PakledHostage

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 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.

Maybe you can help by fixing the inaccuracies in your post? Orbital speed at 15 billion metres altitude above Kerbol is 8821 m/s, not 30000 m/s. I\'ve flown numerous missions outside Kerbin\'s SOI and I\'ve only encountered the Kraken once. I was able to regain control of my spacecraft in that case by turning off the SAS and hand flying. In addition to SunJumper\'s example above, check out the following threads. I think they illustrate quite well that it isn\'t 'nearly impossible' to keep control of your ship in Kerbol orbit: RedDwarfIV\'s \'The Adventures of Solar Observer\' Togfox\'s \'Recording Solar Flares\' challenge BoolyBooly\'s \'Shoot for the Sun!\' challenge

Check out the Mun/Minmus Gravity Slingshot challenge. I submitted two entries for that challenge. The first (faster) trip was a learning experience. The second time, I managed to do what you\'re asking about using only 70 m/s delta-V. I tried to be specific about how I did it so that others could replicate my results if they wanted to try. Hopefully there\'s enough there to answer your question.

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: 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.

Well, coincidentally, I just completed this challenge using a 3.3 km x 3.6 km pre-landing orbit. I don\'t know if that is sufficiently within tolerances to meet the new objectives but I\'ll submit it anyway. Regardless of whether or not it counts, I think my results are interesting. Immediately after starting the scenario, I used my spacecraft and the navball to measure the angle between the Mun and Kerbin. I then made my burn to lower my Munar periapsis by burning while oriented 1/2 way between the two (roughly 25 degrees elevation in the navball\'s brown hemisphere, on a heading of 270 degrees). It cost 6 kg fuel (39.1 m/s delta-V) to lower my periapsis to 4 km. I then waited until reaching periapsis to close down my orbit. I pre-calculated the orbital speed required for the 4 km orbit and burned at full throttle until reaching that speed. My imprecise flying resulted in a 3.3 km by 3.6 km orbit about the Mun. Once inserted into that orbit, I had 200.5kg fuel remaining (net delta-V expended to this point: 330.9 m/s. This includes the 39.1 m/s expended to lower the Pe). From that orbit, I made my DOI burn then coasted along until I came across some suitably high terrain. I had never seen the Mun arch before, so I was quite surprised to see it coming over the horizon. I quickly pressed F5 to save my position then proceeded to violate my own prohibition on 'do-overs' to make my landing with minimal fuel burn, somewhere near the arch. I landed with 114.2 kg fuel remaining after several attempts (net delta-V expended 958.3 m/s). The arch is in the background of the screenshot. I will leave it up to the judges to decide whether to count this entry or not.

Interesting. What happens if you make the 'turn' immediately after starting the scenario? I haven\'t tried it myself, but intuitively you shouldn\'t have to wait until you\'ve crossed the Mun\'s SOI to start the turn. Also, what happens if we just insert into a 3 km x 3 km circular munar orbit rather than landing? For both capture techniques, you need to pass through a 3 km x 3 km circular orbit in order to land. The fuel burn from there to the surface would be the same for both methods.

This sounds interesting... Are you saying that burning retro-grade to lower Pe to 5 km and then landing required roughly the same Delta-V as landing after lowering the Pe to 5 km using the 'turn' method? I haven\'t had a chance to make an attempt at this challenge myself yet, but I\'ll have to make this comparison myself. If it is done correctly, the turn method shouldn\'t add any orbital energy in the same way that turning the handle bars on your bike doesn\'t add energy. Sure you\'ll reach Pe at higher speed, but that is because you have converted potential energy into kinetic energy. The overall orbital energy is the same. Kosmo-not and I discussed this back in April. If the two methods result in roughly the same fuel burn requirements, then I would assume that the only benefit to the 'turn' method would be that it allows us to take greater advantage of the Oberth effect at Pe. I guess the question then would be whether or not the improved efficiency of the higher velocity burn is enough to offset the delta-V 'wasted' on turning (rather than reducing orbital energy). That, in turn, would depend (I\'m sure) on the geometry of the initial Munar transfer orbit...

Nice follow-on challenge to your last one. I will have a go at this when I next get a chance to play the game. It might be a few days, though. I seem to manage to find more time for theorising about rocket science than I do for playing KSP… Probably because my mind drifts to it when I’m bored of doing what I should be doing... (A guy can only take so much of “it is broken, can you make it go?â€)

I can\'t speak to the relative efficiency of my method because I haven\'t compared it analytically with other methods but I would bump my periapsis down to 10 km as soon as I cross the Mun\'s SOI by burning normal to the spacecraft\'s motion but within the plane of the hyperbolic orbit. I would then wait until reaching periapsis to close down my orbit because that would take maximum advantage of the Oberth effect. Intuitively, it makes sense that this is an efficient method. Imagine approaching the Mun from within the Mun\'s orbital plane. Now imagine a 'gate' located on the circumference of the Mun\'s orbital radius at some distance from the Mun\'s centre of mass. Any hyperbolic trajectory originating at the spacecraft\'s current position with a given orbital energy is uniquely defined by the position of such a 'gate'. The hyperbolic Pe distance can therefore be adjusted by 'turning' to pass through one of these uniquely defined 'gates'. Gates that are farther from the Mun\'s centre of mass result in higher Pe, while gates nearer to the Mun\'s centre of mass result in a lower Pe. As we all know, 'turning' is done most efficiently by burning perpendicular to the spacecraft\'s velocity. At large distances, burning in the direction of the Mun\'s orbital motion is roughly perpendicular to the spacecraft\'s trajectory while still in the hyperbolic orbital plane. Also, at large distances, the amount of 'turning' required (and consequently the delta-V required) to change the gate position is very small. Therefore, burning E or W as required to raise or lower the Pe imediately after crossing the Mun\'s SOI, when approaching the Mun from within the Mun\'s orbital plane, would seem to be an efficient method of adjusting the approach trajectory. The orbit could then be closed down to a circular orbit at the hyperbolic Pe to take maximum advantage of the Oberth effect. This is also the method that I would use for trimming my orbital plane during inital approach. For example, if I entered the Mun\'s SOI in the equatorial plane with some random hyperbolic Pe but I wanted a polar orbit, I\'d bump the Pe down to -200 km (i.e. in line with the Mun\'s centre of mass) then orient N or S (as required for the intended orbital direction) and bump the Pe back up to something above the Mun\'s north or south pole. I\'d then close down my orbit upon reaching the hyperbolic Pe. And Tarmenius, I think it would be most objective to start from low Kerbin orbit and fly the whole transfer. As I showed in my comparison between Option 1 and Option 2 in my post above, it is possible to yeild very different munar periapses with very small differences in TMI burn lengths. Expending fuel to adjust one trajectory into another would bias the results in favour of the initial trajectory. That may not be practical though, so feel free to set it up however you feel like. As Closette said, you did an excellent job setting this challeng up. If you want to set up a free-return trajectory for your next challenge, you might want to reference my entry to the free-return trajectory challenge back in January. You can find the details here: http://kerbalspaceprogram.com/forum/index.php?topic=4449.msg70498#msg70498 and here: http://kerbalspaceprogram.com/forum/index.php?topic=4449.msg57476#msg57476

Obviously there are compromises to be made when choosing an orbital altitude for Munar capture, approach and landing. I certainly haven\'t considered all the factors, but I imagine that objectives like minimising communications blackouts (requiring a higher orbit) and minimising fuel burn during the descent to and ascent from the surface (requiring a lower initial orbit) would be key. The ability to choose a suitable landing site and the relative safety of the approach and landing would also be extremely important. Initiating the descent to landing directly from a low Pe flyby trajectory isn\'t very practical, but the analysis of this option seems to illustrate that inserting into a lower initial orbit about the Mun is significantly more fuel efficient than inserting into a higher orbit. Looking at Figure 1 in the article that Tarmenius linked to, a ~60 NM high lunar orbit was used in the Apollo program. When scaled down into the Kerbalverse, that\'s equivalent to a 12-13 km orbital altitude about the Mun. That\'s right in line with your method, Closette. And since setting up for insertion into a non-equatorial orbital plane can be done very economically during the TMI burn, I don\'t think plane change manoeuvres are a critical consideration unless the mission requires them for some reason. It will be interesting to revisit this challenge when docking is finally added to the game. Hopefully it will be possible to build some low delta-V lander and ascent stages, such that efficient trajectory design is actually critical to successful completion of the mission.

You\'re right, but much of the analysis in this thread makes the same assumption. The assumption allows calculation of a rough estimate of the minimum fuel burn required for a given landing trajectory. Interestingly, the top placed entries on the leader board are achieving fuel burns close to those predicted by calculations that use this assumption.

You could also achieve non-trivial fuel savings by bypassing the orbital insertion and making your landing directly from a hyperbolic fly-by trajectory. I did some 'back of the envelope calculations' using patched conics. It is an idealised calculation which assumes that the Mun\'s surface is perfectly spherical with an altitude of 0 m. Also, all burns are assumed to be impulsive except the TMI burn: Starting with the challenge stack in a 100 km altitude circular orbit about Kerbin Option 1: - burn to 3084.1 m/s at 100% throttle. Yields a 100 km munar periapsis at 720.5 m/s (relative to Mun\'s centre of mass). - retro burn to 466.0 m/s for insertion into 100 km munar orbit (-254.5 m/s). - retro burn to 417.4 m/s for DOI (-48.6 m/s). Yields a 1 km x 100 km descent orbit. - retro burn to 0 m/s relative to munar surface at 1 km Pe (-614.0 m/s). - freefall from 1 km altitude to 0 km altitude, followed by impulsive burn to arrest the descent (-57.0 m/s) Net delta-V (post TMI burn) = 974.1 m/s Option 2: - burn to 3082.8 m/s at 100% throttle. Yields a 1 km munar periapsis at 853.0 m/s (relative to munar surface). - retro burn to 0 m/s relative to munar surface at 1 km Pe (-853.0 m/s). - freefall from 1 km altitude to 0 km altitude, followed by impulsive burn to arrest the descent (-57.0 m/s) Net delta-V (post TMI burn) = 910.0 m/s That\'s an improvement of 64.1 m/s for the case where landing is initiated directly from hyperbolic flyby trajectory (65.4 m/s if you include the difference in delta-v between the two TMI burns).

I couldn\'t agree more. Just imagine if some of your favourite games of the past were designed so that everybody could achieve all of the objectives within the first couple of times that they played? How many times would you have played Super Mario if you successfully rescued the princess the third time you tried? A new planet would be a nice progression to the game\'s challenge... You start by trying to make orbit, get better at that then fly to the Mun. Minmus is a little bit harder to get to because of the orbital inclination but you soon master that too. Getting to orbit around another planet is the next challenge. Landing on the new planet and returning to Kerbin is an even greater challenge than that, but the progression keeps you hooked. And finally, I do get a bit frustrated to constantly read that it would be impossible to get to a new planet because of the Kraken or using only the current stock parts. Tim Barret is right. You will be able to get into orbit around any moderate sized new planet using a comparable rocket to what you currently use to get to the Mun. Add a LFT to that and you\'ll be able to get back to Kerbin too. For example, the stack in the image below is enough to get from low Kerbin orbit, down to a 7 million km circular orbit about Kerbol, and back to Kerbin. The same stack could reach orbit around a planet located at 7 million km from Kerbol and get back to Kerbin. In fact, if there was a Kerbin sized planet located at 7 million km, the round trip into orbit around that planet and back to Kerbin could be done with 1/2 a tank less fuel. Sure landing on the surface and returning to Kerbin will take a much larger rocket than just getting to orbit, but someone will be able to pull it off. Check out BoolyBooly\'s Shoot for the Sun challenge and Togfox\'s Recording solar flares challenge for examples of rockets that people have managed to return to Kerbin from Kerbol orbit. It is possible.

Well, here\'s what I would do. (I haven\'t tried it yet but there\'s no reason it shouldn\'t work): An elliptical transfer from 50 km above the Mun to 10 km above Minmus (Pe = 11650 km altitude above Kerbin\'s surface, Ap = 46470 km altitude) takes 75.023 hours. Minmus moves 90.246 degrees in its own orbital plane in that time. As Minmus\' orbit is only inclined at 6 degrees from the Mun\'s, and given the accuracy to which we know the Mun and Minmus\' position at any time, it is reasonable to neglect Minmus\' orbital inclination when calculating its angular displacement from the Mun1. For this transfer orbit, a window occurs every time the Mun is located 89.8 degrees behind Minmus. This first occurs at 1d, 17hr, 51 minutes UT. It occurs again every 44.324 hours after that. In reality, you\'d have a hard time flying that transfer because of the nature of the required Munar escape trajectory (it needs to be elliptical rather than parabolic or hyperbolic). A more realistic transfer is a 11850 km x 46550 km orbit, which would take 75.559 hours from Pe to Ap. Actual transfer time from transfer orbit insertion burn to Minmus intercept is closer to 79 hours due to the time it takes to escape the Mun\'s SOI. The extra transfer time has a significant effect on the timing of the window. To insert into that transfer orbit from a 50 km circular, prograde, equatorial Munar orbit, our Kerbals would have to accelerate to 695.5 m/s starting the instant that the last bit of Kerbin\'s upper limb sets below the Mun\'s horizon, as seen from the 50 km high circular orbit (all described manoeuvers should occur within the Mun\'s equatorial plane). That\'s a delta-V of only 185.1 m/s and results in an elliptical orbit with Ap outside the Mun\'s SOI (Ap = 3050 km). If they flew the transfer orbit insertion burn accurately, then our Kerbals should find themselves in a 11850 km x 46550 km orbit about Kerbin after leaving the Mun’s SOI. That transfer orbit should intercept Minmus with only minor orbital trim manoeuvres. As for intercepting Minmus near the ascending or descending node, you\'d have to wait a long time before a launch window results in a Minmus intercept near one of the nodes. The earliest such option that results in Minmus intercept within 1000 km of the Mun\'s orbital plane occurs at 12d, 19h, 53 min UT (intercept ~850 km below Mun\'s orbital plane). The next one after that occurs at 31d, 7hr, 7min UT (intercept ~275 km above Mun\'s orbital plane). Fortunately though, it doesn\'t take much Delta-V to adjust your orbital plane enroute to Minmus if you start your transfer at one of the other windows, so you should be able to use any of the windows. Just be sure to start your insertion burn at exactly the instant that Kerbin sets, and try to have that burn start coincide as close as possible with the windows that I\'ve given above. I hope this helps. PH. 1I made up a MathCAD spreadsheet that takes the orbital inclination into account, but the difference is insignificant for the purposes of planning this type of transfer. The maximum error that results from assuming that the Mun and Minmus are coplanar when calculating their angular displacement is on the order of 0.16 degrees. This can safely be ignored because it is silly, after all, to measure with a micrometer when you\'re marking with chalk and cutting with a chainsaw... The Mun moves through 0.16 degrees every minute. Minmus moves through 0.16 degrees every 8 minutes.

Only until aerodynamic drag starts to kick in. And you waste a lot of fuel getting to equilibrium if your rocket is underpowered. You want a TWR closer to 2:1. Check out the Mini Challenge: Max Altitude with this supplied spacecraft thread over in the challenges section. There is excellent discussion and analysis in there about optimal ascent profiles.

It can\'t be atmospheric drag if he\'s not moving. I think the sticky pad theory is the best so far. Also, if the TWR is 1.43:1, the rocket\'s initial acceleration will only be 0.43 g\'s because you\'ve first got to overcome Kerbin\'s gravity.

You could look up WxEcho\'s orbit mechanic. I\'ve never used it so I don\'t know its limitations but it might help you. I\'m not sure that there\'s very much we can offer other than that. We\'d need a lot more information about what you\'re trying to do if you want a better answer. The problem is you not only need to know how fast you should be going but in what direction. It is possible to work out the required burn for an efficient transfer (even without resorting to MechJeb), but it will take careful planning.

Yes, exactly. I haven\'t tried it explicitly either, although I\'ve approximated it during my landing burn at periapsis. Thanks for verifying my numbers. Accounting for the fact that the Mun is rotating and that a point on the surface at the equator moves at ~9 m/s, we\'re in the same ballpark. I also used a periapsis of 2.5 km in my calculations because the terrain height where I landed is about 1600 m.

That was easily the best Kerbal video I\'ve ever seen. Having just finished watching it, I find myself feeling strangely melancholy that there hasn\'t been a truly monumental manned mission since the end of the Apollo program... Hopefully Apollo wasn\'t a flash in the pan and we\'ll see a return to the moon within our lifetimes.

Here\'s another entry, this time flown using my joystick. Otherwise, I used the same approach as my earlier entry: Lower my periapsis to about 1 km above terrain then perform my pre-landing burn at periapsis, while allowing the spacecraft to settle towards the ground. When horizontal velocity has been eliminated, turn perpendicular to the surface and arrest the remaining vertical velocity for landing. Edit: By my figures, the spacecraft needs to slow by just shy of 50 m/s to drop its periapsis to 1 km above terrain. It then needs to decelerate by a further ~610 m/s at periapsis, followed by a last little bit of (difficult to quantify) delta-V prior to landing safely. I used a total of 708 m/s delta-V in my best attempt, while Apotheosist used 699 m/s. That doesn\'t leave much room for improvement for this method. Does anyone have any other ideas to try? What about a reverse gravity turn?

Impressive!

Here\'s my entry. I had 139.8 kg fuel remaining upon landing on Mun\'s surface. I went around once from the persistence file\'s start point, and then lowered my periapsis to about 1 km above terrain. I started my landing burn a few seconds before reaching periapsis. I could have probably saved a tiny bit more fuel (and avoided breaking off 3 of the 4 landing legs) if I\'d used my joystick. Just to prove it is possible, I also flew the return to Kerbin. I didn\'t quite make the landing (splashing down at 22 m/s). The Apollo guys didn\'t get 'do overs' so neither do my Kerbals, but I\'m sure someone who\'s a better pilot than me could do it on the first try. Edit: I just noticed that Zephram beat me to it, not only returning to Kerbin but also landing successfully. Nice job.

You certainly stand out as the most methodical of the contributors to this challenge! Thanks for your efforts and for sharing your results. Empirically, I\'ve found that my best results occur when I get set up on a trajectory that gets me above 38 km altitude relatively early in the flight, yet with low enough vertical velocity that I can continue to accelerate down range for an extended period before my apoapsis reaches 75 km. That requires a later pitchover and a more aggressive down range turn than just a gravity turn. A couple of times, I tried modulating the throttle to more closely follow a gravity turn trajectory, but it didn\'t work very well. I am able to reliably get into the 87 kg range, and I once achieved an 88.3 kg flight, but I haven\'t matched the top results yet. That said, I agree with Tarmenius that we are definitely down to the short strokes here. The difference between an 88 kg flight and an 89 kg flight is only an eighth of a second at full throttle!

I\'ve updated the leader board, but that\'ll be it for me until after the weekend. Maybe a new record will be set by the time I get back. Until then, Happy Canada Day to my fellow Canadians and an early 'Happy Independence day' to our neighbours to the south! Canadian National Canthem

On that topic, I was e-mailed a press release the other day about a 'Rocket Science 101' application from NASA that 'lets players build and launch a virtual rocket'. While it is nice that they are making the effort, I think it misses the mark. KSP\'s charm will be far more effective at getting people excited about space and space exploration! Any chance you could convey that to your sister?