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MontBlanc2012 MontBlanc2012 is offline
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Default Low energy Moon-Earth transfer
by MontBlanc2012 11-13-2017, 06:57 AM

Here (below) is a simple challenge designed to test your thinking about ways of making transfers from low lunar orbit to the Earth.

The scenario set out in the attached .scn file is as follows: A Delta Glider starts in a 20 x 20 km lunar orbit with a total 700 m/s of main fuel and 10 m/s of RCS. The challenge is to land anywhere on Earth. For the purpose of this exercise, achieving an Earth approach orbit with an Earth-centric periapsis < 70 km altitude constitutes a 'landing'.

Now, at first blush, this sounds simple enough - after all, it is one of the first 'missions' that budding Orbinauts will set themselves. However, Apollo-style Moon-Earth transfers typically require around 1,000 m/s of fuel. But in this challenge, the DeltaGlider starts with just over two-thirds that amount. In fact, the fuel needed to lunar escape from a 20 km parking orbit is 692 m/s - so, you have barely enough fuel to 'escape' from the Moon let alone set up a low perigee encounter.

Finding a Moon-Earth transfer solution that meets the rather restricted fuel constraints of this scenarion requires a little 'out of the box' thinking. As will all problems of this kind, the solution is quite simple once you frame the problem in the rightway; and devilishly hard if you don't. But once you know what you are doing, it's quite easy to set up in Orbiter using standard tools.

As a hint, you may find that TransX is not of much use in finding a solution.

For those that easily manage to find the solution, well done! - and you may wish to pay appropriate hommage to Messrs. Lagrange and Belbruno.

I'm happy to provide a solution upon request.

As for the challenge scenario itself, here it is:


Code:
BEGIN_DESC
Delta Glider in a 20 x 20 km lunar orbit with a total 700 m/s of main fuel and RCS.  The goal is to land anywhere on Earth.  For the purpose of this exercise, achieving an Earth approach orbit with an Earth-centric periapsis < 70 km altitude constitutes a 'landing'.
END_DESC

BEGIN_ENVIRONMENT
  System Sol
  Date MJD 52006.7542552436
  Help CurrentState_img
END_ENVIRONMENT

BEGIN_FOCUS
  Ship GL-NT
END_FOCUS

BEGIN_CAMERA
  TARGET GL-NT
  MODE Cockpit
  FOV 40.00
END_CAMERA

BEGIN_SHIPS
GL-NT:DeltaGlider
  STATUS Orbiting Moon
  RPOS 1677740.022 -137005.718 -506935.076
  RVEL 477.7254 -65.8104 1598.8449
  AROT 15.788 -10.126 42.772
  AFCMODE 7
  PRPLEVEL 0:0.015058 1:0.004584
  NAVFREQ 0 0 0 0
  XPDR 0
  HOVERHOLD 0 1 0.0000e+000 0.0000e+000
  AAP 0:0 0:0 0:0
END
END_SHIPS
And for those requiring a .scn file, see the attached .zip file.
Attached Files
File Type: zip Moon Earth transfer.scn.zip (1.2 KB, 5 views)
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Old 11-14-2017, 01:27 PM   #2
MontBlanc2012
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And just as a demonstration, here is a link to a YouTube video demonstrating the basic scheme (contains spoilers!):

https://www.youtube.com/watch?v=m4FD...ature=youtu.be

Sadly, I fluffed the step back to real time upon Earth approach and ended up scooting by Earth with a ~ 70 km encounter. But, meh!, not a big deal.

Last edited by MontBlanc2012; 11-23-2017 at 07:23 AM.
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Old 11-17-2017, 05:30 PM   #3
BrianJ
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Hi,
I couldn't resist this one :-)
I managed to land close to Overberg base in SA after a fairly rough reentry (8g at one point) with 220g fuel remaining in RCS.

I used the IMFD "Delta Velocity" + IMFD "Map - Plan View" to plan the escape burn. Set up a Prograde burn of 670m/s and tweak the burn time until you get reasonable solution, then tweak the dV.
Here's my escape burn solution:
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Coming up on Moon-flyby:
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MCC at first Apogee:
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Scraping the tanks for a last Perigee adjust burn before reentry:
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On the ground:
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Thanks for the challenge, I enjoyed it
Tip o' the hat to Mssrs.Lagrange and Belbruno!
Cheers,
Brian
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Old 11-21-2017, 01:51 PM   #4
MontBlanc2012
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Quote:
I used the IMFD "Delta Velocity" + IMFD "Map - Plan View" to plan the escape burn. Set up a Prograde burn of 670m/s and tweak the burn time until you get reasonable solution, then tweak the dV.
A very reasonable solution.

Although we can set up low energy transfers of this kind using IMFD, we can't ensure that these transfers are 'optimal'. However, I suspect that with a bit of tweaking with dates and prograde quantities, we an get to within 5 - 10 m/s of the optimal amount without much effort.

You might want to try the inverse low energy transfer from low Earth orbit to lunar ballistic capture. The trajectory is essentially the time-reversed, mirror image of the Moon to Earth transfer. I haven't tried this myself yet, but I'm sure it is doable in Orbiter.
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Old 08-27-2018, 08:36 PM   #5
Ajaja
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Sorry for necroposting.
I've been having fun with GMAT for last days learning this tool and trying to solve these old Orbiter challenges.
Looks like here we can use Earth gravity to minimize delta-V when we escaping the Moon and then we can use a gravity assist maneuver near the Moon to fly away from the Earth to give the Sun a chance to return us back home.
For example, this is 664.5 m/s trajectory, calculated in GMAT (a single prograde burn at t+3900 sec).

----------
P.S. Don't try to use these numbers in Orbiter with this scenario, I've mixed up frames of reference
It turned out that there are even better solutions without grav.maneures. Eg. 650 m/s at t+530s.
Attached Thumbnails
moon_earth.png  

Last edited by Ajaja; 08-29-2018 at 12:28 PM. Reason: edited
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Old 08-28-2018, 10:35 AM   #6
BrianJ
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Beautiful! Must get GMAT back on board :-)
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Old 08-28-2018, 12:20 PM   #7
MontBlanc2012
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Very nice.

As with BrianJ, it looks as if I'm going to have download GMAT.
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Old 09-26-2018, 06:02 PM   #8
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Thanks MontBlanc2012 for this beautiful challenge. This one definitely deserves to be included in challenges of the next Orbiter update.

I had seen this challenge when it was first posted and in my initial attempts, I barely made it back to Earth using all the fuel at my disposal - even the RCS.

Then I saw Ajaja's post about the GMAT 650 m/s solution, so I had to try again and see if I could reproduce it with IMFD.

After a few minutes of playing around with the prograde dV and time variables, it seemed that 647.18 m/s was the "magic number".
Even 0.01 m/s less than that and IMFD can't find any solution that gets back to Earth.

But the 645.18 - 646 m/s range seems "special". Those solutions end up in a nice 3 Moon encounters trajectory that ends up with a straight path to Earth's core!



I tried it first "just as is". Single burn from the Moon, no corrections, just to see where I'd end up.
The result was very promising, I missed the Earth by "only" ~30M meters in the retrograde direction.

So, in the next attempt, I performed the same burn but this time aimed for a 30M PeA in the prograde direction. Result: smack right in the center of Earth, using a single burn of 645.5 m/s.

"If PeA = 30M Pro" gets me to the center, surely 36M meters will get me where I want, right?" Wrong!
I tried all the burns that started from 35M and ended up to 105M meters above Earth's surface (adding 10M in each attempt) and they all ended up hitting the Earth near the center.
In all the attempts, the difference in the dV was miniscule, but after that 3rd encounter with the Moon, you can see the PeA relative to Earth shrinking until it gets near the center.

Below is a video of one of those attempts and the best dV result I had. (Being extremely careful with the RCS used for orientation and using only a single burn).




Thanks again for this wonderful challenge!
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Old 09-29-2018, 11:27 AM   #9
MontBlanc2012
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Quote:
But the 645.18 - 646 m/s range seems "special". Those solutions end up in a nice 3 Moon encounters trajectory that ends up with a straight path to Earth's core!
I read this while holidaying in Croatia recently. While driving on the freeway from Dubrovnik back to Zagreb to catch my return flight home, I pondered dgatsoulis' solution to the problem. Judging by the IMFD trajectory plan, what I think is going on is actually quite complex with the trajectory consisting of a number of distinct parts. A stylised representation of the trajectory is as follows:

1. An impulsive burn to transfer the vehicle from Low Lunar Orbit to a planar Lyapunov orbit around EM L1 - (i.e., the Lagrange point between the Earth and the Moon). Approach to the Lyapunov orbit is along the stable manifold of the trajectory.

2. Because the Lyapunov orbit is unstable, as it approaches the Lyapunov orbit, it transfers quickly from the stable manifold to the unstable manifold of the same orbit. This unstable manifold enables a transfer to EM L2 - (i.e., the Lagrange point on the far side of the Moon.) This transfer trajectory is ballistic and appears to require approximately three reasonably close passes of the Moon (as per Ajaja's suggestion in another thread.)

3. Approach to EM L2 is roughly along the stable manifold of the L2 point. Again, and because L2 is unstable, as it approaches L2, it quickly transfers to the unstable manifold leading away from the Earth-Moon system. With correct timing, this unstable manifold connects with the stable manifold of the Sun-Earth L1 or L2 (not sure which).

4. Finally, as it approaches SE L1/L2, it transfers to the unstable manifold which again, with correct timing leads back to a close encounter with Earth.

This strategy for solving the problem - i.e. to target EM L1, then a ballistic transfer to EM L2, leading to a transfer to SE L1/L2 and then back to Earth - appears to be the least-cost solution to the problem.

The logic of this transfer mechanism warrants further explanation, but I'll need to assemble some graphics to elucidate the mechanism (and to show why there might be a lower bound to the dV required to implement this strategy.)
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Old 09-29-2018, 11:54 AM   #10
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The outline makes perfect sense of how the flight "felt".
Especially the part leaving the EML2 and coasting to SEL1 (it was the L1 in this case).
It felt like there was a giant funnel in that region of space and I was getting to it at just the right speed to have it grab me and throw me back in towards Earth.

What was really surprising was the range of the different dV and time combinations that produced the exact same result. To someone untrained (like me), it would seem that there is only one extremely special combination of burn variables that get you to that region, but I was happily surprised to find out otherwise.

Looking forward to reading more on this from you when you have the time.
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Old 10-01-2018, 05:10 AM   #11
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Quote:
Looking forward to reading more on this from you when you have the time.
Slowly getting over the jet lag and have had a chance to look more closely at your ('dgatsoulis') solution. When looking at these low-energy solutions, it often makes sense to transform the problem to rotating coordinates in which the position of the Earth and the Moon (in the case of the Earth-Moon system) are fixed. After carrying out the transformation, this is your solution trajectory.



Here, the Earth is located at the coordinate origin (0,0); and the Moon is located at the point (1,0). The two red dots are the collinear Lagrange points EM L1 and EM L2. The green oval is a planar Lyapunov orbit about L2. The blue line is the solution trajectory path starting from Low Lunar Orbit and ending 2.5 M seconds (or 28 days) later as it departs the Earth-Moon system via L2 on its way to SE L1 and then back to Earth.

Having plotted this graph, it seems clear that the planar Lyapunov orbits around EM L1 are not being used here. The key, really is the small planar Lyapunov orbit around EM L2 shown in green. The stable manifold of this orbit traces back in time (via a few laps around the Moon) to a low passage around the Moon. On the other hand, the unstable manifold of this Lyapunov orbit leads to SE L1 and then back to the Earth. In short, the green planar orbit provides the connection between lunar departure and Earth arrival.

Now, in your solution, you don't spend much time 'on' the Lyapunov orbit: as you approach the orbit, you transition quickly to the unstable manifold and depart the Earth-Moon system for SE L1. However, one could imagine using the Lyapunov orbit as a kind of 'parking orbit' where you approach the orbit ballistically. This would look something like:



Once on the Lyapunov orbit, and with the help of some minimal station-keeping, one could then choose a departure time along the unstable manifold leading to SE L1. Of course, with a single burn in Low Lunar Orbit, it isn't possible to carry out this sequence of operations, but using Lyapunov orbits (and other centre manifold solutions) helps provide structure to setting up and executing transfers of this kind.

As for the lower bound cut-off, this is what I think is going on:



In rotating coordinates, there is a conserved quantity (often called the Jacobi constant) for any given trajectory. This functions like conserved energy in inertial coordinates and can be used to define forbidden regions where the vessel cannot go. In the above diagram, the forbidden region for some trajectory the Earth-Moon system is shown in grey. For low enough values of the Jacobi constant, one gets a bottleneck close to L2. Any trajectory with that Jacobi constant has to pass through the bottleneck if it is to make it 'outside' of the Earth-Moon system. The smaller the Jacobi constant, the tighter the bottleneck. At some critical value of the Jacobi constant, the bottleneck closes entirely and access to the 'outside' is forbidden. My guess is that your 645.18 m/s solution is close to the point where the bottleneck closes leaving just a fairly small neck through which the trajectory must pass. Reduce the dV further and any trajectory that does make it 'outside' no longer has the correct timing to reach SE L1.

---------- Post added 10-01-18 at 05:10 AM ---------- Previous post was 09-30-18 at 10:58 AM ----------

Out of curiosity, I took a closer look at my 700 m/s example solution. Plotting in rotating co-ordinates in the same way that I did for dgatsoulis' 645.2 m/s, one gets the following:



And for comparison, here is dgatsoulis' solution again:



My 700 m/s solution just takes advantage of the dynamics around the SE L1/L2 Lagrange point - and makes no use whatsoever of the EM L1/L2 Lagrange points. It's a fair bet that the best low-energy escape roots from Low Lunar Orbit (LLO) back to Earth follow the dgatsoulis model - i.e., a connection from LLO to EM L2; then to SE L1/L2; and then back to Earth rather than my more simple solution of a direct transfer from LLO to SE L1/L2 and then back to Earth.

(Although it is a different solution - and I don't have the details to plot it properly, plot it - Ajaja's 655 m/s solution seems to be much the same as dgatsoulis' in that it too makes use of the dynamics around EM L2.)
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