# Optimal Descent (to the Mun)

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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.

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If anyone is interested I can make a plot of delta-v vs. periapsis altitude and attach it to this post later on.

That sounds really cool, I\'d love to have a look at it (not that it would help me stop from crashing into the ground repeatedly xD).

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Did the mechjeb file and just clicked land. . . 122.1kg of fuel left and one messed up lander. . . it landed me on the side of a crater, so I lost all of my legs and my engine from rolling down a hill. The kerbal comic comes to mind. . . Trying again, watch here. Great challenge by the way!

EDIT: Used Smart ASS for a retrograde, then activated mech jeb at the periapsis, much more efficient and I still have an intact lander this time. 143.4 kgs of fuel left. GOIN HOME BABY!

FINAL EDIT: Not only home, but curtisy of auto-land we are back at KSC after 10 horrible days and burning all but .3kg of gas, I feel pretty damn boss.

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PakledHostage, what do you mean by a 'reverse gravity turn'? Is that where you start out burning horizontally and let the craft slowly drift to vertical?

Yes, exactly. I haven\'t tried it explicitly either, although I\'ve approximated it during my landing burn at periapsis.

48.5 m/s to get the periapsis down to 1 km altitude

623 m/s to null the horizontal velocity

57 m/s to drop down vertically and land

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.

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Okay, I finally got home and just flew my first run. 148.6 kg

I started my burn a little too late and had to nose up to avoid hitting the surface. I will give it another go later.

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i cant seem to open this file wtf?

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ASAS has it\'s own force than can applied without limit (without fuel to power it I mean), like SAS units (else, what would be the point of them?).

Nope, it doesn\'t. It just appears that way since it controls Pod SAS.

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When pod torque gets nerfed ASAS may be less a bit less useful, but right now it\'s ridiculously handy. It\'s way better than a basic SAS for keeping ships in line, since it controls pod torque, gimbaled engines, RCS and control surfaces.

It\'s amazing for hands-off landings; just kill your horizontal velocity, point your lander straight up, and you won\'t have to worry about anything but your rate of descent. I can land without it, but it\'s an annoying luxury to give up.

Oh, and I also tried my hand at this challenge. I know I could land with a kg or two more fuel if I picked an ideal landing spot and descended in a perfect arc, but this will have to do for now:

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Nice results everyone! I\'ve updated the Leaderboards accordingly.

First, this is a very well set-up challenge.

Thanks! And good work with the mathmatical analysis. If you feel like making that chart, I think it would be interesting and helpful for anyone who wants to get the most out of their landers. The deeper into this game I get, the more I want to brush up on my Advanced Algebra. Ten years of not using it has really taken its toll.

On a related note, how hard is it to convert fuel spent into delta-v?

I feel pretty damn boss.

And with good reason! Congrats on performing above and beyond! And with practically only fumes left in the tank, too.

i cant seem to open this file wtf?

No need to open it. First move your existing persistent.sfs file from the 'KSP/saves/default' folder to somewhere outside KSP so you can keep any ongoing missions you may have. Then copy mine and paste it where yours was.

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Thanks! And good work with the mathmatical analysis. If you feel like making that chart, I think it would be interesting and helpful for anyone who wants to get the most out of their landers. The deeper into this game I get, the more I want to brush up on my Advanced Algebra. Ten years of not using it has really taken its toll.

I am just learning it right now (sophmore) and have taken quite a few orbital equations to my Geometry teacher, eventually she asked what for, I showed her so now we get to spend a week working on advanced formula\'s in KSP Mrs. Smiller is amazing!!! Unfortunatly its only the demo, and we have very specific tasks to complete with pre assigned rockets, and we have to show work, but still... video games in class, most assuradly affirmative!

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@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).

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Excellent work as usual, closette; that chart will come in very handy. As for the calculations of delta-v, don\'t worry about not going through the steps. It looks a little beyond what I\'d be comfortable doing anyway. Though I\'ll still be trying to reach for that 162.5kg mark

And to BlazingAngel665, I would also be very interested to hear how it goes using KSP in the classroom. Had this been around when I was in high school, my Applied Physics teacher would have loved it. That was the best class ever.

[EDIT]: In my most recent attempt, I set my Pe to 2500m. I came very close to being a crater on some mountain, so I decided to land. After some pretty sloppy maneuvering, I still touched down with 143.9kg in the tank.

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For a theoretical minimum delta-V of 674.4 m/s, I\'m calculating a minimum 151.8 units of fuel remaining. This assumes landing at an altitude of 0 m. If you landed on a 3 km mountain, theoretical minimum delta-V decreases to 666.0 m/s and 152.9 units of fuel. The landing legs could probably absorb another 10-15 m/s too.

Teacher always said to show your working:

Note that the \'kg\' value seen when right-clicking a fuel tank has absolutely no real meaning as of 0.15.2 and will be refered to as units to avoid confusion.

Initial mass

[table]

[tr][td]Command pod [/td][td]1[/td][/tr]

[tr][td]ASAS[/td][td]0.8[/td][/tr]

[tr][td]Full fuel tank[/td][td]1.25[/td][/tr]

[tr][td]Engine[/td][td]0.5[/td][/tr]

[tr][td]Landing legs[/td][td]0.3[/td][/tr]

[tr][td]TOTAL[/td][td]3.85[/td][/tr]

[/table]

Final mass

The half fuel tank holds 250 units of fuel and the difference between full and empty mass is 1.25 - 0.2 = 1.05 tonnes.

Each unit of fuel is 1.05 / 250 = 0.0042 tonnes.

Final mass of rocket is (3.85 - 0.0042x) tonnes where x is the amount of fuel used.

Exhaust velocity

Exhaust velocity is thrust / mass flow rate.

Thrust = 50 kN

Mass flow rate = unit mass of fuel Ã— burn rate = 0.0042 Ã— 2 = 0.0084 tonne/second

Exhaust velocity = 50 / 0.0084 = 5,952.38 m/s

Delta-V

Using the rocket equation:

Delta-V = 5952.38 Ã— ln (3.85 / (3.85 - 0.0042x))

Solving for delta-V = 674.4, x = 98.2 units of fuel used. Or 151.8 units of fuel remaining.

If you landed on a 3 km mountain and let the landing legs absorb 10 m/s, fuel remaining goes up to 154.3 units.

I tried this (with a healthy amount of quicksaving) by decreasing Ap to 200 m above ground level and executing a reverse gravity turn with full thrust until touch down and had 154.5 units of fuel remaining. The fact this is over the 154.3 units theoretical maximum leads me to three conclusions:

[list type=decimal]

[li]Cheating. I can\'t really prove I didn\'t cheat but the fact pushingrobot got 150.5 units remaining should be indication that I didn\'t edit the cfg files.[/li]

[li]Incorrect calculations. I hope it\'s not this, otherwise I\'ll look like a massive idiot.[/li]

[li]Rounding errors, time-warp errors, phantom forces caused by internal collisions, deep space kraken\'s tamer less violent brother?[/li]

Conclusion

I have too much time on my hands.

Fin.

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Congratulations on taking the lead, JellyCubes! And an impressive show of mathmatical force as well.

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@JellyCubes

For a theoretical minimum delta-V of 674.4 m/s, I\'m calculating a minimum 151.8 units of fuel remaining.

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 Ã— 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!

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Wow, the progress being made is fantastic! It\'s a pleasure reading these posts . I have tried getting a very low periapsis inside a deep crater, to cancel out my horizontal velocity as close as possible to the mun. Hasn\'t gone too well. Now I see landing on a mountain may be more efficient? @Jellycubes, that landing spot looks perfect! Is there any chance of an image or two showing the location you landed at?

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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.

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closette- I\'m trusting you for three reasons: 1: Your reputation on these boards warrants it in my view, 2: PakledHostage (who made the challenge that inspired this one) did as well, and 3: My true goal for this challenge is to establish the most efficient descent method that anyone who wants to can duplicate. Competition and placement on the Leaderboards is of lesser importance. At least, that\'s how I see it.

Have you tried assigning the throttle controls to different keys, freeing up CTRL? I haven\'t needed to re-map any, so I don\'t know if certain functions are locked to certain keys.

On somewhat of a side note, I got real curious about how closely our descent paths matched those of the Apollo missions. I found the NASA Technical Memorandum 'Apollo Lunar Descent And Ascent Trajectories,' March 1970. There\'s some very interesting stuff in there. http://www.hq.nasa.gov/alsj/nasa58040.pdf Of particular relevance, the 2nd and 3rd pages describe the descent.

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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.

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2 more on topic points, someone had something in the right hand upper corner that showed true altitude. . . Can I have that please??

2nd I am trying this vanilla w/out mechjeb and we will see how this works. . . probbably leaving some stranded kerbonauts in a transfer orbit this time. . .

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That\'s pretty awesome. Now I have an urge to watch an internet-cafe full of people play KSP with horribly (or awesomely) Kerbal rockets!

Tosh\'s FPS Pod plugin is what will give you a display of your current actual altitude. You can find it here: http://kerbalspaceprogram.com/forum/index.php?topic=11214.0

closette: A Masters thesis devoted entirely to a Lunar descent? Wow. How awesome is this game that lets us so casually accomplish what so many brilliant minds devoted so much energy toward.

Now, I\'m off to try landing on a Munar mountaintop...

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Wow! So much good stuff in this thread. I\'m impressed!

After reading about the Apollo descent design, mountain tops, and reverse gravity turns, I decided to try again. (Knowing there is so much room for improvement helped too!) So here\'s my updated entry: 148.5 kg kilogunits -ish.*

*I hearby propose we rename the fuel units -ish. For example, 'precisely 148.5-ish fuel remaining'.

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I\'m going to update my score from some attempts I just made. This time I tried using a reverse gravity turn type descent from a low periapsis, and I managed 153.8kg of fuel left. I like the concept of renaming Kerbal units of mass, but I don\'t like 'ish' myself. It is very Kerbal sounding, but I fear the confusion it may cause.

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Ish by all means, I love it!

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Good job guys. I\'ve updated the Leaderboard.

I liked the idea to rename fuel units, but as Apotheosist mentioned, 'ish' might cause some confusion over the accuracy of our numbers. So, after some searching I decided to (tentatively) rename them 'Gubs.' It\'s a shortened version of 'Gubbins' which means 'an object of no value.' I thought it was fitting. If someone has something better, feel free to post it.

I\'ve been trying to stick a mountaintop landing, but mostly I end up cartwheeling off into the distance. I\'d set it up so that my reverse gravity turn ends just above the peak, but in practice it\'s quite difficult. I either add more craters to the Mun, or use too much fuel during vertival descent. I\'m determined to pull it off at least once, though!

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