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

[r_rolo1]

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.

No, it happens when you have a referential that it is not the center of gravity of the rocket + exhaust system ( that is, always ). Gravity has zero to do with that, so don't bring it to the discussion...

Edited November 11, 2014 by r_rolo1

[Starman4308]

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.

There is one non-gravitational reference I can think of where the Oberth effect is meaningful: "George, George, George of the Jungle, strong as he can be, watch out for that tree!"*

*Technically this specific example involves gravity, but if you had something about to collide with you in deep space, gravitational reference frames are not particularly relevant at the moment, but kinetic energy is certainly relevant.

EDIT: Basically, the Oberth effect is relevant whenever kinetic energy is relevant, and that is not constrained to orbital maneuvers.

Edited November 11, 2014 by Starman4308

[Kamuchi]

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.

I understand thanks

I was mainly using EL to build things off Kerbin to avoid having the repeat launches, Minmus is a perfect place to build/assamble larger ships in orbit with a very low dV infinite fuel pump below.

Then use Kerbin to slingshot out into the solar system using Minmus offset orbital plane to match the target, timing would still remain an issue

[arkie87]

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.

I like you description of "talking past each other" as i think it is a good way of explaining what was happening .

I think the "chemical energy" concept works for people who dont want to do the math/kinetic energy balance, and just throw any perceived imbalances into "chemical energy". This is not something exact, but rather, just an analogy/way of explaining where the extra energy comes from at high velocity (or where the extra energy goes at low velocity). The analogy obviously breaks down when the velocity of the craft is much larger than that of the propellant (as you correctly explained), and the only approach that makes any sense is to perform a kinetic energy balance.

[arkie87]

No, it happens when you have a referential that it is not the center of gravity of the rocket + exhaust system ( that is, always ). Gravity has zero to do with that, so don't bring it to the discussion...

What you are saying is a generalization of what I am saying.

[Starman4308]

Then use Kerbin to slingshot out into the solar system using Minmus offset orbital plane to match the target, timing would still remain an issue

Well, if you weren't averse to manual number-crunching, you might try exiting Minmus far ahead of time, and tweaking orbit so that, when the launch window comes (technically a few days before the launch window comes, to provide time to drop down from high orbit), you are directly opposite the burn location. There's some leeway, because the launch window has leeway, but you'd have to set up your orbital period such that you're at the right phase when the launch window comes.

[John FX]

So I looked at the thread to see an easy way to transfer to other planets and what I got was an intense discussion of the Oberth effect.

Any chance the OP could rename the thread to match the change in topic discussion?

[arkie87]

There is one non-gravitational reference I can think of where the Oberth effect is meaningful: "George, George, George of the Jungle, strong as he can be, watch out for that tree!"*

*Technically this specific example involves gravity, but if you had something about to collide with you in deep space, gravitational reference frames are not particularly relevant at the moment, but kinetic energy is certainly relevant.

EDIT: Basically, the Oberth effect is relevant whenever kinetic energy is relevant, and that is not constrained to orbital maneuvers.

EDIT: This is a good way of summing/closing the discussion, and shows many different ways of looking at the Oberth effect.

Thank you.

Edited November 11, 2014 by arkie87

[arkie87]

So I looked at the thread to see an easy way to transfer to other planets and what I got was an intense discussion of the Oberth effect.

Any chance the OP could rename the thread to match the change in topic discussion?

It was not my intention to discuss Oberth effect to death, but you know how debates go in online forums...

I think the first few pages discuss other algorithms for getting to other planets more efficiently without too much calculations/measurements/plugins.

[r_rolo1]

What you are saying is a generalization of what I am saying.

Well, you spoke about gravitational referentials as the only ones where the Oberth effect is not meaningless. That is factually wrong, hence my correction. You are right in saying that you need a external referential, but gravity has nothing to do with the issue. To avoid bringing more misunderstandings to the discussion, it is better to not bring gravity to the fray until needed

[arkie87]

Well, you spoke about gravitational referentials as the only ones where the Oberth effect is not meaningless. That is factually wrong, hence my correction.

Well i didnt intend that gravity is the ONLY reference that makes Oberth effect relevant ever, but rather, the one that is relevant to the discussion at hand. But thanks for your clarification/generalization (not sarcastic).

You are right in saying that you need a external referential, but gravity has nothing to do with the issue.

It does have to do with the issue, since we were discussing Oberth effect in a gravitational well.

To avoid bringing more misunderstandings to the discussion, it is better to not bring gravity to the fray until needed

I only brought gravity into the discussion since we were specifically discussing a case involving the Oberth effect in a gravitational well

Edited November 11, 2014 by arkie87

[GoSlash27]

Well, you spoke about gravitational referentials as the only ones where the Oberth effect is not meaningless. That is factually wrong, hence my correction. You are right in saying that you need a external referential, but gravity has nothing to do with the issue. To avoid bringing more misunderstandings to the discussion, it is better to not bring gravity to the fray until needed

^This. The Oberth effect has nothing to do with gravity wells. Looking at it from that perspective isn't stating the same thing in a different way. It's just that the relationship between velocity and kinetic energy isn't linear, so adding velocity when travelling rapidly adds more kinetic energy than adding velocity when travelling slowly. That's really it, period, full stop.

The original question was why (or indeed if) it was possible to achieve a Jool intercept more cheaply from the bottom of Kerbin's gravity well than the top, and this is why. You have more kinetic energy at the time of firing down there, so you create more kinetic energy for the same fuel expenditure.

Best,

-Slashy

[arkie87]

^This. The Oberth effect has nothing to do with gravity wells. Looking at it from that perspective isn't stating the same thing in a different way. It's just that the relationship between velocity and kinetic energy isn't linear, so adding velocity when travelling rapidly adds more kinetic energy than adding velocity when travelling slowly. That's really it, period, full stop.

Looking at from a different perspective can of course be stating the same thing in a different way. Look at D'Alembert's_principle (http://en.wikipedia.org/wiki/D'Alembert's_principle) which essentially states that F-ma = 0 rather than Newton's Law of F=ma. These are the same thing, but the alternate form/perspective allows one to see inertia as a pseudoforce.

Thus, to some people, one explanation might more "intuitive" sense over another, even though they are equivalent.

From my perspective, it is obvious that kinetic energy is non-linear with velocity, and realizing that gravity wells are energy wells not deltaV wells made the result of the Oberth effect (escaping faster leaves more kinetic energy upon departure) more easily understandable.

And the discussion from the very beginning has been about Oberth effect in gravity wells, so of course gravity is relevant...

Edited November 11, 2014 by arkie87

[arkie87]

The original question was why (or indeed if) it was possible to achieve a Jool intercept more cheaply from the bottom of Kerbin's gravity well than the top, and this is why. You have more kinetic energy at the time of firing down there, so you create more kinetic energy for the same fuel expenditure.

By bottom of the gravity well do you mean in LKO? If so, please stop mentioning it. That misunderstanding was clarified by Cocentric 6 pages ago... not sure why you are still dwelling on that...

What i dont understand is how its possible from the very bottom i.e. on the surface (lets say north pole) where kinetic energy is equivalent to being just outside SOI with same orbit.

[GoSlash27]

arkie87,

Again, it's not escaping faster that does it, but rather burning while traveling at a higher velocity.

So accepting your assertion that you understand the Oberth effect, you should have no problem understanding that 1) the charts are actually correct, 2) the cost of a 2 burn profile is more than just the DV to escape Kerbin's SOI, and 3) The penalty in terms of DV actually increases with distance to the target planet.

This should set to rest all of the questions that you had.

Best,

-Slashy

[GoSlash27]

By bottom of the gravity well do you mean in LKO? If so, please stop mentioning it. That misunderstanding was clarified by Cocentric 6 pages ago... not sure why you are still dwelling on that...

What i dont understand is how its possible from the very bottom i.e. on the surface (lets say north pole) where kinetic energy is equivalent to being just outside SOI with same orbit.

Sorry, didn't see this when I posted my last.

I don't know how being static on the North pole could ever be more efficient than floating in open space. I'd have to chuck some numbers at it and see if that's possible.

Best,

-Slashy

[arkie87]

arkie87,

Again, it's not escaping faster that does it, but rather burning while traveling at a higher velocity.

By escaping faster i clearly meant having a higher velocity at periapsis, which is, essentially, burning at periapsis, as you mentioned.

So accepting your assertion that you understand the Oberth effect, you should have no problem understanding that 1) the charts are actually correct, 2) the cost of a 2 burn profile is more than just the DV to escape Kerbin's SOI, and 3) The penalty in terms of DV actually increases with distance to the target planet.

Who is talking about the charts and the cost of a two burn profile? These things have been resolved pages ago!

My current question, on the other hand, remains, and stating my answer as "wrong" or self-evident by virtue of the tables/charts/trying it in KSP does not help anyone understand why. I am interested in why. If you would like to answer the question i have asked (and not rehash old discussions that were resolved pages ago), please feel free. Otherwise, please stop responding to my posts.

[arkie87]

Sorry, didn't see this when I posted my last.

I don't know how being static on the North pole could ever be more efficient than floating in open space. I'd have to chuck some numbers at it and see if that's possible.

Best,

-Slashy

That is my current question. Every other thing you mentioned is a revival of a resolved topic, and im not sure why you keep bring it up. Please only discuss this question in the future.

That said (back to the discussion), it seems to me that being static on north pole must take more energy than floating in space since both must overcome Kerbol gravitational well, so those cancel, and all that is left is for kerbal on north pole to overcome Kerbin gravity well. I could understand how they could be close, especially when Kerbol gravitational well is much larger than Kerbin's such that not much energy/dV is lost upon change of SOI (from Kerbin to Kerbol). However, it seems to me that it can never take less energy on north pole than in open space... And im pretty sure someone (i think it as you) said it would take 1050 m/s less from Kerbin than from open space. Or maybe that was you still assuming i was talking about LKO not on north pole...

[GoSlash27]

If I understand your current question, my math says that it is not more efficient at any distance to launch from Kerbin's surface at the north pole than it is to do so from a spot on Kerbin's orbit away from it's SOI.

If you got the idea otherwise from me, it was a misunderstanding, as I was referring to LKO.

And I will take this opportunity to stop responding to your posts.

Cheers,

-Slashy

[Caelib]

Here is what I would like to see added - in the little info/"wiki" popup in-game (the one on the right-side of the map view with the "i" icon) when you have selected a planet, it would be awesome if it would show phase-angle for intercept and "launch window" countdown so you could time your launch correctly.

[kzauner]

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?

Without that SOI-boundaries you will reach relative velocity of close to 0 m/s

at an infinite distance. To be precise: you will never be able to 'escape' the

gravitational field of any body as the later is infinite.

But your question was of theoretical one, as the universe is a) finite and

expanding and hencefore messing up the situation anyway.

HTH

Clemens.

[arkie87]

That said (back to the discussion), it seems to me that being static on north pole must take more energy than floating in space since both must overcome Kerbol gravitational well, so those cancel, and all that is left is for kerbal on north pole to overcome Kerbin gravity well. I could understand how they could be close, especially when Kerbol gravitational well is much larger than Kerbin's such that not much energy/dV is lost upon change of SOI (from Kerbin to Kerbol). However, it seems to me that it can never take less energy on north pole than in open space... And im pretty sure someone (i think it as you) said it would take 1050 m/s less from Kerbin than from open space. Or maybe that was you still assuming i was talking about LKO not on north pole...

The original (incorrect) statement i made (6 pages ago) is:

It doesnt make sense that the differences in deltaV's between transfer from LKO-->Jool directly and a two-step burn from LKO-->Kerbol and then Kerbol-->Jool could be larger than the burn from LKO-->Kerbol.

This is clearly wrong since LKO has kinetic energy which allows Oberth effect to be used.

Thus, i revise my statement:

It doesnt make sense that the difference in deltaV's between transfer from North pole-->Jool directly and a two step burn from North pole-> Kerbol and then kerbol-->Jool could be larger than the burn from North-pole-->Kerbol.

I havent checked the charts yet, but i imagine, this is the lower limit on efficiency, and upper limit on wasted energy.

[arkie87]

Without that SOI-boundaries you will reach relative velocity of close to 0 m/s

at an infinite distance. To be precise: you will never be able to 'escape' the

gravitational field of any body as the later is infinite.

But your question was of theoretical one, as the universe is a) finite and

expanding and hencefore messing up the situation anyway.

HTH

Clemens.

This is incorrect. Even though gravity continues to act even infinite distance away, there is an upper limit (finite) on the amount of gravitational energy than can be absorbed. This upper limit can be calculated analytically using calculus (integration).

In fact, the velocity far from the target is non zero and can be calculated from:

V_inf = sqrt(V^2-V_escape^2)

[arkie87]

If I understand your current question, my math says that it is not more efficient at any distance to launch from Kerbin's surface at the north pole than it is to do so from a spot on Kerbin's orbit away from it's SOI.

If you got the idea otherwise from me, it was a misunderstanding, as I was referring to LKO.

And I will take this opportunity to stop responding to your posts.

Cheers,

-Slashy

That's what i thought and have been trying to clarify this entire time.

I think the upper limit on the wasted energy/dV for the two burn rather than direct transfer is the amount of energy wasted transferring from north pole to Kerbol...

Edited November 11, 2014 by arkie87

[arkie87]

I think the upper limit on the wasted energy/dV for the two burn rather than direct transfer is the amount of energy wasted transferring from north pole to Kerbol...

My rationalization/proof is as follows:

(1): dKE(North-pole-->Jool) >= dKE(Kerbol-->Jool)

(2): dKE(North-pole-->Kerbol,Kerbol-->Jool) = dKE(North-Pole-->Kerbol) + dKE(Kerbol-->Jool)

Thus substituting (2) into (1):

Wasted dV = dKE(North-pole-->Kerbol,Kerbol-->Jool) - dKE(North-pole-->Jool) <= dKE(North-Pole-->Kerbol)