Burn efficiency question: Follow the ball or not?

[The_Rocketeer]

ETA: But then again, if you burn at prograde at opposite sites of an orbit, you're burning in opposites directions and it certainly doesn't cancel out, so I'm probably wrong.

You're not completely wrong. The reason it doesn't cancel out is your frame of reference - Kerbin - has stayed fixed i.e. the direction of the gravity field has turned with you/is also reversed. Mathematically, turning to stay on prograde keeps everything exactly the same except your orbital velocity and therefore altitude, so you could do laps and just spiral faster and higher until you escaped Kerbin's SOI.

[Francesco]

Okay, Tavert and Numerobis are directly contradicting each other:

Who's right? This is the core of my question.

They aren't, they are actually saying the same exact thing.

Following the prograde marker ("straight line") is more efficient than following the maneuver marker ("arc"), but it's also less precise - because you're not performing the exact maneuver you planned.

[tfabris]

Yes, but, ignoring the precision question, which of the two things increases my height more, given the same fuel consumption? (or conversely, would doing one or the other require a slightly shorter burn to accomplish the same height change?)

[numerobis]

I suspect prograde increases your semi-major axis more, whereas burning towards the node increases your apoapsis more. But I don't have a proof.

[The_Rocketeer]

The quickest way to gain altitude is to burn radial out. Burning to a maneuver vector will increase your altitude more because it as the same thrust force as burning radial-out at a point 90 degrees further around your orbit. Think of your orbit not as a circle but as an ellipse, with a width and a height. Burning prograde will keep the width and height approximately equal, whereas burning on a maneuver vector will cause the height to increase much more than the width.

What you're not understanding is Delta-V is a change in velocity. Velocity isn't just the speed you're going at now, it's also the direction you're going in relative to something else. So, a ship that burns into an elliptical orbit by using a maneuver vector will still use exactly the same amount of fuel over exactly the same amount of time to reach the same energy state as a ship that burns into a circular orbit using a prograde-vector gravity turn. The only difference will be that one orbit is shaped like a comet's orbit, and the other like a planet's orbit.

[tavert]

So, a ship that burns into an elliptical orbit by using a maneuver vector will still use exactly the same amount of fuel over exactly the same amount of time to reach the same energy state as a ship that burns into a circular orbit using a prograde-vector gravity turn.

That's not quite the case. Any burn component in a non-prograde direction does not change orbital energy, it only changes the direction of your velocity and hence the shape of your orbit (some combination of eccentricity, inclination, LAN, and/or LPE, but not SMA).

Edited October 23, 2013 by tavert

[The_Rocketeer]

I did qualify that by saying:

burns into an elliptical orbit by using a maneuver vector

That does assume a prograde burn. Radial or retrograde burns will obviously have different effects.

Also, the shape of the orbit will be a factor of the energy state. Two orbits with an apoapsis at the same altitude but a periapsis at different altitudes will have different potential energy states. A rocket with a higher apoapsis in an elliptical orbit doesn't necessarily have more potential energy than one with a lower apoapsis in a circular orbit.

Edited October 23, 2013 by The_Rocketeer