How to get one of them fancified Apollo-style free return trajectories...

[capi3101]

Greetings,

Ever since I started going to Mun in the 0.18 Demo, I've pretty much wound up with the same orbital shape every time I've gone; I get a Mun encounter after reaching my apoapsis, it looks like a curly-Q as I go through the Mun encounter, then if I continue on without retroburning into Munar orbit, I get shot back out on a trajectory that takes me reasonably close to Kerbin. This is what I've gotten every time, and I've recently learned that it is a type of free-return trajectory (a "cislunar" trajectory). What I'd like to know is how to get one of those Apollo-style free turn trajectories - a circumlunar trajectory. What I have been doing is pretty much following the "100-degree" rule, burning prograde along a maneuver node at Kerbin's six-o'clock when the Mun is 100 degrees away from Kerbin's twelve-o'clock (Mun's at about 3:20). Where does the Mun need to be for a circumlunar trajectory (i.e. what is the phase angle)?

[Geschosskopf]

Once you have an encounter with Mun from pulling the prograde gizmo, try pulling the gizmo on the inside of the curve. This will slide the encounter forward along Mun's orbit and change its geometry. What you want is to pass directly in front of Mun into an immediate and very low periapsis. This will make you do 1 retrograde orbit of Mun and out the other side back at Kerbin, no other burns required.

It's not hard to do per se, it just takes careful fiddling.

[Scorpi15]

To get a free return trajectory you need to burn more then you usually would for a normal Mun transfer.

[Johnno]

Made a quick tutorial for this, I'm sure there's plenty of ways to go about it, but this is the way I'm used to doing it and I find it easy enough without having to look up phase angles or such.

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[capi3101]

"Horizontal marker"...I assume that's talking about the radial in/out set on a maneuver node (the blue ones).

What about it, guys? Does it take more delta-V to get into a circumlunar trajectory as opposed to cislunar one? It strikes me like it wouldn't.

[rdfox]

The free-return trajectory used on Apollo *did* require more delta-V than a simple Hohmann transfer orbit, but the big reason for that is that it was selected as a *fast* transfer, taking about three days to make the transit, as opposed to the roughly 14 days that would be required for a Hohmann transfer. While it required more delta-V than the Hohmann transfer, the vast reduction of the required consumables allowed a much lower total mass to LEO than a Hohmann transfer would for the Apollo mission.

I don't know if there's any way to do a Hohmann transfer orbit with a free-return trajectory--the only way I can think of would be to make it a long, lazy loop around the far side and thus insertion into Munar orbit would be a prograde orbit rather than the retrograde orbit used by the Apollo program. So more like a big O instead of the Apollo-style figure-8...

[Tw1]

I don't know if there's any way to do a Hohmann transfer orbit with a free-return trajectory--the only way I can think of would be to make it a long, lazy loop around the far side and thus insertion into Munar orbit would be a prograde orbit rather than the retrograde orbit used by the Apollo program. So more like a big O instead of the Apollo-style figure-8...

That's how I did it in my Apollo replication. One of a few mistakes.

It was a cool mission, none the less.

It's not unlike what's described in the OP, just tweaked a bit.

It made it a longish flight, but flight to the Mun is much shorter than to the Moon anyway.

Edited August 5, 2013 by Tw1

[capi3101]

I managed to get a circumlunar orbit over the weekend. Required a correction burn, but I did get the trajectory. Took a second burn to adjust my inclination; I don't count that one, though.

I did make a subsequent attempt without further success, though. I suppose this is just something that's going to require practice.

[Geschosskopf]

"Horizontal marker"...I assume that's talking about the radial in/out set on a maneuver node (the blue ones).

What about it, guys? Does it take more delta-V to get into a circumlunar trajectory as opposed to cislunar one? It strikes me like it wouldn't.

I'm no rocket scientist although I used to be married to one, so take this with a grain of salt .

Getting into a retrograde Munar orbit (circumlunar) seems to be a function of 2 things. 1st is the phasing of your Munar encounter, so you're out in front of it on its orbital path instead of behind or beside it. Second is having enough prograde velocity relative to Mun's orbit that you stay out front as Mun catches up from behind. You have to stay out in front long enough to cross Mun's orbit, after which Mun's gravity whips you around the far side. If you don't have enough prograde velocity, then Mun will suck you into a prograde orbit (cislunar) behind it.

Compared to a conventional Munar approach, the difference in encounter phasing is free (in theory). That's because you can get it simply by moving your maneuver node forward along your orbit back at Kerbin (that is, burning after instead of at Munrise). However, because the stock game doesn't give you any angular readouts, doing this consistently is quite difficult. Thus, you usually need to add some radial velocity to your ejection burn. Thus, in practice, the phasing usually costs you a bit extre delta-V.

The real source of extra delta-V for such an orbit is the need for the higher velocity at the encounter. First off, you have to burn enough prograde to cross Mun's orbit, whereas with a conventional approach you don't quite need to get out to the orbit, relying on Mun's gravity after SOI change to take you the rest of the way. Secondly, you have to arrive with a fair amount of prograde velocity relative to Mun's orbit so you can stay out in front long enough. This usually means you have to add a noticeable amount of inward radial velocity, which you don't need at all with a conventional approach.

Finally, the retrograde circumlunar free return requires more delta-V to get captured out of. This is because you're moving faster than normal at encounter, and have a pretty low Munar periapsis to point you right back into Kerbin's atmosphere, which pulls you in closer and thus increases your speed at Mun. If you didn't do an Apollo 13 so no longer need the free return, you've got a lot of braking to do now .

[Hour Hero Yes]

I don't know if you use Mechjeb or have ever played with the Conics setting in the config file, but different settings change the way your projected orbit looks after changing SOI's. One of them lets you see exactly what it looks like when you make the SOI transfer, I think it's Conics 1?

[Geschosskopf]

I went back and started playing with doing free returns because something was nagging me about what I'd said previously. So here's a better explanation, and also an easier way to set it up.

First off, an example:

As you can see, this burn requires 899.1 m/s. For this, you get a 116km circumlunar retrograde periapsis followed by a 50km Kerbin periapsis, which will eventually bring you to the surface although you have to wait several long orbits nearly back out to Mun.

Lowering the Kerbin periapsis into the atmosphere like this has 2 effects. First, it increases the amount of delta-V required. Second, it increases the Munar periapsis, so if you like to land low and shallow (which I recommend in 0.21 so you can see all the small craters and no land in them), then you have to do a fairly large burn not only to capture but then to come down. If you want a lower Munar periapsis, you might not get back into Kerbin's atmosphere for free, but it won't be much of a de-orbit burn. Perhaps if you have severe OCD and tweak the node enough, you might find a sweet spot where you get a close Munar periapsis and a re-entering Kerbin periapsis, but trying to do this exceeded my patience.

In any case, 899.1m/s is not too much more than the standard 860m/s shown on the Wiki delta-V map as needed for Munar intercept. However, you also have to spend more to get captured and adjust your orbit so pick your poison.

So, how did I set this up, which is what the question was about.

While it's possible to do the orbit as I described above, using the inward radial gizmo, that's actually very difficult to get everything lined up. It's a lot easier to use the outward radial gizmo. Place the node a bit AFTER Munrise and pull the prograde out until you can reach Mun's orbit. Then slide the node around Kerbin until you get an encounter. At this point, start pulling the outward radial gizmo. It shouldn't take much to turn your orbit into a retrograde figure-8 around Mun.

[jozxyqk]

Just a quick question. At first I was really baffled by the idea of a "free return trajectory" for a mun landing mission. To me, this meant you have to land and get back into orbit to catch your ride home VERY quickly (is this even possible without stupid amounts of fuel?). Can someone confirm that a "free return trajectory" is not actually completed for a mun landing mission and that Apollo simply started out with one for two reasons: 1. safety - if something went wrong you just wait and go home again, 2. speed - as previously said to make eating and breathing easier. That once near the moon, a free return trajectory is simply abandoned and the orbit circularized. The idea that you can return to a free return trajectory is false as you've used fuel to do so and hence isn't free.

[Geschosskopf]

Can someone confirm that a "free return trajectory" is not actually completed for a mun landing mission and that Apollo simply started out with one for two reasons: 1. safety - if something went wrong you just wait and go home again, 2. speed - as previously said to make eating and breathing easier. That once near the moon, a free return trajectory is simply abandoned and the orbit circularized. The idea that you can return to a free return trajectory is false as you've used fuel to do so and hence isn't free.

AFAIK you're correct. The free return was in case the had an Apollo 13 on the way out, so they didn't have to send a rescue mission (which they couldn't anyway). But if all was well getting to the moon, they put the CM/SM into a regular orbit for however long the other guys were playing around on the moon.

[rdfox]

Correct; the free-return trajectory was basically a safety feature in case the Apollo CSM's Service Propulsion System engine didn't ignite for the course-correction burn that lowered periapsis to the target 60 miles. (Given the simplicity of the SPS engine, with only four moving parts--two ball valves and two gimbal actuators--and the ability to make the burn successfully even with the gimbal actuators failed, it was felt that if it lit the first time, it would light for both Lunar Orbit Insertion burns and Trans-Earth Injection.)

Technically, the Apollo 13 mission wasn't actually a free-return trajectory; the tank explosion occurred AFTER the course-correction, and thus required some fancy footwork by the crew and Mission Control to get back onto a return trajectory, including the use of the LM's main engine in two separate burns, one at "pericynthion" (lunar periapsis) to basically do the TEI burn without ever actually entering lunar orbit, and one at PC+2 hours to further accelerate the return trajectory above normal and thus help with the issue of finding enough consumables to get back to Earth alive.

The standard Apollo mission did the course-correction burn about a day out from the Moon, to drop down to the planned periapsis, then did LOI1 at periapsis to drop into a nominal 60 x 160 nautical mile orbit, then did LOI2 the next time around to circularize the orbit at 60 miles. They then stayed for a day before undocking the LM and doing the landing; after the LM returned, the CSM would remain in orbit for another day or so before making the Trans-Earth Injection burn to return home. For the final three landings, this adds up to a total stay in lunar orbit of about five days for the CSM!

[karolus10]

Free return transit time is pretty low, ~5 hours. 890m/s ÃŽâ€V isn't much more than with regular transfer.

Edited August 13, 2013 by karolus10

[goldenpeach]

After you circularized the munar orbit,is there a way to go back in the free return trajectory?

[Kimberly]

After you circularized the munar orbit,is there a way to go back in the free return trajectory?

It's not a free return if you're already in a proper orbit around the Mun. So no, there isn't. But you can apply the same principle to get back to Kerbin, sure. You just have to burn at the right spot in your orbit.

[goldenpeach]

How can I know where is that spot?

EDIT:Can someone post a screenshot of a free return trajectory with the mechjeb node editor,I have tried several time to make a free return trajectory but I have never achieve one.

The node editor would show me the delta-v spent in the prograde marker and the delta-v spent in the radial in/out marker,wich will,I hope,help me a lot.

Thank you in advance!

Edited August 13, 2013 by goldenpeach

[karolus10]

Wen You are using free return (and similar) trajectory, You end up on clockwise orbit around mun. This mean that You must burn pro-grade on far side of the mun in parallel to mun orbit.

Also munar orbit insertion is done on far side of the mun.

This stuff is quite easy to do but need a lot of play with maneuver nodes (not really good, when it comes to fine adjustments and "sliding" nodes on orbit ).

Edited August 13, 2013 by karolus10

[capi3101]

Time out for a second...these free return trajectories put you back at Kerbin at the same Kerbin-relative altitude as when you left? If that's the case, I haven't got this down yet at all...the figure-eight is there but my closest point of return is about 700k.

[Frederf]

The return orbit is very sensitive to how you interact with the Mun's gravity. The most extreme case you exit the Mun with exactly zero orbital velocity relative to Kerbin and fall straight down on a degenerate (collapsed) orbit. What you're basically doing on free return is doing a no-assist gravity assist. You'll find that some Mun SOI interactions are gravity assist increasing the size of your orbit and others are gravity brakes decreasing them. A free return is finding a path through the Mun's SOI such that you are neither assisted nor braked and return to the orbital energy you had on interaction. In reality you want the slightest braking to take your original Pe down to something suitable for aerobraking.

I disagree with using radial component on the maneuver node as with different timing there is an entirely prograde solution for the same impulse vector and will be more efficient. When I do it (usually to dispose of the TMI stage) I set up a Hohmann to intersect Mun CoM and then increase prograde until I arrive ahead of the Mun which sets up a gravity brake. I shuffle the node location and reduce prograde dV until I get the minimum deltaV that produces the desired resultant orbit. You'll find if you "cheapskate" on deltaV you will not be able to both clear the Munar surface and brake enough to lower your resultant Pe. In this case more dV is called for to lower your resultant dV somewhat unintuitively.

[Geschosskopf]

Time out for a second...these free return trajectories put you back at Kerbin at the same Kerbin-relative altitude as when you left? If that's the case, I haven't got this down yet at all...the figure-eight is there but my closest point of return is about 700k.

Technically speaking, you actually want to be rather lower back at Kerbin than when you left. You start from say a 100km orbit, but when you get back you want a periapsis far enough down in the atmosphere to make you land without ever having to burn again after you eject. That's a truly "free" return (provided, of course, you don't stop at Mun, just coast through 1 retrograde orbit and escape it again). If you have to do a deorbit burn back at Kerbin, then it's not a "free" return.

It just takes some OCD fiddling with the node, both its position along your Kerbin orbit and how much you're burning in the various directions, to set this up. There are several different methods to set it up. I outlined a simple but somewhat inefficient method above, Frederf explains a different method that's more efficient but requires a bit more on the OCD side I think.

Personally, I don't think there's much point in going to all this trouble except for the specific purpose of recreating an Apollo mission as accurately as possible. This is because, unless your rocket explodes right at the end of the ejection burn and all you have left is a capsule heading away from Kerbin, a free return trajectory does you no good. I mean, you're doubtless going to Mun specifically to land there, not just do a fly-by, and landing on Mun takes you off the free return trajectory. Because the scenario where a free return trajectory actually does you some good is quite unlikely, you end up doing all this OCD fiddling for nothing.

That said, the basic concept can be useful in other ways, although they're not "free". For example, suppose you've built a big interplanetary ship in orbit out of multiple parts and you plan on aerobraking it where you're going, but you want to make sure it can stand the stress before your launch window. So set up an orbit to whip you out around Mun and back, then tweak your Kerbin periapsis on the way back to aerobrake at different altitudes to see what happens.

[Kimberly]

One benefit of free return maneuvers is that you can ditch any empty stages. While you perform your capture burn around the Mun, they'll keep on going and deorbit around Kerbin rather than causing debris.