Why does the oberth effect work?

[peachoftree]

I know that the oberth effect is that you burn more efficiently when moving faster. What i'm confused about is how it gets you a delta-v advantage when you have to spend delta-v to get up to that speed?

Edited November 11, 2014 by peachoftree

[Taki117]

It all comes down to Conservation of Energy. From Wikipedia. "The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy on top of its chemical potential energy"

[HorusKol]

I know that the oberth effect is that you burn more efficiently when moving faster. What i'm confused about is how it gets you a delta-v advantage when you have to spend delta-v to get up to that speed?

The trick is not necessarily in using your rockets to get up to speed.

If you get yourself into a lowish orbit around a body, and then burn a little to lower your periapsis (while keeping you apoapsis at the original orbit), you can then wait until you ship is at periapsis and then make a more efficient burn as you are being slingshot around the planet - that is where the Oberth effect helps. Then you can use a little more fuel to circularise (if you're just after a higher orbit) - or you have saved some fuel getting to another planet.

Wikipedia also states:

So if a spacecraft is on a parabolic flyby of Jupiter with a periapsis velocity of 50 km/s, and it performs a 5 km/s burn, it turns out that the final velocity change at great distance is 22.9 km/s; giving a multiplication of the burn by 4.6 times.

[Geschosskopf]

I know that the oberth effect is that you burn more efficiently when moving faster. What i'm confused about is how it gets you a delta-v advantage when you have to spend delta-v to get up to that speed?

The delta-V to get from Point A to Point B is always roughly the same (taking into account orbit inclinaction and eccentricity). Oberth doesn't change the amount of delta-V required for a given maneuver. Instead, it decreases the amount of fuel burned for the delta-V needed to do the desired maneuver. Of course, this depends a lot on the mass flow-rate of the exhaust. Oberth doesn't do a whole lot of good for ion and nuke engines. The higher the ISP, the lower the mass flow-rate in general, so Oberth does the most good for LFO engines.

But, the ultimate answer is that Oberth is hard-coded into KSP so just roll with it for profit (it's acconted for in the burn times for nodes) or ignore it at your peril.

Edited November 11, 2014 by Geschosskopf
typos

[Starman4308]

One non-mathematical way to think about it for transfer burns: the longer you spend burning, the more time gravity has to claw back some of that velocity, and you have to add additional velocity to compensate for that loss.

The more mathematical way: kinetic energy (minus relativistic effects) is equal to 1/2 m*v^2. Accelerating a 1kg object from 100-200 m/s will give it 15,000 J of kinetic energy. Accelerating that same object from 1000-1100 m/s will give it 105,000 J of kinetic energy. You've spent 100 m/s of delta-V either way, but in the second case, you've gained a lot more energy.

The delta-V to get from Point A to Point B is always roughly the same (taking into account orbit inclinaction and eccentricity). Oberth doesn't change the amount of delta-V required for a given maneuver. Instead, it decreases the amount of fuel burned for the delta-V needed to do the desired maneuver. Of course, this depends a lot on the mass flow-rate of the exhaust. Oberth doesn't do a whole lot of good for ion and nuke engines. The higher the ISP, the lower the mass flow-rate in general, so Oberth does the most good for LFO engines.

But, the ultimate answer is that Oberth is hard-coded into KSP so just roll with it for profit (it's acconted for in the burn times for nodes) or ignore it at your peril.

You are incorrect. Fuel is delta-V: dV = G*Isp*ln(fuelled mass/dry mass): for a manuever of any given delta-V, you must burn fuel equal to e^(dV/G*Isp) * fuelled mass.

The Oberth effect, as mentioned, deals with the fact that you get more kinetic energy at high velocity, because kinetic energy is proportional to v^2.

EDIT: Also, if you are confused about whether this violates conservation of energy: it doesn't. While you're adding kinetic energy far in excess of the fuel's chemical energy, the exhaust gas is also losing a commensurate amount of kinetic energy (with respect to the present orbital body, and presumably all gravitational fields).

Edited November 11, 2014 by Starman4308
Clarifications

[peachoftree]

Ok. From what I seen here, My problem was that I was assuming fuel = delta v so now I see thats not the case and everything makes sense now (on a side note, does this mean that delta-v numbers in kerbal engineer are just rough estimates?)

[Starman4308]

Ok. From what I seen here, My problem was that I was assuming fuel = delta v so now I see thats not the case and everything makes sense now (on a side note, does this mean that delta-v numbers in kerbal engineer are just rough estimates?)

Delta-V in KER is, to my knowledge, exact*, following the Tsiolkovsky rocket equation (dV = G*Isp*ln(fuelled mass/dry mass)). Geschosskopf was simply confused as to how the Oberth effect works.

*Plus or minus a tiny fudge factor to account for imperfections in the KSP simulation: it must apply thrust and consume fuel in discrete quantities, and is subject to the limitations of machine precision, which means rockets will not exactly follow the equation. I doubt the error will come to even 1 m/s in most cases. However, the delta-V maps are all approximations, because the exact amount of delta-V required depends strongly on how you manage your maneuvers.

Edited November 11, 2014 by Starman4308

[Geschosskopf]

You are incorrect

Not at all, sir. Pray, check your math. Oberth has absolutely zero impact on the delta-V required to get from Point A to Point B, which is entirely a function of geometry and various laws of physics starting with conservation of energy and everything derived therefrom. Oberth is solely and totally about fuel-efficiency as it relates to ship speed vs. the mass flow-rate of its exhaust stream. Period. End of story. Please search this forum for a huge number of threads on this very subject, including one back in ancient history where I got schooled on this very subject.

And as I said above, the fuel economy derived from Oberth is hard-coded into the game. When you create a maneuver node, it shows 2 things: burn time and delta-V. The delta-V won't ever change until you discover a way of beating the 2nd Law of Thermodynamics. But the burn time is all about how efficiently the engine converts fuel into the required delta-V. The shorter the burn time, the less the engine runs. The less the engine runs, the less fuel it burns. And the burn time is factored in under the hood, taking Oberth into account. Which, as I said, is a function of mass flow-rate, so does the most good for LFO engines and the least good for ions. Period. End of story.

[Empiro]

Not at all, sir. Pray, check your math. Oberth has absolutely zero impact on the delta-V required to get from Point A to Point B, which is entirely a function of geometry and various laws of physics starting with conservation of energy and everything derived therefrom. Oberth is solely and totally about fuel-efficiency as it relates to ship speed vs. the mass flow-rate of its exhaust stream. Period. End of story. Please search this forum for a huge number of threads on this very subject, including one back in ancient history where I got schooled on this very subject.

And as I said above, the fuel economy derived from Oberth is hard-coded into the game. When you create a maneuver node, it shows 2 things: burn time and delta-V. The delta-V won't ever change until you discover a way of beating the 2nd Law of Thermodynamics. But the burn time is all about how efficiently the engine converts fuel into the required delta-V. The shorter the burn time, the less the engine runs. The less the engine runs, the less fuel it burns. And the burn time is factored in under the hood, taking Oberth into account. Which, as I said, is a function of mass flow-rate, so does the most good for LFO engines and the least good for ions. Period. End of story.

That's not quite correct -- any game that properly simulates Newtonian gravity (i.e. acceleration due to gravity decreases with the square of distance), and properly simulates the 3 laws of motion will automatically simulate the Oberth Effect "for free". In other words, the Oberth Effect is a natural consequence of Newtonian gravity and Newtonian motion, and as long as you get those things right, you don't need to specifically code it into the game.

The easiest way to see the Oberth Effect work is to do the following: use Hyperedit to put yourself into a 100km orbit, and one into a 47,000km orbit (Minmus altitude). Create a maneuver node that ejects you from Kerbin prograde with relation to the sun and puts you at an AP of 100 million km (about Eeloo's AP). You'll find that you take more Delta-V in the second case, even not taking into account the fact that you would have needed to burn quite a bit of fuel just to get into a super high orbit around Kerbin.

[peachoftree]

The way I see it that works in my mind is that the higher kinetic energy you have at periapsis, the higher your apoapsis will be. Since kinetic energy grows with the square of velocity, the faster you are moving, the less change in velocity you need to get the same change in kinetic energy.

Edited November 12, 2014 by peachoftree
Clarifications

[Empiro]

That's basically correct -- for a given rocket, Delta-V remains the same whether you're moving fast or slow, but if you're moving fast, then changing your velocity by the same amount yields a much bigger change in kinetic energy because as you said -- kinetic energy is proportional to velocity squared.

As you stated in your original post -- you need to spend Delta-V to get fast to begin with. However, that's not all there is to it, since you can speed up and slow down without spending fuel due to gravity. If you spend fuel building up speed, and then travel away from Kerbin (or whatever other body), you'll slow down. If you start your engines again, then you're getting less out of the Oberth Effect since you let gravity slow you down. It would have been better if you were able to do your maneuver in a single burn (which isn't always possible, but it's something to keep in mind when planning your mission).

[GoSlash27]

Discussing this in another thread at this very moment!

The reason the Oberth effect works is because orbital mechanics aren't velocity, but rather kinetic energy.

We all think of it in terms of "ÃŽâ€V" in m/sec, but in reality they're "ÃŽâ€Ek" in terms of joules.

Adding velocity imparts a change in energy proportional to the square of the velocity, not linear. How fast you're going when you impart the velocity has a profound effect on how much energy you add to your motion.

If KSP had an "energy" gauge, we would all understand this implicitly and wouldn't even need to discuss it.

You start on the pad with kinetic energy from rotation and orbit. You have potential energy in chemical form in fuel.

Launching converts the chemical potential energy into kinetic energy, with losses due to gravity and drag.

Your burns then convert chemical energy into kinetic energy and it all suddenly makes sense; converting energy is more efficient when you're going fast.

Best,

-Slashy

Edited November 11, 2014 by GoSlash27

[Srpadget]

Explanation #1 (plain Engish, no math; uses reasoning to illustrate, but doesn't 'prove' anything):

Rockets do 2 things simultaneously. They add energy to the vehicle and push it forward, AND they add energy to the propellant and push it backward. If the vehicle is moving slowly, then a great deal of the total energy is "wasted" shoving the exhaust backward at tremendous speed. HOWEVER, if the vehicle is already moving very fast, then even after being shoved out the back of the vehicle the exhaust is still actually going FORWARD...and thus lots more of the total energy went into the spacecraft instead.

Explanation #2 (uses math; less accessible to many people, but actually proves the point rather than merely armwaving about it with a woefully-inadequate appeal to "common sense"):

Energy and velocity are DIFFERENT THINGS, and they do not scale linearly. Kinetic energy (the energy bound up in a moving body) is given by: KE=1/2mv^2. Note that the equation contains velocity SQUARED rather than just velocity. We can get mass out of this explanation by noting that the KE *per ton of vehicle* is just 1/2v^2. (That's called "specific kinetic energy", for those who care about such things. It's got units of energy per mass, or equivalently length-squared divided by time-squared. I'll be calling it m2/s2 from here on out, though it probably makes more intuitive sense to call it Joules per kilogram.)

So. Let's assume that your vehicle is sitting still (initial v = 0, KE also = 0) and you give it a boost of 100 m/s delta-v. Now your KE per ton of vehicle is 1/2 * (100)^2. 100 squared is 10,000; divide by 2 for 5000 m2/s2.

With me so far? Good.

Now for comparison, let's do the same thing again... but this time let's do it to an object that's already moving at 2000 m/s. That means its initial KE is 2,000,000 m2/s2. Now when we give it an additional dv of 100 m/s, its new velocity is 2100 m/s, for a KE of... 2100 squared is 4,410,000, divided by 2 is 2,205,000 m2/s2. Which means we have now ADDED 205,000 m2/s2 of specific kinetic energy. Go look at the first case, now look at this one. We just got FORTY-ONE TIMES as much energy added to our vehicle with THE EXACT SAME FUEL BURN.

Yeah. Looks like magic, doesn't it?

Here's the secret: the square of those two velocities added together in the second example is NOT the same as squaring each of the velocity contributions and adding them together. Or, in math-speak:

(a+^2 = a^2 + 2ab + b^2; it is NOT a^2 + b^2. That "+2ab" in the middle there? THAT is where Oberth lives.

[Starman4308]

Oberth has absolutely zero impact on the delta-V required to get from Point A to Point B, which is entirely a function of geometry and various laws of physics starting with conservation of energy and everything derived therefrom.

Only in maneuver node land, where burns are instantaneous, is this true. In practice, because burns are not instantaneous, you will always be a bit short of the target orbital energy, and will need to perform a correction burn to get you back on course: this correction will require more delta-V for a low TWR vehicle.

Oberth is solely and totally about fuel-efficiency as it relates to ship speed vs. the mass flow-rate of its exhaust stream. Period. End of story. Please search this forum for a huge number of threads on this very subject, including one back in ancient history where I got schooled on this very subject.

In an indirect way, Oberth is about fuel flow rate as said. The faster you can thrust, the better you can exploit the Oberth effect, because you're performing more of your burn at high velocity. I am, however, completely certain you are misunderstanding what the Oberth effect is, because the Oberth effect is all about the velocity through the course of the burn.

And as I said above, the fuel economy derived from Oberth is hard-coded into the game.

Yes, because KSP simulates Newton's laws of motion in a system with Newtonian gravity. The Oberth effect is an orbital mechanic, and KSP simulates orbital mechanics.

When you create a maneuver node, it shows 2 things: burn time and delta-V. The delta-V won't ever change until you discover a way of beating the 2nd Law of Thermodynamics.

You're living in maneuver node land. In maneuver node land, all burns are instantaneous and make maximum effect of the Oberth effect. In the real world and KSP, if you perfectly complete a maneuver node, you will find yourself not quite having reached the target orbit, because the burn did take a finite amount of time. If your acceleration was low, you will be particularly far off-course in a significantly lower-energy orbit than the target, because it did not maximally exploit the Oberth effect.

But the burn time is all about how efficiently the engine converts fuel into the required delta-V.

No, burn time is a function of thrust, vessel mass, and some dependence on specific impulse, being mostly a function of how quickly the engine converts fuel into the requisite delta-V. The only dependence on efficiency is a negative one: if you have a very efficient engine, you will burn through fuel slowly, so your acceleration will not increase as much, prolonging your burn time. Given two vessels identical in every way but TWR, the higher TWR one will get a slightly better boost out of the Oberth effect, by dint of throwing an identical amount of fuel out the back more quickly.

The shorter the burn time, the less the engine runs. The less the engine runs, the less fuel it burns.

Patently false. To make a concrete example out of the above, imagine you had two identical vessels, with one having its thrust limiter set to 10%. Delta-V = Isp*G*ln(full/empty mass): there is no dependence on thrust or acceleration anywhere. Both can manage the exact same burn using the exact same amount of fuel, but the 10% one will take 10x longer: it is applying 1/10 of the force for 10x longer, performing it in slow motion. When you integrate force over time (which is basically the rocket equation) for a momentum engine (a rocket) to get a change in velocity (after dividing by mass), it doesn't matter how quickly you thrust, it only matters how fast the exhaust is, and what your full/empty mass ratio is.

And the burn time is factored in under the hood, taking Oberth into account.

I have never seen this. Again, to the best of my knowledge, maneuver nodes are calculated based on instantaneous acceleration. Or, perhaps more accurately, the projected orbit from a maneuver node is calculated based on instantaneous acceleration.

Edited November 11, 2014 by Starman4308

[Vanamonde]

This sounds like a question for the Science Labs. Moved.

[peachoftree]

So based on the the oberth effect, would it be a really good Idea to build a refueling base out on minmus because you can get goind crazy fast around kerbin pe and exploit the oberth effect for fun and profit?

can you find my super sneaky pun?

[-Velocity-]

So based on the the oberth effect, would it be a really good Idea to build a refueling base out on minmus because you can get goind crazy fast around kerbin pe and exploit the oberth effect for fun and profit?

can you find my super sneaky pun?

Yea, that should work. By burning right when you get to periapsis, you can preserve much of your orbital velocity around Kerbin by escaping its gravity quickly, before it can act to slow you down- so you can obtain large velocities relative to Kerbin. I used a similar concept for a three-stage ion probe I made- I launched it out to around Eeloo, then reduced its periapsis till it was like 1 solar radii. I did not cheat and use massless batteries- I used solar panels, which work even better near the Sun. As I approached periapsis, I did like a five hour ion "burn" to escape the Sun's gravity. I was able to preserve most of my solar orbital velocity at periapsis, plus a bunch of added delta-V from the ion engines. My final anti-solar velocity was just in excess of 70 km/s. 100 km/s would probably have not be hard to achieve, with more ion stages and better optimization (this was my first try), but this was just a simple afternoon "project".

Sobering thought- even at 70 km/s, and assuming that all distances in the Kerbal universe are 1/10th that in the real universe, it would still take my "high speed" probe 1800+ years to reach the nearest star to Kerbin, 0.43 light-years away. That's 1 week of continuously running the game at 100000X time compression.

Edited November 11, 2014 by |Velocity|

[GoSlash27]

So based on the the oberth effect, would it be a really good Idea to build a refueling base out on minmus because you can get goind crazy fast around kerbin pe and exploit the oberth effect for fun and profit?

can you find my super sneaky pun?

It will save you some DV in the long run, but it makes your transfer windows awfully tricky. Minmus needs to be in the correct phase relative Kerbin (it usually isn't) and you have to account for the time it takes to fall to Kerbin. Plus you have to correct for the inclination change at Kerbin periapsis

Kind of a big hassle for the little gain in kinetic energy, so I don't bother.

Best,

-Slashy

[GoSlash27]

The way I see it that works in my mind is that the higher kinetic energy you have at periapsis, the higher your apoapsis will be. Since kinetic energy grows with the square of velocity, the faster you are moving, the less change in velocity you need to get a bigger change in kinetic energy. If you relate this to conservation of momentum you will find that you need to expel less mass (fuel) to get the same change in kinetic energy because your change in velocity is lower. Is this correct or am i wildly mistaken?

This is correct. All the stuff about escaping gravity wells/ etc is a distraction from this, and you worded it very well.

Getting a desired apoapsis (or periapsis) is not a matter of change in velocity, but rather a change in kinetic energy. Since velocity is not linear with respect to kinetic energy, it stands to reason that a fixed quantity of velocity will add more kinetic energy when you're moving quickly than it does when you're moving slow.

This is the Oberth effect.

Sounds like you've got it pegged.

Best,

-Slashy

Edited November 11, 2014 by GoSlash27

[rkman]

So based on the the oberth effect, would it be a really good Idea to build a refueling base out on minmus because you can get goind crazy fast around kerbin pe and exploit the oberth effect for fun and profit?

I'd think that depends on where you get the fuel. makes quite a difference in overall delta-v cost whether it is mined on Minmus or has to be transported to Minmus from Kerbin.

converting energy is more efficient when you're going fast.

Just to be excruciatingly sure about this:

A given amount of fuel yields more delta-v the higher the initial velocity is, and a desired amount of delta-v requires less fuel the higher the initial velocity is?

[andrewas]

No. The Oberth effect doesn't change your dV in the slightest. It changes the kinetic energy you gain or lose from your change in velocity. The faster you are going, in the frame centered on the body you are orbiting, the more effecting your velocity change is.

[Concentric]

I'd think that depends on where you get the fuel. makes quite a difference in overall delta-v cost whether it is mined on Minmus or has to be transported to Minmus from Kerbin.

Just to be excruciatingly sure about this:

A given amount of fuel yields more delta-v the higher the initial velocity is, and a desired amount of delta-v requires less fuel the higher the initial velocity is?

No. Delta-v for a given amount of fuel (relative to the total mass of the craft) is fixed. However, you gain more kinetic energy per part fuel, as the rocket engine produces useful work from the kinetic energy of the fuel, which is higher at higher velocity. As it's energy-levels that get you places, you can travel further/get a greater orbit energy difference with the same fuel quantity if your initial velocity is higher.

Also, one reason you might want to refuel in higher orbit/Minmus orbit is that if you have the same ship starting full up there and send it down, it'll be more full when it arrives at its final destination, having used less delta-v/fuel than if it started full in LKO. The cost of getting the fuel up there might not be relevant to the particular mission.

[rkman]

No. The Oberth effect doesn't change your dV in the slightest. It changes the kinetic energy you gain or lose from your change in velocity.

Does more kinetic energy (with a given vessel mass) not mean more speed for that vessel?

The faster you are going, in the frame centered on the body you are orbiting, the more effecting your velocity change is.

More effecting in what way? Higher apoapsis compared to a burn at higher periapsis (less Oberth effect) - an apoapsis that would have required more fuel if the burn would have taken place at higher periapsis?

If more Oberth effect verse less Oberth effect does not save fuel (or gets you more speed for the same amount of fuel) then why use it at all?

[Starman4308]

Does more kinetic energy (with a given vessel mass) not mean more speed for that vessel?

No. Rockets operate on momentum and force (velocity and acceleration, once you divide the mass out). The Tsiolkovsky equation is dV = Gm*Isp*ln(full/empty mass): they can change their velocity on a function dependent on mass ratio and engine efficiency. At no point does kinetic energy factor into rocket delta-V. Kinetic energy is a consequence, not a determining factor.

The size of your orbit is dependent on kinetic energy and the strength of the gravitational field. The whole point of the Oberth effect is that you get more kinetic energy if you are already traveling at a high speed: going from 0-100 m/s gives you 5000 J/kg of specific kinetic energy. Going from 1000-1100 m/s, an identical expenditure of delta-V (100 m/s), gives you 105,000 J/kg of specific kinetic energy. Despite costing your rocket an identical amount of delta-V, the second maneuver has given you far more kinetic energy.

More effecting in what way? Higher apoapsis compared to a burn at higher periapsis (less Oberth effect) - an apoapsis that would have required more fuel if the burn would have taken place at higher periapsis?

If more Oberth effect verse less Oberth effect does not save fuel (or gets you more speed for the same amount of fuel) then why use it at all?

Oberth won't change how much delta-V you get.

Oberth will change how much you can get out of your delta-V. A Jool transfer manuever node will use a certain amount of delta-V. You will always need to spend a bit more delta-V than displayed, however, because that maneuver node assumes an instantaneous burn, taking 100% advantage of the Oberth effect. Since real rockets cannot change velocity instantaneously, they cannot take 100% advantage of the Oberth effect, and will have to perform a correction maneuver to get them back on course.

Edited November 12, 2014 by Starman4308

[GregA]

If I were sitting on an asteroid falling towards periapsis, then right at periapsis I pushed myself of of the asteroid, would I pick up kinetic energy greater than my strength in pushing off the asteroid?

I think I am fundamentally misunderstanding this somehow. Does the energy from the Oberth effect come from leaving mass behind at periapsis, that I carried towards periapsis?