oberth effect

[Red Iron Crown]

Exhibit A. If that were true, there would be no consequence of the Oberth effect because the total amount your craft can accelerate would be constant regardless of circumstance. However, your engine is more effective at higher speeds - meaning that if you start at a higher velocity your fuel is worth more delta-V. It's NOT the same.

No, no, no! Your fuel is worth exactly the same delta-V at higher speeds, it just produces more useful energy!

This confusion goes away when you stop thinking of delta-V as a number that MechJeb throws at you as if it were a physical metric of your craft. Delta-V is not an intrinsic property of your craft just as the miles you can drive is not an intrinsic property of a car. It's a potential that is subject to change based on conditions... one of which being how fast you're already going.

=Smidge=

Your delta-V does not change with conditions. I'll ask you to review the rocket equation and point to where "conditions" affect the calculation.

"Miles you can drive" would be an intrinsic property of a car if fuel economy was constant. Which for rockets, it is.

Edited February 26, 2014 by Red Iron Crown
Typos, typos...

[Kryxal]

Rather, DVA and DVE are the same, DVO is not just 0, but nonsensical. Oberth is not "free" delta-v, it's spending delta-v in a place where it will have a greater effect, giving greater speed LATER ON (or lesser, in the case of a retrograde burn). A burn of 100 m/s does not give you 110 m/s, or 200 m/s, or anything like that just because it's near an object, but where you were approaching at a certain speed 6 hours ago, you're leaving again at MORE than 100 m/s greater once you hit the same altitude as the previous measurement.

New thought experiment: car on a hill, stopped, with an identical hill on the other side, with no air resistance and no friction (except the brakes are on for now).

Let h be the height of the car, ma the mass of the car, g the force of gravity, with initial h=100 m and g=9.8 m/s^2

PE = g*ma*h

PE(initial) = 980*ma

Now, release the brakes and the car rolls down to the bottom of the hill

PE(bottom) = 0

KE(bottom) = 980*ma = (ma*v(bottom)^2)/2

v(bottom)^2 = 1960

v(bottom) = 44.27 m/s

Mass, of course, cancels right out. Now apply, say, 10 m/s of delta-v:

v(boost) = 54.27 m/s

v(boost)^2 = 2945

KE(boost) = 1472.6

and see what the speed is once the car is back at a height of 100m

PE(end) = 980*ma

KE(end) = 492.6*ma

v(end)^2 = 985.2

v(end) = 31.4 m/s

So that 10 m/s push gave it over 30 m/s at the end, once it was back to the same height as before. It only sped it up by 10 m/s at the time, though.

Honestly, "frictionless car at the top of the hill" is probably a fair analogy for orbital energy states, especially going with the usual convention of PE(infinity) = 0 and your PE value is negative as you get closer...

[Rhomphaia]

Exhibit A. If that were true, there would be no consequence of the Oberth effect because the total amount your craft can accelerate would be constant regardless of circumstance. However, your engine is more effective at higher speeds - meaning that if you start at a higher velocity your fuel is worth more delta-V. It's NOT the same.

The total amout your craft can accelerate is constant regardless of circumstance if you have a 10ton craft 1ton of which is fuel that has an engine sith a specific impulse of 390. that craft will always be able to accelerate by 403.5m/s

if you were travelling at 100m/s you would be able to accelerate to 503.5m/s, if you were travelling at 1000m/s you would be able to accelerate to 1403.5m/s

What changes is not how much you can change your velocity, but what effect the velocity change has.

Edited February 26, 2014 by Rhomphaia

[Kasuha]

• DVR: Delta-V Required for the planned maneuver. This number is the immutable amount of velocity change necessary for the planned maneuver, as shown by the maneuver node system in the game. This is immutable because it's totally vector addition. At the point where you place the maneuver node, your ship will have a certain velocity vector. The burn at the maneuver node is another vector, DVR. DVR adds to the ship's starting vector to sum up to the vector the ship will have in its new orbit after the burn. Thus, DVR is analogous to the physical straight-line distance between 2 points.
  
• DVC: Delta-V Capacity of the ship, based on its mass, engine, and fuel alone, ignoring Oberth. DVC is consumed during burns. IOW, DVC is the available delta-V number reported by MechJeb and KER. For purposes of this discussion, however, DVC is not applicable because we will assume we always have enough fuel in the tank for the planned maneuver. I mention it only make sure nobody think's I'm talking about it here.
  
• DVA: Delta-V Applied. The amount of velocity change applied to the ship in the course of making a burn. During the burn, DVA < DVR. At the end of the burn, DVA = DVR. If Oberth really has any effect in the game, it supplies some of DVA because it can't have any effect on DVR. Thus, we also have:
  
• DVE: Delta-V from the Engine. This is the portion of DVA due solely to the engine burning fuel without considering Oberth. DVE happens over time throughout the burn and is a function solely of the engine's thrust compared to the sihp's mass and how that changes due to the engine's fuel consumption.
  
• DVO: Delta-V from Oberth Effect. This is the portion of DVA due solely to the Oberth Effect, taking into account the ship's velocity and how that changes throughout the burn.
  

For the purpose of understanding Oberth effect, these are completely irrelevant. You may imagine all dv of a single maneuver is applied instantly, exactly how KSP maneuver nodes assume, and you can still see effects of Oberth effect.

Oberth effect is about where in your current orbit you perform that maneuver.

If you're on elliptic orbit and want to escape SOI of the body you're orbiting, you can burn for escape at any point on your orbit and you will escape it eventually. Oberth effect means that it will cost you least dv to do so from Periapsis, and most dv if you do it at Apoapsis.

The matter is, when you travel from Kerbin to Eeloo, you're not going to spend certain dv. You're going to travel the Kerbin/Sun/Eeloo gravity potential wells and you need to change your orbital energy in a way which will eventually get you from point A (Kerbin orbit) to point B (Eeloo orbit). Amount of dv you will spend to do so depends on how smart you are with deciding when and where to perform changes of your orbital parameters. Oberth effect just means that when you need to increase or decrease your orbital energy substantially, it's best done at the point where your kinetic energy, i.e. velocity is the greatest - and subsequently at the place of the lowest gravitational potential of your current trajectory.

[Smidge204]

No, no, no! Your fuel is worth exactly the same delta-V at higher speeds, it just produces more useful energy!

Seems very straightforward to me that if your fuel is producing more useful energy, then it's worth more delta-V.

Answer me this: Where does that extra useful energy go if not into your craft's kinetic energy, and therefore manifest as extra velocity?

Your delta-V does not change with conditions. I'll ask you to review the rocket equation and point to where "conditions" affect the calculation.

Okay. Here's the rocket equation:

*points to v_e - the exhaust velocity term*

The exhaust velocity relative to the rocket is constant. The faster you go, the slower the exhaust velocity will be relative to some inertial frame. The slower the exhaust velocity, the less energy it has. Where does that energy go? Into your rocket!

Oberth doesn't reduce the dV required for a very specific maneuver, such as "raise the apoapsis of a 100km circular orbit to 1000km". It reduces the dV requirement for larger goals, such as "get from the surface of Kerbin to Duna's SOI". The first case is too specific, by defining the starting orbit we've defined the starting orbital energy, and thus the speed at which the burn will be made, which defines the Oberth effect! The second case gives us the flexibility to choose the speed at which we make our burn and harness the Oberth effect fully.

The goal of "get from the surface of Kerbin to Duna's SOI" is made of several specific maneuvers. If it does not apply to specific maneuvers then it's not at all clear how it would apply to a collection of maneuvers.

Here's an experiment: Build a ship and HyperEdit it into a a few given, nearly perfectly circular orbits around Tylo and see what the total change in velocity it can do in each situation. The ship will comprise the following components:

0.8000t - Command Pod Mk1

0.5625t - FL-T100 fuel tank (full)

0.5000t - LV-909 Engine

Total start mass: 1.8625t

Total empty mass: 1.3625t

Engine Isp(vac): 370 seconds

Calculated dV = 370 * 9.81 * ln(1.8625/1.3625) = 1134.64 m/s

The engine will burn for 9.07 seconds at full throttle.

We will test the claim that the ship has a fixed delta-V. If true, we would expect that whatever our velocity is to start with, our final velocity will be Vo + 1134.6m/s higher when we're done in every case.

If I'm correct, however, we should expect our final velocity to be larger and dependent on our initial orbital velocity.

I'm choosing Tylo because I can get very low and fast without worrying about atmosphere, maximizing the range of starting orbit altitudes. In practice it should not matter, but experimentally it should be easier.

Edit: Okay, Tylo wasn't as ideal as I thought it would be... SOI is actually rather small and it's hard to keep it straight with such a small radius. Going to the Sun!

Solar Orbit 1: 101000 meters at 63124.0 m/s. Final velocity: 64292.0 m/s. Change: 1,168.0 m/s.

Solar Orbit 2: 10119100000 meters at 10627.0 m/s. Final Velocity: 11794.7 m/s. Change: 1,167.7 m/s

Solar Orbit 3: 1012400000000 meters at 1076.0 m/s. Final velocity: 2243.6 m/s. Change: 1,167.6 m/s

Solar Orbit 4: 10200000000000 meters at 339.0 m/s. Final Velocity: 1506.6 m/s. Change: 1,167.6 m/s

Trying to go much further out results in a Kraken encounter. The effect, though, is small but clear.

=Smidge=

Edited February 26, 2014 by Smidge204

[Superfluous J]

I just want to take a break from the drudgery here and thank everybody in this thread for remaining calm and civilized. I'm fascinated by the conversation and appreciate the level headed discussion that's been going on for far longer than most threads of this nature could without people going insane and calling each other names.

I have this weird sort of gut understanding of the Oberth Effect and when I see huge equations my eyes gloss over, so reading all these different examples is really helping cement the ideas in my mind. So thank you all as well for that

Okay, back to talking about stuff I can barely understand!

[Kasuha]

Seems very straightforward to me that if your fuel is producing more useful energy, then it's worth more delta-V.

Forget about fuel. Tsiolkovsky equation and Oberth effect are two completely different things.

Your ship can produce certain amount of dv. That's a given. But depending on where it applies that dv, the energy of the ship changes more or less.

Delta-v of your ship is not equal to change of your ship's energy you can get from it. Your kinetic energy is proportional to square of your speed.

[Geschosskopf]

DVA will always equal DVE, because...

...this isn't a thing. The Oberth effect does not change the amount of speed gained (or lost) from a given burn. A burn that produces 10m/s of DVE will accelerate the craft by exactly 10m/s, no matter what its speed is when the burn is made.

Again, I disagree. If Oberth doesn't change the ship's velocity, then it does nothing. Oberth is expressed in terms of work and energy. You get more Ek by having more velocity. You do more work by having the force move a greater distance in the same time, which again means higher velocity. So in terms of what we can measure, Oberth results in more velocity than the ship would have without it.

Besides, if you read to the bottom of the wiki Oberth page, you'll see that Oberth's net effect is multiplying the impulse provided by the engines by some factor based on its orbital parameters at the time. Impulse is the time integral of force. Force is the product of mass and acceleration. If mass is constant or nearly so, then any multiplication of impulse mostly applies to multiplying acceleration. Acceleration is changing velocity. So again, Oberth is measurable by a change in velocity over and above what the engines provide. And because of this, Oberth really is a source of extra thrust, to create the extra acceleration, which results in the extra velocity. This doesn't violate conservation because this is being sucked out of the exhaust stream.

Here's where I think you're running into trouble. When you set a maneuver node and it calculates the effect of that burn, it takes Oberth into account because it is an inherent characteristic of Newtonian physics. It knows what your speed will be at the node, thus it can accurately calculate how much kinetic energy will be added by the burn.

Don't you see, though, that what you're saying here is really that the delta-V displayed when you create a node is DVR - DVO = DVE. Which means DVO really is a thing, a specific quantity the game is using to modify the raw DVR derived solely from vector math .

But I don't think this is how the game works. If it is, then the way the game handles Oberth is flawed. This is because while the Oberth Effect in general really is "an inherent characteristic of Newtonian physics", it does not affect all ships equally. The Oberth effect relies largely on TWR so doesn't help low-thrust, high-Isp engines as much as it does gas-guzzlers, even though they have the same velocity.

Now, the game knows nothing of a ship's engines until you run them. If you change to another ship, it forgets what it knew about the 1st ship's engines until you come back to it and run its engines again. Yet if you create a node for the ship before running its engines, the game still displays a delta-V number, which is either DVR or (DVR - DVO = DVE). But the burn time says "N.A.", indicating the game doesn't remember anything about the engines. At least it doesn't know the TWR, upon which both the burn time and the magnitude of the Oberth effect depend. So, if the burn delta-V displayed is in the form of DVR - DVO, either the game has a generic DVO value it uses regardless of engine type, or it defaults an unknown DVO to zero and the number displayed is raw DVR so now you'll make too big a burn. Either way, you get the wrong result.

Thus, it would make more sense if the what you're seeing is raw DVR and the DVO is figured into the burn time once you remind the game what type of engine you've got. Of course, there's no guarantee Squad thought of this when they wrote the code so it could well have the problems outlined above. But I figure they had to have known Oberth requires knowledge of TWR, so I think the displayed delta-V is DVR and the DVO is part of the calculated burn time.

The thing is, though, no matter how you slice it, there's a DVO term being taken into account, it's just on 1 side of the equation or the other. Either DVR = DVE + DVO or DVR - DVO = DVE.

To really see Oberth's effect, take an elliptical orbit, burn some amount of dV at Ap and calculate the orbital energy afterwards, then revert to before the burn. Burn the same dV at Pe and calculate the orbital energy, it will quickly become apparent that the burn at Pe gained you more orbital energy. I'm at work now, so I can't perform the experiment, give me a few hours and I'll whip up something and share the results.

I know this is the example in the wiki but it's just vector math combined with the exponential change in the scalar speed value. Also, it would be way more accurate and also less confusing to say "impulse" instead of "dV". So think about this:

How, given the lack of instrumentation for this purpose, would you go about creating equal impulse burns in the game? You have to apply the same impulse, not the same amount of DVE/DVA/DVwhatever. This means you can't use either value (magnitude or duration) of the burn shown by the in-game maneuver node system. If the "delta-V" of a node is displayed in the form (DVR - DVO), then you can't use that at all because DVO at Ap and Pe varies with the orbit's eccentricity. Thus, a display of Xm/s at Ap is not the same impulse as Xm/s at Pe. But OTOH, if the displayed "delta-V" is raw DVR, then the burn time and not the "delta-V" includes the Oberth effect, so if you burn for the specified time, you've overdone Oberth. Thus, the only way to do this is to point prograde, start burning, and time it with a stopwatch so both burns have the same impulse.

Oberth is in the game, it has to be. It's not some mystical effect that needs to be coded separately, it is the observation that, because speed and kinetic energy are not linearly related, expending dV at higher speeds gains more kinetic energy. It is a consequence of the most basic kinetic energy calculation, without which a game like KSP would not be possible.

I have to disagree. Oberth depends on the TWR of the rocket; otherwise it would be of equal benefit to them all. So it has to be coded separately, and thus either was or wasn't, and if it was, there are several ways that could have happened.

Edited February 26, 2014 by Geschosskopf

[MockKnizzle]

Geschosskopf: Do we both agree that the formula for calculating kinetic energy is .5*m*v²?

Good.

Let us do some math:

A rock is sliding on some ice that is frictionless. It has a mass of 1kg. It is currently moving at a constant speed of 10m/s. It is accelerated to a new speed of 20m/s. What was the velocity change of the rock, and what is the change in kinetic energy?

Delta-V = V₂ - V₁

20m/s - 10m/s = 10m/s

.5*m*V₁² = KE

KE₁ = .5*1kg*(10m/s)² = 50J

KE₂ = .5*1kg*(20m/s)² = 200J

KE₂ - KE₁ = 150J

In this case, increasing the velocity of our rock by 10m/s increased our kinetic energy by 150J.

But wait! our rock is accelerated again, to a new speed of 30m/s! What is the velocity change, and what is the change in kinetic energy?

Delta-V = V₂ - V₁

30m/s - 20m/s = 10m/s

.5*m*V₁² = KE

KE₁ = .5*1kg*(20m/s)² = 200J

KE₂ = .5*1kg*(30m/s)² = 450J

KE₂ - KE₁ = 250J

HOLY SPACE KRAKEN the same velocity change resulted in a larger energy change! This is where that "extra" energy goes.

[maccollo]

Seems very straightforward to me that if your fuel is producing more useful energy, then it's worth more delta-V.

Answer me this: Where does that extra useful energy go if not into your craft's kinetic energy, and therefore manifest as extra velocity?

The additional energy is cancelled out by the exponential nature of kinetic energy, resulting in a linear increase in velocity. When we're talking about the oberth effect we are usually interested in the reverse. How much additional energy do we get for the same acceleration when we move faster?

Think about this.

The final increase in velocity is simply the avarage acceleration* time.

The acceleration is the force divided by the mass.

For what you say to be true, if it goes faster, either the exhaust velocity of the engine must go up and thrust increase

or

the exhaust velocity must increase and the fuel flow decrease.

If I'm correct, however, we should expect our final velocity to be larger and dependent on our initial orbital velocity.

I'm choosing Tylo because I can get very low and fast without worrying about atmosphere, maximizing the range of starting orbit altitudes. In practice it should not matter, but experimentally it should be easier.

=Smidge=

I've played KSP for a long long time now so I know you are wrong from experience, but I will do your experiment to show you that you are wrong.

Here is... Tony probe

Tony probe has about 794 m/s of vacuum delta V (Kerbal enginer says 793 but when I do control tests it gets an extra m/s, presumably because of phantom forces).

Here the probe takes of from the moon at a shallow angle to minimize gravity losses.

Velocity increase: 792

Same burn in low earth orbit. The reason the orbital velocity is so high is because I'm using RSS, but the faster the better right?

Velocity increase 794

Same burn at periphrasis in a highly elliptical orbit around the sun.

The gravitational field this close to the sun is extremely strong, so getting no gravity losses is nearly impossible.

velocity increase 791.4

Edited February 26, 2014 by maccollo

[Geschosskopf]

For the purpose of understanding Oberth effect, these <the various delta-V species I defined> are completely irrelevant. You may imagine all dv of a single maneuver is applied instantly, exactly how KSP maneuver nodes assume, and you can still see effects of Oberth effect.

Besides disambiguation, I was just applying the scientific method. When you suspect something may or may not be there, or want to measure its effect, you create a variable for it and slot it into the appropriate place in the equation. Then you can tell whether it exists or not, and what effect it has if it does. That's all I was doing.

The matter is, when you travel from Kerbin to Eeloo, you're not going to spend certain dv. You're going to travel the Kerbin/Sun/Eeloo gravity potential wells and you need to change your orbital energy in a way which will eventually get you from point A (Kerbin orbit) to point B (Eeloo orbit). Amount of dv you will spend to do so depends on how smart you are with deciding when and where to perform changes of your orbital parameters. Oberth effect just means that when you need to increase or decrease your orbital energy substantially, it's best done at the point where your kinetic energy, i.e. velocity is the greatest - and subsequently at the place of the lowest gravitational potential of your current trajectory.

Suppose you travel from Kerbin to Moho. If you make a quick trip, you will encounter Moho at a large angle to its orbital path. Now, to have this occur, you're going very fast relative to every central body in the game so Oberth should be at his best. But the DVR you see displayed to capture at Moho is staggering, like 7000-8000m/s, despite your huge speed and plotting the Pe for the capture burn just skimming Moho's mountaintops. You couldn't make Oberth happier unless you were just a hair above Kerbol's incineration altitude. So where's Oberth when you need him?

The reason this capture burn at Moho is so huge is because of vectors. Your vector is crossing Moho's vector with a significant perpendicular component. You have to kill off all that perpendicular component first, just like killing off horizontal velocity for a lander on top of worrying about gravity. Speed here, at least in a disadvantageous direction, is actually a bad thing. And then you have the actual capture burn, which deals with the parallel component of your velocity vector.

OTOH, if you get to Moho slowly, you meet it closer to parallel to its orbital path and only need 1000-2000m/s to capture. This is because you have a very small perpendicular velocity component relative to Moho At the same Pe, you're going way slower but the capture burn is way smaller. Oberth doesn't like this, but you still burn less fuel than the other way.

This is an example of why it's necessary to separate Oberth out from the sheer mechanics of vector math, at the very least to see if he provides something extra that's not just a result of vector math. Hence, my definition of different species of delta-V. If the results are indistinguishable, then Oberth is not independent. If they are different, then Oberth is a thing unto himself.

But the thing is, the pre-Oberth rocket scientists, who declared interplanetary travel impossible due to the huge energy expenditures involved, were fully versed in vector math and Newtonian orbital mechanics. You can't have one without the other. So they certainly knew of examples like the above. Thus, they had to be missing something, which Oberth showed them. This and the fact that Oberth is a function of TWR also indicate Oberth is a separate thing.

[Claw]

Okay, I've been trying to stay out of this one because it is rapidly blooming, but like the torque question I feel pulled in. I feel like some people are in on this but it's getting mixed up with some of the examples.

Please stop using the car example. While it is an excellent example of conservation of momentum, it is poorly constructed as an analogy for interstellar transfer. The car neither achieves escape velocity from the road, nor does it observe exchanges in potential and kinetic energy from the road's gravity well. Also, the frame of reference is wrong.

I will avoid using numbers for the conceptual part, because numbers simply provide a specific example to confuse the issue. (For those who feel comforted by numbers, the example at the end will include them.)

Parabolic example and detailed proof from wiki. It seems simple enough right?

Specific energy is

So if I'm exchanging chemical energy for dV, then I must be getting free dV because the equation has Velocity!

Well, not exactly. Here is why, and also why many of the examples are lacking a definitive demonstration of the effect.

The Oberth effect is primarily obvious during inter-body transfer. If you are trying to execute an oberth effect around a planet (or the sun) and simply changing the shape of your elliptical orbit, you will not see an effect. To benefit from Oberth, you must leave what KSP calls an SOI (in reality, the "gravity well"). Hence why all the Oberth internet references talk about parabolic or hyperbolic trajectories. Those trajectories enter and then leave an SOI.

Here's a conceptual example:

Scenario 1:

Imagine you're in a 75x75km orbit around Kerbin. So now the PE/AP debate doesn't really matter because it's the same all around. Suppose I do a burn prograde to Kerbin's orbit around the sun. The burn is such that the craft just ever so barely manages to escape to the forward edge of Kerbin's SOI so that it is orbiting the Sun just in front of Kerbin. It's no longer orbiting Kerbin, so from Kerbin's perspective it has zero kinetic energy. Also, it isn't capable of falling on Kerbin, so from Kerbin's perspective it has zero potential energy.

Now, if I am at this point and I burn directly away from Kerbin, my orbit around the sun will change by the amount of dV that I apply. Also, from Kerbin's perspective the speed away is the dV I applied and the kinetic energy is 1/2 * (dV)^2.

Scenario 2:

Imagine you're back in that 75x75km orbit around Kerbin. Now if I do the same "Kerbin orbit relative prograde" burn with enough dV to just escape, plus the dV that I applied once interstellar, I've expended the same amount of dV as I did in the above scenario. My fuel expenditure has NOT changed in any way (i.e. I've used the exact same amount of liquid fuel and oxidizer as scenario 1). As I'm leaving Kerbin's SOI and enter the Sun's SOI, my speed isn't simply the extra dV that I applied. It's the converted kinetic energy. So from Kerbin's perspective, I have no potential energy (because I'm incapable of falling back on Kerbin), but my kinetic energy is higher (because in Scenario 1 at SOI crossing my KE is zero).

Now, that extra dV that I applied interstellar in Scenario 1 has been multiplied. This is because I transferred more kinetic energy with the LKO burn, which has NOW been converted into velocity. How much higher is the escape speed?

Practical Example

"I don't believe you!" Well, for those who like experiments and numbers, here's an example you can do in KSP. Make a ship, any ship that's capable of escaping Kerbin's SOI will do. I made a ship that had almost 4000dV in a 75x75km orbit around Kerbin.

Scenario 1:

Fire the engines while on the back side terminus of Kerbin so that you leave the SOI going forward relative to Kerbin's orbit. Getting an exactly perfect escape velocity is difficult because it's such a narrow speed range, but the burn takes about 935m/s dV. Speed around Kerbin is now about 3200 m/s.

This results in leaving Kerbin's SOI (for my test anyway) at less than 50m/s at the edge of the SOI (took several days). The orbital speed around the Sun was about 9340m/s. I then applied another 300 dV which brought my orbital speed around the Sun up to 9640 m/s.

So total of 1235 dV spent, and a solar orbital speed of 9640 m/s...

Scenario 2:

Fire the engines past the back side terminus of Kerbin. This ensures you're leaving the SOI at the forward edge of Kerbin's orbit. How much to fire the engines? At precisely 1235 dV worth. End result, the speed around 3500 m/s. So a gain of about 300m/s from the Scenario 1 burn. Makes sense.

Except this time when the craft leaves Kerbin's SOI, it was at 1426 m/s. Solar orbital speed is 10708 m/s.

So total of 1235 dV spent, and a solar orbital speed of 10708 m/s. Nearly 1100 m/s higher than Scenario 1.

Back to the theory:

So now if you go back and do the math.

sqrt {1 + ([2 * 3200]/300)} = 4.72

300 dV * 4.72 = 1420 m/s

If you compare the final solar orbital velocity for scenario 2 (10708 m/s) with the solar orbital velocity for a (just barely) Kerbin escape (9340 m/s), you'll see that it's 1368 m/s. So not quite the 1420 expected. What's the difference, the 9340 m/s escape velocity included about another 56 m/s. So add that on to the 1368 m/s and you get 1424 m/s. Pretty good test of the theory considering the lack of real instrumentation.

Conclusion

So yes, in a manner of speaking a craft does get free dV. But it isn't because of some magical change in fuel efficiency, Isp, or slingshot effect. It's because it's conserving Total Energy (which includes Kinetic, Potential, and Chemical energy). This is why it originally confused rocket scientists. Because kinetic exchange during the SOI transfer didn't make sense when you only consider the chemical energy of the rocket and not the conversion of kinetic energy.

Edited February 27, 2014 by Claw

[maccollo]

Specific energy is http://upload.wikimedia.org/math/7/c/9/7c9acb63207d58c5946e35b0a2b69ccf.png

So if I'm exchanging chemical energy for dV, then I must be getting free dV because the equation has Velocity!

The equation doesn't have velocity. It has velocity squared, which is kinetic energy, not velocity.

[Claw]

The equation doesn't have velocity. It has velocity squared, which is kinetic energy, not velocity.

Well, I think you skipped over some of my explanation and just jumped onto a point I was arguing against.

[maccollo]

Well, I think you skipped over some of my explanation and just jumped onto a point I was arguing against.

You're right, I did, apologies

But I wouldn't say you need to escape the body you are orbiting in order to benefit from the oberth effect.

If you just want to raise orbit it's most efficient to do so at periapsis. If you want to cut travel time to the moon but you don't want to take the full 7 days you just need 40 or so m/s more in low earth orbit to cut the transfer time in half.

Edited February 27, 2014 by maccollo

[Claw]

You're right, I did, apologies

But I wouldn't say you need to escape the body you are orbiting in order to benefit from the oberth effect.

Even if you just want to raise orbit it's most efficient to do so at periapsis. If you want to cut travel time to the moon, a slight increase in velocity close to the earth when you are moving

This is true, but it's much less due to Oberth effect. It's inefficient to change your orbit by circularizing at intermediate steps. So if you burn at other than the PE, you're basically spending some amount of dV on circularizing.

Just like most efficient place to raise your PE is at your AP. Oberth is lower at your AP then anywhere along your orbital path, but it's still efficient here because your potential energy relative to the body isn't going away like when you leave an SOI.

[maccollo]

Just like most efficient place to raise your PE is at your AP. Oberth is lower at your AP then anywhere along your orbital path, but it's still efficient here because your potential energy relative to the body isn't going away like when you leave an SOI.

You still gain more orbital energy by burning at periapsis than apoapsis, which is why the most efficient way to increase the size of the orbit is to burn at periapsis first, and then apoapsis.

As an example, say we are in an eliptical orbit around Kerbin that is 75 km x 3500 km.

We want to turn this into a 50 000 km circular orbit.

If we do the first burn at 3500 km and the second at 50 000 km this will cost 930 m/s.

If we do the first burn at 75 km and the second burn at 50 000 km it will cost 438 m/s.

*edit

I should also add that If we do the first burn halfway between apoapsis and periapsis it will cost 705 m/s, so that is still more efficient that doing it from apoapsis.

However, if we do this we will suffer much higher gravity losses since we are burning away/towards Kerbin.

Edited February 27, 2014 by maccollo

[SSSPutnik]

I just want to take a break from the drudgery here and thank everybody in this thread for remaining calm and civilized. I'm fascinated by the conversation and appreciate the level headed discussion that's been going on for far longer than most threads of this nature could without people going insane and calling each other names.

I have this weird sort of gut understanding of the Oberth Effect and when I see huge equations my eyes gloss over, so reading all these different examples is really helping cement the ideas in my mind. So thank you all as well for that

Okay, back to talking about stuff I can barely understand!

Lol I almost lost my temper last night but breathed a few times, and was OK, amazing thread and a credit to all that it hasn't descended into ranting so far.

[SSSPutnik]

You're right, I did, apologies

But I wouldn't say you need to escape the body you are orbiting in order to benefit from the oberth effect.

If you just want to raise orbit it's most efficient to do so at periapsis. If you want to cut travel time to the moon but you don't want to take the full 7 days you just need 40 or so m/s more in low earth orbit to cut the transfer time in half.

Oberth definitely happens whether you leave SOI or not.

Real space does not have SoI.

Scott Manley has a video of this, (he does actually leave SoI but the effect is clearly visible before he does.)

Edited February 27, 2014 by SSSPutnik

[maccollo]

Oberth definitely happens whether you leave SOI or not.

Scott Manley has a video of this, (he does actually leave SoI but the effect is clearly visible before he does.)

That's what I was arguing.

Although perhaps I should have said "You definitely don't" instead to be more clear.

"Real space does not have SoI."

Indeed, but they are a pretty good approximation.

If I can do an apollo mission in the RSS mod with nearly the exact same delta V as the Saturn V had, and set up the same free return trajectory, it's good enough for our purposes

Edited February 27, 2014 by maccollo

[SSSPutnik]

Morning

[MockKnizzle]

I finally figured out how to do the math I've been wanting to do for days now! We're gonna show that burning the same amount of dV lower in the gravity well of a planet increases the specific energy of our orbit more, for the case of an elliptical orbit.

We have two ships in orbit around Kerbin. One (orbit A) is in a 200km circular orbit, and the other (orbit is in a 70km-330km elliptical orbit. Both of these orbits have the same specific orbital energy, since they have the same semi-major axis (800km).

In this case, that energy is -2.20725 MJ/kg for both orbits.

We can also find the velocity at periapse for each orbit.

For orbit A, the periapse velocity is just the circular velocity, which is 2101.071m/s.

For orbit B, the periapse velocity is 2475.397m/s.

Now, we're gonna add 100m/s to our ship's velocity at periapse in each case (doesn't matter where we add it for the circular orbit).

For orbit A, our new velocity is 2201.071m/s.

For orbit B, our new velocity is 2575.397m/s.

By speeding up at periapse, we have now increased our apoapse.

For orbit A, our apoapse has changed from 200km to 372.764km.

For orbit B, our apoapse has changed from 330km to 536.713km.

Since we applied our dV at periapse, those values don't change.

Now, we can go back and calculate the specific orbital energy of each orbit again!

For orbit A, the new specific orbital energy is -1.99214 MJ/kg.

For orbit B, the new specific orbital energy is -1.95471 MJ/kg.

And now calculating the change...

For orbit A, the change in specific orbital energy is 0.21511 MJ/kg.

For orbit B, the change in specific orbital energy is 0.25254 MJ/kg.

There it is! We added the same amount of dV - 100m/s - and we got different values for the change in specific orbital energy, because in the case of orbit B we burned when we were traveling faster.

Edited February 27, 2014 by MockKnizzle

[Red Iron Crown]

I've had a couple of conversations going with different people in this thread, some of whom are arguing the same things, so I'd like to consolidate my responses into this single post. If I missed some point that you feel needs addressing, my apologies and please point it out to me.

I think we can all agree that KSP uses Newtonian physics. One of the most basic equations in Newtonian physics is the kinetic energy equation:

E_k = 0.5*m*v²

Once this equation is programmed into the game engine (an early, early point), the Oberth effect is there. The Oberth effect is, in I think its clearest form:

To maximize energy change, change speed at the highest speed possible.

There is no extra speed change. The extra energy is contained in the speed that was just added.

We've kicked around a bunch of analogies, so I thought I'd add one more in which I've tried to keep the math to a minimum while still showing how "proportional to the square of" works(and it's the only thing better than a car analogy, a money analogy):

You have a golden goose that lays one egg a month. You can only sell it once, so you want to get the best price for it. Kerbonomicists say that the price is always equal to the square of tomorrow's day of the month minus the square of today's day of the month. Or, in an equation:

Price = DayOfMonth_Tmw² - DayOfMonthToday_Tdy²

When would you sell your egg to get the best price?

Even if you're not mathematically inclined, I'm sure you would take the time to work out which day is best:

[table=width: 300, align: center]

[tr]

[td]Day[/td]

[td]1[/td]

[td]2[/td]

[td]3[/td]

[td]4[/td]

[td]5[/td]

[td]6[/td]

[td]7[/td]

[td]8[/td]

[td]9[/td]

[td]10[/td]

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[td]14[/td]

[td]15[/td]

[td]16[/td]

[td]17[/td]

[td]18[/td]

[td]19[/td]

[td]20[/td]

[td]21[/td]

[td]22[/td]

[td]23[/td]

[td]24[/td]

[td]25[/td]

[td]26[/td]

[td]27[/td]

[td]28[/td]

[td]29[/td]

[td]30[/td]

[td]31[/td]

[/tr]

[tr]

[td]Price[/td]

[td]3[/td]

[td]5[/td]

[td]7[/td]

[td]9[/td]

[td]11[/td]

[td]13[/td]

[td]15[/td]

[td]17[/td]

[td]19[/td]

[td]21[/td]

[td]23[/td]

[td]25[/td]

[td]27[/td]

[td]29[/td]

[td]31[/td]

[td]33[/td]

[td]35[/td]

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[td]61[/td]

[td]63[/td]

[/tr]

[/table]

Obviously, the egg is much more valuable near the end of the month, so you would sell it on the highest numbered day of the month.

Similarly:

You have a 2 kg rocket with an SRB with 1m/s of delta-V. You can only fire it once, so you want to get the most kinetic energy change from it. You are in an eccentric orbit that varies from 1m/s to 30m/s of orbital speed. Rocket scientists say that the kinetic energy change is equal to the square of final speed minus the square of the initial speed. Or, in an equation:

Kinetic Energy Change = V_f² - V_i²

At what speed would you fire your rocket?

The amount of kinetic energy gained would follow the chart above, with day being speed and price being the kinetic energy change. The golden egg is your 1m/s of dV, and note that selling your egg later in the month doesn't get you any extra eggs, just more money for the same egg. Similarly, increasing speed at higher speeds does not get you any extra dV, it just gets you more kinetic energy for the same amount of dV.

Or more generally:

To maximize energy change, change speed at the highest speed possible.

Edited February 27, 2014 by Red Iron Crown

[Claw]

You still gain more orbital energy by burning at periapsis than apoapsis, which is why the most efficient way to increase the size of the orbit is to burn at periapsis first, and then apoapsis.

As an example, say we are in an eliptical orbit around Kerbin that is 75 km x 3500 km.

We want to turn this into a 50 000 km circular orbit.

If we do the first burn at 3500 km and the second at 50 000 km this will cost 930 m/s.

If we do the first burn at 75 km and the second burn at 50 000 km it will cost 438 m/s.

*edit

I should also add that If we do the first burn halfway between apoapsis and periapsis it will cost 705 m/s, so that is still more efficient that doing it from apoapsis.

However, if we do this we will suffer much higher gravity losses since we are burning away/towards Kerbin.

Sure, I will get on board with this. And I also agree, you certainly don't have to be out of the SOI to see the results of burning at PE before leaving the SOI. Mostly I was trying to avoid adding qualifier after qualifier and delving into the equations, because when people have been referring to changing parameters of an orbit, there starts to become confusing about potential energy vs. kinetic energy vs. chemical energy. When considering the effect for inter-body transfers, it becomes clear that crossing the SOI (when potential energy hits zero) nets a gain in speed that is higher in the new reference frame than the dV applied.

Some of the examples I have been seeing simply focus on "why doesn't the speed or fuel efficiency increase when burning at the PE."

Also I modified my post. I said interplanetary at first, but the conceptual exercise works equally as well leaving the Mun's SOI and returning to Kerbin's SOI. Burn lower at the Mun and you'll have more (or less) orbital velocity around Kerbin when you emerge from the Mun's SOI. You'll also have steadily higher velocity as you leave the Mun's SOI than if you did the same burn from a higher altitude.

To defy the intuition of your burn to 50km example, another weird phenomenon...

-From that 75x75km orbit (orbital velocity about 2300m/s), burn prograde 500 dV. You end up with an AP of about 1.35Mkm

-At the APof 1.35Mkm (orbital velocity about 950m/s), burn another 500 dV. The craft ends up with a 1.35Mkm x 2.23Mkm orbit.

So at this point, 1000 dV has been added.

-At the AP of 2.23Mkm (orbital velocity about 1000m/s), burn 500 dV retrograde. The PE ends up inside Kerbin's lithosphere.

So how the heck did I add 500 dV at 2.23Mkm (at a speed of 1000 m/s) and it lowered my 1.35Mkm PE way below my original 75km orbit (where I added 500 dV at more than twice the speed)? - (NOTE: I'm not looking for an answer to that.)

My point is, this is why I dislike trying to use examples of Oberth inside an SOI. I see wacky examples like this (not necessarily in this thread), but there's so many things going on that it doesn't explain Oberth at all. In fact, if you just look blindly at this example, it looks like Oberth is a complete lie, which it clearly isn't. One has to be careful about setting up experiments based on a piece of an equation/theory, without understanding the assumptions behind it. That's why my explanation focuses on intra-SOI transfers, since everyone quoth the wiki, and that's what the wiki assumes.

Edited February 27, 2014 by Claw

[Claw]

By the way, the Oberth effect is not dependent on TWR. Oberth effect is a theory which relies on orbital velocities and KE exchange. TWR can affect it, but Oberth is not dependent on it.

If your velocity isn't rapidly changing, then changes in TWR are going to have little impact. It does have to do with how long it takes you to apply those changes (TWR), but relative to how fast the velocity is changing. If the velocity is changing very slowly (i.e. a circular orbit), then TWR is much less significant.