What is the most efficient way to do high energy plane changes?

[Loren Pechtel]

At what point, if ever, do you gain by raising your orbit before doing them?

[bewing]

Below 45 degrees, it doesn't help. Say you are in a low circular orbit, at speed X. A 45 degree plane change will cost you an additional X. But an X/2 burn will get you to the SOI boundary, where your plane change cost is zero. But then you need to come back down again, for an additional cost of X/2. So there's your cost of X again.

For a plane change of more than 45 degrees, it starts to be energetically beneficial.

Edited February 6, 2017 by bewing

[Reusables]

18 minutes ago, bewing said:

Unless you need to modify your orbit, it's usually a wash? Say you are in a low circular orbit, at speed X. A 45 degree plane change will cost you an additional X. But an X/2 burn will get you to the SOI boundary, where your plane change cost is zero. But then you need to come back down again, for an additional cost of X/2. So there's your cost of X again.

Doesn't it cost sqrt(2-sqrt(2))X for the direct plane change, and nearly (2sqrt(2)-2)X for SOI boundary plane change? Which are 0.765X and 0.828X each. If you meant energy cost, the energy cost per delta v is dependent on speed, thus kinda redundant.

[FancyMouse]

16 minutes ago, bewing said:

an X/2 burn will get you to the SOI boundary

The burn is in fact of a factor sqrt(2)-1 (since escape velocity = sqrt(2) of circulr orbit velocity). But 1/2 is a good approximation that accounts for the actual nonzero plane change. I'm ok with that for simple calculation.

But 45 degree isn't quite the breakeven point - I would need a v*sqrt(2-sqrt(2)) budget for a direct plane change (I burn to the target circular orbit, not along initial normal direction), which is just about 3/4 of v, less than 2(sqrt(2)-1)v~0.8v to burn to high elliptical. In fact, if you use v as an approximate for the high elliptical plan, then the breakeven point is exactly 60 degrees (equilateral triangle ftw).

[Loren Pechtel]

29 minutes ago, bewing said:

Below 45 degrees, it doesn't help. Say you are in a low circular orbit, at speed X. A 45 degree plane change will cost you an additional X. But an X/2 burn will get you to the SOI boundary, where your plane change cost is zero. But then you need to come back down again, for an additional cost of X/2. So there's your cost of X again.

For a plane change of more than 45 degrees, it starts to be energetically beneficial.

I didn't know where the breakpoint was but it sure seemed like it cost less to climb first.  Thanks!

Edited February 6, 2017 by Loren Pechtel

[bewing]

53 minutes ago, Abastro said:

Doesn't it cost sqrt(2-sqrt(2))X for the direct plane change, and nearly (2sqrt(2)-2)X for SOI boundary plane change? Which are 0.765X and 0.828X each. If you meant energy cost, the energy cost per delta v is dependent on speed, thus kinda redundant.

There are those of us who do all our orbital calculations via approximation in our heads.

 

[Superfluous J]

1 hour ago, bewing said:

There are those of us who do all our orbital calculations via approximation in our heads.

Those people have one on me. I make exploratory maneuver nodes and add up the numbers in my head, rounding to the nearest 100.

[GoSlash27]

 

Just so happens we had a very interesting discussion about this problem just last year. I was afraid that it would be hard to track down, but this is it. The result is there is a definite function that correlates a plane change with the amount the Ap should be raised in order to minimize DV.

HTHs,
-Slashy

[Aegolius13]

22 hours ago, bewing said:

Below 45 degrees, it doesn't help. Say you are in a low circular orbit, at speed X. A 45 degree plane change will cost you an additional X. But an X/2 burn will get you to the SOI boundary, where your plane change cost is zero. But then you need to come back down again, for an additional cost of X/2. So there's your cost of X again.

For a plane change of more than 45 degrees, it starts to be energetically beneficial.

On atmospheric bodies, I would think you can save some of the recirculaization cost by aerobraking, which would mean it's potentially cost effective for somewhat under 45 degrees.   Does that make sense?

Also, it wasn't addressed in the question, but if you're entering a new SOI, you can generally change planes for very cheap.  Just do the minimum possible capture burn to get your orbit inside the SOI, then change at apo.  This is also useful for splitting up long burns if you have a low TWR.

 

[bewing]

2 hours ago, Aegolius13 said:

On atmospheric bodies, I would think you can save some of the recirculaization cost by aerobraking, which would mean it's potentially cost effective for somewhat under 45 degrees.   Does that make sense?

Makes excellent sense.

2 hours ago, Aegolius13 said:

Also, it wasn't addressed in the question, but if you're entering a new SOI, you can generally change planes for very cheap.  Just do the minimum possible capture burn to get your orbit inside the SOI, then change at apo.  This is also useful for splitting up long burns if you have a low TWR.

 

Yup.