closette

The mass units in KSP are a bit of a mess right now. For now I would modestly propose that the fuel unit displayed as 'kg' be a kerbal-gram, while the mass of parts be measured in 'pods' since a Mk1A command pod has a mass shown in the VAB as 1.0. By comparing the empty and full weight of a fuel tank, one finds that 1 fuel unit (displayed as 'kg') equals (1.25-0.2 pod units) / 250 fuel units = 0.0042 of these 'pod' units. Some places, such as the KSP Wiki http://kspwiki.nexisonline.net/wiki/Category:Default_Part, refer to the pod mass as 1 kilogram, but a more reasonable guess is that it has a mass of 1000 kilograms or 1 metric ton. It\'s been a while since I looked into it, but if I recall correctly, 1000 kilograms per pod means that the Solid Rocket Booster part works out to have a ~realistic density, and will explain why a full one will sink in Kerbin\'s ocean while an empty one will float (try it!). That would make 1 'kg' kerbal-gram fuel unit equal to 4.2 terrestrial kilograms. But one 'pod' unit of mass could be 600 Earth kilograms for all we know, or some other number in between, with other units (fuel, thrust) scaling along with it. It gets worse when dealing with the atmosphere: its density, and drag vs. frontal area of parts etc. but I\'m enjoying the break from thinking about the atmosphere in this thread so I won\'t derail it! (And of course we happily accept that Kerbin, Mun and Minmus are all denser than lead!). I understand that parts may undergo re-scaling of their dimensions for the upcoming 0.16 version, so that the Kerbals can make reasonable EVAs. So there may be some more unit confusion coming our way, but I hope it can be resolved soon after 0.16 comes out.

You are very kind Tarmenius. That Apollo reference is a good one, and kind of confirms the approach (pun intended) that some of us are applying to the idealized problem - Hohmann transferring the craft from an initial 100km circular orbit down to a highly eccentric 'orbit' with the spacecraft at apoapsis hovering just above the Mun\'s surface. A while ago I found some other papers on descent profiles for post-Apollo missions. I\'ll dig around and attach the best ones to this post. One of them was an entire Masters thesis on lunar descent! Most had some optimal control theory and none of them could be directly applied to our situation (at least not by me) to generate an optimal descent profile. One constraint they had/have is that the crew should be able to eyeball the landing site on approach, so window placement (and markings) can be driven by the profile chosen. Another I guess is that abort-to-orbit should be quickly available at any time prior to touchdown, requiring a more 'nose up' attitude.

141.9 kg remaining according to the quicksave file on my latest attempt, sort of a 'direct' method in that I set periapsis just _below_ ground level and did a white-knuckle gravity turn as I came in. I still can\'t right-click to show fuel status. On a Macbook, right click is usually CTRL-click but the CTRL key is bound to the throttle so that doesn\'t work. You could (1) trust me or (2) measure pixels on the fuel indicator in the screenshot below. Sorry. Anyone else using a Mac found a way around this? I tried landing back on Kerbin but only had a sliver of fuel left after re-entry, not enough to slow down from 100 m/s. I\'m all the more impressed by those who made it back safely.

@JellyCubes I think you mean a maximum of 151.8 units of fuel remaining. (Yes it\'s confusing that they are referred to as 'kilograms' in the display). Thanks for showing your work and checking on mine. I usually write up my calculations but ran out of time yesterday. It looks like I can\'t add, since I used a different initial mass in my calculation (3.35 pod units on my Post-it, so I may have forgotten the engine mass). Checking your numbers, I agree with them, so the fact that you have beaten the idealized 3-impulse model shows that one of my assumptions is wrong. The Mun\'s rotation saves you ~ 9 m/s, and careful choice of terrain can save you a bunch of deltaV_vert which can be tens of m/s (see my reply #25 above). And as you mention, the lander legs (and if needed, ASAS module) can absorb some delta-v on impact landing. By comparison, near the end of the descent I find that each 'kilogram' of fuel provides a delta-V of 7.3 m/s. I believe this is what Tarmenius was asking about above. Derivation here: Take the derivative of your rocket equation Delta-V = 5952.38 m/s × ln (3.85 / (3.85 - 0.0042x)) with respect to the fuel kilograms used 'x' and evaluating at the landing value of x ~ 98 kg. OR you can go back to basics and do conservation of momentum in the rocket\'s frame with the spacrcraft mass of 3.44 pod units expelling 1'kg' of fuel (=0.0042 pod masses) at the effective exhaust speed of 5952.4 m/s. So it turns out that the leaderboard has some very-close-to-optimal piloting! The gravity turn path may also save some delta-V over the simple impulsive model I was using - in the literature it seems to be the common method of descent on to the lunar surface. I\'m still most impressed that Zephram was able to get back to Kerbin safely even though he had less fuel remaining on landing than some of the other entries!

@BlazingAngel665 - you have a good teacher. Please report back on the forums your (and your classmates\') experiences with KSP as an educational aid. Even the free version offers a lot of sciencey potential. @Tarmenius - I updated my post (#25 in this thread) with the chart of delta-V vs. intermediate periapsis altitude. You asked how to convert delta-V to mass remaining, and in this case it\'s a bit more tedious because one has to convert between 'fuel units' (which display as kg) and what I have taken to call 'pod' units (where the default command pod has mass 1.0), and use the Tsiolkovsky rocket equation. (I cheated and used this calculator: http://www.strout.net/info/science/delta-v/ with the effective exhaust velocity of 5952.4 m/s for your craft\'s engine+fuel tank). EDIT - I made an error on the mass calculation below - see JellyCubes post below for better numbers: I\'ll give you the short answer: for the theoretical minimum delta-V of 674 m/s to land an idealized spacrcraft, you would end up with 168.2 kg of fuel remaining, which again I will tentatively predict will be the hard limit for this challenge. (For a free fall from 1km periapsis which a few people have used, the best you can get should be 162.5 kg, which might actually be attainable by some awesome pilot). Sorry I don\'t have time right now to go through all the steps of that calculation, which I hastily scribbled down Kerbal-style on Post-it notes while I was supposed to be doing something else. If the terrain allowed, a near-optimal landing could be achieved by setting as low a periapsis as you dare, then after coasting to it, nulling all horizontal velocity just in time to settle down on a mountain-top conveniently placed at just the right height! For the more realistic case of final descent from a safe periapsis, there is a lot of literature about the optimal path, and of course one of the earliest arcade games Lunar Lander was all about this problem. Reverse gravity turn (mentioned by PakledHostage above) and 'linear tangent steering law' descents seem to come up a lot in the literature. It can also be shown that the fuel-optimal final descent is the same as the fastest final descent (less time for gravity to change the craft\'s momentum). This basically means you should free-fall until you\'re knuckles turn white, and then apply full thrust just in time to prevent a crash. P.S. My own attempt - undocumented since I can\'t right-click to get fuel state to appear on my MacBook, is 140.1 kg remaining (read from the quicksave.sfs file).

First, this is a very well set-up challenge. My own attempts so far have fallen well short of optimal but I\'ll keep trying. A brief note on the 'direct' vs. 'periapsis' methods. I did some simplified calculations of the delta-V required to land on a non-rotating Mun with an 'idealized' constant-mass spacecraft that could apply near-infinite thrust (i.e. change its velocity instantaneously). For this spacecraft, a 'direct' landing from 100km orbit by first killing orbital speed, and then killing all vertical speed just above the surface, required a delta-v of 931 m/s. Interestingly, for this altitude the 'stop' delta-V and 'drop' delta-V contributed exactly equally to this total. See the graph I made in this thread earlier this year: http://kerbalspaceprogram.com/forum/index.php?topic=6484.msg96021#msg96021 On the other hand, a 3-stage orbital transfer maneuver to the surface required: deltaV_orbital = 48.5 m/s to get the periapsis down to 1 km altitude deltaV_horiz = 623 m/s to null the horizontal velocity deltaV_vert = 57 m/s to drop down vertically and land for a total delta-V of 728 m/s. Some of you have already beaten that limit! These separate contributions to delta-V are in close agreement with PakledHostage\'s in-game estimates above. If you throw reality away completely and drop periapsis down to 0.000 km (so that the craft is just skidding above the surface) the smallest possible delta-V is 674 m/s, which I tentatively predict will the the 'hard limit' for this challenge. (I could well be wrong though, since the spacecraft in the challenge is losing mass as it descends). Basically, the lower the periapsis after your de-orbit burn the better, although for the spacecraft used in this challenge you obviously can\'t go too low, since it takes time to change its velocity. If anyone is interested I can make a plot of delta-v vs. periapsis altitude later on have posted a plot of delta-V vs. intermediate periapsis altitude, showing the three contributions, below.

Thanks for the update and insightful comments. I am still left with a lingering question of 'What is the optimal ascent profile?', which would take some gnarly optimal control theory to solve. Are we all missing something that could put us in orbit with 90kg of fuel remaining? It\'s a tough problem to solve. Most of the papers I\'ve read are quick to make a 'flat Earth' assumption, which does not work for a small planet like Kerbin, and of course they use a different drag law. Also most of them have constraints that we do not, such as on the allowed deviation of the velocity and pitch attitude for mechanical reasons. I believe there is MATLAB code out there for solving the ascent problem (as a two-point boundary value problem) which could be adapted, if we have any MATLAB experts on the forum...

Well a recently discovered favorite (or 'favourite' since it\'s British) - the Clangers, sort-of 1970s knitted versions of the Kerbals: The narrator\'s prologues are simply amazing: '...This calm serene orb, sailing majestically among the myriad stars of the firmament...' when you consider that it was watched by 3-6 year olds! Compared to kids\' shows today that insult even a preschooler\'s intelligence, this is TV gold.

82.1 kg trying to follow the ascent profile in jqhullekes\' video, and I got the orbit in range for once at 75.9x75.5km (so the superscript can be removed from my leaderboard entry), and I\'m in the 80-something kg club at last! But that\'s not why I\'m posting. A few things I noted: 1. During the coasting phase in one of my attempts, I saw the apoapsis 'decay' from 75500 m down to below 73000 m. I then had to decide whether to (a) re-boost while in the atmosphere (at ~30 000m) to an apopapsis above 76000m and let it decay back down, or ( wait until I was above 70 000m and boost then, or © something in between, i.e. wait until the atmosphere got thinner, then reboost. Not sure what I did, but I was already out of contention on that attempt. Ideas as to which option would have been best, and why? 2. On my 82.1kg ascent I tried to follow the video closely, including the advice to keep the heading slightly to the left of the edge of the yellow ball. This means a slightly nose-up attitude in flight, perhaps more so than a real rocket can do due to aerodynamic stresses. Technically, a gravity turn 'follows the ball' exactly so that the most kinetic energy is gained per delta-v, which is what I was doing before, but this is clearly not optimal. 3. The initial pitchover at 8km has to happen fast and be 'just right' to set up the rest of the ascent profile. I suspect that may be the discriminating factor between the top few entries on the leaderboard. Comments anyone?

Kind of you to incorporate my suggestions - it\'s a great craft, and exciting to fly. I also appreciate having the extra fuel for messing around trying to get to the perfect spot on the Mun and Minmus.

The drag force on each part of your craft is given by FDrag = 1/2maximum_drag x mass x density x speed2 where the density as a function of height in this equation is given by 1.223125 x 0.008 x Exp[-height/5000 meters], as described by PakledHostage above. and the total drag is just the sum of the drag forces on all parts. For the Goddard challenge rocket flown at close to optimal ascent speed, by modeling the forces I found that 'max Q' occurred during the initial ascent phase at only ~ 250 m altitude, and decreased slowly from there. That would be approximately true for any rocket ascending vertically at a fuel optimal speed. Model flight profile and .craft file attached if you are interested, and I have reposted a graph of drag and velocity vs. altitude for the close-to-optimal ascent profile and unpowered fall back to Kerbin. Please take a look at the ascent profile, which would be similar in shape for most spacecraft up to 18km. After the initial boost at 100% throttle, the combined effects of (1) decreasing mass and (2) especially the rapidly decreasing density with altitude, more than compensate for the increasing velocity-squared term in the drag equation, so that the total drag force decreases smoothly all the way to MECO. On the fall back down to Kerbin\'s surface (here from 34500 m, but similar for re-entry from orbit) most of us have seen the rapid deceleration that occurs around 8-10 km altitude due to the exponentially thickening atmosphere.

Good point. No spacecraft has ever lifted off another solar system body with an atmosphere (like a Mars sample return or manned mission would have to). Of course the devs can construct whatever conditions they want, such as an inner planet with a thinner atmosphere (and similar temperature to Kerbin\'s). That would make rocket launches easier but parachute-only or parachute-assisted landings harder - until we get airbags!

Ok tried it some more and have some feedback. - The parachutes on the engines are a nice touch but they are destroyed anyway, so not necessary. - As you promised, there was plenty of fuel left over once in orbit. It took me a while to make a Minmus-matching orbit, making lots of incorrectly oriented burns, but I still had some main engine fuel left until final descent. Speaking of which... - Controlling the lander was a pain, in fact I crashed on Minmus. I would relocate the ASAS module to the lander stage above the tank. That should damp out any (my) control inputs to keep the landings vertical. - I would also relocate the topmost stack decoupler to below the RCS tank, not above it. That way you can get back to Kerbin with a pod+RCS 'lifeboat'. As long as you retrofire or land in the ocean when arriving home, the crew will survive. - I might add a second RCS tank since I used up most of the RCS fuel doing maneuvers in orbit, since the rotational inertia of the main engine stage is quite large (i.e. slow to change direction without RCS help) Otherwise, as I said above, a very nice introduction to the advantages of ramjets as low-altitude boosters! If you would like to make some of these suggested changes, please reply with the new .craft file. And post your craft to the 'stock' spacecraft exchange thread in the Exchange forum if you want more people to try it.

As for Kerbin residing outside Kerbol\'s 'habitable zone', there may be an explanation in a strong greenhouse-effect contribution from the atmosphere, see this discussion: http://kerbalspaceprogram.com/forum/index.php?topic=5623.msg92110#msg92110 Without the atmosphere (so this applies to the Mun and Minmus) I find surface temperatures at 0.09 A.U. from Kerbol would be in the chilly range of 100-150 K (depending on albedo). Kerbin still lies within the tidal locking radius though, so its fast spin could be explained by either its recent Creation, or its perfect spherical shape. (I think Kerbals might find consider this as evidence for Intelligent Design...and they would be right!).

I know what you mean, thorfinn, but you just described Mars! More distant gas-baggiant planets could have large Titan-like moons with an atmosphere but that\'s a while off I bet. An inner Venus-like planet with an atmosphere would be fun to try and get to and land on with parachutes, and I hope/presume something like that is being worked on...

Why methane hydrate in particular? Am I missing something? Solid methane + some impurities works fine for me. I just did a quick (well not so quick) calculation of the effective temperature of a greenhouse-effect-free planet at the 0.09 A. U. distance from Kerbol, which I assumed had a red dwarf luminosity equal to that of Proxima Centauri (if anything it would be a little bit less, but Kerbol\'s stellar astrophysics doesn\'t make sense in any case). I used a relation found here: http://en.wikipedia.org/wiki/Effective_temperature#Planet ... and got numbers between 100 K and 190 K, depending on surface albedo. Someone please check my calculations. So you see Kerbin really needs that greenhouse effect! (One inconsistency - Kerbin\'s rotation should be tidally locked to its orbit around Kerbol, but either it is a new planet recently placed there by HarvesteR-God, or it has no tidal bulges). It also took some digging but I found a phase diagram of pure methane, and sure enough it would be a solid, but probably slowly sublimating away on the sunny side of the planet. Not enough to give you the 50-bar pressure of a methane atmosphere though! For other objects, perhaps a Venus or Titan analogue, I do like the idea of surface fog, which would make eyeballing a landing more interesting! One could do this on the Mun already with dust clouds kicked up by rocket engines - '..picking up some dust' (Buzz Aldrin, Apollo 11 landing - http://www.hq.nasa.gov/alsj/a11/a11.landing.html

It looks like frozen methane is HarvesteR\'s working hypothesis: http://kerbalspaceprogram.com/forum/index.php?topic=13530.msg219016#msg219016 The sulphur hexafluoride component in Kerbin\'s atmosphere was my suggestion, based on the molecular weight derived from the scale height (5000 m) and the need for a greenhouse gas to keep its oceans and grass, e.g. http://kerbalspaceprogram.com/forum/index.php?topic=2062.msg106155#msg106155 But there are quite a few inconsistencies with 'real world' expectations for all the objects in KSP. The new KSP wiki gives Minmus\' density as 46.8 times that of water, somewhere between Kerbin\'s and the Mun\'s density: http://kspwiki.nexisonline.net/wiki/Minmus so it\'s certainly not methane all the way through, or 'cometary' material. It must have at least some of the super-dense (denser than Osmium) material that is a component of the Mun or Kerbin.

Nice! What video capture software did you end up using? I see you were very careful with the circularization burn(s), I\'ll be more careful too.

Make that 6 downloads and I like it - a nice introduction to ramjets for those who just upgraded from the free version. It seems to be fairly efficient on ascent as well. Not knowing much about ramjets, I\'m guessing that you arranged for these to run out of fuel right around the altitude where they lose effectiveness anyway? (About 18000m I think).

Hi DayBlur, I finally managed to look at your flight data (thanks!) to see how your optimal ascent strategy differs. As you predicted 'not by much' is the answer. The C++ code I was using to predict the maximum altitude is designed to 'chase' the terminal velocity (i.e. provide enough thrust to reach the terminal speed). Yes, I do include the changing gravity, and maximum_drag factor as fuel is used up, and a rudimentary centripetal term, as does the MechJeb implementation I believe. (Although Mechjeb does so using an arbitrary 'falloff' factor which I don\'t quite understand). it sounds like your method explicitly calculated the acceleration required 'right now' (= v dv/dy) assuming that your ship is already going at the terminal speed. This means that your thrust profile is slightly more aggressive, resulting in a higher speed at burnout (which also happens a little earlier than for mine since you burn fuel a bit faster). As you say, the differences are only just above the 'simulation noise' of time-steps, time warps etc. but they worked well in your favor! I do hope you can work on the 2-D ascent problem - there are a couple of related Challenges going on right now (just follow my posts). I have attached a line-by-line comparison of our ascent methods. Columns beginning with a 'C' are for the 'Closette' model ascent. Those beginning with a 'D' are the 'DayBlur' measured profile.

I\'m trailing behind in the non-MechJeb leaderboard, but I bettered my own record to get 79.02683 kg remaining, with a 79.0 x 79.4 km orbit. Yes I overshot again, which seems to be easier for me than going under 75km (where one barely scrapes out of the atmosphere at a shallow angle before doing the circularization burn). Sorry I can\'t right-click to verify that fuel status on my MacBook, but I\'ve nothing to gain from fooling anyone!

Any chance you could upload a video of your ascent, please? A 'boring' one with no changes to the camera angle, so that we can see how the pitchover angle varies with time. (And in fact there might be a way to measure it from the video).

That\'s a big improvement Tarmenius, I bet it will be hard to beat. Regarding TWR and drag, we know that for a vertical ascent, the fuel-optimum speed vs altitude results in an equal 'gravity loss' vs. 'drag loss', and I\'ve seen it suggested that the same 'equipartition' might also hold for an optimum 2D ascent. So could I suggest that some of us non-Mechjeb leaderboard winners try the ascent with a MechJeb module installed and the 'Ascent stats' turned on, so that we can see how gravity loss and drag loss compare on these near-optimal paths?

That would be a perfect 'mouseover' punchline (a la xkcd)

I was not much of a fan of these (to each their own) but this is the best one yet, since many of us can relate to it, and thank you for referring to them as kerbOnauts in the cartoon.