not and closette. 0.2kg separating the two of you. Very exciting!

PakledHostage, you made me Google 'nadir.' Good work. :D That graph shows an interesting range of deviation from Retrograde where fuel consumption is hardly affected. Also, it appears you can almost tell what direction Kerbin was in by the shape of the curve. And thanks for recommending OpenOffice. I\'ll have to dust off my spreadsheet skillz.

After I retry the Low-Fuel Lander, I\'m planning on trying different bi-elliptic transfers from the original Mun Pe of 900-something km. It\'ll be intersting to see how those stack up.

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Low-fuel lander: 99.78 kerbalgram (kg) remaining (from the quicksave file, I forgot to bring the borrowed 2-button mouse to school), I wasted a lot of fuel trying to get the 3.4x3.5km orbit within range.

Landed with 14.48 kg remaining but toppled over and disconnected pod - typical for me these days!

Why so few contenders in this challenge? It\'s very relevant to everyday game play, and just trying without winning is fun and makes one a better navigator overall.

Perhaps changing the title to 'Optimal Munar orbit insertion challenge [stock]' might help?

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Good work closette. You\'ve done better than I have with both Landers. My best so far with the Low-Fuel version is 86.2kg. And I still can\'t manage to get it to the ground with any fuel left.

Thanks for the suggestion to change the challenge\'s name. I\'d been wondering about why there are so few entries myself. I had been chalking it up to low forum activity overall, but you may be onto something with the title.

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Why so few contenders in this challenge? It\'s very relevant to everyday game play, and just trying without winning is fun and makes one a better navigator overall.

Rest assured I\'ve been following this thread with great interest, and I do intend to do this challenge soon. The launch has been delayed several times do to work schedule awkwardness and such.

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Zephram, I knew you\'d be doing this challenge sooner or later. BlazingAngel665 as well, I\'m sure. My guess is that with 0.16 getting closer, others are reluctant to start a challenge that may need to be re-made in a couple weeks. Myself, I\'m looking forward to seeing how the new parts and balancing affect this and the other two related challenges. I plan to revisit them once I get a feel for the changes.

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202.7 kerbalgrams (found the mouse!) and a nice shot of the Arch. Someone should try landing there from this challenge.

I\'m pretty much doing the same thing each attempt, spending most of my time in low orbit waiting to reach one of the apses for a circularization burn. I\'m going to try some other strategies to see if there is another way to save more fuel and blow everyone else away.

Landed intact for once at 108.3 kg remaining. Should be possible to do much better though...

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Here\'s my latest entry: 202.9 kerbalgrams fuel remaining upon achieving a 2.95 km x 3.45 km orbit.

delta-V used to lower periapsis to 3 km: 29.9 m/s

delta-V used during Munar capture burn: 284.5 m/s

Total delta-V used: 314.4 m/s

UBSKD.png

Dz4IO.png

When I next get a chance to play, I will leave the spacecraft in the free return trajectory until entering the Mun\'s SOI. I will then compare a couple of different orbital insertions starting from near the Mun\'s SOI boundary. After all, if the free-return trajectory was used to ensure that our brave Kerbals could still make it home in the event of an inflight problem, then it doesn\'t make sense to divert from that free-return trajectory early in the flight. I am most curious if the 'turn' method or the retro-burn method works better once we\'ve crossed into the Mun\'s SOI.

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I am most curious if the 'turn' method or the retro-burn method works better once we\'ve crossed into the Mun\'s SOI.

...or some combination thereof! From your systematic investigations so far, I am confident you\'ll pin down the best strategy. Nice job on nudging into the lead. Calculating the theoretical best-case impulsive delta-V is a bit harder for this challenge as it crosses an SOI boundary but I\'ll have a go. Your step-by-step delta-V breakdowns are very helpful.

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Congratulations, PakledHostage! It\'s incredibly close between the three of you. I don\'t think I\'ve seen such a narrow margin between the top spots in a challenge like this. I\'m very impressed.

When I did my comparison of the two methods just inside the Mun\'s SOI, I found that I used less fuel when burning directly toward Kerbin as opposed to retrograde. I also tried a couple bi-elliptic transfers, but those ended up being less efficient than my posted results. Unless I didn\'t perform them correctly. I\'ll have to do some more and actually record some data so we can have something tangible to compare.

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Soft landing, standard lander: 111.9kg remaining. (Still not as good fuel-wise as my 'hard landing' with 116.1 kg, Reply 18).

I couldn\'t beat PakledHostage or Kosmo-not on the 3x3km orbit part although I tried a new strategy - making my initial burn not completely retrograde, but trying to 'turn inwards' to reduce the patched-conic-projected Pe to under 5km as efficiently and as quickly as possible. Perhaps I did not do it right as I ended up in an inclined orbit around the Mun. I ended up with 202.3kg remaining in a 3x6km orbit so gave up on that part of the challenge.

BUT from there with a better gravity turn profile than I\'ve used before, I was able to soft land with 111.9kg left in the tank.

I\'m beginning to like the multi-part challenge with more-than-1 leaderboard. If I miss one part I can still continue with the other. (Tarmenius, I know it means more work for you to keep track of it all).

Also Tarmenius, you are not being fair to yourself. Your 4th test in Reply 19 shows you made orbit with 202.6kg remaining - that should be up there on the leaderboard! (You also have PakledHostage on there twice).

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Nice job on the landing, closette. I wonder how far you\'d be able to get from the surface with that 111.9kg...

...you made orbit with 202.6kg remaining - that should be up there on the leaderboard! (You also have PakledHostage on there twice).

That\'s what I get for updating at 2 in the morning :D Also, when I posted those tests I hadn\'t planned on putting the results on the Leaderboard because I hadn\'t taken screenshots. But I suppose that since I\'ve done it for others, I should allow it for myself as well. It\'s just too bad the game froze before I could achieve the estimated 204.1kg from the third. Anyway, thanks for pointing those out.

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With 111.9kg I was able to return to Kerbin\'s upper atmosphere, and aerobrake 4-5 times around before going in, but I did not land safely - I think it was an 80 m/s impact after running out of fuel at ~200m altitude. But it seemed do-able with some skill, luck and patience.

You belong on that leaderboard. I just wish there were more new names doing this challenge. Despite the closely packed leaderboard, I have a hunch that we haven\'t found the optimum entry path yet...

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I ran three tests from within the Mun\'s SOI this afternoon. None of them improves on the current records, but I\'m sharing them because they are potentially interesting:

Test 1:

- Burn retrograde at 2150 km altitude above Mun.

- 209.2 kg fuel remaining after Pe lowered to 3.2 km. (delta-V expended = 271.3 m/s)

- 179.1 kg fuel remaining after closing down orbit to 2.8 km x 3.2 km (delta-V expended = 208.8 m/s)

- Net delta-V expended 480.1 m/s

Test 2:

- Burn normal to spacecraft\'s velocity (but in the Mun\'s orbital plane) at 2150 km altitude above Mun.

- 224.5 kg fuel remaining after Pe lowered to 3.3 km. (delta-V expended = 168.0 m/s)

- 179.8 kg fuel remaining after closing down orbit to 3.0 km x 3.4 km (delta-V expended = 307.2 m/s)

- Net delta-V expended = 475.2 m/s

Note: I am not using MechJeb so I didn\'t do a perfect job of holding a burn heading normal to the spacecraft\'s velocity. This may have affected the results of this test.

Test 3:

- Burn towards Mun\'s centre of gravity, in the plane normal to the view direction towards the Mun (i.e. towards 90 degrees on the navball) at 2150 km altitude above the Mun.

- 227.9 kg fuel remaining after Pe lowered to 3.6 km. (delta-V expended = 145.3 m/s)

- 188.4 kg fuel remaining after closing down orbit to 3.3 km x 3.5 km (delta-V expended = 269.5 m/s)

- Net delta-V expended = 414.8 m/s

I would summarise my observations from this and the other challenges that we\'ve all participated in as follows:

An efficient transfer from Kerbin to Munar orbit and landing starts with a well planned TMI burn. If contingency for in-flight failure includes initial insertion into a free-return trajectory, then that free-return trajectory should have as low a Munar Pe as possible.

If the Munar orbit insertion requires lowering the Munar Pe, then the burn to lower the Pe should be made towards the Mun\'s centre of mass, in the plane normal to the current view direction towards the Mun (i.e. heading of 90 degrees on the navball)1. The earlier this burn is made after entry into the Mun\'s SOI, the less fuel will be required. The Munar orbit insertion burn should occur as late as possible however, as this maximises the window during which a failure could occur and still allow the mission to take advantage of the Munar flyby abort trajectory. Due to the conflicting pressures of minimising fuel burn and maximising safety contingencies, a balance must be found. Presumably the MOI burn will occur within the Mun\'s SOI.

The Munar orbit insertion burn should then occur as close as possible to Pe to maximise the benefits of the Oberth effect. Orbital trim manoeuvres should then be carried out as required to circularise the Munar orbit.

The DOI burn should be done 180 degrees from the planned landing site, and Pe of the descent trajectory should be as close to the surface as is safe. Final approach to landing should be done using a reverse gravity turn.

Any comments about or additions to the above?

1 I realise it is a bit of a stretch to draw this conclusion based on just three tests, but intuitively it makes sense that this is the most efficient means of lowering the hyperbolic trajectory\'s Pe. Maybe I\'ll try to find some time to do the analysis to verify my intuition. But for the time being, the experimental evidence suggests that it is an efficient method.

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Very interesting, indeed. I hadn\'t considered burning directly toward the Mun\'s center of gravity. I still haven\'t been able to find a more efficient method (not sure I\'m doing the bi-elliptic transfers correctly though), so I\'ll give that one a try and see how it goes for me.

Your summary sounds right and makes me wonder just how big of a difference the right TMI burn would make. I see togfox just posted, and I would also be interested in diagrams :)

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202.8 kg remaining for the standard lander, 3.9x3.9km orbit, despite numerous attempts to get 203 kg. I\'m level with Kosmo-not! (Not for long I\'m sure).

Landed with 109.1kg remaining, and what is that in the landed screenshot? The MunArch!

@PakledHostage:

burn to lower the Pe should be made towards the Mun\'s centre of mass, in the plane normal to the current view direction towards the Mun (i.e. heading of 90 degrees on the navball)

I think I know what you mean, but for newcomers a rephrasing might help clarify (as well as a diagram). I think you mean 'in the plane of the hyperbola, turn towards the Mun'.

My hunch is that in fact it would be better to turn more - 90 degrees to your velocity, not towards the Mun, since that would reduce periapsis by reducing both energy and angular momentum, and burning radially does not change the latter. I have a scribbled list of orbital equations that I need to work through though when I can get some quiet time...

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Having thought about it a bit more since yesterday’s post, I think that there’s some further testing that could be done to follow up on my three tests.

First of all, Test 2 could be done again using MechJeb or by someone who’s a better pilot than me to determine if the difference between Test 2’s results and Test 3’s results are due to experimental error or are a real effect. And even if it isn’t a real effect, then it would be interesting to compare the two methods accurately using MechJeb. Maybe the method used in Test 3 is just more practical in that it is easier to achieve using the instruments that we’ve got available?

Some more tests could also be done, burning at a range of angles in the plane of the hyperbolic trajectory, between retrograde and normal to the spacecraft’s velocity. If we had data from that range of tests, then we could plot it like I did in my post from last week. Maybe there’s an optimum point that hasn’t been found yet? I\'ll try to fit in a bit of experimenting.

And finally, I agree that my description of Test 3’s burn direction wasn’t very good. I struggled to come up with something that wasn’t too wordy. I ended up leaving out a couple of critical ones. I should have just posted a picture:

1Y18g.png

Better late than never… I can sketch up a diagram too if people think it would help.

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So I went ahead and made a persistent.sfs using the Standard Lander with MechJeb onboard. It\'s also on a free-return starting at the MCC1 position. I\'m not going to add it to the original post, though since it\'ll probably only be used by the group of us who are following this thread anyway. :)

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I ran four more tests, all of them burning towards 90 degrees (like in Tests 1, 2 and 3 and as shown in the image above), except I varied the angle of the heading reticle in the navball from 65 degrees above the horizon (retrograde) to 30 degrees below the horizon (roughly normal to the inital velocity vector). I\'ve plotted the data below:

gT6gq.png

Test 4:

- Burn toward 30 degrees above horizon starting at 2150 km altitude above Mun.

- 224.2 kg fuel remaining after Pe lowered to 3.6 km.

- 189.0 kg fuel remaining after closing down orbit to 2.9 km x 3.4 km

- Net delta-V expended 410.6 m/s

Test 5:

- Burn toward 15 degrees above horizon starting at 2150 km altitude above Mun.

- 227.2 kg fuel remaining after Pe lowered to 3.5 km.

- 189.5 kg fuel remaining after closing down orbit to 2.9 km x 3.1 km

- Net delta-V expended 407.1 m/s

Test 6:

- Burn toward 15 degrees below horizon starting at 2150 km altitude above Mun.

- 227.3 kg fuel remaining after Pe lowered to 3.6 km.

- 185.4 kg fuel remaining after closing down orbit to 3.2 km x 3.3 km

- Net delta-V expended 435.8 m/s

Test 7:

- Burn toward 30 degrees below horizon starting at 2150 km altitude above Mun.

- 225.2 kg fuel remaining after Pe lowered to 3.4 km.

- 180.5 kg fuel remaining after closing down orbit to 2.9 km x 3.3 km

- Net delta-V expended 470.2 m/s

Note that burning normal to the spacecraft\'s velocity results in a burn that starts at about 35 degrees below the horizon and ends at about 20 degrees below the horizon. The variation during the burn is as a result of the spacecraft turning.

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Another awesome job, PakledHostage. It\'s curious that burning 15o above the horizon would yield the most efficient results. Is it because that burn is doing two jobs at once (being just about inbetween Retrograde and Normal)?

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