How much dV do you lose with a higher LKO really?

[Snark]

Lots of wordage below, but my points boil down to,

• Yes, @Red Iron Crown's arguments are correct.
• @Geschosskopf, you seem to have a misunderstanding about what Oberth effect is.
• I would love for @OhioBob to weigh in on this matter.  I've got a physics degree, and have written software that accurately deals with this stuff, so I'm pretty confident I know what I'm talking about, but for the final word on physicsy, sciencey, mathy stuff, Bob's the guy, as far as I'm concerned. 

Oberth "effect" isn't an "effect" that has to be "modeled" by software, any more than "centrifugal force" does.  It falls out naturally from Newtonian mechanics.  For example, this sort of statement,

10 hours ago, Geschosskopf said:

So unless you add an algorithm to calculate what the Oberth effect should be and apply it, you have no Oberth effect.

No.  This is simply and completely wrong.  If you could see into the KSP source code, you wouldn't find anything in there that says "...and now let's apply the Oberth effect."  I say this not only as a physics major, but as a modder who has written mods that deal with dV and KSP physics, and who has written orbit-simulator programs.  If, for example, you use BetterBurnTime and you believe its numbers, then you're believing code that came out of the same head that's giving you this advice now.  If I'm wrong, then BetterBurnTime is wrong. 

Isp has nothing to do with Oberth effect, not directly.  You can demonstrate the Oberth effect completely with pencil and paper, using orbit geometry alone without any reference to rocket characteristics.

Let's say you're in circular LKO at 100 km altitude, traveling 2245 m/s, and you want to eject from Kerbin's SoI with 1000 m/s of excess velocity.  How much dV do you need?  Let's explore two possible approaches.

• Approach #1:  Do a single burn in LKO.
• Approach #2:  Do a single burn in LKO that boosts your Ap up to, say, 40,000 km.  Coast up to that Ap, then do another burn to eject.

Approach #1 requires a single burn of 1084 m/s.

Approach #2 requires one burn of 903 m/s to get your Ap up to 40,000 km, then a second burn of 1030 m/s at that altitude to get your ejection from Kerbin.  Thus, a total dV of 1933 m/s.

In other words:  if you do all your burn at low altitude, you need 1084 m/s of dV to accomplish the desired ejection.  If you do a lot of your burn at high altitude, you need 1933 m/s.  Therefore, you get more bang for your buck at the low-altitude burn.

That's it.  That's what the Oberth effect is.  That's all there is to it.

Note that at no point in the above scenario description do I mention anything at all about the rocket itself:  not its size, not its Isp, not its TWR, nothing.  That's because all of that is irrelevant to Oberth effect.

All of the above discussion, and the example I give, is just a more verbose way of stating what the Oberth effect's Wikipedia page says in the opening paragraph:

Quote

In astronautics, a powered flyby, or Oberth maneuver, is a maneuver in which a rocket falls into a gravitational well, and then accelerates when its fall reaches maximum speed.^[1][2] The resulting maneuver is a more efficient way to gain kinetic energy than applying the same impulse outside of a gravitational well. The gain in efficiency is explained by the Oberth effect, which is that the use of a rocket at high speed generates greater mechanical energy. In practical terms, this means that the most energy-efficient time for a spacecraft to burn its engine is at the lowest possible orbital periapse, when its orbital velocity (and so, its kinetic energy) is greatest.

Gonna re-quote Red Iron Crown for truthiness:

1 hour ago, Red Iron Crown said:

This is nonsense. The Oberth Effect is a classical newtonian effect, as explained and mathematically demonstrated above. I don't know how to put this more clearly:

The Oberth Effect is entirely a consequence of E_k = 0.5*m*v² and constant thrust engines, not a separate phenomenon.

^ This.  This right here.  This summarizes pretty much everything I'm saying.  Every word here is correct.  If this statement were wrong, not only would all of my physics classes be incorrect, but none of the software I've written for modeling orbits and maneuvers would have worked.

Now, it's true that TWR can make a difference at how practical it is to use Oberth effect to one's advantage.  The above scenario assumes that the "single burn" of scenario #1 happens entirely while the ship's at (or close to) the minimum altitude.  If a ship has a very low TWR, so that it needs a really long burn, then it can't accomplish that within the relatively short period that the ship's at periapsis.  In that case, yeah, it's hard to take advantage, unless you're willing to do multiple passes with a burn at Pe each time.  And even that has a limit, since if you want to eject from the planet with a lot of excess velocity, the final burn has to be a big one and you can't come around for another pass.

That is why ion ships have trouble with Oberth benefits-- it's just an outgrowth of their low TWR, and has nothing to do with their Isp per se.  If you were to mod the Dawn engine so that it gives a thrust of, say, 200 kN instead of 2 kN, but leave all its other parameters (including Isp) intact, it would get just as much Oberth benefit as an LFO engine would.

And that's what the Oberth effect's wikipedia article is talking about when it says this:

Quote

The Oberth effect is strongest at a point in orbit known as the periapse, where the gravitational potential is lowest, and the speed is highest. This is because firing a rocket engine at high speed causes a greater change in kinetic energy than when fired at lower speed. Because the vehicle remains near periapse only for a short time, for the Oberth maneuver to be most effective, the vehicle must be able to generate as much impulse as possible in as short a time as possible. Thus, the Oberth effect is much more useful for high thrust rockets like liquid-propellant rockets, and less useful for low-thrust reaction engines such as ion drives, which take a long time to gain speed.

Hope that clears things up-- would love to hear OhioBob weigh in on this. 

 

[OhioBob]

11 hours ago, Geschosskopf said:

The net result of the Oberth effect is that the rocket gains more energy per unit of fuel burned when travelling faster than it does burning the same amount of fuel when travelling slower. 

The result of the Oberth effect is that a rocket gains more energy per unit of Δv when traveling faster than it does when traveling slower.  That is, adding 1 m/s when traveling 2000 m/s adds more energy to the rocket than adding 1 m/s when traveling 1000 m/s.  Change in velocity comes from acceleration * time, and acceleration comes from thrust / mass.  As long as the physics model correctly computes thurst, mass, acceleration, time, and kinetic energy, the Oberth effect is there whether you want it to be or not.  There is no need to add a separate algorithm to compute Oberth effect.  In fact, we don't "compute" Oberth effect at all, it is simply a consequence of the equations that are fundamental to any proper physics model.  Furthermore, Oberth effect doesn't care where the Δv comes from.  1 m/s is 1 m/s regardless of whether is comes from a Mainsail or an ion engine.

It is true that propellant mass flow rate and specific impulse are components in the calculation of thrust.  In KSP thrust is computed as follows,

F = ṁ * I_sp * g_o

where ṁ is the propellant mass flow rate, I_sp is the specific impulse, and g_o is the standard acceleration of gravity.  In KSP, I_sp is provided by a floatCurve in the engine's configuration file, where it is a function of ambient air pressure.

So mass flow rate and specific impulse do have an effect on the Δv produced by a particular system, but to relate that to the Oberth effect is not correct.

 

Edited August 22, 2016 by OhioBob

[Geschosskopf]

2 hours ago, Red Iron Crown said:

No no no! You do NOT get more dV from the same fuel with Oberth. You get MORE energy from the SAME dV/fuel. Thrust and Isp don't enter into it. There's no "thrust buff" or "Isp buff" at all.

Hehehe, you even said the key thing about the Oberth Effect but you're missing the point, so I'll try to explain it again....

You are totally correct in saying that the result of the Oberth Effect is that "You get MORE energy from the SAME ... fuel".  But how do you measure that extra energy?  By a change in the rocket's velocity, aka dV.  So, because of the Oberth Effect, you get MORE dV from the SAME fuel.  Can't have one without the other.  And that is, at the bottom line, indistinguishable from an improvement or "buff" to the engine's thrust and/or Isp.

But, wait a minute, you say....The engine has constant stats of thrust and Isp, and a given amount of fuel has a constant amount of potential energy, so the engine should always produce the SAME amount of energy/dV from the same amount of fuel.  So how can the rocket ever get more energy than that?

The answer is, yes, the engine is always producing the same energy, limited by the potential energy of the amount of fuel burned.  However, as the rocket's speed increases, more of that energy is applied to pushing the rocket forward instead of pushing the exhaust backwards.  THAT is the Oberth Effect, this imbalance in where the total amount of energy released by the burn ends up. Wikipedia even says so:

Quote

It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's kinetic energy is balanced by a relative decrease in the kinetic energy the exhaust is left with (the kinetic energy of the exhaust may still increase but it does not increase as much).^[3]:204 Contrast this to the situation of static firing where the speed of the engine is fixed at zero. This means its kinetic energy does not increase at all and all the chemical energy released by the fuel is converted to the exhaust's kinetic energy (and heat).

At very high speeds the mechanical power imparted to the rocket can exceed the total power liberated in the combustion of the propellant; this may also seem to violate conservation of energy. But the propellants in a fast moving rocket carry energy not only chemically but also in their own kinetic energy, which at speeds above a few km/s actually exceed the chemical component. When these propellants are burned, some of this kinetic energy is transferred to the rocket along with the chemical energy released by burning. This can partly make up for what is extremely low efficiency early in the rocket's flight when it is moving only slowly. Most of the work done by a rocket early in flight is "invested" in the kinetic energy of the propellant not yet burned, part of which they will release later when they are burned.

So, there you have it.  Due to the Oberth Effect, the rocket is not only being pushed by the chemical energy of the fuel, but also by absorbing some of the kinetic energy of the exhaust stream.   Because the rocket is receiving more energy than it would otherwise get just by burning the fuel, the net result of the Oberth Effect is an increase in the rocket's fuel efficiency.  The faster it goes, the more dV it gets per drop of fuel burned, even though its engine's stats are constant.

Now note that burns are done to move the rocket to a specific point in space, so have a finite dV requirement.  You burn the engine until you have changed your velocity by the needed amount.  For 2 ships orbiting a planet at different altitudes, the dV required to go to the same distant place is essentially the same.  However, the one going faster to start with will burn less fuel making this burn because of the Oberth Effect.  It will thus have more fuel remaining when it reaches its target, so can do more stuff there.  That is why the Oberth Effect is a useful thing in rocketry.

Bottom line:  The Oberth Effect does not actually change the stats of an engine, but it results in a situation indistinguishable from that happening.  The engine effectively has more thrust and/or Isp because of the Oberth Effect, so from the POV of the player, the Oberth Effect is a "buff".

This is why I've been going on about exhaust streams.  In real life, the Oberth Effect grows out of the exhaust stream, as the repository for the extra energy the rocket absorbs.  But KSP doesn't track exhaust streams at all---the rocket's mass simply is not conserved and all we see is a particle effect.  This is why you need additional code to model the Oberth Effect in KSP, to add the extra energy to the rocket artificially.  And that, according to Squad's own information, is what KSP does.

 

1 hour ago, Snark said:

@Geschosskopf, you seem to have a misunderstanding about what Oberth effect is.

No, I understand it quite well.  I've had it explained to me in great and gory detail by many knowledgeable people for a number of years.  The Oberth Effect is all about fuel efficiency, a way to wring more dV out of a given tank of fuel.  It cannot be explained simply by Ek = 1/2 (mv^2).  It is expressed in terms of work.

[Red Iron Crown]

Just now, Geschosskopf said:

You are totally correct in saying that the result of the Oberth Effect is that "You get MORE energy from the SAME ... fuel".  But how do you measure that extra energy?  By a change in the rocket's velocity, aka dV.

You measure it by a greater magnitude change in semi-major axis, not a greater change in speed. The speed change is the same, the delta-V is the same, etc. This thinking that more speed is gained for the same dV expenditure is the root of your misunderstanding here. 

Quote

<More stuff about exhaust>

The exchange of energy with the exhaust is the mechanism by which Oberth works with rocket engines while not violating conservation. It is not what causes the Oberth effect, that happens whether there is exhaust or not. For a real world example, a solar sail craft experiences Oberth, it gains more kinetic energy for the same speed change if that speed change is done at higher speed, even though there is no carried propellant or exhaust. See the first worked example in my longish post earlier, even if there is no mass change the energy gained is greater.

[OhioBob]

1 hour ago, Geschosskopf said:

This is why I've been going on about exhaust streams.  In real life, the Oberth Effect grows out of the exhaust stream, as the repository for the extra energy the rocket absorbs.  But KSP doesn't track exhaust streams at all---the rocket's mass simply is not conserved and all we see is a particle effect.  This is why you need additional code to model the Oberth Effect in KSP, to add the extra energy to the rocket artificially.  And that, according to Squad's own information, is what KSP does.

No, we don't need extra code.  We know the thrust of the engine and the mass of the vehicle, from which we compute the acceleration.  We then integrate over time to get the change in velocity, from which we get the change in energy.  We don't need to know anything about the exhaust stream and we don't need to add anything artificially.

(edit)
Note that all we need to know about the exhaust stream is provided to us by the thrust.  Know the thrust and we can compute everything else.  There is nothing more about the exhaust that we must consider.
 

Edited August 22, 2016 by OhioBob

[qoonpooka]

So, I'm a PhD student in Law and Public Policy.  This is decidedly not my field, I just really enjoy building and flying rockets.

But it is a thing of beauty to watch y'all discuss this stuff and I'm learning a tremendous amount from it.  So thank you all for having this discussion in front of me.

That is all.

[Laie]

The first part of your quote:

1 hour ago, Geschosskopf said:

It may seem that the rocket is getting energy for free, which would violate conservation of energy.

...already points out that the following is a rationalization for why Oberth doesn't violate the conservation laws. It's no longer about Oberth proper, which is that a linear increase in velocity is giving you a nonlinear increase in kinetic energy. Which seems like a free lunch but isn't.

But if it helps: fuel also has a disproportional change in kinetic energy, because it's expelled along the same axis, though in the other direction. Again, velocity of expulsion matters; it's not mass alone that makes it work.

[Geschosskopf]

4 hours ago, Red Iron Crown said:

You measure it by a greater magnitude change in semi-major axis, not a greater change in speed. The speed change is the same, the delta-V is the same, etc. This thinking that more speed is gained for the same dV expenditure is the root of your misunderstanding here. 

Um.......no.   Now you're confusing instantaneous kinetic energy with orbital potential energy,which is not time-dependent.  You've quoted the equation for Ek enough now that you know it has no term in it for SMA.  Orbital energy is just another way of expressing potential energy, the same as saying a rock on the edge of a high cliff has more potential energy than a rock that has already fallen.  A rocket's kinetic energy is not the same thing as its orbital energy.  If it's orbit is not perfectly circular, a rocket's velocity, and therefore its kinetic energy, varies at every point around the orbit, whereas the orbital energy remains the same until the rocket does another burn.  Ek and orbital energy are apples and oranges.

Besides, the Wikipedia article expressly states that the "extra" energy gained by the rocket is mechanical, kinetic energy, not potential orbital energy.  Because the rocket's mass is decreasing with fuel consumption, its Ek can only increase if its velocity increases.  If the velocity increases, then you have obtained dV from the kinetic energy.  QED.

Tell you what.  Go rewrite the Wikipedia article to reflect your views.  If it withstands peer review, I'll change my opinion   And then we'll both tell Squad they're doing it wrong, because they've stated they do things as I've been describing.

 

3 hours ago, OhioBob said:

No, we don't need extra code.  We know the thrust of the engine and the mass of the vehicle, from which we compute the acceleration.

And that totally ignores the extra thrust provided by the kinetic energy soaked up from the burning propellant, so you get no Oberth Effect.  Look, just read the Wikipedia page.  It's pretty easy to understand.  The rocket is getting what is effectively extra thrust from the kinetic energy of the exhaust, over and above that provided by the combustion alone, and it's the combustion alone that determines the thrust rating of the engine.  If you don't take the contribution of the exhaust kinetic energy into account, then you're playing with billiard balls, not rockets.

Edited August 22, 2016 by Geschosskopf

[Red Iron Crown]

1 hour ago, Geschosskopf said:

Um.......no.   Now you're confusing instantaneous kinetic energy with orbital potential energy,which is not time-dependent.  You've quoted the equation for Ek enough now that you know it has no term in it for SMA.  Orbital energy is just another way of expressing potential energy, the same as saying a rock on the edge of a high cliff has more potential energy than a rock that has already fallen.  A rocket's kinetic energy is not the same thing as its orbital energy.  If it's orbit is not perfectly circular, a rocket's velocity, and therefore its kinetic energy, varies at ever point around the orbit, whereas the orbital energy remains the same until the rocket does another burn.  Ek and orbital energy are apples and oranges.

Kinetic energy is a component of orbital energy. Orbital energy = kinetic energy + potential energy. As the craft moves through an elliptical orbit, kinetic energy is converted to potential energy as it rises to apoapsis, then converts back into kinetic energy as it falls towards periapsis, but the sum of the two always remains the same while in freefall. That's the secret here, it is easier to build kinetic energy with unvarying thrust when speed is high, since more of the orbital energy is not "locked away" as potential energy where it can't be built upon as easily.

1 hour ago, Geschosskopf said:

Besides, the Wikipedia article expressly states that the "extra" energy gained by the rocket is mechanical, kinetic energy, not potential orbital energy.  Because the rocket's mass is decreasing with fuel consumption, its Ek can only increase if its velocity increases.  If the velocity increases, then you have obtained dV from the kinetic energy.  QED.

Specific orbital energy, i.e. orbital energy per unit mass, is the really critical measure, because all orbits with the same specific orbital energy have the same SMA and vice versa. Change the specific orbital energy and you change SMA. Change the SMA and you've changed specific orbital energy. Does a vessel that splits in half suddenly shrink its orbit because its kinetic energy just got halved? No, because it's specific energy hasn't changed, even though its gross kinetic energy has.

1 hour ago, Geschosskopf said:

Tell you what.  Go rewrite the Wikipedia article to reflect your views.  If it withstands peer review, I'll change my opinion   And then we'll both tell Squad they're doing it wrong, because they've stated they do things as I've been describing.

I don't have to, the wiki page already agrees with me (as does Squad's implementation, please show me where they ever mention implementing Oberth as some separate function, rather than just thrust and E_k=0.5*m*v²). You are interpreting the wiki article incorrectly.

Are you ever going to provide some math to support your assertions? You have not commented on the examples I posted earlier. Please find the error in them if you can and I will happily retract.

1 hour ago, Geschosskopf said:

Look, just read the Wikipedia page.  It's pretty easy to understand.

The irony is awfully thick with these two statements.

[OhioBob]

1 hour ago, Geschosskopf said:

And that totally ignores the extra thrust provided by the kinetic energy soaked up from the burning propellant, so you get no Oberth Effect.  Look, just read the Wikipedia page.  It's pretty easy to understand.  The rocket is getting what is effectively extra thrust from the kinetic energy of the exhaust, over and above that provided by the combustion alone, which is determines the thrust rating of the engine.  If you don't take that into account, then you're playing with billiard balls, not rockets.

No, we are not getting extra thrust.  Thrust is equal to the rate that mass is ejected out the back of the rocket times its velocity relative to the rocket.  This relative velocity is constant regardless of the speed the rocket is traveling, therefore the thrust is constant (ignoring the possible effects of changing ambient air pressure).

 

[OhioBob]

Example #1

A rocket has an initial mass of 10,000 kg.  Its engine produces a thrust of 100,000 N with a specific impulse of 350 s.  When operating, the engine’s mass flow rate is,

ṁ = 100000 / (350 * 9.80665) = 29.135 kg/s

The rocket is traveling in gravity free space at an initial velocity of 1000 m/s.  We burn the engine for 10 seconds, resulting in a final mass of,

m_f = 10000 – 29.135 * 10 = 9708.65 kg

The change in velocity is,

Δv = 350 * 9.80665 * LN( 10000 / 9708.65) = 101.5 m/s

The final velocity is,

v_f = 1000 + 101.5 = 1101.5 m/s

And the spacecraft’s change in kinetic energy is,

ΔE_k = 0.5 * 9708.65 * 1101.5² – 0.5 * 10000 * 1000² = 889,763,445 joules

Example #2

A rocket has an initial mass of 10,000 kg.  Its engine produces a thrust of 100,000 N with a specific impulse of 350 s.  When operating, the engine’s mass flow rate is,

ṁ = 100000 / (350 * 9.80665) = 29.135 kg/s

The rocket is traveling in gravity free space at an initial velocity of 2000 m/s.  We burn the engine for 10 seconds, resulting in a final mass of,

m_f = 10000 – 29.135 * 10 = 9708.65 kg

The change in velocity is,

Δv = 350 * 9.80665 * LN( 10000 / 9708.65) = 101.5 m/s

The final velocity is,

v_f = 2000 + 101.5 = 2101.5 m/s

And the spacecraft’s change in kinetic energy is,

ΔE_k = 0.5 * 9708.65 * 2101.5² – 0.5 * 10000 * 2000² = 1,438,166,420 joules

 
@Geschosskopf,

As you may have noticed, everything in these two examples is exactly the same except for the initial velocity of the rocket.  The thrust is the same, the specific impulse is the same, the mass flow rate is the same, and the burn duration is the same.  As you can see, the second example results in a much greater increase in kinetic energy than the first example. I didn’t require any “additional code” or perform any mathematical shenanigans to artificially add energy to the rocket.  There is no extra thrust involved.  I’m just performing straightforward calculations like we do in this forum all the time.

If you think there is something missing from these calculations, what is it?  What additional code would you add?  If my method doesn’t take into account the Oberth effect, then why is the change in kinetic energy greater in the second example than in the first?

 

Edited August 23, 2016 by OhioBob

[GoSlash27]

@Geschosskopf,

 The Oberth effect is not hard to understand if you look at it logically:

 It does not exist.

It is merely a construct to help explain why you are not seeing the results you expect to see. It's like the Doppler Effect; the train moving toward you doesn't actually raise it's horn's frequency, just as the Oberth Effect doesn't actually give you free DV. It just *seems* that way because orbits don't work like we implicitly expect them to work.

Every orbit exists because the object carries energy. Mostly kinetic at periapsis and mostly potential at apoapsis, but always the same total energy. You get to where you're going not because you have velocity, but because you have energy.

 Going back to high school physics, the kinetic energy of an object is directly proportional to it's mass, and directly proportional to the square of it's velocity. The relationship between energy and velocity is not linear, but rather exponential.

 We tend to think of it as linear simply because everything we do is measured in DV, but that is incorrect and that's why the Obert Effect seems to exist (and why you are able to "cheat" DV maps if you play your cards right).

A fixed change in velocity adds more kinetic energy to an orbit at higher velocities than lower velocities. This is simple to prove mathematically.

Let kinetic energy equal k(constant)v^2

 Our change in energy in response to a change in velocity is therefore k(v+delta)^2-kv^2

We can ignore "k" since it's equal in both.

Foiling out (v+delta)^2 yields v^2+2vdelta+delta^2-v^2.

Cancelling the v^2 leaves 2vdelta+delta^2

The delta is common to both arbitrary examples, yet that "2vdelta" remains.

*That's* where your "Oberth Effect" is coming from.

Accordingly, the same change in velocity makes a different change in apoapsis depending on how fast you're going at the moment of burn. Same goes for any Hohmann transfer, or even interplanetary intercept (which is really just a Hohmann transfer that extends beyond the SoI). If you're moving quickly, your DV has a larger effect on the apoapsis. If you're moving slowly, it has less effect.

It has nothing to do with what engine you use, or the mass of it's exhaust, or any of that stuff. It is simply math behaving as it ought. Accordingly, it does not need to be modeled or accounted for in the game engine. So long as we follow Newtonian physics, it will appear on its own like it's supposed to.

But remember: It doesn't actually exist. It's just an illusion you *think* you see because physics doesn't work quite like what you expect.

HTHs,
-Slashy

PS Don't argue orbital mechanics with OhioBob. He's right.

*edit* Ninja'd...

 

 

 

Edited August 23, 2016 by GoSlash27

[OhioBob]

Quote

At very high speeds the mechanical power imparted to the rocket can exceed the total power liberated in the combustion of the propellant; this may also seem to violate conservation of energy. But the propellants in a fast moving rocket carry energy not only chemically but also in their own kinetic energy, which at speeds above a few km/s actually exceed the chemical component. When these propellants are burned, some of this kinetic energy is transferred to the rocket along with the chemical energy released by burning.

This is why the change in kinetic energy of the rocket is greater in example #2 than in example #1.  Thrust comes from the chemical energy released by burning, which is why relative exhaust gas velocity and thrust are constant.  However, it is the transfer of the propellant's greater kinetic energy in example #2 that accounts for the rocket's greater increase in kinetic energy.  This results just falls out of the calculations, as demonstrated in my examples.  There is no need to add or fudge anything.  (As Slashy said, "it will appear on its own like it's supposed to.")

Edited August 23, 2016 by OhioBob

[OhioBob]

23 minutes ago, GoSlash27 said:

PS Don't argue orbital mechanics with OhioBob. He's right.

I do make mistakes occasionally, so don't ever hesitate questioning me or calling me out on something.  You just might be right.

[Blaarkies]

17 hours ago, Geschosskopf said:

No, I'm saying you do not understand how Oberth works so did not design your program to take it into account properly.  This is an honest mistake, not a deliberate deception.

That's a pretty bold statement for someone who does not provide an experiment like I asked, or provides any maths to support his claim, or has dabbled in simulating physics to see how programming this stuff really works. You are following a very "tinfoil hat" approach to this argument by dismissing counter-arguments(made by some very prominent member of this forum) without disproving their claims.

There is a reason why the majority does not agree with your point...it's the same reason why the majority of the science community does not agree with the EM drive. Not because it is ridiculous, but because there is not enough evidence for the claim. Did you actually read all the posts in this thread? You know how you find it hard to believe that my simulation actually exhibits the Oberth effect without it specifically being programmed into...I feel the same way about you continually jumping over people's explanations, without telling us why they are wrong.

You claim that a ship with fuel for 100m/s dv will only be able to achieve 100m/s dv in deep space...but deep in a gravity well it somehow will achieve 100+ m/s dv?(please record the change in speed within 10 seconds of the burn, since the escape of SOI will be greatly influenced by the effect that the rest of us are trying to explain to you).
This is because of the "ISP buff" that the Oberth effect induces, right?

I do not believe you are a fool, and neither do i believe you are trying to embarrass yourself...but you are definitely trolling this thread big time, I have never seen such a successful troll EVER! Well done!

Edited August 23, 2016 by Blaarkies

[OhioBob]

Here’s another way of looking at the problem.  Let’s go back to our 10,000 kg rocket traveling at an initial velocity 1000 m/s.  Let’s first compute the initial momentum of the rocket,

P_initial = 10000*1000 = 10,000,000 kg·m/s

Let’s say we now expel 1000 kg of reaction mass out the back of the rocket at 3000 m/s.  Since the rocket is traveling +1000 m/s and the exhaust is traveling -3000 m/s relative to the rocket, the exhaust velocity is -2000 m/s as seen from a stationary reference frame.  Therefore the momentum of the exhaust is,

P_exhaust = 1000*(-2000) = -2,000,000 kg·m/s

We know that linear momentum must be conserved, therefore the final momentum of the rocket plus the momentum of the exhaust must equal the initial momentum.   Or, in other words, the final momentum of the rocket equals the initial momentum less that of the exhaust,

P_rocket = 10,000,000 – (-2,000,000) = 12,000,000 kg·m/s

We can therefore calculate the final velocity of the rocket,

V_final = 12,000,000 / 9,000 = 1333.333 m/s

We see that the Δv is 333.333 m/s.

Let’s now compute the initial kinetic energy and the final kinetic energy,

KE_initial = 0.5*10000*1000^2 = 5,000,000,000 J

KE_final = KE_exhaust + KE_rocket

KE_exhaust = 0.5*1000*(-2000)^2 = 2,000,000,000 J

KE_rocket = 0.5*9000*1333.333^2 = 8,000,000,000 J

KE_final = 10,000,000,000 J

We see that the system gained 5,000,000,000 J of kinetic energy, which is the energy released by the burning of the propellant.  (It is this energy that gives our reaction mass its 3000 m/s exhaust velocity.)

Let’s now work the same problem but with an initial velocity of 2000 m/s.

P_initial = 10000*2000 = 20,000,000 kg·m/s

P_exhaust = 1000*(-1000) = -1,000,000 kg·m/s

P_rocket = 20,000,000 – (-1,000,000) = 21,000,000 kg·m/s

V_final = 21,000,000 / 9,000 = 2333.333 m/s

KE_initial = 0.5*10000*2000^2 = 20,000,000,000 J

KE_exhaust = 0.5*1000*(-1000)^2 = 500,000,000 J

KE_rocket = 0.5*9000*2333.333^2 = 24,500,000,000 J

KE_final = 25,000,000,000 J

Again we see that the Δv is 333.333 m/s, and again the energy gained by the system is 5,000,000,000 J.  This is what we should expect because we’re burning the same amount of propellant, releasing the same amount of energy, and expelling the same amount of reaction mass at the same exhaust velocity.

However, we see that the energy gained by the rocket, KE_rocket-KE_intial, in the first case is 3,000,000,000 J, and in the second case it is 4,500,000,000 J.  Also note that we don't have to consider the energy of the exhaust at all to compute the change in energy of the rocket.  Computing the energy of the exhaust here is only a convenience to show that net energy change of the system is the same in both examples.  All we need to know to compute the energy change of the rocket is its change in velocity, which we get from analyzing momentum.  Here we calculated the momentum of the exhaust in order to determine the change in momentum of the rocket, but we normally don’t have to do that.  Thrust gives us what we need to know because thrust is simply the rate of change of momentum.

Note that the examples above include some mathematical simplifications.  (For explanation, see hidden contents below.)  This was done to demonstrate the basic concept without overcomplicating the math.  In real life the math is a bit more complicated and the resulting numbers will be different, but the concept is the same.  My previous examples are a better demonstration of the real life calculations.

Spoiler

First, I assumed that the reaction mass is expelled instantaneously, which is not how it works in real life.  Second, you may have noticed that I computed the velocity of the expelled mass relative to the initial velocity of the rocket.  I then computed the final velocity of the rocket from conservation of linear momentum.  This sequence of operations resulted in the expelled mass having a velocity of -2000 m/s and the rocket having a velocity of +1333.333 m/s (first example).  This means that velocity of the exhaust relative to the rocket is actually 3333.333 m/s, not 3000 m/s as stated in the problem.

 

Edited August 23, 2016 by OhioBob

[mhoram]

1 hour ago, Blaarkies said:

I do not believe you are a fool, and neither do i believe you are trying to embarrass yourself...but you are definitely trolling this thread big time

After reading his latest posts, I came to the same conclusion.

Everything about the Oberth effect has already been said in more detail and from more perspectives than I have ever seen in a single thread on this forum, and still it seems to be futile to try to convince him. Will bookmark this thread for future reference.

[foamyesque]

Here's a way to picture Oberth that doesn't even consider thrust. Suppose you're skiing, and you enter a half-pipe. You go in with 10m/s horizontal velocity; the half-pipe is 12.75m deep. Assuming friction to be negligable, if one were to accelerate the skier by 1m/s at any point in traversing the half-pipe, at what point would applying the change in velocity get you the highest exit velocity, and what would it be?

 

The answer is: at the trough of the half-pipe, and you would come out moving at 11.4m/s -- an efficiency gain of 40% over doing so at the top.

[taniwha]

On 23/08/2016 at 7:50 AM, Red Iron Crown said:

I don't have to, the wiki page already agrees with me (as does Squad's implementation, please show me where they ever mention implementing Oberth as some separate function, rather than just thrust and E_k=0.5*m*v²). You are interpreting the wiki article incorrectly.

As far as I know (there may be something in aero heating), other than calculating a vessel's SMA and eccentricity, KSP doesn't directly implement E_k=0.5*m*v² even: just no need. I've done physics (and orbit, and even rocket) simulations myself and it's simply a_grav = -μ(r.r)^-3/2r where both a_grav and r are vectors. You then integrate with respect to time to get velocity and again to get position. Without doing anything else, you will get correct orbital motion (depending on the quality of your integrator (*glares at PhysX*)). No calculation of kinetic or potential energy, or angular momentum, or centrifugal force or anything else.

If you want a rocket in there, you just do a_total = a_grav + a_thrust (again, vectors). If you calculate a_thrust correctly (engine thrust (vector) and vessel mass), and vary mass with fuel consumption, you get the rocket equation for free.

There is no need to do any calculations for the Oberth effect because there's no need to do any calculations for energy (kinetic or potential). And one of the biggest reasons you don't need to do any energy calculations is energy is (I believe) a fictitious thing that makes it easier to work with physics mathematically (and conceptually).

Now, a bit about rockets and Isp and thrust. The starting point is F = m_e v_e (or really F = m a, but meh) where m_e is your mass flow rate (usually kg/s) and v_e is your exhaust velocity (usually m/s). Do all the math and you get Δv = v_e ln(m₁/m₀) where m₁ is your full mass and m₀ is your empty mass. I_sp is just a way of setting the argument over m/s vs ft/s and so Ve = g₀ I_sp where I_sp is in seconds and g₀ is 9.81m/s²always (doesn't matter where you are). KSP uses Isp to calculate your mass flow rate from your thrust (prior to 1.0, KSP held thrust constant and thus flow rate varied with atmospheric pressure, but since 1.0, KSP holds flow rate constant and thus thrust varies with atmospheric pressure, but calculates the base flow rate from vacuum thrust and I_sp). However, much to the disappointment of many, KSP does not actually calculate Δv = v_e ln(m₁/m₀) (MechJeb and Kerbal Engineer do, though): it doesn't need to as it just falls out from the integration of the basic equations.

So, in summary: no calculation of Δv, no calculation of energy (except to calculate orbit parameters), and certainly no calculation of Oberth Effect (and I agree, it really is a misnomer: it should have been called Oberth's Rocket Law or something like that, as that's what it is: an observation of what is, thus a (scientific) law): they all show up uninvited.

Edited August 26, 2016 by taniwha

[Superfluous J]

1 minute ago, taniwha said:

and I agree, it really is a misnomer: it should have been called Oberth's Rocket Law or something like that, as that's what it is: an observation of what is, thus a (scientific) law

I think the name's fine. It's an effect. A result of things being set up a certain way, not a cause. The wrong word would have been affect. Law works just fine though too.

"In this space ship we obey the Oberth Law."

[OhioBob]

3 hours ago, taniwha said:

So, in summary: no calculation of Δv, no calculation of energy (except to calculate orbit parameters), and certainly no calculation of Oberth Effect: they all show up uninvited.

I've done plenty of space filght simulations and I agree with you.  All that we really need to do is generate a force diagram (gravity, thrust, and drag), compute the accelerations, velocities, and displacements, and then repeat.  We integrate with respect to time because those force vectors are constantly changing.  That's really all there is to it.

Many of my simulations have reproduced the results of well documented real life space flights with negligible error (such as here and here), thus proving that my methods are sound.  If I'm failing to account for Oberth effect by leaving out some required code, then the real life universe must not have gotten the memo either.

[FullMetalMachinist]

On 8/21/2016 at 9:17 PM, Red Iron Crown said:

All that is required for Oberth is kinetic energy increasing with the square of speed and thrust not changing with speed

You said this a couple times in this thread. Why does Oberth care if thrust is constant if the only thing that matters is adding to your velocity when you're already going fast? 

[Red Iron Crown]

Just now, FullMetalMachinist said:

You said this a couple times in this thread. Why does Oberth care if thrust is constant if the only thing that matters is adding to your velocity when you're already going fast? 

I mention this because it is so different than most Earthbound propulsion, which is part of why Oberth is so counterintuitive. Most propulsion is relatively constant power (energy/second), so acceleration decreases as speed increases (i.e. more fuel is needed for the same amount of speed change if the initial speed is higher). Oberth doesn't apply to such propulsion. Rockets in vacuum are fairly unusual in this regard, in that they deliver the same acceleration no matter what the initial speed. It is this unusual characteristic that enables them to harness the Oberth effect.

[FullMetalMachinist]

11 minutes ago, Red Iron Crown said:

Rockets in vacuum are fairly unusual in this regard, in that they deliver the same acceleration no matter what the initial speed. It is this unusual characteristic that enables them to harness the Oberth effect.

That makes sense. Thanks 

[dire]

11 hours ago, Red Iron Crown said:

I mention this because it is so different than most Earthbound propulsion, which is part of why Oberth is so counterintuitive. Most propulsion is relatively constant power (energy/second), so acceleration decreases as speed increases (i.e. more fuel is needed for the same amount of speed change if the initial speed is higher). Oberth doesn't apply to such propulsion. Rockets in vacuum are fairly unusual in this regard, in that they deliver the same acceleration no matter what the initial speed. It is this unusual characteristic that enables them to harness the Oberth effect.

It is also this effect that acceleration decreases as speed increases, which is responsible for the speed of light being the universal speed limit. Even rockets aren't magic.