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Oberth effect... optimizing.


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Ok, deep stuff for the math geeks out there.

I'm looking for some clarity on calculating the benefits of Oberth effect, and using this to best effect.

Specifically for the very simple case.

One gravitational source, my rocket in a circular orbit around it. No other gravitational influences, just me _ rocket equation + oberth.

Given a set delta-v, and the goal to achieve the best possible escape velocity, how do I optimize my burn?

Running a few tests in KSP, i find some very interesting results!!

Starting from an orbit around the sun, virtually identical to Kerbin's orbit but outside Kerbin SOI...

(for purpose of this test, i "define" excess escape velocity as the amount of retrograde burn i can apply, when crossing Jool orbit heading out, that will still give a apogee date >> 50 years)

Starting with about 20896m/s delta-v

If I burn pure prograde, my ship reaches excess escape velocity of 22012m/s

If i "waste" 7495.2m/s of my deltav, to drop sun periapsis to 10km, and then burn the remaining 13400m, I get excess escape velocity of 44500m/s!

but!!

If i burn prograde for 2000 thus stretching my orbit to almost Dres, then retro for 3382, i get the same perigee. But using 2113m/s less, *and* moving a fraction faster at peri.

Using this, I can get excess escape speed up to 50190m/s!

{image below, to illustrate what I mean}

So, counter all intuition, the more efficient flight plan is to first expand my orbit, then shrink it, then expand it again!?

Ok.. So has anyone out there got the mathematics (mathemagics?) to explain and quantify this?

Obviously the best strategy would change, depending on initial orbit and available delta-v.

so, just what would the *optimal* burn be, for a ship with 20000m/s delta-v, in circular sun orbit at Kerbin orbital radius, to maximize escape speed?

(maybe I should make this a challenge... and forbid entering any SOI other than the sun... hmmm)

((disclaimer))

the numerical figures above are accurate, but as per maneuver node predictions.

Due to burn times, the best I could achieve with my 3-burn strategy was about 47000m/s excess. (that's about 52400 m/s, when crossing Jool orbit)

Still, not bad, considering I only started with 9300 orbital velocity and 20896 in the tank, and had to climb up to Jool.

rlKRsja.png

Start d-v of 20896 available, Kerbin-type orbit.

burn1: pro 2000 immediate

burn2: retro 3382.25 at apo

burn3 pro 15513 at peri

"burn4" is virtual, it is used to measure speed while crossing Jool orbit. In this case, -50190, thus speed = 55590 as escape at Jool = 5400m.s.

Edited by MarvinKitFox
image to explain
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The first two burns you did - extending your orbit above Dres, then lowering your periapsis from there - are example of bi-elliptic transfer to low Sun periapsis. You could save more dv by extending your initial apoapsis higher. Theoretically dv-optimal transfer is bi-parabolic, where you raise your velocity to exactly escape speed, coast to infinity, then at infinity perform correction for 0 m/s for arbitrarily small Sun periapsis, and then at that periapsis spend the rest of your fuel. Of course coasting to infinity makes no sense so you need to stop somewhere earlier.

Edited by Kasuha
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Kasuha's right, you've stumbled independently onto biecliptic transfers. They're dV efficient in most cases but consume a lot of time. Not a factor in KSP where kerbals reside happily in a cramped pod for years, but IRL you'd basically never do it with a manned ship.

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The first two burns you did - extending your orbit above Dres, then lowering your periapsis from there - are example of bi-elliptic transfer to low Sun periapsis. You could save more dv by extending your initial apoapsis higher. Theoretically dv-optimal transfer is bi-parabolic, where you raise your velocity to exactly escape speed, coast to infinity, then at infinity perform correction for 0 m/s for arbitrarily small Sun periapsis, and then at that periapsis spend the rest of your fuel. Of course coasting to infinity makes no sense so you need to stop somewhere earlier.

So absolutely best delta-v achievable in this scenario is:

(Starting dv - dv to minimal escape) * Achievable Oberth factor?.. that makes it easier.

My initial orbit is at 9300m/s, v at periapsis is about 92800m/s,

This acts on my remaining 17045 d-v

so sqrt(1+Ve/dv) = 3.448 as Oberth factor.

I should be getting 3.448*17045= 58771 over escape velocity.

Nodes prediction yields 55590, practical flying yields 52400m/s.

I'd say that's in agreement with the math.

(The math just got a *lot* easier when I'm using 'infinite' drop to sun apo, i.e. exactly equal to escape velocity!)

So back to my original puzzle, at how much delta-v does one gain benefit from diving to the sun for Oberth, when departing on interstellar trip...

it is when (dv-3280) * 3.448 > dv, thus when dv >= 13275

I wonder if I can achieve my goal of 100000m/s interstellar speed.

Would only need a ship of 33200 deltav in LKO, and good TWR for the sun boost.

hmmmm... thats not so bad.

Scratch that last paragraph, im an idiot.

The more you burn, the less the relative effect of Oberth is..

to get 100k interstellar, i need...

{bashes calculator mercilessly}

67700??? eww!

Edited by MarvinKitFox
im an idiot
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Kasuha's right, you've stumbled independently onto biecliptic transfers. They're dV efficient in most cases but consume a lot of time. Not a factor in KSP where kerbals reside happily in a cramped pod for years, but IRL you'd basically never do it with a manned ship.

Combining the two optimizations... Bi-elliptic transfers, and maximal Oberth boost.

its *fun* getting 50K speed out of a 20K d-v rocket!

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