oberth effect

[SSSPutnik]

I think the main issue with low TWR, is by the time you finish your burn, periapsis was an hour ago and you've lost significant speed as you climb back up the gravity well and Oberth has started thinking about going for an early beer.

To get maximum Oberth effect, you want to do as much of the burn as possible at the highest speed you can. You can't do an instantaneous burn though, (an explosion is close but still takes some time).

If it takes 2 minutes to do the burn you want, you start it one minute before periapsis, so at halfway through you're at peak speed.

@claw - The wiki doesn't talk SOI at all... It talks about hyperbolic orbits, which are quite different. SOI is just an artefact of KSP physics. I think Wiki and Rho project mention them due to, in the real world, it's what you would be using Oberth for.

But... There is nothing stopping you from using Oberth to raise or lower a closed orbit. It's just in real life you don't do this very often which is why the articles fail to mention it.

Oberth boils down to dumping high kE fuel in lower orbits, (or higher orbits if burning retrograde) giving you a helping hand.

This matters, otherwise according to some people, dumping a stage/engine/kerbal at periapsis would give you a boost, which it doesn't. The item just follows with you in the same orbit.

[Red Iron Crown]

Oberth boils down to dumping high kE fuel in lower orbits, (or higher orbits if burning retrograde) giving you a helping hand.

This matters, otherwise according to some people, dumping a stage/engine/kerbal at periapsis would give you a boost, which it doesn't. The item just follows with you in the same orbit.

Oberth makes retrograde better at lower orbits, too. Remember our exploding ship example from a few pages back.

Dumping a stage does give a boost, it is the same as expending fuel (only at a much, much lower Isp). The delta-V is very, very small though. But you will see that even with the weakest decoupler the two objects drift further and further apart as they follow their very slightly different orbits.

[Claw]

SOI is a real assumption. It's a way to simplify orbital dynamics, and it's considered a valid assumption. Call it a gravity well if you wish, it's the same idea. It's an artifact of scientific assumption.

You can't have a hyperbolic trajectory that doesn't leave the SOI. That's the definition of a parabolic/hyperbolic curve. It runs off to infinity. Now that doesn't mean you have to fly the whole trajectory.

Also, I said I agree that you can use Oberth inside an SOI. I didn't use that as an example because the math is more complex, so it makes using a conceptual example overly complicated for posting on a text forum. Also, the OP said he read the wiki and didn't understand. So I was explaining the wiki and focusing on it's explination in an attempt to not get bogged down in details.

TWR affects how much Oberth effect you can harness, but Oberth is not dependent on your TWR. It's a subtle distinction. If I somehow had a very high TWR and a very long burn time, the ability to harness Oberth is effected in a similar way as if I had a low TWR and the same burn time. The assumption of TWR effecting Oberth is making a secondary assumption that the same impulse is being applied, which wasn't stated.

So if I said, "For a given dV burn, having a higher TWR means I can harness more Oberth" that would be more accurate. This is implicit in the theory, but that's because the theory assumes an instantaneous application of a given dV.

But here's another thought. Assume a circular orbit such that orbital speed is unchanging (PE = AP). If the engine is fired at any arbitrary point, ship speed increases. Does this increase my Oberth?

Edited February 27, 2014 by Claw

[Claw]

Oberth makes retrograde better at lower orbits, too. Remember our exploding ship example from a few pages back.

Dumping a stage does give a boost, it is the same as expending fuel (only at a much, much lower Isp). The delta-V is very, very small though. But you will see that even with the weakest decoupler the two objects drift further and further apart as they follow their very slightly different orbits.

Your exploding ship example is right on. In fact, it isn't even that the retrograde portion is experiencing Oberth. It is "because of" the retrograge portion that you have Oberth. You're litterally stealing it's kinetic energy. (And the opposite is also true.)

Again, this is why I dislike examples that stay inside an SOI and discuss things like AP retro burns. They defy intuition and simply confuse the issue of Oberth.

And agreed. Dumping a stage in orbit does give a change if any amount of force was used in the separation (i.e. a decoupler).

[Geschosskopf]

OK, OK, OK.

I will accept that there is no such thing as DVO. The reason I defined it was to see if it really existed and if so, how big it was. Mission accomplished.

But this means orbital maneuvering is totally a function of vector addition. Current vector + burn vector = new vector. Because burns consume fuel, the bigger the burns you make, the fewer burns you can do, which limits your ability to make further maneuvers. Therefore, we always place our nodes to minimize the size of the burn vector needed to go from Orbit A to Orbit B. To do this, we look for where the burn vector has the most "leverage" to "pry" our current orbital path from where it is now into alignment with the desired orbital path. This intuitively tells us to put Ap-changing burns at Pe and vice versa, and inclination-changing burns at the most remote ascending or descending node.

This is just plain common sense. You can do this without ever having heard the term "Oberth Effect", and in fact many people do. This thread demonstrates that applying the mysterious name "Oberth Effect" to the simply process of looking for vector "leverage" is a source of unnecessary confusion. If you understand the vector nature of spaceflight, then the word "effect" naturally makes you look for some external force that changes the underlying vector math. But apparently there isn't one. Thus, "Oberth Effect" IMHO is a poor choice of words.

Thus, after all this discussion, I return to the place I started in maintaining that there is no need to worry about Oberth at all. Just keep doing what you're doing in terms of seeking to minimize the burn vectors you use, and you get the benefit of Oberth. Because they're exactly the same thing.

Still, I have this nagging doubt. I learned orbital mechanics from the Newtonian vector side without ever hearing of Oberth until I came to this forum. I have since read that the pre-Oberth guys, who certainly knew all about Newtonian vectors, couldn't figure out how to make maneuvers work on realistically available amounts of fuel. This I can't understand, because Oberth as defined here is nothing but a natural result of how the vectors work. So how did the pre-Oberth guys miss something so obvious? This creates in me the suspicion that Oberth has to be more than just saying "put your burn vectors where they're the smallest".

But even if that is the case, it still makes no difference to the practical matter of maneuvering your ship so, again, there's no need to worry about Oberth.

[Red Iron Crown]

But here's another thought. Assume a circular orbit such that orbital speed is unchanging (PE = AP). If the engine is fired at any arbitrary point, ship speed increases. Does this increase my Oberth?

The Oberth effect is equal anywhere along a circular orbit because orbital speed is constant.

[Claw]

The Oberth effect is equal anywhere along a circular orbit because orbital speed is constant.

But once I start firing engines, my speed increases. That's the key point of this thought exercise. Does increasing my speed over time improve my ability to harness Oberth?

[Claw]

This is just plain common sense. You can do this without ever having heard the term "Oberth Effect", and in fact many people do. This thread demonstrates that applying the mysterious name "Oberth Effect" to the simply process of looking for vector "leverage" is a source of unnecessary confusion. If you understand the vector nature of spaceflight, then the word "effect" naturally makes you look for some external force that changes the underlying vector math. But apparently there isn't one. Thus, "Oberth Effect" IMHO is a poor choice of words.

SNIP...

Still, I have this nagging doubt. I learned orbital mechanics from the Newtonian vector side without ever hearing of Oberth until I came to this forum. I have since read that the pre-Oberth guys, who certainly knew all about Newtonian vectors, couldn't figure out how to make maneuvers work on realistically available amounts of fuel. This I can't understand, because Oberth as defined here is nothing but a natural result of how the vectors work. So how did the pre-Oberth guys miss something so obvious? This creates in me the suspicion that Oberth has to be more than just saying "put your burn vectors where they're the smallest".

But even if that is the case, it still makes no difference to the practical matter of maneuvering your ship so, again, there's no need to worry about Oberth.

Yes. You don't need to do anything special to harness Oberth. There are ways to go out of your way to harness it, but for the most part isn't necessary.

The Newtonian vectors work fine enough in/around Earth when you ignore Oberth. What was throwing people off was that original pure vectors only considered chemical energy, and not the exchange of Kinetic. So they proclaimed you needed vast amounts of dV to get inter-planetary. If you include KE in the vectors, then it's a non-factor.

AND YES!!

KSP factors this all in. So put your nodes down and enjoy the fact that KSP calculates it for you.

[Jouni]

Still, I have this nagging doubt. I learned orbital mechanics from the Newtonian vector side without ever hearing of Oberth until I came to this forum. I have since read that the pre-Oberth guys, who certainly knew all about Newtonian vectors, couldn't figure out how to make maneuvers work on realistically available amounts of fuel. This I can't understand, because Oberth as defined here is nothing but a natural result of how the vectors work. So how did the pre-Oberth guys miss something so obvious? This creates in me the suspicion that Oberth has to be more than just saying "put your burn vectors where they're the smallest".

The history of science is full of ideas that are obvious in retrospect. As far as I understand, people before Oberth assumed that the thrust of a rocket was somehow constrained by the chemical energy in the fuel. The faster the rocket went, the less thrust it was assumed to produce - as seemed to happen with every other kind of vehicle. Rocketry was also a rather marginal hobby until the early 20th century, so people just didn't think things like that too much.

[Kasuha]

But once I start firing engines, my speed increases. That's the key point of this thought exercise. Does increasing my speed over time improve my ability to harness Oberth?

TWR and time needed to perform a burn are not relevant to understanding of Oberth effect.

Assuming you start the burn at the place most suitable to do the burn and it takes long time, yes, Oberth effect will be the reason why you'll find the burn was not as effective as if you were able to deploy the whole impulse instantly. You can imagine that you applied the burn as a series of instant impulses evenly distributed over the time interval and because you're moving, only the first impulse is at the most suitable place and every subsequent later impulse is at slightly less optimal place. That's also why Oberth effect is not considered important for probes using ion engines.

But to understand Oberth effect, assuming all burns are instaneous is better. Because then you can see that where you do the burn is the matter.

[SSSPutnik]

Oberth makes retrograde better at lower orbits, too. Remember our exploding ship example from a few pages back.

Dumping a stage does give a boost, it is the same as expending fuel (only at a much, much lower Isp). The delta-V is very, very small though. But you will see that even with the weakest decoupler the two objects drift further and further apart as they follow their very slightly different orbits.

Yes agreed. I meant more detaching rather than eject. ( let's say you just unscrewed the bolts).

I did some testing tonight with a highly elliptical orbit around Jool.

Periapsis I was over 9000ms and at apoapsis around 350ms.

Using MJ I did 100dV burns and I can confirm speed does not change. Whoever was saying this is indeed 100% correct and it IS just orbital energy changing.

Burning 100dV changed my speed 100ms, prograde, retrograde, periapsis or apoapsis.

It can easily be seen using Mechjeb that changing your semi major axis at apo or periapsis makes a huge difference in dV requirement. At periapsis it told me I needed 16dV, at apoapsis 700dV, more than 10x different. (For the same semi major axis change).

Further reading online says that in many cases it is actually more efficient to do a direct Hohmann transfer than to set up Oberth maneuvers.

Anyway, its been a great learning experience!

Edited February 27, 2014 by SSSPutnik

[Red Iron Crown]

Still, I have this nagging doubt. I learned orbital mechanics from the Newtonian vector side without ever hearing of Oberth until I came to this forum. I have since read that the pre-Oberth guys, who certainly knew all about Newtonian vectors, couldn't figure out how to make maneuvers work on realistically available amounts of fuel. This I can't understand, because Oberth as defined here is nothing but a natural result of how the vectors work. So how did the pre-Oberth guys miss something so obvious? This creates in me the suspicion that Oberth has to be more than just saying "put your burn vectors where they're the smallest".

Oberth's confusing because it only works for reaction drives that carry all their reaction mass with them (i.e. engines that accelerate by ejecting mass from the craft) and furthermore it only works in the prograde and retrograde directions.

The Newtonian physics we are used to with non-reaction drives are different. Most surface-bound non-reaction engines have fixed power (J/s) with varying thrust. The faster you are going, the more energy it takes to change speed by a given amount, so acceleration slows as you go faster because the engine is adding energy at a fixed rate. This is a consequence of kinetic energy being proportional to the square of speed.

Reaction drives are unique in having fixed thrust with varying power. This is because we expend fuel at the same rate, but the fuel has more energy when burned at a higher speed. So a given amount of fuel burned adds more kinetic energy when going faster.

Oberth only works in the prograde or retrograde direction because those are the cases where we're trying to change the amount of kinetic energy. For a radial or normal burn, kinetic energy doesn't change, just the direction of the velocity vector (the inclination or eccentricity of the orbit, not its total energy). In fact, the reverse is true for normal or radial burns, they are more effective the slower the craft is going.

Edited February 27, 2014 by Red Iron Crown

[Red Iron Crown]

But once I start firing engines, my speed increases. That's the key point of this thought exercise. Does increasing my speed over time improve my ability to harness Oberth?

For an instantaneous burn, the Oberth effect is constant because all acceleration occurs at the same speed (this is a contradiction, part of why instantaneous burns are impossible).

For burns with a non-zero duration, i.e. all real burns, things change. A prograde burn experiences more Oberth effect because some of the acceleration occurs at a faster than initial speed. Conversely, a retrograde burn experiences less Oberth effect because some of the acceleration occurs at a slower than initial speed.

[SSSPutnik]

Yes your fuel has more kinetic energy but remember you are gaining that energy by dropping your fuel into a lower energy orbit, or pushing it into a higher energy orbit (retrograde burn).

Oberth comes into effect by balancing this with an orbital energy addition or subtraction (retrograde burn).

[Red Iron Crown]

Yes, the amount of energy exchanged between craft and fuel is greater the faster it is going.

[Kasuha]

For burns with a non-zero duration, i.e. all real burns, things change. A prograde burn experiences more Oberth effect because some of the acceleration occurs at a faster than initial speed. Conversely, a retrograde burn experiences less Oberth effect because some of the acceleration occurs at a slower than initial speed.

I disagree with this.

Instaneous prograde burn of X m/s will give you more energy than instaneous retrograde burn of the same X m/s will take from you.

Effectivity or ineffectivity of non-instaneous burns only depend on how far you get from the optimum point while you're burning.

[Red Iron Crown]

I disagree with this.

Instaneous prograde burn of X m/s will give you more energy than instaneous retrograde burn of the same X m/s will take from you.

Effectivity or ineffectivity of non-instaneous burns only depend on how far you get from the optimum point while you're burning.

You're right, I should clarify.

The Oberth effect is dependent on the difference between the square of final speed and the square of initial speed. So an instantaneous prograde burn gains more energy than a retrograde one loses, because final speed is higher.

But it is also true that a non-instantaneous burn changes things even without accounting for closeness to the optimum. If you split a 2m/s prograde burn into two 1m/s prograde burns, the second burn experiences more Oberth effect because both initial and final speed are greater. Similarly, for a retrograde burn split the same way, the second burn experiences less Oberth effect because initial and final speed are lower.

Edit: On further reflection, the second part of this post is true for instantaneous burns, too. You're absolutely right Kasuha, thanks for the correction.

Edited February 27, 2014 by Red Iron Crown
Eating humble pie.

[Claw]

TWR and time needed to perform a burn are not relevant to understanding of Oberth effect.

Yes, I fully agree. Hopefully my thought exercise didn't imply otherwise. It was more to illustrate how you can't peanutbutter spread a theory around.

But again, if I could employ a partial but instantaneous impulse of dV, my orbital speed just increased. So if I can immediately apply another partial impulse of dV, did I maximize my Oberth?

I'm not saying this is correct, I'm trying to illustrate examples that seem intuitive but are not accurate representations of the theory. Because as you said, you're not at the optimal place for the new orbit (after the first impulse) even though you're going faster than you were.

[Claw]

Muahaha. I was ninja'd while posting and hopefully I've made my point about intuition and theory.

[Red Iron Crown]

But again, if I could employ a partial but instantaneous impulse of dV, my orbital speed just increased. So if I can immediately apply another partial impulse of dV, did I maximize my Oberth?

No, it would be the same as if you did a single instantaneous burn.

[Vanamonde]

If this thread ever was a gameplay question, it has long since mutated into a science discussion. So by burning at periapsis, its orbit has shifted to Science Labs, gaining kinetic energy but not speed in the transition, or something like that.

[Yasmy]

For a radial or normal burn, kinetic energy doesn't change, just the direction of the velocity vector (the inclination or eccentricity of the orbit, not its total energy).

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.

Edited February 27, 2014 by Yasmy
Edited to quote just the portion I was objecting to.

[SSSPutnik]

Ahh, but your exhaust has to lose or gain orbital 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 in return.

Edited February 27, 2014 by SSSPutnik

[Rakaydos]

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.

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.

[Yasmy]

I disagree with this.

Instaneous prograde burn of X m/s will give you more energy than instaneous retrograde burn of the same X m/s will take from you.

Effectivity or ineffectivity of non-instaneous burns only depend on how far you get from the optimum point while you're burning.

I'm not sure this is universally true. At this point in the thread, such statements should be qualified or backed up with equations.

So let's look at pure prograde/retrograde instantaneous burns, using dm fuel to change the rocket velocity by dv:

K_i = 1/2 m v²

K_f = 1/2 (m-dm) (v+dv)²

Conservation of momentum for a prograde burn: m v = (m-dm) (v+dv) + dm (v - v_e), where v_e is the exhaust velocity g I_sp.

Simplifying gives dm = m dv / v_e. (Note that integration of this equation gives the rocket equation: Delta-v = ve ln(m_i/m_f.)

For a retrograde burn, just let dv be negative, but you must explicitly replace v_e with -v_e. So let's use dm = +/- m dv / v_e, where the top sign is for prograde, the bottom for retrograde.

So the change in kinetic energy is dK = K_f - K_i:

dK = 1/2 (m-dm) (v+dv)² - 1/2 m v²

dK = 1/2 m (1 -/+ dv/v_e) (v + dv)² - 1/2 m v²

dK = m v dv -/+ 1/2 m v² dv/v_e

dK/dv = m v (1 -/+ v/v_e)

Note that this is the statement of the Oberth effect: The change in kinetic energy due to a burn is proportional to your velocity (or momentum): dK/dv = m * v * stuff. The bigger your velocity v, the bigger dK/dv. That's it.

dK/dv_pro = m v (1 - v/v_e)

dK/dv_retro = m v (1 + v/v_e) (dv is negative, so the change in kinetic energy dK = m v (1 + v/v_e) dv is negative.)

Conclusions:

1) You get a larger change in kinetic energy per delta-v spent in the retrograde direction than prograde direction. This is basically because for a retrograde burn the change in v² is in the same direction as the change in m (both get smaller): (m-dm) * (v-dv)^2 vs (m-dm) * (v+dv)^2.

2) Here's a fun one: For prograde burns, if your rocket velocity is greater than your exhaust velocity, you actually lose kinetic energy.

You gain speed, but (1 - v/v_e) is less than zero, so you lose kinetic energy. Cute.

3) You do of course gain specific orbital energy, which is really what matters. When not burning, your mass is irrelevant, so it makes more sense to look at your energy divided by mass.

dE_pro = 1/2 (v+dv)² - 1/2 v² = 2 v |dv| + dv²

dE_retro = 1/2 (v-|dv|)² - 1/2 v² = -2 v |dv| + dv²

Note that I've used -|dv| for retrograde because earlier I chose to let the sign of dv indicate prograde or retrograde. Here taking the sign out of dv makes it clear that the two changes in specific energy are different.

Now clearly prograde burns add more specific orbital energy than retrograde burns subtract specific orbital energy.

So, I guess I don't disagree with Kasuha so much as caution people to be specific. (No pun intended.)

Edited February 28, 2014 by Yasmy