Orbital inclination change: Efficiency question.

[Guest]

Just a quick question about changing the inclination of an orbit. Is there an optimal way to do it? As in, would having a highly elliptical orbit and changing the inclination at either apogee or perigee provide any deltaV savings over a circular orbit?

Are there any tips and tricks for efficiently  changing inclination?

Edited June 11, 2019 by Guest

[Incarnation of Chaos]

1 minute ago, Dale Christopher said:

Just a quick question about changing the inclination of an orbit. Is there an optimal way to do it? As in, would having a highly elliptical orbit and changing the inclination at either apogee or perigee provide any deltaV savings over a circular orbit?

Are there any tips and tricks for efficiently  changing inclination?

You change inclination at your ascending or decending nodes; decending node is most efficient. Changing at apogee or perigee is wildly inefficent and likely won't change the inclination as much as you want; also the cost for a plane change is mostly based on the degree of inclination you wish to change. Going from 120 Degree inclination to 0.0 is gonna hurt; even if you're doing it "Efficiently". Always try to launch into the inclination desired; or carry the extra delta-v to achieve the desired inclination.

[HebaruSan]

1 minute ago, Incarnation of Chaos said:

You change inclination at your ascending or decending nodes; decending node is most efficient.

No, there's no difference between the ascending and descending nodes.

1 minute ago, Incarnation of Chaos said:

Changing at apogee or perigee is wildly inefficent

Not so; a high apoapsis is very efficient for plane changes, assuming you've aligned it with your ascending or descending node.

[Incarnation of Chaos]

Just now, HebaruSan said:

No, there's no difference between the ascending and descending nodes.

Not so; a high apoapsis is very efficient for plane changes, assuming you've aligned it with your ascending or descending node.

Really? I just had a Eve probe that needed 400m/s less delta-v to perform the same inclination change at the decending node (Which was also CLOSER to Eve). Also aligning the nodes requires more dv; and does not make that statement false. Because most times the nodes won't be aligned with apogee or perigee; they're going to be somewhere inbetween. 

[HebaruSan]

1 minute ago, Incarnation of Chaos said:

Really?

Yes, really. The distinction is arbitrary; if you swapped the north and south poles, the AN would become the DN and vice versa, but the details of the orbits would not change.

1 minute ago, Incarnation of Chaos said:

I just had a Eve probe that needed 400m/s less delta-v to perform the same inclination change at the decending node (Which was also CLOSER to Eve).

The delta V will vary between the AN and DN, depending on which is higher, but it's not always going to be less at one or the other. But if you think that a lower node was more efficient, then you missed or misunderstood something. Got a screenshot?

[Incarnation of Chaos]

1 minute ago, HebaruSan said:

Yes, really. The distinction is arbitrary; if you swapped the north and south poles, the AN would become the DN and vice versa, but the details of the orbits would not change.

The delta V will vary between the AN and DN, depending on which is higher, but it's not always going to be less at one or the other. But if you think that a lower node was more efficient, then you missed or misunderstood something. Got a screenshot?

I don't but i'm actually going to end up sending that probe again so gimme a few days and i'll get it; it was pretty hilarious since the AN was several million KM higher but the DN ended up being the most efficient.  And i'm well aware the positions of them are arbitary; but that's just what iv'e gathered from my experience+ online materials. But KSP was acting very strange that day since it had been up for ~3 days in a row; so it may very well have been a glitch.

[Kerbart]

5 minutes ago, Incarnation of Chaos said:

Really? I just had a Eve probe that needed 400m/s less delta-v to perform the same inclination change at the decending node (Which was also CLOSER to Eve). Also aligning the nodes requires more dv; and does not make that statement false. Because most times the nodes won't be aligned with apogee or perigee; they're going to be somewhere inbetween. 

Adjusting the apoapsis to coincide with an ascending/descending node takes very little DV

Changing inclination, on the other hand, does; as you're adjusting the direction of your velocity vector.  At 2000 ^m/_s, changing inclination from 0° to 180° will take, obviously, 4000 ^m/_s, but raising your apoapsis to 10,000 km will take far less than ¹/₄ of that and reversing your orbit at that point will only take a few 100 ^m/_s. Then you have to drop your apo (same cost) but you end up doing the same maneuver for a little over half the cost.

But don't take my word for it; look at real life spaceflight. That's usually how inclination changes are done; raise apo, change plane, drop apo. Much cheaper than doing it in low orbit.

[Incarnation of Chaos]

Just now, Kerbart said:

Adjusting the apoapsis to coincide with an ascending/descending node takes very little DV

Changing inclination, on the other hand, does; as you're adjusting the direction of your velocity vector.  At 2000 ^m/_s, changing inclination from 0° to 180° will take, obviously, 4000 ^m/_s, but raising your apoapsis to 10,000 km will take far less than ¹/₄ of that and reversing your orbit at that point will only take a few 100 ^m/_s. Then you have to drop your apo (same cost) but you end up doing the same maneuver for a little over half the cost.

But don't take my word for it; look at real life spaceflight. That's usually how inclination changes are done; raise apo, change plane, drop apo. Much cheaper than doing it in low orbit.

in KSP how would you align the nodes then? Radial out or in?  

[Kerbart]

6 minutes ago, Incarnation of Chaos said:

in KSP how would you align the nodes then? Radial out or in?  

Prograde/retrograde, at 90° from apo/peri. Or if you have a very elliptical orbit, raise periapsis first (that's cheaper than lowering apoapsis)

“Even in a very elliptical orbit?”

Especially in a high elliptical orbit, if the alternative is a plane change at low altitude (because of vis viva that automatically means high velocity)

[Incarnation of Chaos]

2 minutes ago, Kerbart said:

Prograde/retrograde, at 90° from apo/peri. Or if you have a very elliptical orbit, raise periapsis first (that's cheaper than lowering apoapsis)

“Even in a very elliptical orbit?”

Especially in a high elliptical orbit, if the alternative is a plane change at low altitude (because of vis viva that automatically means high velocity)

Ah i see; you burn prograde/retro at apo/peri until the node aligns and then perform the inclination change desired?

[KerikBalm]

Consider the case of a 180 degree inclination change from LKO.

Orbital velocity is about 2350. In this case, there is no An/Dn (or they are everywhere, however you want to look at it) because the desired orbit and target orbit are coplanar. To reverse your orbit in LKO, you'd need to burn 2350 to kill your orbital velocity, and then another 2350 to reestablish orbit in the other direction: 4,700 m/s for the plane change....

or...

You burn 900 m/s and get your apoapsis way out... at apoapsis, its only a few hundred m/s to reverse direction, lets say your orbital velocity at Ap is 200 m/s, so to change inclination 180 degrees is 400 m/s.

Then you burn 900 m/s again at Pe, total cost: 2200... less than half changing it in LKO. 

You always want to change your inclination when you are moving slowest. However, you can't change your inclination to your desired inclination at any point in your orbit: ie when you are over the north pole, there is no burn that will put you in an equatorial orbit, you must first cross the equator before a burn can put you in an equatorial orbit. You can only insert into the desired orbit from the An/Dn. 

For small plane changes, it doesn't make sense to put your An/Dn very high, but for big inclination changes, it does.

Back to the LKO example, if instead of a 180 degree change, we did a 90 degree one, (starting from equatorial orbit with kerbin's rotation) we'd need to kill our "east velocity" component, and give a north or south velocity component of 2350 m/s. We could do this with a 45 degree burn, which would require 2350/sin(45) = 3,323 m/s... still a lot more that boosting Ap (at an An/Dn), and then doing the plane change.

Now lets say a 45 degree change... then our forward velocity decreases by 1-1/sqrt(2), and our upward one increases by 1/sqrt(2). Ie "X": velocity from 2350 to 1661, Y velocity from 0 to 1661. deltav_X is 689, dV Y is 1661. We could do this with a single burn (pythagorean theorem) of sqrt( 689^2+1661^2)= 1798... call it 1,800 m/s... vs which is what it would take to boost Ap way up, then bring it down (a 45 degree change wouldn't take 400 m/s at Ap, which was estimated for the 180 change... lets say 100 m/s instead).

So... the two become relatively equal at about 45 degrees. If you need to change your inclination more than 45 degrees, boost your Ap way up at the An or Dn. If you need to do a change of less than 45 degrees, just pick the An or Dn that you are moving slowest at.

[Kerbart]

8 hours ago, Incarnation of Chaos said:

Ah i see; you burn prograde/retro at apo/peri until the node aligns and then perform the inclination change desired?

Almost, if you burn at apo/peri their positions don’t change. You want to burn in between them to alter their positions. As @KerikBalmmentions below, if there’s a node reasonably close by (where velocity is low) you don’t need to. It’s all about changing inclination when your speed is low.

[Guest]

Thanks a lot ppls! This is great infos! Thanks to your help we are one more step towards a bright future of Kerbal kind venturing into the heavens, boldly flying in the face of the destiny, to rise above the potential of mere little green men... to grab hold of the cosmos, and to smack the mouth of god! :3

[Nich]

6 hours ago, KerikBalm said:

Consider the case of a 180 degree inclination change from LKO.

Orbital velocity is about 2350. In this case, there is no An/Dn (or they are everywhere, however you want to look at it) because the desired orbit and target orbit are coplanar. To reverse your orbit in LKO, you'd need to burn 2350 to kill your orbital velocity, and then another 2350 to reestablish orbit in the other direction: 4,700 m/s for the plane change....

or...

You burn 900 m/s and get your apoapsis way out... at apoapsis, its only a few hundred m/s to reverse direction, lets say your orbital velocity at Ap is 200 m/s, so to change inclination 180 degrees is 400 m/s.

Then you burn 900 m/s again at Pe, total cost: 2200... less than half changing it in LKO. 

You always want to change your inclination when you are moving slowest. However, you can't change your inclination to your desired inclination at any point in your orbit: ie when you are over the north pole, there is no burn that will put you in an equatorial orbit, you must first cross the equator before a burn can put you in an equatorial orbit. You can only insert into the desired orbit from the An/Dn. 

For small plane changes, it doesn't make sense to put your An/Dn very high, but for big inclination changes, it does.

Back to the LKO example, if instead of a 180 degree change, we did a 90 degree one, (starting from equatorial orbit with kerbin's rotation) we'd need to kill our "east velocity" component, and give a north or south velocity component of 2350 m/s. We could do this with a 45 degree burn, which would require 2350/sin(45) = 3,323 m/s... still a lot more that boosting Ap (at an An/Dn), and then doing the plane change.

Now lets say a 45 degree change... then our forward velocity decreases by 1-1/sqrt(2), and our upward one increases by 1/sqrt(2). Ie "X": velocity from 2350 to 1661, Y velocity from 0 to 1661. deltav_X is 689, dV Y is 1661. We could do this with a single burn (pythagorean theorem) of sqrt( 689^2+1661^2)= 1798... call it 1,800 m/s... vs which is what it would take to boost Ap way up, then bring it down (a 45 degree change wouldn't take 400 m/s at Ap, which was estimated for the 180 change... lets say 100 m/s instead).

So... the two become relatively equal at about 45 degrees. If you need to change your inclination more than 45 degrees, boost your Ap way up at the An or Dn. If you need to do a change of less than 45 degrees, just pick the An or Dn that you are moving slowest at.

Forgot to mention leaving the SOI.  I once managed a 167 degree inclination change around the Mun with 12 Dv but that was a very special case.  If you are boosting your AP to change inclination you generally want to do some of the change while boosting and unboosting (if airless body) as well but the savings are small (5% at most).  The only way I have found for a minimum Dv is numerically solving.

[Gargamel]

Moved to GamePlay Q's

[Incarnation of Chaos]

Alright; tested all of this last night. And literally saved thousands of delta v performing plane changes around duna. Thanks for all the input; and the humility of realizing I don't know jack after hundreds of hours in ksp xD