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Is "Low-energy transfer" possible in KSP?


Nefrums

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I read an article on "Low-energy transfer" and wondered if this is possible in KSP.   And if they would require less dV than standard Hohmann transfers in KSP?

https://en.wikipedia.org/wiki/Low-energy_transfer

WSB transfer requires 3-Body simulation that we do not have in KSP, but I could not really figure out if the Low-energy transfers depended on that.

 

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1 hour ago, Veeltch said:

So it's another name for... gravity assist? Except somehow fancier? Like when you get a retrograde Tylo encounter and free orbital insertion?

No. KSP uses a patched-conics two-body approximation of gravity, and these low-energy transers require the influence of additional massive bodies to affect the gravitational field itself. It's the same phenomenon that creates Lagrange points.

Like @regex said, the Principia mod intends to incorporate an "N-body" Newtonian gravitational model to replace the Keplerian one we've got.

Edited by pincushionman
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Yes after further reading it is quite obvious that this requires n-body.

However I wonder if a mun transfer could be done cheaper with a kerbol gravity assist, in the 2-body world of KSP. It should be possible to exit Kerbin SOI and shortly after fall back into it.

But I suspect that a simple flyby of the mun before a second encounter would be simpler and cheaper dV wise.

 

 

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17 minutes ago, Nefrums said:

Yes after further reading it is quite obvious that this requires n-body.

However I wonder if a mun transfer could be done cheaper with a kerbol gravity assist, in the 2-body world of KSP. It should be possible to exit Kerbin SOI and shortly after fall back into it.

But I suspect that a simple flyby of the mun before a second encounter would be simpler and cheaper dV wise.

I don't see how that makes sense. A Kerbin eject trajectory costs more dV than a Mun intercept. Even if you manage to come back at a "nicer" angle, you will be going faster than escape velocity when you eventually intercept the Mun, and theres no way this can save you some dV. The only way to reduce or eliminate capture costs with Keplerian physics (disregarding aerobraking) is through the influence of a third child body to the one you want to be captured around. You can't work it out by going to a parent body.

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I disagree.    The size of the capture burn depends on the relative velocity of your ship and the body you what to be captured by.

In a hohmann transfer to the mun, when you intercept the mun at your AP, you ship is traveling much slower than the mun.  If you where to intercept the mun at your PE when coming in from an Kerbin escape trajectory your ship would be traveling faster than the mun but relative velocity would be lower,  (I think.. haven't actually done the math)

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3 hours ago, Nefrums said:

Yes after further reading it is quite obvious that this requires n-body.

However I wonder if a mun transfer could be done cheaper with a kerbol gravity assist, in the 2-body world of KSP. It should be possible to exit Kerbin SOI and shortly after fall back into it.

But I suspect that a simple flyby of the mun before a second encounter would be simpler and cheaper dV wise.

 

 

I've worked out a lower-dV-than-Hohmann way of getting from LKO to low Munar orbit. It's a bit of a simulation of the low-energy transfer method you mention in the OP, or at least the best you can do without 3-body path calculation. This way goes from LKO just high enough to touch Mun's SOI, Mun then pulls your path up just enough so you get a 2nd Mun encounter that pulls you up the rest of the way. To save even more dV a third pass of Mun then throws you out almost to the edge of Kerbin's SOI, where a deep-space manuever is made to raise your periapsis to Mun's orbit, you then finally approach Mun at a much lower speed than you would from a Hohmann transfer. The total savings is only about 5% over a Hohmann though. (Hohmann: 856+273=1129m/s, mine:844+16+209=1069m/s) In real life of course the Moon's 'SOI' goes all the way to Earth so you can use properly timed little bitty tugs from the Moon to slowly pull your orbit up from much lower.

Here's an album showing the approach (I then went on to Minmus, I was trying to land a stock Kerbal-X on both Mun and Minus and return to Kerbin without refueling).

I wonder if L1 could be simulated in a 2-body program by treating it as a small mass with a zero diameter and a small SOI, you'd have to force it to orbit Kerbin more slowly than it should though. I don't know if the game engine would allow that.

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20 minutes ago, TotallyNotHuman_ said:

@tutike2000 Whoa. Enlighten me if I'm wrong but is this ship encountering Laythe in such a way that Laythe captures the ship into a stable orbit?

How is that even possible?

That is what the picture is showing. I'm not 100% sure how it is possible, but If the relative velocities were very low, and the body's gravity bent the trajectory of the spacecraft's orbit just enough, a perfect orbit could be achieved without any insertion burn. This would, of course, require an intercept window with almost no margin of error, and an orbit that had a velocity very close to the target body.

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5 hours ago, LordKael said:

That is what the picture is showing. I'm not 100% sure how it is possible, but If the relative velocities were very low, and the body's gravity bent the trajectory of the spacecraft's orbit just enough, a perfect orbit could be achieved without any insertion burn. This would, of course, require an intercept window with almost no margin of error, and an orbit that had a velocity very close to the target body.

It is theoretically possible, at least with 2 body physics, N-Body I'm not sure whether it is, but in this situation the gravitational force starts instantly when you enter the SOI of laythe and the only thing that keeps it from getting ejected out of orbit is the lack of Jool's influence that had previously been there, and by it switching SOI the craft defies the laws of physics and creates phantom energy, which happens everytime you switch SOIs but here it was so close that the small phantom energy was just enough to place it in a stable orbit.

Huh.

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2 hours ago, nosirrbro said:

It is theoretically possible, at least with 2 body physics, N-Body I'm not sure whether it is, but in this situation the gravitational force starts instantly when you enter the SOI of laythe and the only thing that keeps it from getting ejected out of orbit is the lack of Jool's influence that had previously been there, and by it switching SOI the craft defies the laws of physics and creates phantom energy, which happens everytime you switch SOIs but here it was so close that the small phantom energy was just enough to place it in a stable orbit.

Huh.

Not saying you're wrong, but you may want to consider the floating point errors in the tiniest decimals of the orbital parameters which I think have a bigger influence here and in almost all subtle physics behaviors in KSP.

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That doesn't really look like a glitch or floating-point error, just a weird consequence of SOI-physics. The ship and Laythe are already in very similar orbits. The encounter is near the ship's apoapsis, where its orbit gets high enough to cross into Laythe's SOI, and its relative velocity at that point happens to be one that gives it a stable orbit. (And since Jool ceases to exist for the ship at that point, nothing else matters.)

Normally, encounters have the ship on a very different orbit from the target, so it has a much higher relative velocity at the encounter point, way too fast for a stable orbit. This can happen in real life (captured asteroids) though obviously not in the same way (probably has something to do with lagrange points...).

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12 hours ago, kotomikun said:

That doesn't really look like a glitch or floating-point error, just a weird consequence of SOI-physics. The ship and Laythe are already in very similar orbits. The encounter is near the ship's apoapsis, where its orbit gets high enough to cross into Laythe's SOI, and its relative velocity at that point happens to be one that gives it a stable orbit. (And since Jool ceases to exist for the ship at that point, nothing else matters.)

Normally, encounters have the ship on a very different orbit from the target, so it has a much higher relative velocity at the encounter point, way too fast for a stable orbit. This can happen in real life (captured asteroids) though obviously not in the same way (probably has something to do with lagrange points...).

 

"It is theoretically possible, at least with 2 body physics, N-Body I'm not sure whether it is, but in this situation the gravitational force starts instantly when you enter the SOI of laythe and the only thing that keeps it from getting ejected out of orbit is the lack of Jool's influence that had previously been there, and by it switching SOI the craft defies the laws of physics and creates phantom energy, which happens everytime you switch SOIs but here it was so close that the small phantom energy was just enough to place it in a stable orbit"

Pretty much exactly what I said.

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There isn't any "phantom energy"... changing SOI doesn't change your velocity, just your frame of reference, and which object's gravity affects you. That doesn't defy any laws of physics, but it's an approximation, so in some cases it does weird things like this that don't really make sense.

But "someone said something I already said on the internet" is a close companion of "someone is wrong on the internet," so I expect the second one to appear shortly...

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