Algorithm for Making Interplanetary Transfers (almost) as Easy as Getting to The Mun!

[arkie87]

*Sigh*

Shouldn't have bothered. I have tried to explain to you in detail why your underlying assumptions are incorrect and how the math really works. But you're convinced that everybody who came up with this stuff is wrong because it "doesn't make sense" to you.

I'm not trying to "out-nerd you" but rather going out of my way to help you out by explaining it to you, but you don't want to listen, so have it your way. Not worth my time and not appreciated by you.

Screwit,

-Slashy

Concentric's explanation made sense to me. Yours didnt because you kept telling me i was wrong or insisted i didnt understand some concept, rather than addressing the question i had asked.

As i said, i am not concerned with who is right, but rather, why.

[arkie87]

*Sigh*

Shouldn't have bothered. I have tried to explain to you in detail why your underlying assumptions are incorrect and how the math really works. But you're convinced that everybody who came up with this stuff is wrong because it "doesn't make sense" to you.

I'm not trying to "out-nerd you" but rather going out of my way to help you out by explaining it to you, but you don't want to listen, so have it your way. Not worth my time and not appreciated by you.

Screwit,

-Slashy

I think you will agree that when something doesnt make sense to you you need someone to explain it to you in a way the hones in on your misunderstanding, rather than repeating things you already know. Apparently, we dont think alike enough to do that.

No hard feelings.

EDIT: I'm sure it has happened to you that you asked someone a specific, detailed, technical question (usually by email) with an elaborate explanation and laying out the things you know etc..., and they respond by answering a simpler/more obivous question than you asked or repeating stuff you've already established etc... I feel like that is what happened here.

Edited November 11, 2014 by arkie87

[GoSlash27]

Concentric's explanation made sense to me. Yours didnt because you kept telling me i was wrong or insisted i didnt understand some concept, rather than addressing the question i had asked.

As i said, i am not concerned with who is right, but rather, why.

You're not listening, you *don't* understand the concept, and I'm done explaining. Your defensiveness ain't my problem, so maybe Concentric will have better luck.

Good luck,

-Slashy

[arkie87]

You're not listening, you *don't* understand the concept, and I'm done explaining. Your defensiveness ain't my problem, so maybe Concentric will have better luck.

Good luck,

-Slashy

Still at it, so thanks for proving my point.

If the escape velocity of a planet is 1 km/s, and you accelerate up to 1.1 km/s, how fast will you be moving infinitely far from the planet?

And yes, this IS the Oberth effect, though it might be a different way than you are used to looking at it.

[GoSlash27]

I think you will agree that when something doesnt make sense to you you need someone to explain it to you in a way the hones in on your misunderstanding, rather than repeating things you already know. Apparently, we dont think alike enough to do that.

No hard feelings.

EDIT: I'm sure it has happened to you that you asked someone a specific, detailed, technical question (usually by email) with an elaborate explanation and laying out the things you know etc..., and they respond by answering a simpler/more obivous question than you asked or repeating stuff you've already established etc... I feel like that is what happened here.

*Grumble...*

If you "already knew" this stuff, you wouldn't have this confusion.

Forget what you think you "know" about the Oberth effect and orbital mechanics and pay attention, and *maybe* you'll actually understand it if somebody's willing to take the time to explain it.

Honestly... I don't know why I'm willing to give you another shot at this point, but I am.

Act like you don't know anything about this, and I'll help you.

What don't you get? How getting out of the bottom of a gravity well can be cheaper than operating in free space?

[arkie87]

*Grumble...*

If you "already knew" this stuff, you wouldn't have this confusion.

Forget what you think you "know" about the Oberth effect and orbital mechanics and pay attention, and *maybe* you'll actually understand it if somebody's willing to take the time to explain it.

Honestly... I don't know why I'm willing to give you another shot at this point, but I am.

Act like you don't know anything about this, and I'll help you.

What don't you get? How getting out of the bottom of a gravity well can be cheaper than operating in free space?

Eh, why should i act like i dont know anything about this? What kind of a person asks that of someone they are trying to help? And we already resolved my confusion pages ago, but you still seem to think i dont know anything.

Please just answer the question i posed above:

If the escape velocity of a planet is 1 km/s, and you accelerate up to 1.1 km/s, how fast will you be moving infinitely far from the planet?

The answer is clearly not 100 m/s, and this is due to the fact that gravity requires an energy well that absorbs a certain amount of kinetic energy, not velocity. I have shown this pages ago, and this IS the oberth effect. If you review the equations, you will see how they are actually the same thing, and yet, for some reason, you told me I am wrong and i dont know what the Oberth effect is....

Edited November 11, 2014 by arkie87

[GoSlash27]

Eh, why should i act like i dont know anything about this? What kind of a person asks that of someone they are trying to help? And we already resolved my confusion pages ago, but you still seem to think i dont know anything.

Please just answer the question i posed above:

If the escape velocity of a planet is 1 km/s, and you accelerate up to 1.1 km/s, how fast will you be moving infinitely far from the planet?

The answer is clearly not 100 m/s, and this is due to the fact that gravity requires an energy well that absorbs a certain amount of kinetic energy, not velocity. I have shown this pages ago, and this IS the oberth effect. If you review the equations, you will see how they are actually the same thing.

Sorry, wrong answer.

You are too proud to accept instruction, so I can't help you (even though I want to and I know what's going on here).

Good luck looking elsewhere and I honestly hope you find the answers you're looking for.

-Slashy

[Superfluous J]

Are you also claiming that starting from the surface of Kerbin requires less dV/energy than starting on the other side of Kerbol i.e. point (?

This wasn't directed at me but:

If Kerbin had no atmosphere AND you were orbiting right at the surface (essentially skimming over it like a meter off the ground) then yes. It would be more efficient. Being stopped on the surface puts you at a HUGE disadvantage and having to carve through the atmosphere just adds to it.

The reason the surface launch is expensive has nothing to do with the reason the orbit-to-transfer burn is the most efficient.

[arkie87]

Sorry, wrong answer.

You are too proud to accept instruction, so I can't help you (even though I want to and I know what's going on here).

Good luck looking elsewhere and I honestly hope you find the answers you're looking for.

-Slashy

Lol. Wow dude. If you know what's going on, the please explain it. It seems to me that you just realized you were wrong and now cannot admit it.

"even thought i want to and I know whats going on here"

So please, explain away. This is a very simple problem...

[Specialist290]

Alright, I'm calling a time-out on this discussion before this goes any further. Watch this space.

EDIT: Alright, time-out is over. After some deliberation, we of the moderation team have decided to reopen this discussion. However, please remember that condescending and / or cutting remarks aimed at other users are not tolerated on these forums. Any further remarks questioning any participant's intelligence and good faith, rather than the topic at hand itself, will be construed as user harassment. If you can't respond in a constructive and civil fashion, don't respond at all.

Edited November 11, 2014 by Specialist290

[Starman4308]

Okay, perhaps I can take a crack at this. As mentioned, the Oberth effect favors burns at high speed. One way to think about it: if you escape Kerbin very quickly, Kerbin's gravity has less time to claw away and try to eliminate your escape velocity. If you just barely escape LKO into the same orbit as Kerbin, the amount of orbital velocity lost to Kerbin gravity is at a maximum. If you burn straight from LKO to, say, Jool, you will be escaping much quicker, so you will lose less velocity during Kerbin escape. Both ways of going to Jool will have the same velocity exiting Kerbin orbit, but the two-burn strategy loses some because you dawdled in Kerbin's SOI and lost more velocity to gravity.

That is why Oberth favors transfers direct from LKO* in delta-V terms: you lose some velocity escaping Kerbin, and the faster you escape Kerbin, the less velocity you lose to gravity. In kinetic energy terms, GoSlash is almost exactly right: because a direct transfer lets you add the velocity from orbiting Kerbin, you get the same kinetic energy for less change in velocity. The goof: I am pretty sure Oberth doesn't have much to do with the chemical energy of the rocket: conservation of energy is maintained because your fuel is losing kinetic energy equal to the energy your rocket is gaining. I have no idea where the chemical energy is going (probably heat, for the most part), but it is patently ridiculous to conclude that you are getting "more out of your chemical energy", because as you increase velocity, you will swiftly reach the point where kinetic energy per kilogram of propellant burned exceeds that kilogram's chemical energy.

There's also the little bit that energy should be conserved no matter what reference frame you're in. There's probably some fancy mathematical proof which states that, for a momentum engine (a rocket), energy gained between all potential energy fields (gravity wells, etc) and reference frames (magic spaceships traveling 0.99c in the other direction) is equal to energy lost by the exhaust, again in all potential energy fields and reference frames. Maybe there's a bit of leeway, but it would have to be strictly less than the useful energy gained from burning the rocket fuel, because of conservation of energy.

*Or possibly straight from the pad. I've always set myself up in a parking orbit before interplanetary transfer.

Edited November 11, 2014 by Starman4308

[arkie87]

Okay, perhaps I can take a crack at this. As mentioned, the Oberth effect favors burns at high speed. One way to think about it: if you escape Kerbin very quickly, Kerbin's gravity has less time to claw away and try to eliminate your escape velocity. If you just barely escape LKO into the same orbit as Kerbin, the amount of orbital velocity lost to Kerbin gravity is at a maximum. If you burn straight from LKO to, say, Jool, you will be escaping much quicker, so you will lose less velocity during Kerbin escape. Both ways of going to Jool will have the same velocity exiting Kerbin orbit, but the two-burn strategy loses some because you dawdled in Kerbin's SOI and lost more velocity to gravity.

That is why Oberth favors transfers direct from LKO* in delta-V terms: you lose some velocity escaping Kerbin, and the faster you escape Kerbin, the less velocity you lose to gravity. In kinetic energy terms, GoSlash is almost exactly right: because a direct transfer lets you add the velocity from orbiting Kerbin, you get the same kinetic energy for less change in velocity. The goof: I am pretty sure Oberth doesn't have much to do with the chemical energy of the rocket: conservation of energy is maintained because your fuel is losing kinetic energy equal to the energy your rocket is gaining. I have no idea where the chemical energy is going (probably heat, for the most part), but it is patently ridiculous to conclude that you are getting "more out of your chemical energy", because as you increase velocity, you will swiftly reach the point where kinetic energy per kilogram of propellant burned exceeds that kilogram's chemical energy.

*Or possibly straight from the pad. I've always set myself up in a parking orbit before interplanetary transfer.

There are many equivalent ways to look at Oberth effect:

Your explanation is correct (that gravity has less time to pull back at you, so you end up going faster if you leave faster)

The chemical energy explanation that more of the rocket gasses energy goes into kinetic energy rather than chemical

The simplest explanation (to me, at least) is that gravity is a energy well, and requires a certain energy, rather than velocity to escape, and since kinetic energy increases quadratically, escaping faster than necessary results in higher velocities after escaping... in fact, the Oberth effect ONLY applies in gravitational fields, since otherwise, a change in kinetic energy is meaningless (since gravitational wells are an absolute point of reference)

That said, i'm not sure which point you were trying to clarify, since there is no quote...

[Starman4308]

That said, i'm not sure which point you were trying to clarify, since there is no quote...

I thought you were having issues with trying to understand how a burn straight from LKO was more efficient than escaping, and then going somewhere?

In any case, I am 99% sure the "chemical energy explanation" is bunk, because conservation of energy has to be true no matter what silly reference frame you choose (magic rocket going the other way at 0.9c, for example), and the differences in KE are greatly exaggerated in some reference frames, leading me to suspect that's not exactly what's going on.

[Kamuchi]

*snip*

Wait, my nub rocket scientist moment

With what you`re saying and using my Mun/Minmus base/orbit fuel depo in mind:

Actually burning trajectory from Kerbin in some cases would require less fuel then plummeting from Mun/Minmus to Kerbin for a slingshort+burn?

Lets just throw something else into the mix as I saw a Manley vid some time ago (forgot wich one) where he actually got more dV out of a Kerbin slingshot going clockwise (against rotation) then when you go anti clockwise (with rotation) for specific escape trajectories to other planets.

He`s a rocket scientist and I`m not, so I`m still scratching my head abit wondering why&when you do something that appears insane?

Edited November 11, 2014 by Kamuchi

[GoSlash27]

In kinetic energy terms, GoSlash is almost exactly right: because a direct transfer lets you add the velocity from orbiting Kerbin, you get the same kinetic energy for less change in velocity. The goof: I am pretty sure Oberth doesn't have much to do with the chemical energy of the rocket: conservation of energy is maintained because your fuel is losing kinetic energy equal to the energy your rocket is gaining. I have no idea where the chemical energy is going (probably heat, for the most part), but it is patently ridiculous to conclude that you are getting "more out of your chemical energy", because as you increase velocity, you will swiftly reach the point where kinetic energy per kilogram of propellant burned exceeds that kilogram's chemical energy.

There's also the little bit that energy should be conserved no matter what reference frame you're in. There's probably some fancy mathematical proof which states that, for a momentum engine (a rocket), energy gained between all potential energy fields (gravity wells, etc) and reference frames (magic spaceships traveling 0.99c in the other direction) is equal to energy lost by the exhaust, again in all potential energy fields and reference frames. Maybe there's a bit of leeway, but it would have to be strictly less than the useful energy gained from burning the rocket fuel, because of conservation of energy.

*Or possibly straight from the pad. I've always set myself up in a parking orbit before interplanetary transfer.

Starman,

The "lost" energy you're talking about is going into potential energy relative to Kerbin. That energy does nothing to aid a trip to Jool, and is thus wasted.

The important part is that adding velocity at a higher velocity creates more kinetic energy than adding velocity at a lower velocity.

That's why it's cheaper to make a trip to Jool from LKO than it is to make it from Kerbin's orbit around Kerbol even though it's in the bottom of a gravity well.

Kamuchi,

Your practice of refueling at high orbit and then slingshotting is completely sound. Harder to match your launch windows, but your potential energy is returned to kinetic just before the burn and your burn is even more efficient because you're doing it at higher speed.

Best,

-Slashy

Edited November 11, 2014 by GoSlash27

[Kamuchi]

Thanks,

Feel less stupid for understanding half of what is said here and that I`m on track

[arkie87]

I thought you were having issues with trying to understand how a burn straight from LKO was more efficient than escaping, and then going somewhere?

I was having trouble understanding how a burn to Jool from LKO could require less energy than a burn from the same apoapsis/periapsis outside of Kerbin's SOI... but that was back on page 2 or 3... Cocentric has since helped me resolve the issue by pointing out the incorrect assumption (LKO carries with it 2 km/s of velocity, so the difference between methods can be more than just the extra needed escape velocity)...

My current un-answered question is how is it possible that someone on the surface of Kerbin could require less energy/dV to get to Jool (assuming zero rotation of Kerbin) than someone in Kerbin apoapsis/periapsis but outside Kerbin SOI, since the former has to overcome two gravitational wells (Kerbin's and then Kerbol's) whereas the latter only has to overcome one (Kerbol's) and by the exact same amount as the former...

In any case, I am 99% sure the "chemical energy explanation" is bunk, because conservation of energy has to be true no matter what silly reference frame you choose (magic rocket going the other way at 0.9c, for example), and the differences in KE are greatly exaggerated in some reference frames, leading me to suspect that's not exactly what's going on.

"Chemical energy" explanation means that any remaining energy is assigned to "chemical energy" when computed in different reference frames such that conservation of energy still holds. It is just a way of understanding how a rocket can produce a different amount of power depending on location/velocity, and isnt quite so scientific. It is like the difference between enthalpy and internal energy, where all additional energy is said to be in "internal" energy.

But you are right that kinetic energy, and particularly differences in kinetic energy, vary wildly with reference frame, which is why the Oberth effect only matters in gravitational fields, where a reference frame is given.

Personally, I don't even see a reason for labeling the "Oberth effect". Understanding that gravitational wells absorbs a certain amount of energy (rather than dV) when escaping will make its effect clear.

Edited November 11, 2014 by arkie87

[GoSlash27]

Personally, I don't even see a reason for labeling the "Oberth effect". Understanding that gravitational wells absorbs a certain amount of energy (rather than dV) when escaping will make its effect clear.

Arkie,

The Oberth effect isn't about gravity wells absorbing energy. It's about the nonlinear relationship between velocity and kinetic energy.

Kinetic energy is proportional to the square of velocity. adding velocity is (V+D)^2, or v^2+2VD+D^2. It's that "+2VD" part that increases your efficiency the faster you're going.

Best,

-Slashy

[arkie87]

Arkie,

The Oberth effect isn't about gravity wells absorbing energy. It's about the nonlinear relationship between velocity and kinetic energy.

Kinetic energy is proportional to the square of velocity. adding velocity is (V+D)^2, or v^2+2VD+D^2. It's that "+2VD" part that increases your efficiency the faster you're going.

Best,

-Slashy

It really is the same thing and/or its really about both... when calculating velocity infinitely far from the planet, you always subtract the escape kinetic energy:

V_final = sqrt(V_total^2 - V_escape^2) = sqrt((V_escape+dV)^2-V_escape^2) = sqrt(2*V_escape*dV + dV^2)

V_final/dV = sqrt(1+ 2*V_escape/dV)

The non-linear relationship between deltaV and energy is what makes V_final larger than just deltaV i.e. V_final/dV > 1.

Thus, either they are two ways of looking at the same thing, or, it really is about both...

Edited November 11, 2014 by arkie87

[r_rolo1]

Yeah, gravity has zero to do with the Oberth effect. It is simply a fancy way to say that, as you go faster the kinetic energy of the system rocket + exhaust tends to be more on the side of the rocket than on the exhaust ( some simple math with high school physics will show you that ). OFC that is quite visible in lower parts of gravity wells ( because you necessarily need to go fast there, otherwise you would not be in orbit ), but it is not exclusive of those locations ...

[Starman4308]

Starman,

The "lost" energy you're talking about is going into potential energy relative to Kerbin. That energy does nothing to aid a trip to Jool, and is thus wasted.

I suspect we're talking past each other. My point is that I am pretty sure the rocket's kinetic energy gains out of the Oberth effect are not because "you are getting more out of the chemical energy of the fuel". This is clearly ridiculous, because if you apply thrust at 0.9c, the amount of kinetic energy your rocket gains is vastly in excess of the chemical energy from the fuel. Instead, I am pretty sure your kinetic energy boosts can be explained by the fact that the exhaust gases are losing a commensurate amount of kinetic energy, possibly with a wiggle factor no greater than the useful work generated by burning propellant.

The important part is that adding velocity at a higher velocity creates more kinetic energy than adding velocity at a lower velocity.

I am not disagreeing.

Kamuchi,

Your practice of refueling at high orbit and then slingshotting is completely sound. Harder to match your launch windows, but your potential energy is returned to kinetic just before the burn and your burn is even more efficient because you're doing it at higher speed.

As mentioned, this should work. It might burn more fuel overall in some cases*, but I am pretty sure it'll let you get away with smaller fuel tanks on your vehicle, because the amount of dV spent going from a circular high orbit to an elliptical high orbit should be less than the amount of velocity you can add to the slingshot.

*For example, if your target is "just barely escape Kerbin", you're burning almost enough fuel to escape Kerbin, refuelling, burning a bit more to set up the slingshot, and then burning a substantial fraction of the fuel necessary to escape Kerbin: this is in addition to the fuel it took to set up the fuel depot in the first place. More overall fuel expenditure, though neither maneuver (getting to the fuel depot, or setting up the slingshot plus slingshot) uses as much fuel as a direct burn.

Edited November 11, 2014 by Starman4308

[Kamuchi]

It really is the same thing and/or its really about both... when calculating velocity infinitely far from the planet, you always subtract the escape kinetic energy:

V_final = sqrt(V_total^2 - V_escape^2) = sqrt((V_escape+dV)^2-V_escape^2) = sqrt(2*V_escape*dV + dV^2)

V_final/dV = sqrt(1+ 2*V_escape/dV)

The non-linear relationship between deltaV and energy is what makes V_final larger than just deltaV i.e. V_final/dV > 1.

Thus, either they are two ways of looking at the same thing, or, it really is about both...

Is that the equationfor *moar boosters*?

[JimmyAgent007]

It would be interesting if the Kerbin Orbit path was highlighted different colours for the launch windows to different planets. Say Kerbin is passing through an optimal Duna launch window, you go to the map and see Kerbin passing through a red section of its orbit. Maybe label it in case two or more planets overlap. That way you can, without launching or even doing heavy math, be able to see when the best time to launch the mission is. Also if your ship is around Duna then its orbit gets highlighted for windows. Same for if you orbit the star, then its your ships orbit that gets broken down into the windows. This would save people from having to do the math, check window calculators, and be very friendly to new players. As with anything it can be toggled for those who like doing things the hard way.

[arkie87]

Yeah, gravity has zero to do with the Oberth effect. It is simply a fancy way to say that, as you go faster the kinetic energy of the system rocket + exhaust tends to be more on the side of the rocket than on the exhaust ( some simple math with high school physics will show you that ). OFC that is quite visible in lower parts of gravity wells ( because you necessarily need to go fast there, otherwise you would not be in orbit ), but it is not exclusive of those locations ...

Without a gravitational reference, the Oberth effect is meaningless, since i could be going 0.9c or stationary (and the amount of kinetic energy increase for a 1 m/s burn will vary wildly)... It is only when there is a gravitational reference that speaking about the Oberth effect makes any sense, since the reference frame is constrained: all velocities are measured with respect to the gravitational well.

[arkie87]

It would be interesting if the Kerbin Orbit path was highlighted different colours for the launch windows to different planets. Say Kerbin is passing through an optimal Duna launch window, you go to the map and see Kerbin passing through a red section of its orbit. Maybe label it in case two or more planets overlap. That way you can, without launching or even doing heavy math, be able to see when the best time to launch the mission is. Also if your ship is around Duna then its orbit gets highlighted for windows. Same for if you orbit the star, then its your ships orbit that gets broken down into the windows. This would save people from having to do the math, check window calculators, and be very friendly to new players. As with anything it can be toggled for those who like doing things the hard way.

I'm glad to get back on topic of the thread.

A nice gui like that would be pretty nice, and maybe it could be enabled/disabled with difficulty settings.

- - - Updated - - -

Is that the equationfor *moar boosters*?

No, adding moar boosters is like multiplying by infinity