Orbital Maneuver Question

[SYDWAD]

I would like to try make a low fuel orbit reversing maneuver and i would like to know if its possible, i made a crude drawing of my question. 

or Jule or whatever

[Cpt Kerbalkrunch]

You can get a pretty good boost from Tylo and Jool. If you have some serious power, so much the better. I think I would try Eeloo at it Ap, though. But low fuel? No way. Unless somebody knows some moves I've never heard of.

[regex]

The further out you are (where orbital velocity is the lowest) the easier it is to make a plane change, which is probably the lowest fuel maneuver to accomplish what you want done. You could also potentially use a planet to sling you most of the way through the plane change.

[Red Iron Crown]

Moved over to Gameplay Questions.

[Spricigo]

 

The deltaV requirement for inversion of orbit will be twice your orbital velocity at the maneuver node. Certainly gravity assists may help alleviate the cost, but a single flyby around Duna is out of question for sure.

Jool can help more since is farther away from Kerbol (lower orbital velocity) and is more massive. But reaching Jool has its own cost.

[Aegolius13]

If I understand this right,  this is equivalent to a "figure 8" maneuver around the moon/Mun, to reverse one's orbit around earth/Kerbin.  EDIT: not quite a figure 8 actually; that's more a standard free return trajectory that does not reverse direction. 

So yeah, that's possible in theory,  but you need a powerful gravity assist relative to your orbital velocity.   Jool seems like the best bet,  but I'm not sure if the math works out our not. 

Edited April 19, 2017 by Aegolius13

[Streetwind]

10 hours ago, SYDWAD said:

I would like to try make a low fuel orbit reversing maneuver and i would like to know if its possible

As other have said: a single flyby won't give you nearly enough of an assist to turn you around. Remember that every planet also moves counterclockwise around Kerbin. How can something that moves counterclockwise push your ship in a clockwise direction? It only works for small amounts of push.

However, how much time do you have? If you are willing to invest a lot of flight time, you can make your orbit reversal very cheap by executing what's called a bi-elliptic transfer. First, raise your apoapsis very very high. Past Eeloo kind of high. Then travel to apoapsis. Once out there, you will only be going a hundred m/s or so. That means you only have to spend 200 m/s or so to reverse your orbit around the sun. Set your desired periapsis, then travel there. Finally, burn at periapsis to circularize.

It may seem counterintuitive, considering that you must do two huge burns to first raise your apoapsis and then bring it back down again, but this approach is actually cheaper than a straight unassisted reversal. The more eccentric your transfer orbit is (in other words, lower down you start and the higher up you push your apoapsis), the greater the cost advantage gets.

The only downside is that you'll be traveling for years on end just to get to AP and back.

[Spricigo]

5 minutes ago, Streetwind said:

As other have said: a single flyby won't give you nearly enough of an assist to turn How can something that moves counterclockwise push your ship in a clockwise direction? It only works for small amounts of push.

Nit-picking : if you are passing in front it's pushing in the opposite direction of the movement. I think you meant that changing direction relative to planet may be not enough to change direction relative to the star,  given the planet's orbital speed.  Anyways, just want to clarify. 

[Reusables]

Impractical with single gravity assist. It will only change direction of relative velocity to the planet, flipping it at best. So you need to approach the planet with 2 times the orbital velocity of the planet to go retrograde. That costs more dv to plot those trajectories for inner planets. (They can't deflect your orbit much with the speed)

Shooting Jool could do it in single assist, but it should be better to use Tylo (or Laythe).

Edited April 19, 2017 by Reusables

[Kryxal]

Back when BTSM was still reasonably current, I accidentally managed a retrograde orbit by a close pass on Laythe, though I forget the details now.  I'd not be surprised if I was close to Jool then caught Laythe on the outbound leg, skimming prograde.

[Snark]

On 04/18/2017 at 10:36 AM, SYDWAD said:

I would like to try make a low fuel orbit reversing maneuver and i would like to know if its possible, i made a crude drawing of my question. 

A few things:

You're not going to be able to make a simple U-turn maneuver at any planet, because physics.  Details in spoiler section:

Spoiler

A couple of reasons contribute to that:

First, the only way to make a an escape trajectory around a planet where the direction reverses 180 degrees is if the ship's going right exactly at escape velocity.  (If it were going slower, it would be a captured elliptical orbit; and if it were going significantly faster, the ship's path would bend less than 180 degrees relative to the planet.)  But in that case, it means that the ship's velocity relative to the planet will be very very low-- close to zero-- at the SoI boundary.  Which would mean that the maneuver wouldn't give you any significant benefit.  If you enter the SoI going, say, 50 m/s solar-prograde relative to the planet, and leave it going 50 m/s solar-retrograde relative to the planet, you've only bought yourself 100 m/s.  Hardly worth the bother.

Second, you'd never be able to get such an encounter anyway.  Why?  Because, for you to enter the planet's SoI at such a low relative velocity, it would mean you would have had to have a near-perfect match to the planet's orbit already, before entering the SoI-- which excludes any transfer from another planet's orbit, unless you did a horribly Oberth-inefficient, fuel-wasting maneuver to match orbits beforehand.

Certainly it's possible to get a lot of dV from a pass by a planet, but a "useful" pass (i.e. which contributes a lot of dV) won't be anything like a 180-degree U-turn; you'd have to be crossing the orbit non-tangentially if you were trying to optimize for reversing to a solar-retrograde orbit.

And there's going to be an upper limit to how much dV you can pull off, this way.  Certainly you'd never be able to use Duna to reverse your orbit.  Why?  Because its orbital speed around the Sun is around 7500 m/s, whereas its escape velocity is a measly 1372 m/s.  I'm not sure what the optimum possible dV would be that you could get by a slingshot maneuver (would have to grind through a bunch of math), but broadly speaking, I expect it's going to be around the planet's escape velocity (perhaps some constant multiple thereof).

You'd have much better luck with Jool, since 1. it has a much higher escape velocity  (9700 m/s), and 2. it has a slower solar orbital velocity (around 4100 m/s).  I'm not saying that what you want is necessarily possible with Jool-- like I said, would take some significant math cranking-- but I'd say that's the only planet where something like what you want to do is even vaguely possible.

On 04/18/2017 at 8:46 PM, Streetwind said:

How can something that moves counterclockwise push your ship in a clockwise direction?

By flying in front of it, so it's pulling you backwards.

[Spricigo]

8 hours ago, Snark said:

And there's going to be an upper limit to how much dV you can pull off, this way...

... I'm not sure what the optimum possible dV would be that you could get by a slingshot maneuver (would have to grind through a bunch of math)... 

I have this vage hope someone will show up to explain that maths. 

The sentence between parenteses made the hope fade a bit

 

[Kryxal]

The possible delta-v from a slingshot is going to be less than twice your relative velocity just after crossing the SoI boundary.  A double slingshot is going to complicate the issue there, of course ... but also remember that the faster you approach, the harder it's going to be to change your vector a great deal.

[Reusables]

On 2017. 4. 21. at 0:09 AM, Spricigo said:

I have this vage hope someone will show up to explain that maths. 

The sentence between parenteses made the hope fade a bit

I think it's not too complex, with a bit of help from wikipedia.

https://en.m.wikipedia.org/wiki/Hyperbolic_trajectory

To gain deflection angle of 2D,

Needed Eccentricity: e=1/sinD.

Closest Approach Distance: r=(-a)(e-1).

Semi-Major Axis: -a=GM/{v_inf}^2.                               ------------ (Definition)

Escape Velocity :  {v_esc}^2=2GM/r                             ------------ (On Closest Approach)

 

Let's compare the relative speed on SOI entry with the escape velocity, as it gives reasonable measure of how fast the entry speed is for the planet.

{v_inf/v_esc}^2 = -r/2a = (e-1)/2, e=2{v_inf/v_esc}^2+1.

D=asin (1 / [2 * {v_inf/v_esc}^2 + 1])

This formula looks horrible.. Let's just put some numbers here.

 

With Eve gravity assist, escape velocity can get up to approx. 4.85km/s.

For the last assist just before reaching circular retrograde orbit, the relative velocity(entry speed) should be 20km/s approximately!

In this case, sinD=0.028 and D~1.6deg. So you can barely get 3.2 degrees of deflection here, which shouldn't be meaningful. Literally, it was too fast for such a small(?) and hot planet.

 

With Kerbin, that's 3.43km/s and 18.6km/s, sinD=0.167, D~1deg. Even worse! (2 degrees deflection)

With Duna, escape velocity is smaller than 1.4km/s(!) while relative velocity should still be 15km/s. No need to bother with this one.

 

Now, with Jool:

It has escape velocity over 9.7km/s, while its orbital speed is under 4.4km/s. Relative velocity is now up to 8.8km/s. So even in the worst case, sinD=0.378 and D~22deg. That's 44 degrees of deflection!

...Actually, that's not so much. It means you were already going retrograde(clockwise) before the swingby.

 

Let's push a bit more. There is one more benefit of Jool: even with tiny amount of retrograde velocity after gravity assist, the Pe should be sufficiently high. 5.0km/s relative velocity could be feasible, and that gives D~41deg. Now this gives 82 degrees of deflection, which is enough for this kind of gravity assist.

Still, it's hard to get into retrograde circular orbit; It'll cost either bunch of time or major burn.

 ( - As Jool is your only solution for retrograde circularization. Other planets gives what, 4 degrees?)

 

EDIT: Cleaned up a bit.

Edited May 3, 2017 by Reusables

[Grounder]

On 18/04/2017 at 11:46 PM, Streetwind said:

If you are willing to invest a lot of flight time, you can make your orbit reversal very cheap by executing what's called a bi-elliptic transfer. First, raise your apoapsis very very high. Past Eeloo kind of high. Then travel to apoapsis. Once out there, you will only be going a hundred m/s or so. That means you only have to spend 200 m/s or so to reverse your orbit around the sun. Set your desired periapsis, then travel there. Finally, burn at periapsis to circularize.

Sorry to bust into this thread but I just came across this Bi-Elliptic transfer in the Wiki, claiming it saves delta V compared to the Hohmann method.  My question is, has anybody tried this in KSP to see if its confirmable?

[Spricigo]

9 minutes ago, Grounder said:

Sorry to bust into this thread but I just came across this Bi-Elliptic transfer in the Wiki, claiming it saves delta V compared to the Hohmann method.  My question is, has anybody tried this in KSP to see if its confirmable?

Corfirmed both by mathematical demonstration and practical examples a few times. (need to dig the forum a bit to find) 

However, as @Streetwind points, the flight time will be longer.  Also there is the complexity factor  

[Grounder]

51 minutes ago, Spricigo said:

Corfirmed both by mathematical demonstration and practical examples a few times. (need to dig the forum a bit to find) 

However, as @Streetwind points, the flight time will be longer.  Also there is the complexity factor  

Spricigo,  I can never claim to excel at the math but the concept is interesting since, at first glance, it looks counter-intuitive.

Much thanks, I think i'm gonna play with this in game.

[Streetwind]

10 hours ago, Grounder said:

Sorry to bust into this thread but I just came across this Bi-Elliptic transfer in the Wiki, claiming it saves delta V compared to the Hohmann method.  My question is, has anybody tried this in KSP to see if its confirmable?

Note that the bi-elliptic transfer is not always cheaper than a Hohmann transfer. There needs to be a sufficiently large difference between origin and destination (a solar orbit reversal obviously being a very large difference), and you need sufficient room to either lower or raise your orbit a whole lot in one direction. So for example, going from Kerbin to Minmus, the bi-elliptic method wouldn't work out for you. Not only is the Hohmann transfer fairly cheap to begin with, but you also don't have room above Minmus to raise your apoapsis very much. You'll quickly hit Kerbin escape velocity.

[Zeiss Ikon]

18 hours ago, Grounder said:

Sorry to bust into this thread but I just came across this Bi-Elliptic transfer in the Wiki, claiming it saves delta V compared to the Hohmann method.  My question is, has anybody tried this in KSP to see if its confirmable?

I've done bi-elliptic reversals around Minmus -- orbital velocity is so low it's easy.  Pretty small burn to raise apoapsis near the edge of the SOI, then a pretty small burn to reverse the orbit, and another pretty small burn to circularize.  I've also done it with Kerbin orbit, when I was trying to intercept an asteroid during the rock's encounter -- again, with a highly eccentric orbit that peaks beyond Minmus, the velocity is so low at apoapsis that it takes little dV to reverse the orbit.

Others have explained pretty well that the upper limit for the dV you can get from a planet gravity assist is no more than twice the planet's escape velocity; in practice, you can't get even that much.  Now, if you were to add an Oberth-enhanced maneuver at a very low periapsis, your combined dV total could be close to twice escape velocity plus several times the actual maneuver dV, but possible ejection angles from this kind of maneuver rule out the 180 degree turn the OP illustration seems to be looking for.

Edited April 22, 2017 by Zeiss Ikon
typo

[Red Iron Crown]

3 hours ago, Zeiss Ikon said:

Others have explained pretty well that the upper limit for the dV you can get from a planet gravity assist is no more than twice the planet's escape velocity; in practice, you can't get even that much.

Minor correction: Twice the body's orbital velocity, not its escape velocity. Not sure why people are on about escape velocity, it is not relevant at all.

[Human Person]

I'm going to attempt this with a powered gravity assist from Jool. I got the feeling this is possible, I'll do it with infinite fuel and keep track of the delta-v expenditure with mechjeb. In the same mission I'll search for a new gravity-assist path to Jool, using only Eve. I want a direct Eve-Jool route so the eccentricity of my solar orbit is as high as possible before doing the reverse-assist.

This might get us in a highly eccentric, reversed solar orbit.

 

Edit: I did it and it worked. Unfortunately I have no pictures, but her're some rough numbers:

 

LKO-Eve: 1200 m/s

Powered Eve-assist: 1500 m/s (I could have encountered jool with much less effort, but I thought I should encounter it with decent speed (apohelion way bejond jool's Orbit) to get a better Jool-assist.

I entered Jools SOI at 3.6 km/s

at Jools PE, it only took another 1000 m/s to get into a (highly eccentric) reversed solar Orbit.

 

After that, I messed around a bit and encountered EVE again. I entered its atmosphere at 26 km/s

 

I'll do more work on this to increase overall Efficiency.

Edited April 23, 2017 by Physics Student

[Kryxal]

On 2017-04-22 at 6:31 PM, Red Iron Crown said:

Minor correction: Twice the body's orbital velocity, not its escape velocity. Not sure why people are on about escape velocity, it is not relevant at all.

I was thinking of your relative velocity to the planet.  Without spending fuel, the best you can do is go back the way you came, but you'll have the same speed at the same altitude.

[Lelitu]

58 minutes ago, Kryxal said:

I was thinking of your relative velocity to the planet.  Without spending fuel, the best you can do is go back the way you came, but you'll have the same speed at the same altitude.

which is a change of twice the velocity. Velocity includes direction. at 1000m/s prograde, you need 2000m/s deltav to get to 1000m/s retrograde.

[Nicias]

I think it feels like it should be twice your escape velocity since the best you could do it turn around with the same speed. However, it then is not twice your escape velocity, it is twice you excess velocity at escape. The higher this is the straighter your orbit is. So you want to maximize this number times the sine of half the angle your orbit turns you. This results in the optimization procedure @Abastro described.

[Red Iron Crown]

The theoretical limit is twice the planet's orbital velocity, assuming no fuel spent. You will enter and leave the SoI with the same speed relative to that body, but your speed relative to the grandparent body can be changed by no more than twice the orbital speed of the slingshot target. Imagine you enter the SoI exactly retrograde and exit exactly prograde, your speed relative to the grandparent body will have increased by exactly twice the orbital velocity of the slingshot target. Escape velocity and excess escape velocity are not relevant aside from affecting how closely your trajectory can approach that ideal.