Apoapsis plane change manoeuvre Δv

[Supernovy]

I'm using "apoapsis plane change manoeuvre" to refer to the three burn manoeuvre (starting from a circular orbit) of increasing apoapsis, performing a plane change at apoapsis, then re-circularising at periapsis.

If I'm correct, the total Δv for this manoeuvre is:

Δv_T = 2[(μ(2/a₁ - 1/a₂))^1/2 - (μ/a₁)^1/2] + (μ(2/(2a₂-a₁)-1/a₂)^1/2(sin(Δi)/sin(90-(Δi/2)))

Whereas the Δv for a simple plane change is:

Δv = (μ/a₁)^1/2(sin(Δi)/sin(90-(Δi/2)))

μ is the gravitational parameter, a₁ is the circular orbit SMA, a₂ is the elliptical orbit SMA and Δi is the desired change in inclination (in degrees here). If either of these equations are wrong or not fully simplified, please correct me. I can post the derivation if needed.

I am interested in finding the "break even point" - the a₂ for a constant Δi, a₁ and μ at which the Δv for the apoapsis plane change manoeuvre equals the Δv for the simple plane change. I hope to use this to find a simple (enough) equation that determines a good a₂ for a given Δi, a₁ and μ.

So far, I've figured out that the difference between the simple plane change Δv and the apoapsis plane change manoeuvre Δv must be equal to twice the prograde burn Δv (the total Δv of the apoapsis raising burn and the recircularisation).

(μ/a₁)^1/2(sin(Δi)/sin(90-(Δi/2))) = 2[(μ(2/a₁ - 1/a₂))^1/2 - (μ/a₁)^1/2] + (μ(2/(2a₂-a₁)-1/a₂)^1/2(sin(Δi)/sin(90-(Δi/2)))

=>

(μ/a₁)^1/2(sin(Δi)/sin(90-(Δi/2))) - (μ(2/(2a₂-a₁)-1/a₂)^1/2(sin(Δi)/sin(90-(Δi/2))) = 2[(μ(2/a₁ - 1/a₂))^1/2 - (μ/a₁)^1/2]

Any help in simplifying or solving this expression would be greatly appreciated.

Edited February 8, 2017 by Supernovy
restoring broken symbols ( ÃŽâ€� -> Δ , μ -> μ)

[Odielthen]

Have you tried putting this into Wolfram Alpha? Even if you can't type in the entire equation at once, it can at least simplify it for you.

[Geschosskopf]

Why are you changing plane at Ap instead of the AN or DN?

[Taki117]

Have you tried putting this into Wolfram Alpha? Even if you can't type in the entire equation at once, it can at least simplify it for you.

Wolfram hates it.

Wolfram|Alpha doesn't understand your query

Showing instead result for query: (2/(2x-y)-1/x)1/2

[alexmun]

Your basic plane change ÃŽâ€v formula can be simplified to:

ÃŽâ€v = 2v sin(ÃŽâ€i/2)

Second, let me take a tangent to solve a related problem. There is a threshold of ÃŽâ€i at which the bi-elliptic method of plane change becomes more efficient. Below that threshold it doesn't matter how high you raise your apoapsis it will never be more efficient than doing the plane change from your original orbit. We can find that threshold fairly easily by solving for the case where your a₂ = infinity (i.e. a parabolic orbit where the apoapsis is at infinity). At infinity, the your orbital velocity is zero and therefore your plane change costs zero ÃŽâ€v and we can calculate the total ÃŽâ€v as just the ÃŽâ€v to raise your apoapsis to infinity and then lower it again. The velocity required to put your apoapsis at infinity is just the escape velocity, so that ÃŽâ€v is twice the difference between the escape velocity and your orbital velocity. So we can solve:

v = sqrt(μ/r)

v_esc = sqrt(2μ/r)

2(v_esc - v) = 2v sin(ÃŽâ€i/2)

ÃŽâ€i = 2 asin(v_esc/v - 1)

If your ÃŽâ€i is above that threshold, it will be more efficient to raise your apoapsis, do the plane change, then recircularize than to do the plane change directly from your circular orbit (well, unless it's really really close to the threshold and you'd have to put your apoapsis outside of the SoI). The higher you can raise your apoapsis, the more efficient the overall maneuver will be.

Edit: Assuming I haven't made any errors (which may be a big assumption), I think your original equation simplifies down to:

a2 = a1 * (sin(theta / 2)^2 + 1) / (4 * sin(theta / 2) + 2) Nope, this was wrong.

Edited February 15, 2014 by alexmun

[Kosmo-not]

Assumption: starting from and returning to a circular orbit.

First, the term for the ÃŽâ€v needed to change your apoapsis twice.

2*sqrt(μ/r₁)*(sqrt(2*r₂/(r₁+r₂))-1)

Then, your pick from either of the next two terms for changing inclination.

For inclination change by pointing normal to your velocity vector for the burn duration:

v₂*ÃŽâ€i

or

For inclination change involving pointing in a single direction for the burn duration (most efficient, and what maneuver nodes do):

sqrt(2*v₂^2*(1-cos(ÃŽâ€i)))

Add those two terms together and voila.

List of variables:

μ = we all should know this one

r₁ = periapsis height

r₂ = apoapsis height

v₂ = velocity at apoapsis

*edit* facepalm. Kosmo-not, learn to read before replying.

Edited February 15, 2014 by Kosmo-not

[Supernovy]

alexmun said:

If your �i is above that threshold, it will be more efficient to raise your apoapsis, do the plane change, then recircularize than to do the plane change directly from your circular orbit (well, unless it's really really close to the threshold and you'd have to put your apoapsis outside of the SoI). The higher you can raise your apoapsis, the more efficient the overall maneuver will be.

So I suppose I was asking the wrong question? Is there no case for which a higher apoapsis results in less Δv than than a lower apoapsis (aside from your case) for the overall manoeuvre?

Also thanks alexmun and Kosmo-not for the math.

Geschosskopf: In the case where you're going from an inclined orbit to equatorial, you'd put the apoapsis at either the ascending or descending node. For going into a more inclined orbit, it doesn't really matter I believe.

Edited February 8, 2017 by Supernovy
restoring broken symbols ( ÃŽâ€� -> Δ )

[alexmun]

So I suppose I was asking the wrong question? Is there no case for which a higher apoapsis results in less ÃŽâ€v than than a lower apoapsis (aside from your case) for the overall manoeuvre?

Actually, I just realized that this equation:

ÃŽâ€i = 2 asin(v_esc/v - 1) = 2 asin(sqrt(2) - 1) = 48.9 degrees.

So that threshold is constant regardless of your starting orbit. If you are making a plane change of greater than 48.9 degrees, it is most efficient to raise your apoapsis as high as possible, do the plane change, then recircularize. The higher you raise your apoapsis in that case, the more efficient the maneuver will be. If you are making a plane change of less than 48.9 degrees it is most efficient to just do it in your existing orbit.

[Kosmo-not]

So I suppose I was asking the wrong question? Is there no case for which a higher apoapsis results in less ÃŽâ€v than than a lower apoapsis (aside from your case) for the overall manoeuvre?

Near and before the critical angle, the delta-v savings grow, peak, and then goes negative further out (I could be in error if my spreadsheet is wrong). I think I will study this further.

Edited February 15, 2014 by Kosmo-not

[alexmun]

Ok, I worked on this some more. The solution to your original equation is:

Apoapsis = 2 * r * sin²(θ/2) / (2 - (sin(θ/2) + 1)²)

Where r is your initial circular orbital radius and θ is the angle of inclination change. That equation will give you the apoapsis where raising your orbit to that apoapsis, doing the plane change, then circularizing costs exactly the same ÃŽâ€v as doing the plane change directly.

However, what we actually want to know is what is the optimal apoapsis to raise our orbit to for a minimal ÃŽâ€v plane change. That equation turns out to be:

Apoapsis = r * sin(θ/2) / (1 - 2 * sin(θ/2))

You will find that for angles less than 38.9 degrees, that formula gives you an apoapsis less than your current orbital radius (which is non-sensical). So for those angles, doing the plane change without changing your apoapsis is optimal. At 60 degrees, the formula gives an apoapsis of infinity, so for angles greater than 60 degrees raising your apoapsis to just inside the SoI and doing the maneuver is optimal. For angles between 38.9 degrees and 60 degrees, raise your apoapsis to the radius given by the formula, do your plane change maneuver there, then re-circularize. For plane change angles less than 50 degrees or so, you're going to save less than 5% ÃŽâ€v compared to just doing a simple plane change in your original orbit, so for small plane change angles this method is probably not worth it. For large plane change angles, however, the savings become significant, getting to about 40% by 90 degrees and up to a max of about 60% for a 180 degree plane change.

Edited February 15, 2014 by alexmun

[metaphor]

Ok, I worked on this some more. The solution to your original equation is:

Apoapsis = 2 * r * sin²(θ/2) / (2 - (sin(θ/2) + 1)²)

Where r is your initial circular orbital radius and θ is the angle of inclination change. That equation will give you the apoapsis where raising your orbit to that apoapsis, doing the plane change, then circularizing costs exactly the same ÃŽâ€v as doing the plane change directly.

However, what we actually want to know is what is the optimal apoapsis to raise our orbit to for a minimal ÃŽâ€v plane change. That equation turns out to be:

Apoapsis = r * sin(θ/2) / (1 - 2 * sin(θ/2))

You will find that for angles less than 38.9 degrees, that formula gives you an apoapsis less than your current orbital radius (which is non-sensical). So for those angles, doing the plane change without changing your apoapsis is optimal. At 60 degrees, the formula gives an apoapsis of infinity, so for angles greater than 60 degrees raising your apoapsis to just inside the SoI and doing the maneuver is optimal. For angles between 38.9 degrees and 60 degrees, raise your apoapsis to the radius given by the formula, do your plane change maneuver there, then re-circularize. For plane change angles less than 50 degrees or so, you're going to save less than 5% ÃŽâ€v compared to just doing a simple plane change in your original orbit, so for small plane change angles this method is probably not worth it. For large plane change angles, however, the savings become significant, getting to about 40% by 90 degrees and up to a max of about 60% for a 180 degree plane change.

Interesting. I wouldn't have thought there was an optimal apoapsis for certain angles.

Related question/comment: if you use aerobraking to circularize after the plane change maneuver instead of a propulsive burn at periapsis, the apoapsis plane change maneuver becomes even more efficient since you're effectively getting half the delta-v for free. I think the cut-even point for that is around 25 degrees (for any plane change of more than 25 degrees it's more efficient to raise your apoapsis high, perform the plane change there, then aerobrake back down), but I wonder if there's an optimal apoapsis for that too.

[alexmun]

Interesting. I wouldn't have thought there was an optimal apoapsis for certain angles.

Yes, it surprised me too.

Related question/comment: if you use aerobraking to circularize after the plane change maneuver instead of a propulsive burn at periapsis, the apoapsis plane change maneuver becomes even more efficient since you're effectively getting half the delta-v for free. I think the cut-even point for that is around 25 degrees (for any plane change of more than 25 degrees it's more efficient to raise your apoapsis high, perform the plane change there, then aerobrake back down), but I wonder if there's an optimal apoapsis for that too.

If you only have to pay half the ÃŽâ€v to raise then re-circularize, the formula becomes:

Apoapsis = r * sin(θ/2) / (0.5 - 2 * sin(θ/2))

In that case, the critical angles are 19.2 degrees and 28.96 degrees. Below 19.2 degrees do the maneuver in your original orbit. At or above 28.96 degrees raise your apoapsis to the SoI. In between use the formula.

And, of course, if you don't have to pay any of the ÃŽâ€v to raise then re-circularize (e.g. you're aerobraking in from an interplanetary transfer) then you always want the apoapsis for your plane change maneuver to be as high as possible. For example, aerobrake to a very high apoapsis, do your plane change maneuver, then aerobrake to a circular orbit.

[Supernovy]

alexmun said:

Ok, I worked on this some more. The solution to your original equation is:

Apoapsis = 2 * r * sin²(θ/2) / (2 - (sin(θ/2) + 1)²)

Where r is your initial circular orbital radius and θ is the angle of inclination change. That equation will give you the apoapsis where raising your orbit to that apoapsis, doing the plane change, then circularizing costs exactly the same Δv as doing the plane change directly.

Which answers my first question.

alexmun said:

However, what we actually want to know is what is the optimal apoapsis to raise our orbit to for a minimal Δv plane change. That equation turns out to be:

Apoapsis = r * sin(θ/2) / (1 - 2 * sin(θ/2))

You will find that for angles less than 38.9 degrees, that formula gives you an apoapsis less than your current orbital radius (which is non-sensical). So for those angles, doing the plane change without changing your apoapsis is optimal. At 60 degrees, the formula gives an apoapsis of infinity, so for angles greater than 60 degrees raising your apoapsis to just inside the SoI and doing the maneuver is optimal. For angles between 38.9 degrees and 60 degrees, raise your apoapsis to the radius given by the formula, do your plane change maneuver there, then re-circularize. For plane change angles less than 50 degrees or so, you're going to save less than 5% Δv compared to just doing a simple plane change in your original orbit, so for small plane change angles this method is probably not worth it. For large plane change angles, however, the savings become significant, getting to about 40% by 90 degrees and up to a max of about 60% for a 180 degree plane change.

Which answers my second question.

alexmun said:

If you only have to pay half the Δv to raise then re-circularize, the formula becomes:

Apoapsis = r * sin(θ/2) / (0.5 - 2 * sin(θ/2))

In that case, the critical angles are 19.2 degrees and 28.96 degrees. Below 19.2 degrees do the maneuver in your original orbit. At or above 28.96 degrees raise your apoapsis to the SoI. In between use the formula.

And, of course, if you don't have to pay any of the Δv to raise then re-circularize (e.g. you're aerobraking in from an interplanetary transfer) then you always want the apoapsis for your plane change maneuver to be as high as possible. For example, aerobrake to a very high apoapsis, do your plane change maneuver, then aerobrake to a circular orbit.

And this is a bonus question answered. Thank you very much, alexmun and metaphor. I think I can set this thread to "answered" now. This is a valuable piece of knowledge added to the Kerbal knowledge base - when and how to save Δv on plane changes by raising apoapsis and doing the change there.

In summary, for less than 38.9 degrees, don't raise apoapsis. For 60 degrees or more, raise it as much as possible. In between those, raise it as given by the second equation. If aerobraking, these angles become 19.2 degrees and 28.96 degrees respectively, and in between use the third equation.

I'm very surprised to see that it ends up independent of μ - but then again, so is the phase angle equation. I suppose the dependence on orbital velocity gets eaten up by the plane change term.

Edited February 8, 2017 by Supernovy
restoring broken symbols ( ÃŽâ€� -> Δ , μ -> μ , θ -> θ )