oberth effect

[Yasmy]

I'm pretty sure it changes the net vector, but if your prograde/retrograde speed is zero on your new heading, oberth applies as if your speed was zero.

It's why inclination changes are so Deltav expensive- despite being close to a mass, you dont get ANY oberth to help you out.

I edited my post to indicate that I was explicitly referring to the fact that he said that you could make a burn perpendicular to your velocity without making a kinetic energy change.

I removed one sentence from the quote. Please see my quote in the new context.

I did not speak one way or another about when you do or do not get the benefit of the Oberth effect. I'm not sure why you responded to me.

But you are wrong that you don't get any Oberth effect when making inclination changes. You can only make very small inclination changes by burning perpendicular to your initial prograde. If you want to make a serious inclination change, without extending your orbit, you have to make your final speed equal to your initial speed. That means you have to burn somewhat along your initial retrograde direction. See Scott Manley's Orbital Mechanics tutorial on inclination changes. He goes through the trig to pick the correct burn direction for an inclination change which doesn't extend your orbit. It ain't perpendicular to your initial prograde.

Orbitial Mechanics 2

Edited February 27, 2014 by Yasmy

[Yasmy]

Ahh, but your exhaust has to lose kinetic energy... If you burnt at right angles, your exhaust orbital energy doesn't lose or gain... So you don't lose or gain any orbital energy.

Of course it does. If your velocity is v, and you burn by dv at a right angle, your new velocity is sqrt(v^2 + dv^2). Your new orbital energy is approximately 1/2 (m-dm) dv^2 greater.

It's much smaller than for a prograde burn, where the change in energy is approxiamtely (m-dm) v dv, but it's not zero.

Edited February 27, 2014 by Yasmy

[SSSPutnik]

Of course your kE changes, agreed, but you get zero Oberth "boost".

I updated my Oberth for Dummies pic based on what I've learnt.

Edited February 28, 2014 by SSSPutnik

[maccollo]

This is incorrect. Any radial or normal burn (or mixed radial and normal) burn will change the magnitude of your velocity, thus your kinetic energy. It's plain old Pythagorus. Initially, K = 1/2 m v^2. Finally, K = 1/2 (m-dm) (v^2 + dv^2).

So for a radial or normal or mixed radial/normal burn, dK = 1/2 (m-dm) (v^2 + dv^2) - 1/2 m v^2 = 1/2 (m - dm) dv^2 - 1/2 dm v^2

If you want to do a pure plane change maneuver, resulting in the same shape orbit but at a different inclination, you need to burn away some of your initial prograde velocity. Even then, your kinetic energy (and total energy) will be lower in the end, because you reduced your mass by the amount of fuel spent.

The "Oberth only works in the prograde or retrograde direction" part is correct, precicely because of Pythagoras theorem.

Or rather, you get increasingly reduced oberth boost as you rotate the acceleration vector towards 90 degrees.

the change in kinetic energy (ÃŽâ€k) is half the square of the final velocity (v2) minus half the square of the initial veloicty (v1):

ÃŽâ€k=v2^2 - v1^2

2

To get the final velocity when burning in line with the motion vector we simply add the initial velocity with the total acceleration V2=v+ÃŽâ€v

ÃŽâ€k=(v+ÃŽâ€v)^2 - v^2

2

However, like you said, for a for a burn that is radial to the motion vector we have to use Pythagoras to get the final velocity said V2=sqrt(v^2+ÃŽâ€v^2).

So we insert Pythagoras theorem into the equation to solve the final velocity:

ÃŽâ€k=(sqrt(v^2+ÃŽâ€v^2)^2) - v^2

2

The square root of the Pythagoras theorem cancels out the square of the final velocity, therefore:

ÃŽâ€k=v^2+ÃŽâ€v^2 - v^2

2

+v^2 cancels out -v^2, therefore:

ÃŽâ€k=ÃŽâ€v^2

2

The conclusion is that if you burn at a right angle to the motion vector you get precisely 0 benefit from the oberth effect, because the initial vecloity does not factor into the change in kinetic energy.

*edit

Also massively ninjad. My post is completely redundant XD

I should add a few qualifiers:

The above only concerns burns at exactly 90 degrees to the motion vector.

As Yasmy said, proper inclination changes are not 90 of the motion vector, and you therefore get some small amount of boost from the oberth effect

Edited February 28, 2014 by maccollo

[SSSPutnik]

It's a good post though and you back it with math.

And yes good ol Pythagoras means if you say, burnt at 45 degrees, you would get Oberth, just less of it.

[maccollo]

Of course your kE changes, agreed, but you get zero Oberth "boost".

You only get 0 oberth boost if you burn at 90 right angles to the motion vector.

If you burn 45 degrees prograde, you acceleration in the prograde direction is the acceleration*cos45, and you get the oberth boost for that acceleration.

As Yasmy stated, a proper inclination change isn't at a right angle to the motion vector.

[SSSPutnik]

LOL ninja'ed you again. (And he did say "right angle").

maccollo, its good to see a lot of people learnt about Oberth, I certainly was mistaken in large parts of the theory but now have a solid grasp.

Also again, a credit to everyone that it didn't reduce to ranting!

Edited February 28, 2014 by SSSPutnik

[maccollo]

Since we're touching on the issue of burning off motion vector, there is one riddle that must be solved.

Is this an efficient way of doing an escape burn? I see people do this pretty often. The burn is so long that the acceleration vector points 45 degrees away from the prograde marker.

My gut says "HELL NO, you should point prograde all the way, like this!"

But I haven't done the math to prove it to myself.

Edited February 28, 2014 by maccollo

[SSSPutnik]

That's a big dV burn. You're pointing that way because you are doing a long burn. (LV-N) isn't it? In T- 9m 43s you will be pointing prograde.

KSP is keeping your burn on the same vector is all.

Gotta be careful doing those, I've ended up in the atmosphere a number of times...

And you're off topic!

[maccollo]

And you're off topic!

Well not quite. I brought it up because the argument I've heard people is that is efficient because you're "using the oberth effect more by not falling out from the gravity well as fast so you loose less kinetic energy".

But since I now know that all radial acceleration gets no oberth boost, this can't be true.

Edited February 28, 2014 by maccollo

[Superfluous J]

Well not quite. I brought it up because the argument I've heard people is that is efficient because you're "using the oberth effect more by not falling out from the gravity well as fast so you loose less kinetic energy".

But since I now know that all radial acceleration gets no oberth boost, this can't be true.

No, that's not Oberth. That's LV-Ns having crappy Isp

It's more efficient to burn prograde in almost every instance (when you actually want to turn or are in atmosphere - or both are a couple of the exceptions). The problem with burning prograde when you've got a maneuver node set up that's supposed to be prograde but isn't (because it's a 10 minute burn in LKO, for instance), the problem is you won't end up going where you wanted to go.

[Red Iron Crown]

This is incorrect. Any radial or normal burn (or mixed radial and normal) burn will change the magnitude of your velocity, thus your kinetic energy. It's plain old Pythagorus. Initially, K = 1/2 m v^2. Finally, K = 1/2 (m-dm) (v^2 + dv^2).

So for a radial or normal or mixed radial/normal burn, dK = 1/2 (m-dm) (v^2 + dv^2) - 1/2 m v^2 = 1/2 (m - dm) dv^2 - 1/2 dm v^2

If you want to do a pure plane change maneuver, resulting in the same shape orbit but at a different inclination, you need to burn away some of your initial prograde velocity. Even then, your kinetic energy (and total energy) will be lower in the end, because you reduced your mass by the amount of fuel spent.

You would be correct if the radial burn were instantaneous, or if by radial burn I meant "perpendicular to the initial prograde direction for the entire burn." Radial burns are one case where instantaneous burns are not equivalent to non-instantaneous burns.

A real radial burn is not instantaneous, nor does it stay perpendicular to the initial prograde direction. It changes to stay perpendicular to the current prograde direction. So it ends up burning away the initial velocity's kinetic energy at exactly the rate it adds it in another direction.

Imagine that our craft is traveling in a straight line at 10m/s (or, for the truly pedantic, in a circular orbit large enough that a straight line is a reasonable approximation for the time of the maneuver). Our ship's engine has an absurdly high Isp, so the mass of propellant spent for a burn is insignificant, just to keep things simple. Now, say we want to change the direction of our velocity by exactly 90 degrees while keeping the magnitude the same.

If our craft could perform instantaneous burns, it is intuitive that we could make the change in two burns. One burn of 10m/s in the retrograde direction to cancel our original velocity, then another burn of 10m/s at 90 degrees to the initial prograde direction to add the new velocity. Seems simple.

But...we can combine those two burns into one and save dV. If, instead, we burn at 135 degrees from initial prograde (which is 45 degrees from retrograde) we can cancel our initial velocity and add our new velocity all in one go. Your friend Pythagoras will tell you that the burn will be sqrt(2)*10m/s, or about 14.14m/s. So by combining our two burns into one we've saved almost 6m/s of dV for the same vector direction change. Our orbital energy hasn't changed, as our speed and altitude are the same as before (and we're assuming mass hasn't changed). Our course plot would look like a right angle.

Now, what happens when we do the same maneuver, but the burn is not instantaneous? Again, we could do one 10m/s burn to cancel our initial velocity, then another 10m/s burn at 90 degrees to add our new velocity for a cost of 20m/s of dV to complete the maneuver.

What if we want to combine the two burns into one to save dV? We need to combine burning to cancel velocity with burning to add it in a new direction. So we start a maneuver where we continuously rotate to stay headed at 90 degrees to current prograde for the duration of the burn. So at the beginning of the burn we are exactly 90 degrees from initial prograde and at the end we will be 180 degrees from initial prograde. This will combine adding our new vector with cancelling our old vector, and will do so in such a way that we're adding kinetic energy to the new one at exactly the same rate that we are subtracting it from our old one. The dV cost for the maneuver is about 14.14m/s. (The last two statements require calculus to show mathematically, which I'd rather not get into.) Our course plot would look like a rounded right angle. Both speed and orbital energy remain the same throughout the entire maneuver.

Oddly, TWR doesn't affect this. Whether it takes a second or an hour to do the maneuver doesn't matter, as long as the time is not zero and the craft can rotate quickly enough to keep up with the changing vector.

***

Your point about orbital energy changing during a plane change maneuver because we are shedding mass is true. Where we were talking about the Oberth effect and the velocity component of kinetic energy, I tried to make mass irrelevant by ignoring the mass change for clarity. Properly speaking, for a craft losing mass during a burn, I should have said specific orbital energy rather than orbital energy, i.e. orbital energy per unit mass. Thanks for the correction.

Edited February 28, 2014 by Red Iron Crown

[Yasmy]

Deleted for wrongness. Details in my next reply.

Edited February 28, 2014 by Yasmy
eating crow

[SSSPutnik]

No, that's not Oberth. That's LV-Ns having crappy Isp

You mean crappy thrust.

[Superfluous J]

You mean crappy thrust.

Crap. I actually meant to change that but failed. You are correct. Great Isp, crappy thrust.

But the basic point still stands.

[maccollo]

What if we want to combine the two burns into one to save dV? We need to combine burning to cancel velocity with burning to add it in a new direction. So we start a maneuver where we continuously rotate to stay headed at 90 degrees to current prograde for the duration of the burn.

That is not correct.

You just point 135 degrees from the initial prograde and then you burn.

The cost of the maneuver will be the exactly the square root of 200, about 14.1421. It doesn't matter if you use point acceleration or not.

Your way of doing it will cost more delta V.

How much more it's going to cost can be aproximated by simply calculating how much it would cost to change some fraction of the 90 degrees, like one percent of a degree, and then extrapolating to the full range.

The equation for determining the delta V cost of a plane change with equal initial and final velocity is: ÃŽâ€V=2V*sin(°/2)

were V is the initial and final velocity and ° is the amount of plane change we want in degrees.

2x10*sin(0.01/2)= 0.00174532924977908918851394674284

So that's the cost of 100:th of a degree. Now we can just multiply that with 9000 to see what it would cost to make a 90° inclination change one 100:th a degree at the time.

And we get: 15.70796324801180269662552068556

About 1.5658 m/s more expensive.

Now there's probably some way to calculate how much of a boost we get from the oberth effect by pointing in one direction the whole burn. I'm gonna bet it will account for the difference exactly.

Edited February 28, 2014 by maccollo

[Red Iron Crown]

Hmmm, you both might be right, let me think about it for a bit and do some testing. It is entirely possible that my understanding of inclination changes is wrong.

[maccollo]

I think figured out how to calculate exactly how expensive your maneuver is.

ÃŽâ€V=V2*À/(360/°)

In your example it is:

ÃŽâ€V=10*2*À/(360/90)

And we get exactly:

ÃŽâ€V=15,707963267948966192313216916398

So while the delta V cost of a proper inclination change can be represented by the length one side in a triangle where the two other sides is V the angle between V1 and V2 is the desired plane change. Your way can be represented by a fraction of the circumference of a circle where the radius is V, and the fraction is 360 divided by the desired plane change.

I'm shocked that there was a solution that was that simple O_o.

...

Still gotta try to solve for the oberth effect tho.

Edited February 28, 2014 by maccollo

[Yasmy]

Bingo, maccollo. In the limit that the angle dphi goes to zero, dV = 2 V sin(dphi/2) -> V dphi. Then integrating from 0 to final inclination phi gives ÃŽâ€V = V phi.

That's in radians. In degrees, ÃŽâ€V = V phi (pi/180).

For phi = 90 degrees, ÃŽâ€V = pi/2 V ~= 1.5708 V.

[maccollo]

Well That's a little embaressing. I could easily have avoided that if I had just payed attention a little better. Maximum derp XD.

Sorry guyzes I did a 0 degree plane change.

Result was Kraken

Edited February 28, 2014 by maccollo

[SSSPutnik]

Brain singularity.

[Yasmy]

So Red Iron Crown said you can change your inclination without changing your specific orbital energy by burning continuously normal to your current prograde.

I thought Red Iron Crown was wrong, but he is pretty much right, and I was wrong. Sorry about that.

Let x be your initial prograde direction, and y be your initial normal direction.

Burning normal to your velocity (vx, vy) means you burn in the direction (-vy,vx) or (vy,-vx):

dv_x/dt = - a v_y(t) / sqrt(v_x(t)² + v_y(t)²)

dv_y/dt = a v_x(t) / sqrt(v_x(t)² + v_y(t)²)

A perfectly good solution to this system of equations is:

v_x(t) = v cos(a t / v)

v_y(t) = v sin(a t / v)

So your total velocity (and thus specific orbital energy) is constant at all times: v_x(t)² + v_y(t)² = v².

Your delta-v is the time integral of your constant magnitude acceleration: sqrt((dv_x/dt)² + (dv_y/dt)²) = a

delta-v = a t

To change your orbital inclination to phi degrees, wait until v_x = v cos(phi * pi / 180).

This happens when (a t / v) = phi * pi / 180. And we already determined that delta-v = a t.

Thus delta-v = phi * pi/180 * v, exactly like maccollo found. So thanks for teaching me something, Red Iron Crown.

The following is wrong, but retained for thread continuity:

Note that I assumed constant thrust. If you can make the whole maneuver with essentially zero change in your thrust to weight, a continuous normal burn will maintain your specific orbital energy. In practice your thrust to weight increases (if you are not staging) while burning. So, if for the ease of it, you want to do a continuous normal burn plane change, keep your acceleration (thrust to weight) somewhat constant by decreasing your thrust over time.

Edited February 28, 2014 by Yasmy
lots of typos and an embarrassing mistake

[Red Iron Crown]

Yes, I did some testing last night and I was right about speed (and thus specific orbital energy) remaining the same during the rotating normal burn, but tremendously wrong about it being more efficient than doing a straight line burn. Which I find frustrating because I've been doing manual inclination changes this way since, well, forever, and was disappointed that maneuver nodes didn't allow for that type of maneuver. Guess my future missions will make better use of their dV. Thanks for the lesson.

Also embarrassingly, I was conflating radial and normal burns in some of my previous posts. Radial for eccentricity changes, normal for inclination changes, is what I should have written.

Yasmy, if your TWR is increasing during the rotating burn maneuver, the vector direction changes more quickly and the rate of rotation increases to keep pointed normal, but the specific orbital energy remains the same. Doesn't matter how fast or slow the maneuver is completed. Though I guess I'm prepared to drop discussion of the maneuver now that it's been shown to be inefficient.

Edited February 28, 2014 by Red Iron Crown

[Yasmy]

Doesn't matter how fast or slow the maneuver is completed.

Right again, Red.

This time I'll stop before doing the general Fourier series solution for arbitrary thrust profile.

You would think I hadn't seen simple harmonic motion a hundred times in the past 25 years...

[Chobotnatec]

I've read nearly all this forum and have seen a lot of really complicated and unpractical examples so here is my own actually useful example:

When you have mining and refueling station at Minmus orbit and want to continue for example to Jool. Than it is better to at first lower you periapsis to 71km near Kerbin and at this point start to thrust for Jool.

Why? Because while you need to spend some dV around 200m/s to lower your Pe then you get much faster at Pe(in order to get to Jool) and it means that gravity of Kerbin has shorter time to slow you down than it had for accelerating you when you've been going towards Pe and because difference in time is quite large you not only get back yours 200m/s but even more.

And that is Oberth Effect which explains why longer road(by distance) is actually shorter in terms of dV. Another Similar situation can be met when going from Layethe back to Kerbin