PLAD

A very interesting challenge. For a quick opener I would like to submit my SSTO to Eeloo entry for this challenge as well, though it shows the earlier and later parts of the trip. I entered Eve's SOI at the 13th frame and this challenge would end when I entered Eeloo's SOI. If I understand the scoring right it would go like this: Entered Eve's SOI +20 points Entered Kerbin's SOI. +20 points Passed through Mun's SOI. +20 points Left Kerbin's SOI. Re-entered Kerbin's SOI 213 days later. +10 points Passed through Minmus' SOI. +20 points Passed through Jool's SOI +20 points Entered Eeloo's SOI +20 points Total dV used between entering Eve's SOI and entering Eeloo's SOI: 27m/s. Round up to 100m/s. Total score: 130/100 =1.30 I agree with pds314, the key to a high score is Jool and it's 5 moons. There's up to 222 points to be found there and then Tylo or Laythe could fling you to another planet. I have a suggestion for a mercy addition to the rules. You might put a total time limit on the mission, say 15 or 20 'Earth years'. I imagine the craft's electronics are degraded by exposure to the interplanetary radiation levels or something. This would prevent someone from being tempted to circle the sun for 100 years waiting for a chance encounter with another planet. After all as long as you are near the plane of a planet there will always be another encounter eventually.

I took a Kerbal to the surface of Eeloo and back on 510 units of fuel, using a 12-ton SSTO that passed through the SOI of Mun, Minmus, Eve, and Jool on the way, BUT I used Mechjeb for the last 300 meters of the landings and for the takeoff from Eeloo and to execute the maneuver nodes. So I can't enter it. It suggests what is possible for a better pilot than I though. Wouldn't a trip to Eeloo score about 90 billion points for its distance from Kerbin?

Oh jeesum, now I noticed that the box in the lower right says 1901 "Flybys found" even though they aren't flybys when it's a 2-body porkchop plot. Maybe it should say "paths found" or "solutions found" instead. Something I might also not make sufficiently clear for all is that is still only works with 24 hour days/ 365 day years, like KSP v0.22 and earlier, but now KSP has the option for 6 hour days/ 426 day years. Does anyone have a feel for what percentage of users use each format? I stuck with 24/365 because I memorized the orbital and synodic periods for most of the bodies in that format and switching would confuse my sometimes-overwhelmed mind, but if a lot of people go 6/426 then I can add the option. We'll have endless confusion in discussions of flight windows though.

Thanks for the pictures, now I'm sure it's the right version. Pick the planet selection drop-down for the "2nd encounter". The bottom-most choice should be "none". Pick that and now hit "begin search". You should get the pork-chop for the first two planets (in this case Kerbin to Eve). I completely re-did the 'slide show' introducing the features of FF 0.60 in the first post. If you are already an expert with the program (and I know you are!) you might not have reread that. It covers the new tricks. You're right, if you're looking at the interface it is not obvious what to do.

Doh! I just downloaded it from the link in the first post and I got the latest version, 0.60 (and ran it to be sure). It is possible you downloaded it in the time between when I started changing the post and the time I set the link. The new version is named Flyby060.exe. I'm going to delete the 0.50 links and remove them from the dropbox folder to be sure they aren't getting mixed up. Please let me know if it works. Sorry about this!

I've just released version 0.60, it's in the first post in this thread, along with an all-new primer on using it. No double flybys yet. It's going to be a while on that. It does do a 2-body porkchop for the first two bodies now, and I increased the search grid to 100x100 from the earlier 80x80, the maximum flybys found to 4500 from 3000, and a few other small changes. You still pretty much have to have a maneuver node editor in order to do flybys well, I recommend Mechjeb. -PLAD

A good question. I've hit a wall in writing something that works AND is still reasonably fast. The problem is that there are many possible ways to do a double flyby. There are 2 key parameters- how many orbits does the planet make between the flybys, and how many orbits does the ship make between flybys. In doing experiments with the Lambert spreadsheet I found that useful flights could be found where those numbers were as high as 5 and 5, and the algorithm I have in mind needs to search 19 of those 25 possibilities, where for a standard flyby it would only search one. Ugh! So I've been playing the game a while and using v0.50 to see what is most useful and what needs to be improved, and hoping for a revelation. Then I got to noticing that it would be really nice to allow multi-orbit flybys with at least 2 orbits in between since this would double the stuff you would find, and deep-space maneuvers dramatically improve your solution field but I can't think of an algorithm for that! So I'm in a bit of a funk right now. I should release version 6 soon, but it won't have double-flyby. It will do a 2-body porkchop, increase from an 80x80 to a 100x100 search, increase solution limit from 3000 to 10000, and make it a little easier to find the departure orbit. But these are small things. I need to work on double-flyby. Doh!

A few months back Metaphor found a way to get from Kerbin's surface to Moho's surface for 7217m/s total. It's here. Since he used about 4400m/s to get to LKO from the surface, it was about 2850 from LKO to landed on Moho's surface. The trick was doing a flyby of Eve when Eve was in the plane of Moho's orbit- this means Eve both eliminates the plane-change expenditure and it lowers your transfer orbit apoapsis and thus arrival speed at Moho. He used some other hardcore tricks to lower the Moho capture to an astonishingly low value, but even without those you could do the whole surface-to-surface trip for under 8000m/s. Getting back could be done using Eve also, but I haven't seen anyone try that yet.

With an heroic rescue and feats of daring showing the minimum dV required for a Mun mission. The Kerbal X rocket which comes with KSP was improved when version 0.24 came out and can now do a Munar land-and-return (MLAR) mission without too much trouble. However the older versions had about 500m/s less and could not quite make it to a Munar landing and back. See this thread for details and a demonstration flight with the .242 Kerbal X. The .235 version has about 6818m/s of total dV. I'd done the mission with 6869m/s. I made only 3 changes to the stock .235 Kerbal-X: 1. Added Mechjeb. I can't do precise nodes without it, or repeatable ascents and landings. 2. Changed when the gantry towers release the rocket. They originally release after the main engines ignite, I made them release at the same time as ignition. 3. Removed the default monopropellant from the Mk1-2 capsule. A Kerbal-X doesn't need it! This raised the total vacuum dV to 6854m/s. I bet 3 Kerbal's lives that I could do a Munar mission with that. I was wrong.... 6809m/s. A personal record -but why wasn't it enough? I forgot that not all of the 6809m/s is expended in a vacuum! In the atmosphere the Isp of the motors are lower than in a vacuum. Doh!! The mission was doomed in the first 2 minutes. I do think that I have proven that the old Kerbal-Xs cannot do an MLAR. The theoretical limit for an MLAR is around 6716m/s. But rounding errors, SOI changes, control inefficiencies (limited maximum thrust, inaccurate pointing and dV)and the like will add dozens of m/s to that. The practical limit is probably around 6780. Definitely more than the 6717 (vacuum equivalent) the old Kerbal-X can give.

Have you seen my Flyby Finder? TOT gives more details of an individual flyby (I particularly like the orbit diagrams), and it allows for a thrust during the flyby, but FF shows what's possible over a wide area. I'd never seen that 150-or-so-day free return Kerbin-Duna-Kerbin path before, that is cool.

4350 to LKO is quite difficult, my best ever was 4349 and I only did it once, generally I can do 4356 reliably with Mechjeb and the right ship. But since I'm assuming everything is perfect for a theoretical minimum I used it. The 762m/s from sitting on the Mun is the bare minimum to escape Mun's SOI. Your VSOI would be 1m/s, which is, need I say, less than what will get you back to Kerbin directly. But using repeated Munar flybys and small deep space maneuvers you could theoretically build up enough dV to get back to Kerbin. In practice I can make it back to Kerbin's atmosphere from a 5.5x5.5km orbit with a 207m/s thrust, it gives 125m/s at the SOI, then a 16m/s deep space manuever and another Munar flyby gets you home.Since circular V at 5500m is about 563m/s that means 786m/s is already practical. Enough being mysterious, here's how I did 6869m/s with the 24.2 Kerbal X. This will be my entry in a minimum dV Mun challenge if there is interest:

I just noticed that the stock Kerbal X rocket has about 500m/s more dV then it has in 23.5 and earlier versions of KSP. To be precise (according to Mechjeb): Version........total vacuum dV............total weight on launchpad 22..............6832 m/s......................131.67 tons 23.5...........6818............................131.44 24.2...........7333............................131.24 The differences between 22 and 23.5 are due to the small parts that became weightless (ladders, solar panels) and the addition of monopropellant to the capsule. The huge difference in 24.2 is due to the Mainsail's higher Isp and the lower weight of the Poodle (2.0 versus 2.5 tons earlier). What does this matter? It means the Kerbal X can now easily do a Mun land-and-return mission! I recall an old challenge where the goal was to land the Kerbal X on Mun with the most fuel still in the tanks, or even try to get back to Kerbin. No one made it back. Recently I've been trying to do a Kerbin-Mun land-Kerbin mission for as little dV as possible and was down to 6863 m/s and decided to give the X a try in 24.2. Piece of cake! Even with the reduced T/W of the X over my test craft I did the mission with 6869m/s and so had plenty of fuel left over. I'd been thinking of another challenge but it would have to be with version 22 or 23.5 and so would not be very inclusive. Maybe a challenge could be done that requires 24.2 and is just to try to have the lowest total dV or the most fuel left over after returning to Kerbin. Extra note: my take on the theoretical minimums for a Mun landing mission: 75x75km Kerbin orbit to Mun surface pass: 840m/s Mun pass to landing (in an instant explosion of thrust on the 5500m mountain) : 762m/s Mun surface to Munar escape: 762m/s Add in 4350 for Kerbin surface to 75x75 orbit and the mission total is about 6720m/s.

Sorry to wait so long on this, I hadn't built a successful SSTO before and the flight took a while. I hope the thread isn't considered dead yet. Mission summary: SSTO from Kerbin surface to Eeloo surface and back. Stock aerodynamics, all stock parts except for Mechjeb. 11 years 277 days total mission time. Ship: 12.059 tons on the launchpad, 62 parts. No clipping except for the ladders, little girders, and one slightly misplaced air intake. 3207m/s available once in LKO. 7.5 intakes per jet engine. Flight profile: K-Mun-Eve-K-K-Jool-Eeloo out, then Eeloo-Jool-Kerbin back. Additional non-useful flybys of Minmus and Mun. 1955m/s from LKO to Eeloo surface and 942m/s from Eeloo surface to parachutes opening on Kerbin. (Softened landing with final thrust, would have landed at 17m/s without it.) LKO to Eeloo surface and back to Kerbin for 2897m/s!!

I did not know that and I've used MechJeb for 6 months. So I looked and found a description of the feature on the web, it is not in the Mechjeb documentation. Just tried it now, and oh lord, this would have saved me so much time...

It struck me that it is a bit tricky to edit this spreadsheet from scratch, so here is some detail on how I used it to find when Kerbin crosses the plane of a planet, Eve in this example. I added some colors to it- I use green to show the cells you are supposed to change and orange or yellow for outputs you need to check. And to answer the original question, here is when Kerbin crosses the orbital planes of these planets, given in 24-hour UT days: Moho: Day 21.74 and 75.00. Eve: Day 5.47 and 58.73. Dres: Day 30.62 and 83.88. Eeloo: Day 15.82 and 69.08. Add Multiples of Kerbin's orbital period (106.5225 days) to these numbers to find more plane-crossings unto infinity. I didn't do Duna or Jool because their orbits have such low inclinations.

The Wiki gives us the orbital parameters for the planets, including their anomalies on Day 1 at 0:00:00. But as someone pointed out there is nothing we can refer to as '0 degrees'. For instance Kerbin starts with an anomaly of 3.14 radians, or about 180 degrees on day 1. That's great to measure from on day 1, but after that you have to calculate what angle Kerbin is at at a given time before you can use it as a reference. No good. So I do it like this (For reference I'm looking from above the plane of Kerbin's orbit, defined as the direction from which Kerbin appears to be traveling counter-clockwise around Kerbol. I also do all calculations using Earth-time, 24 hour days.): Moho first passes through the plane of Kerbin's orbit moving up (so ascending node) on day 13.128. Defining "180 degrees" as the position Kerbin is at on day 1:00:00:00, Moho's ascending node is at 70.00 degrees. It passes through descending node on day 25.09, at angle 250.00 degrees. From this we can compute when Kerbin passes through the plane of Moho's orbit,, since we know when Kerbin was at 180 degrees and we know how many degrees it travels in a (24-hour) day. This is what I think you want to know- Kerbin passes through the plane of Moho's orbit on day 21.74 and day 75.00, and every 106.5225 days after those times. You can't say that Kerbin is always in the plane of Moho's orbit on a given Kerbal-time (6-hour days) day, because a Kerbal year is an integer number of Kerbal days, but Kerbin's real orbit is not an integer number of days, so as the years go by you'll be off by more and more. I used my 'Lambert E' spreadsheet to figure out this data, get it here. The page 'Calc positions' gives the x, y, and z positions of a planet at a given time as well as the anomaly as measured with the 'Kerbin-starts-at-180-degrees method. Look for the times that the z-position of a planet passes through 0 meters and see what 'approx absolute angle' it's at on that day. In summary I can give you the 1st UT day that Kerbin passes through the ascending and descending nodes of a planet, you would have to make a little calculator that adds Kerbin years to the 1st pass to determine all later passes.

Using the Kerbin-Eve-Eve-Moho route can get you from Kerbin's surface to Moho's surface for well under 8000m/s total. For details on the route check out this challenge. Even skipping the Mun flyby and Metaphor's Messenger-style braking passes can still get you there for around 8000. Using Moho-Eve-Eve-Kerbin can get you home for under 2500m/s but it's a rare window, I only know of one that leaves Moho on Y2 D339 (Earth-time).

LKO to Eeloo surface: 1934m/s. I figured out a way to avoid the 230m/s burn when flying by Jool on the way to Eeloo. I used the atmosphere. I knew the VSOIJ I would have when coming from Kerbin and the VSOIJ I needed to leave with for an efficient path to Eeloo and calculated the equivalent energy reduction at 135km above Jool (so just in its atmosphere). It worked out to 150m/s, and 3 trial shots were enough to determine the necessary periapsis for the atmosphere to do that- 134150m. Once I set up a Jool flyby with a 134km periapsis I put a node there and a 'fake' 150ms retrograde burn so I could fiddle with what to do after that. I finally found a good place to do a burn to correct for the directional error that using the atmosphere caused. Enough yacking- here it is: Of course now we know there's a way to leave Kerbin for about 100m/s less, but for the rest of the flight I think this is pretty close to the minimum possible. Maybe 50m/s could be shaved by flying by Jool at a better point in its orbit, since I used 298m/s to brake into a 10x10 orbit around Eeloo, and about 247m/s looks like the lowest possible when coming from Jool. Note that if your ship has different drag from mine then it would need a different Jool periapsis. For those who haven't seen it, this mission uses my 1011m/s to Jool flight up to just after the 2nd Kerbin flyby.

Vector- KLO to Eve surface for 873m/s has got to be an all-time record. That is spectacular. All previous low-dV records can now be reduced by at least 100m/s with this trick. My god, you could probably get to Jool for less than 900m/s with care. Thank you for putting the saves up there, I ran one and found I could leave Kerbin after 2 Mun passes and get into a 6:5 Kerbolar orbit (with a bonus Minmus flyby no less) but that would not reduce the ultimate dV total to Eve at all. I am thinking that we need to have a general 'Lowest dV to'... challenge that includes lowest dV to all bodies in the Kerbol system. The lowest dV paths to Moho and Jool will start with a flight to Eve, and the lowest to Eeloo and Dres will then continue from Jool, so making this incredibly low-dV flight to Eve could lead to record flights to 4 other worlds! The concept could take you directly to Duna too, of course. None of them would be easy, but we do these things not because they are easy, but because they are hard... In the meantime, my Eeloo and Jool record flights still stand.

That is a very impressive series of Munar flybys Vector did. He didn't even encounter Mun at the same point in its orbit each time. Now some egghead analysis... I think Vector's Mun flyby mission proved 2 things and gave them some numbers as well. The first 3 flybys showed the benefit of multiple Mun flybys to adjust the Mun departure vector to be tangential to Mun's motion and thereby as efficient as possible. It looks like when he entered Kerbol orbit he was moving at 8709m/s, Kerbins Vcircular is 9284m/s so his V leaving Kerbin's SOI (I'll call it VSOIK) must have been about 575m/s. This equals a V over Vcirc in a 75x75km Kerbin orbit of 985m/s, since he started with 862 the total gain from the 3 Mun flybys was 123m/s. Just 42m/s more (at the 75x75 orbit start) and that would be enough to get to Eve all by itself. Probably only a 30m/s or so gain over just doing one really low Mun flyby on the way to Eve though. The 4th flyby was much more powerful. He went from a 7:6 resonance-with-Kerbin orbit (SMA about 12.272 million km) to a Kerbin-Eve Hohmann transfer orbit (SMA 11.716 mkm). The V apoapsis of those are 8768 and 8505m/s respectively. This means he entered SOIK at 516m/s and left at 779m/s. The 4th Mun flyby gave him 263m/s VSOIK! To be fair, at a Kerbin distance of 12000km (where Mun is) the speed change (relative to Kerbin) is 878 to 1054m/s or only 176m/s but that's still a lot. And I agree it should be possible to do that again and again, although the faster you flyby the less energy you can steal from Mun so it would be diminishing returns. There's also the pain of many-year coasts between encounters. It would be nice to see the VSOIK is when entering and leaving around a Munar flyby, if possible, to better quantify the effect. I'm missing something, leaving SOIK at 575m/s and doing a 42m/s burn suggests returning at 533m/s but I calculate 516m/s. I'm missing 17m/s somewhere, probably from bad transfer orbit SMA guesses.

Vector-That is an interesting idea. I'm a big fan of Munar flybys. I'm particularly proud of this mission where I got from LKO to Jool orbit for 1011m/s. I saved 87m/s by using a Munar flyby, and I thought that was the best that could be done. But your suggestion of using multiple Munar flybys could blow that out of the water. Normally if you leave Kerbin's SOI and just coast around back into Kerbin's SOI without flying by anything else or making any deep space maneuvers then you will enter with the same speed you exited with. All you can change is the direction you leave Kerbin at. But going Kerbin-Mun-Kerbin-Mun could get around this. Do you have record of a mission where you did this? I see why you should try to get in a solar orbit that has a period of exactly 1 Kerbin year if you used a minimum-energy flight to Mun (856m/s from a 75x75km Kerbin orbit) since you won't have the energy to get in an orbit that gets back quickly (for instance a 10:9 resonance orbit would take 9 years to get back and be a major pain). What could the savings be? I figure the minimum V you have to depart Kerbin with to get a useful Eve flyby works out to a start boost at 75x75 of 1044m/s, 1 Munar flyby can reduce that by about 90m/s, so about 954m/s is the minimum with 1 Munar flyby. The minimum speed to get to Mun is about 856m/s so potential savings are about 98m/s minus maybe a 10m/s correction budget per flyby. Enough to win a challenge if everything else is optimal. Gawd will it be tricky to execute though.

allmhuran- That is an impressive trip to Eeloo (and everywhere else). I think I saw you start the deboost burn into Eeloo orbit at around 950m/s? That suggests something around 1000m/s or a bit more depending on gravity losses total for going from Eeloo SOI to landing. With the other known burns that puts your LKO to Eeloo surface at Round 2200m/s but I can't say exactly because the effect of your maneuverings around Jool system are hard to be sure of. I think you beat my score. You figured out how to cut the Jool periapsis burn out- you aerobraked into orbit around Jool and took the time to set up a Tylo slingshot to Eeloo. That may be the best possible way to do it. I can't make this a formal entry though but I'll put an honorable mention in the listings. Great video- my favorite parts were the takeoff from Bop (the soundtrack made it) and the timing on running out of fuel around Ike. Talk about luck!

That is an excellent discovery on using Wine to run FF on a Mac. Thank you for posting that! Vz is the normal component of your velocity when leaving the start planet. If you start from an orbit with the right inclination you will not need to use a z component in the burn (relative to your orbital velocity). It's a bit hairy to describe, check out this challenge where I use the Vz and start inclination numbers that the FF detail box would give me (though here I put them in LambertE) to leave Kerbin in the most optimal way possible. FF finds paths where you only have to do one big boost right at the beginning. If KSP had 30-digit accuracy in everything you probably could do the whole flight with only the start boost in Kerbin orbit, but in practice rounding errors force you to correct your path every now and then on the way.

Sorry, no, I don't have the software or the hardware to do that right now.