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True Anomaly & Burn Efficiency Equation


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When entering the SOI of a planet, we'd like to start our retrograde burn as close to periapsis as possible and with a low periapsis. For craft with low TWR, this will result in inefficiency, because you must start the burn pretty far away. My suspicion is that the greater you are from the true anomaly of 0 (Pe = True Anomaly of 0 degrees) the less efficient it is.

What is the equation that gives the efficiency of a finite burn compared to an equivalent impulse burn?

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2 hours ago, Wcmille said:

When entering the SOI of a planet, we'd like to start our retrograde burn as close to periapsis as possible and with a low periapsis. For craft with low TWR, this will result in inefficiency, because you must start the burn pretty far away. My suspicion is that the greater you are from the true anomaly of 0 (Pe = True Anomaly of 0 degrees) the less efficient it is.

What is the equation that gives the efficiency of a finite burn compared to an equivalent impulse burn?

It's going to be ugly and not so simple, because you're traveling along the line of an ellipse and you're also going to have varying acceleration as you go.

If I may ask-- is there a particular reason you want this?  i.e. is it just out of academic curiosity?  'Coz I'm having trouble thinking of a practical in-game reason why you'd need an actual number.

If all you want is to maximize your efficiency, that's not the same thing as having an equation.  :)  The answer is basically "do as much of your burn at as low an elevation as possible."  Which means spread it out around periapsis.

The most efficient way to capture with a very-low-TWR craft (assuming that gravity assist or aerobraking aren't available) is to split up the burn.  Set up as low a Pe as possible, then do enough of a burn to just barely capture (i.e. Ap is still really high)-- typically you'll split the difference of that burn around Pe.  Then coast around, do another burn at Pe that's not too long, and keep repeating with a Pe burn each time until you get your Ap down to where you want it.  That way, with the possible exception of the initial burn, you're basically doing all your burning right at Pe and you're getting the max possible benefit from Oberth effect.

Edited by Snark
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I don't mind assuming constant acceleration.

Here are the reasons:

  • If you are coming in at really high speed (e.g. Kerbin to Moho), the capture dV is going to be very high. You must pay the capture cost in a single pass.
  • After capture, the orbital period is pretty high, so you will spend days on the first few burns if you do a split burn. Exactly how much can you spend on each pass, so that you make as few passes as possible?

 

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If I had to hazard a guess, the instantaneous loss is probably involves the sine of true anomaly (how bad is the angle?) and the difference of the squares of velocity at the present altitude versus the Pe from the loss from the Oberth effect, but I'd like to work it out more closely than that.

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2 minutes ago, Wcmille said:

I don't mind assuming constant acceleration.

Here are the reasons:

  • If you are coming in at really high speed (e.g. Kerbin to Moho), the capture dV is going to be very high. You must pay the capture cost in a single pass.
  • After capture, the orbital period is pretty high, so you will spend days on the first few burns if you do a split burn. Exactly how much can you spend on each pass, so that you make as few passes as possible?

 

The answer is going to be really ugly and will involve differential equations.  Even if you assume constant acceleration, the shape of your path is going to be changing as you slow down.  I don't know if an analytical solution (as opposed to a numeric one) is even possible.

I generally approach such situations by taking the total required burn time to capture and splitting it 50/50 across the periapsis.  Then for the subsequent passes, I try to limit my burn to about plus-or-minus 10 degrees from periapsis, and just do as many passes as it takes.

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8 hours ago, Wcmille said:

If I had to hazard a guess, the instantaneous loss is probably involves the sine of true anomaly (how bad is the angle?) and the difference of the squares of velocity at the present altitude versus the Pe from the loss from the Oberth effect, but I'd like to work it out more closely than that.

I assume by efficiency, you mean in terms of delta-v.  You haven't specified.  You also have not specified what you mean by "finite burn compared to an equivalent impulse burn". Equivalent in what manner? Without specifying the details of your finite burn, there is no way to know. You stipulated that constant acceleration is an OK assumption. (While not always realistic, it is an OK starting place.) Now fill in the rest of the details of the circularization process.

Once you do define the problem, you still won't find an equation for your efficiency measure.  The equations of motion for a rocket undergoing arbitrary thrust in a gravitational field are not analytically solvable. You have a couple options:  1) simulation.  You could do a bunch of tests in KSP or you could numerically integrate the equations of motion a bunch of times for different thrust profiles. The equations of motion are pretty simple to specify in the restricted two body system of KSP:  dv/dt = (1/m(t)) (FThrust + FGravity). You can solve this in two dimensions if the thrust is always in the plane of the initial trajectory. Bold quantities are vectors. m(t) is mass as a function of time. You need to specify the direction and magnitude of FThrust as a function of time. 

2) approximation: Instead of a single impulse burn, approximate the finite burn as a small number of impulse burns. This is a bit hairy, but doable. Instead of the equations of motions, you use the vis-viva equation and equations for the geometry of conic sections. In practice, simulation is more accurate, and in fact, easier, than doing the approximate calculation.  (I have been working on a related calculation in my spare time, and it's seriously ugly. Many pages of equations, even without showing the intermediate algebra. It took me way less time (< 1/10th) to write a trivial numerical simulator and run some quick tests.) 

Edited by Yasmy
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