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[LordSlowpoke]

one does not simply maneuver nodes

one does not maneuver nodes at all

[forsaken1111]

one does not simply maneuver nodes

...I do. Even if I dont have mechjeb I can get pretty much anywhere by messing with nodes.

[Twi]

one does not simply maneuver nodes

...I do. Even if I dont have mechjeb I can get pretty much anywhere by messing with nodes.

Manuver nodes are pretty darn useful, yeah.

[Dutchling]

I can't do shit without nodes.

[BigFatStupidHead]

Things were pretty rough before nodes, yes.

[jocan2003]

Things were pretty rough before nodes, yes.

Things were pretty rough before internet ( research wise, hours and hours at the local library ) Now, its easier, so you dont use it? why not use tools at your disposal? Dont you use calculator sometime? Bah enough of that XD

[BigFatStupidHead]

Things were pretty rough before nodes, yes.

Things were pretty rough before internet ( research wise, hours and hours at the local library ) Now, its easier, so you dont use it? why not use tools at your disposal? Dont you use calculator sometime? Bah enough of that XD

Huh? Nodes are a great thing. They make the game a lot better.

[jocan2003]

Im for the nodes, but some purist says they shouldnt be in or something... I dont think i understand fully here, but thats what my french head understand.

[BigFatStupidHead]

S'alright. I have a way with words that can even confound native english speakers. Nodes = good.

[OREOSOME]

Kerbal warmachines have never been so amazing.

[DrPoo]

Kerbal warmachines have never been so amazing.

Old

Spoiler (click to show/hide)

but gold

[Twi]

Kerbal warmachines have never been so amazing.

Also old and awesome, but...
I see your two minutes and raise you one hour. With Touhou music!

[da_nang]

Interesting... I was plotting a maneuver to fully reverse the orbital plane (of a circular orbit) from 180º to 0º by raising the apoapsis, reverse the planes at the new apoapsis and finally lower the apoapsis to its original height and I wanted to find the minimum apoapsis where the maneuver would be more efficient than a straight retrograde burn on the original orbit. Turns out, in KSP any higher apoapsis seems to do. I also found the minimum amount of delta-v required for this maneuver to be roughly 41.4% of the delta-v for the simple retrograde burn. In other words, you save at most 58.6% delta-v doing this maneuver, the higher the better. (Or in KSP's case, aim for an apoapsis close to SoI.) Furthermore, with an apoapsis at least 8 times higher than the circular orbit, you'll save at least 50% delta-v.

Example, if you're orbiting Eve (Radius 0.7 Mm) at 1 Mm (1 Mm + 0.7 Mm = 1.7 Mm), you can save 50% delta-v with a "work" apoapsis at 12.9 Mm (1.7 Mm * 8 - 0.7 Mm).

Spoiler: Any higher apoapsis will do (click to show/hide)

Assume circular orbit.

Initial velocity (circular orbit): v
Initial orbital height (calculated from planet center): R

"Work" apoapsis: h
Apoapsis speed: u

Planet gravitational parameter: μ

Delta-v for initial transfer burn:
v = sqrt(μ/R)

v₁ = sqrt(μ/R)sqrt(2h/(R+h))

|DV₁| = v₁ - v = sqrt(μ/R)(sqrt(2h/(R+h)) - 1)

Delta-v for "reverse" burn:

|DV₂| = 2u = 2*sqrt(μ/h)sqrt(2R/(R+h))

Delta-v for circularizing:

|DV₃| = |DV₁| = sqrt(μ/R)sqrt(2h/(R+h)) |Initial orbit with reverse direction

Total delta-v:

DV_tot = |DV₁| + |DV₂| + |DV₃| = 2*|DV₁| + |DV₂|
= 2*[sqrt(μ/R)(sqrt(2h/(R+h)) - 1)] + 2*sqrt(μ/h)sqrt(2R/(R+h)) = 2*sqrt(μ)[sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h))]

Delta-v for "simple" retrograde burn:
|DV_rev| = 2v = 2*sqrt(μ/R)

Minimum height h:

DV_tot < |DV_rev|

2*sqrt(μ)[sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h))] < 2*sqrt(μ/R) |/[2*sqrt(μ)]

sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h)) < sqrt(1/R) |*sqrt(R)

sqrt(2h/(R+h)) + R*sqrt(1/h)sqrt(2/(R+h)) < 2 |*sqrt(h)

h*sqrt(2/(R+h)) + R*sqrt(2/(R+h)) < 2*sqrt(h) |*sqrt((R+h)/2)

h + R < 2*sqrt(h)sqrt((R+h)/2) |Both sides positive, square both sides

(h + R)^2 < 2h(R+h)

(h + R)^2 - 2h(R+h) < 0

(R+h)(R-h) < 0 |R+h > 0

h > R

Q.E.D.

Spoiler: Max efficiency is 58.6 % (click to show/hide)

Assume h>R for a given R>0. (Specifically above atmosphere...)

DV_tot(h) = 2*sqrt(μ)[sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h))]

Note that for R > 0, h > 0, the function g(h) = sqrt(2h/(R+h)) - 1, g(h) > 0, h>0 is always increasing, which is easily seen from a first order derivative test.
Thus the function DV_tot(h) should also always increase as its second part, sqrt(1/h)sqrt(2R/(R+h)), is never negative.
Therefore, an extrema can only be found in the end intervals of h according to Fermat's theorem. For h = R, the DV_tot(R) = 2*sqrt(μ/R) which is the same as DV_rev. The upper interval is at infinity: lim_h→∞ DV_tot(h) = lim_h→∞ 2*sqrt(μ)[sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h))] = 2*sqrt(μ/R)(sqrt(2) - 1).
Thus the maximum efficiency is 100% - [2*sqrt(μ/R)(sqrt(2) - 1)/2*sqrt(μ/R)]*100% = 100%*(2-sqrt(2)) ~ 58.6 %

Q.E.D.

Spoiler: 50% efficiency at h = 8R (click to show/hide)

2*sqrt(μ)[sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h))] = 0.5*2*sqrt(μ/R) |/[2*sqrt(μ)

sqrt(1/R)(sqrt(2h/(R+h)) - 1) + sqrt(1/h)sqrt(2R/(R+h)) = 0.5*sqrt(1/R) |*sqrt(R*h)

h*sqrt(2/(R+h)) + R*sqrt(2/(R+h)) = 1.5*sqrt(h) |*sqrt((R+h)/2)

h + R = 1.5*sqrt(h)*sqrt((R+h)/2) |Square both sides

(h+R)^2 = (9/8)*h(R+h) |/(R+h)
h+R = (9/8)h
h/8 = R
h = 8R

Q.E.D.

[Mookzen]

I opened the spoilers expecting pretty pictures but alas, *gasp* there's only math... 

[da_nang]

I opened the spoilers expecting pretty pictures but alas, *gasp* there's only math... 

I suppose you want some empirical evidence?

Spoiler: Simple retrograde (click to show/hide)

Spoiler: Maneuver (click to show/hide)

Total DV for maneuver is 881.96 m/s compared to the simple 1748.8 m/s. The apoapsis is a bit lower than "50%-efficiency" height (50% was outside SoI at the time). Actual efficiency was at 49.7%.