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Free-return trajectory


Aiyel

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Launch a craft on Munar intercept, jettison everything but your capsule and a parachute, and make it safely back to kerbin with no thrust expended after your TMI burn.

Due to the nature of the assignment, any parts pack is available, so long as you\'re flying a vanilla mk. 1 capsule.

Screenshots of craft and map screen required at TMI cut-off, upon entering munar SOI, munar peri, re-entering of Kerbin\'s SOI, and atmospheric entry.

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I don\'t think it\'s possible to use the mun\'s gravity to slingshot you back to earth, the gravity isn\'t strong enough.

I did this today granted i didn\'t eject everything i had a full rocket because i was planning on burning but it is possible, the return trajectory i happened onto put me within 10km of kerbin on the low side which was more than enough to catch my ship.

It seems to simply be dumb luck though at this point.

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It\'s possible, but *very* difficult if you intend to have a survivable landing.

On Apollo, the free-return trajectory window was ~0.1 meters per second wide at TLI. The Saturn V simply couldn\'t inject the vehicle onto such a precise trajectory, and Houston couldn\'t maintain such a precise path, anyway. (Frank Borman, on Apollo 8, was trying to demand precision to within three feet per second, or slightly less than 1m/s, and Houston couldn\'t even provide that.) This is why all the 'free return' trajectories involved the use of some course-correction burns after swinging around the Moon, using either the SPS or RCS engines to tweak the trajectory and keep it steep enough to avoid skipping off the atmosphere, but shallow enough to avoid salsifying the crew or disintegrating the CM.

The only time we ever tested the free-return trajectory was Apollo 13; ironically, the O2 tank explosion occurred AFTER LOI1 took the vehicle off of its initial free-return trajectory, and required the use of the LM descent engine to get back onto a free-return trajectory. (The LMDE was then used for a pericynthion-plus-two-hours burn meant to change it to a true free return trajectory, and then at least once more for a midcourse correction burn; I believe there was a second midcourse correction done with the LM RCS because it was a small one.)

So it\'s possible, yes. But to make a survivable free-return to Kerbin without ANY burns after TMI? That\'s gonna be *hard*, man...

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  • 2 weeks later...

It\'s possible, but *very* difficult if you intend to have a survivable landing.

On Apollo, the free-return trajectory window was ~0.1 meters per second wide at TLI. The Saturn V simply couldn\'t inject the vehicle onto such a precise trajectory, and Houston couldn\'t maintain such a precise path, anyway.

[snip]

So it\'s possible, yes. But to make a survivable free-return to Kerbin without ANY burns after TMI? That\'s gonna be *hard*, man...

I was intrigued by this problem so I ran some numbers. It turns out that you are quite right, rdfox, even though it isn\'t as sensitive a problem in KSP as it sounds like it is in the real world. I found that a difference in speed of plus or minus 1.3 m/s, post TMI burn, was enough to make the difference between being too high for re-entry and re-entering too steeply ... Here\'s a plot showing the two cases I tried.

SEdiG.png

Both cases were for the spacecraft configuration shown below, with 100% full LFT and 100% full RCS tank at the start of the TMI burn, and orbiting in a circular orbit at 152.0 km. In both cases, the burn would start the instant that the centre of the Mun\'s disk is on the horizon, and would use only 33% throttle. All of this is critical because the starting fuel mass, throttle setting, starting altitude and TMI burn location affect the rate of acceleration and the resulting trajectory\'s orientation relative to the Mun.

index.php?action=dlattach;topic=4575.0;attach=7710;image

If you were to nail the 89.4 second burn, you\'d expect to end up in a 153.3 km x 13550 km orbit that would set you up for a 240 km high Munar periapsis once you transitioned into the Mun\'s SOI. Your return trajectory would intercept Kerbin\'s atmosphere (I\'m defining the atmospheric boundary somewhat arbitrarily as 48 km) with an angle of 21 degrees below the horizon. If you keep your heading centred in the prograde recticle throughout the TMI burn, your post TMI burn speed should be 2971.5 m/s at an altitude of 159.1 km.

Similar precision flying the 89.3 second burn would find you in a 153.3 km x 13280 km orbit, set up for a 189 km high Munar periapsis. In this case though, your return trajectory would only take you within 112 km of Kerbin. MECO would be at 2970.2 m/s. That\'s a difference of only 1.3 m/s between the two cases.

I\'ve used the numerical model that I used for these calculations to plan four different missions (my Kerballo 8, 10 and 11, plus a mission to geo-synchronous orbit directly over the KSC), and I\'ve found it to be accurate. The main limitation is my flying ability.

If anyone wants to try flying these trajectories, I\'d recommend setting your throttle to 33% before activating the TMI stage. That way you get the throttle setting you need just by hitting the space bar. I\'d also recommend retarding the throttle at 2965 m/s in the 89.3 second burn case and at 2966 m/s in the 89.4 second burn case. By the time you get the throttle down, you\'ll be within a m/s or two of the intended speed. I usually trim my orbit post-TMI burn using the RCS. It usually doesn\'t require more than a couple of m/s delta V to achieve the correct apoapsis.

I\'ll try flying these profiles myself too, but I probably won\'t have time for a couple of weeks. Somehow I seem to have more time for calculating trajectories and planning flights than I do for flying...

PH

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  • 2 weeks later...

Did you model ksp single gravitational body physics? The Mun has zero pull untill you enter its SOI, then kerbin has no effect. I\'m not sure how the SOI is defined though. Or how large the difference is on a free return trajectory.

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Did you model ksp single gravitational body physics? The Mun has zero pull untill you enter its SOI, then kerbin has no effect. I\'m not sure how the SOI is defined though. Or how large the difference is on a free return trajectory.

I first created my numerical model as a restricted three body model (one where only the parent body and natural satellite have significant masses; the spacecraft\'s mass is assumed to be zero) and later added an option to enable a 'KSP' mode. In the KSP mode, only Kerbin\'s gravity affects the spacecraft while it is outside the Mun\'s SOI. I then 'turn off' Kerbin\'s gravitational effects and “turn on†the Mun’s once the spacecraft enters the Mun\'s SOI. I have compared the two models and found them to be similar, but with a few notable caveats and exceptions. I posted a plot in another thread that compares free-return trajectories in the two models.

http://kerbalspaceprogram.com/forum/index.php?topic=4615.msg54462#msg54462

Unfortunately, direct comparison of the two models is difficult. Not only because the Mun and Kerbin\'s gravity do slightly affect the spacecraft whether it is inside or outside the the Munar SOI (the SOI only defines the region where one body\'s gravity dominates), but also because there\'s a notable inconsitency in the Mun\'s orbital speed in the game.

There is an equation that calculates the speed of a satellite of significant mass orbiting its parent body in the restricted circular three body case. That equation predicts that a satellite having the Mun\'s mass and orbiting a parent body with Kerbin\'s mass in a circular orbit of 12000 km radius, should be moving at 547.4 m/sec not the 542.5 m/sec used in the game. My three body model independently corroborates the 547.5 m/sec value, but the 4.9 m/s difference makes it impossible to directly compare the two models because the Mun is in a different place at a given time in one model vs. the other.

In my 'KSP' 3-body model, I artificially restrict the Mun\'s speed and position to move in a 12000 km radius circle 'on rails' at 542.5 m/s, so that it more correctly reflects the game\'s physics. This was a relatively easy change to implement once I had the restricted three body model working. I used the “KSP†3 body model to calculate the trajectories I described earlier in this thread.

I hope this helps.

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  • 2 weeks later...

I actually managed a free return orbit by pure, dumb luck. And of course I didn\'t think to record any evidence of it... Once I realized what was happening, I let it swing by the moon and back to Kerbin, where I entered the atmosphere at what seemed like a reasonable angle and landed safely. No idea if the angle would have caused my capsule to burn up if heating had been in the game, though.

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Interesting that this thread was revived today. Coincidentally, I just completed this challenge the other evening and was going to post about it tonight... At least now I won\'t get critcized for necrobumping! What is it with those criticisms anyway? I kind of like the fact that some old threads are dredged up and dusted off from time to time. Often they\'re quite interesting. The drag race to 100 km altitude thread is a good example... But I digress.

Anyway, in fairness, it has been a couple of weeks since this thread was current. I got distracted by the Other Planet Orbit - rendevous (and optional return) thread. For anyone who hasn\'t checked that thread out, it is an interesting challenge. Especially the returning to Kerbin part.

Back on topic, I planned the trajectory for this mission a little differently than I described in my earlier post in this thread. I planned it so that they\'d re-enter a little more steeply than my earlier plan. I wanted a little margin for error because I wanted to succeed on the first try; I only get one or two chances a week to fly a longer mission. And in the game\'s current incarnation, it doesn\'t seem to be possible to turn your Kerbals into goo or to burn them up during reentry anyway.

I flew this challenge real-time. I\'ve found that you end up in a slightly different location when orbits are propagated normally vs. when warping on rails. I have a sneaking suspicion that there\'s a numerical imprecision bug in KSP or one of its supporting libraries because large orbits seem to 'grow' slightly while propagating in 1x or 2x warp. A tiny imprecision the forces acting on the spacecraft could explain this. As I understand it, no forces act on the spacecraft while warping at 5x or above. But I digress once again... The fact of the matter is that the game\'s physics are only so precise, there is a slight difference if you use warp or not, so I stuck to one option: That of not using warp. It worked out and my Kerbals made it home after making a lap of the Mun.

Here are my screen shots:

LytLQh.jpg

nsjTnh.png

rMbIUh.jpg

Yeah, I\'d scream too!

9QUyFh.jpg

Outbound:

rYpC1h.png

Munar flyby:

jQnv0h.png

Return:

sTYWSh.png

Predicted as 278.5 km at TMI+ 5:34 (actual 288.19 km at TMI+5:30)

ZFHdj.png

Kerbol, Kerbin, Mun

HGLuh.png

Predicted at TMI+ 11:23 (actual at TMI+11:24)

XW2WE.jpg

9n05Oh.png

Taking inspiration from F. Abilleira at JPL\'s 2011 Mars Science Laboratory Mission Design Overview article, I also tried to predict my Kerbonaut\'s reentry location. If a numerical integration is good enough for planning Curiosity\'s mission trajectory, then why can\'t the technique also be accurate enough for Kerbal? (I recently overheard a friend commenting about me to another of my friends that 'I like to do recreational math'. What\'s wrong with that? Everybody\'s got to have a hobby...)

xBYSHh.png

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I think most of the imprecision is from the floating-point numbers used for physics calculations. At high speeds, a seemingly negligable margin of error becomes pretty noticable.

/Originally wrote 'margarine of error'; I assume it\'s a Kerbin staple :V

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-snip-

Slightly related question: how do you accurately measure and determine lead-lag angles? The orbit mechanic happily calculates these angles for me, but I don\'t really know how to apply them.

Oh, and when trying a timed burn: does the engine shut-off time matter? How do you deal with that?

(KSP really needs an engine on-off button)

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