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

[GoSlash27]

I understand what Oberth effect is. But i still maintain that these charts make no sense:

Consider a spacecraft in the same orbit as kerbin but on opposite side of Kerbol such that it isnt in Kerbin SOI. It would take X deltaV to get from there to Jool.

That value is roughly equivalent to the amount of deltaV required for a spaceship that just barely left Kerbin SOI to get to Jool (maybe larger, since eccentricity might be aiming the right direction so some of the deltaV wont be wasted).

Thus, the maximum wasted energy is the ~1 km/s wasted deltaV needed to just barely get out of Kerbin SOI. The oberth effect allows you to double count your transfer burn and Kerbin SOI burn...

Can anyone explain why this is wrong without a generic "oberth effect rah rah rah"?

I think we're trying to

The charts aren't incorrect, your assumptions about how engines convert chemical energy into kinetic energy are incorrect.

The reason it's cheaper to hit a transfer from low orbit (and even cheaper from the pad) is "oberth rah rah rah". It's not a matter of "double- counting your burn", but rather the engines being more efficient when they're moving faster. They spend less of their chemical energy spitting exhaust out the back and more of it moving the vehicle forward. Your vehicle is moving faster at low orbit than high orbit, so your engines are more efficient down there. That's the Oberth effect.

Best,

-Slashy

[Kamuchi]

Wouldn`t it then be more efficient to start the trajectory from the Mun or Minmus (if you have a refueling base there) to plunge towards Kerbin the fastest speed you can get without losing the gravitational slingshot trajectory by re-entry and do the burn at PE?

I suck at science and that`s most likely why I always have to much fuel, or run out short

[arkie87]

I think we're trying to

The charts aren't incorrect, your assumptions about how engines convert chemical energy into kinetic energy are incorrect.

The reason it's cheaper to hit a transfer from low orbit (and even cheaper from the pad) is "oberth rah rah rah". It's not a matter of "double- counting your burn", but rather the engines being more efficient when they're moving faster. They spend less of their chemical energy spitting exhaust out the back and more of it moving the vehicle forward. Your vehicle is moving faster at low orbit than high orbit, so your engines are more efficient down there. That's the Oberth effect.

Best,

-Slashy

I appreciate you trying to, but its not answering my question (its just repeating about the Oberth effect).

The oberth effect is:

Let's say it takes 1 km/s to escape a planets SOI from low orbit. If you burn 1 km/s in the right direction, you will end up with 0 m/s velocity. If, instead, you burn 1.1 km/s, you will have more than 100 m/s velocity left over since gravitational wells require a certain amount of energy to escape them rather than deltaV. In fact, you will have (1100^2-1000^2)^(1/2) = ~ 458 m/s left over (an additional 358 m/s). THIS is the Oberth effect. That additional velocity can be used towards raising apoapsis towards Duna etc... In addition, if you were to burn 10 km/s instead of 1.1 km/s, you would escape SOI with 9,950 m/s and as the burn approaches infinity, the dV required to escape SOI approaches zero... However, you will notice that the velocity you have upon escaping SOI is never larger than the initial burn (obviously) and the velocity difference is never larger than the amount of dV needed to escape SOI i.e. 1 km/s....

That said, please explain the following paradox:

I understand what Oberth effect is. But i still maintain that these charts make no sense:

Consider a spacecraft in the same orbit as kerbin but on opposite side of Kerbol such that it isnt in Kerbin SOI. It would take X deltaV to get from there to Jool.

That value is roughly equivalent to the amount of deltaV required for a spaceship that just barely left Kerbin SOI to get to Jool (maybe larger, since eccentricity might be aiming the right direction so some of the deltaV wont be wasted).

Thus, the maximum wasted energy is the ~1 km/s wasted deltaV needed to just barely get out of Kerbin SOI. The oberth effect allows you to double count your transfer burn and Kerbin SOI burn...

Can anyone explain why this is wrong without a generic "oberth effect rah rah rah"?

[Concentric]

I understand what Oberth effect is. But i still maintain that these charts make no sense:

Consider a spacecraft in the same orbit as kerbin but on opposite side of Kerbol such that it isnt in Kerbin SOI. It would take X deltaV to get from there to Jool.

That value is roughly equivalent to the amount of deltaV required for a spaceship that just barely left Kerbin SOI to get to Jool (maybe larger, since eccentricity might be aiming the right direction so some of the deltaV wont be wasted).

Thus, the maximum wasted energy is the ~1 km/s wasted deltaV needed to just barely get out of Kerbin SOI. The oberth effect allows you to double count your transfer burn and Kerbin SOI burn...

Can anyone explain why this is wrong without a generic "oberth effect rah rah rah"?

Okay. Bringing up the Hop's Blog link I edited in earlier...

Basically, with the spacecraft on the opposite side, you're producing the entirety of your kinetic energy by thrusting, many times your escape requirement (2750m/s rather than 950m/s). However, close in, in Kerbin orbit, you can take advantage of Kerbin's v_inf and your orbital speed, needing only to increase your escape by (in this case) 965m/s over the 950m/s to increase your specific orbital energy by the same amount. What you're doing is taking advantage of the squaring: (u+v)² = (u² + v²) + 2uv. The bolded term is what's making the difference. With your existing escape velocity u, you don't need nearly as high a v to get the same energy if you add before squaring rather than after - which means doing the burn at the same time as you do the escape burn. Additionally, as you leave Kerbin's SoI faster, you lose less of your orbital velocity to Kerbin's gravity, so you make more use of your escape burn.

The Hop's Blog link is talking about Earth-Mars relationship, specifically figuring out the required difference over your escape burn to perform the transfer if you perform them simultaneously. It's much, much smaller than the burn that would be needed if performed after you barely escaped, precisely because of this root-of-sum-of-squares - the critical difference being when you sum and when you square, which is entirely about when you burn.

Wouldn`t it then be more efficient to start the trajectory from the Mun or Minmus (if you have a refueling base there) to plunge towards Kerbin the fastest speed you can get without losing the gravitational slingshot trajectory by re-entry and do the burn at PE?

I suck at science and that`s most likely why I always have to much fuel, or run out short

What you're talking about there is the two-burn method. A quick look seems to say that that's more effective in some situations, but not necessarily all the time. Additionally, you'd have to get your rocket up to whatever starting point in the first place, which may cost you more overall.

Edited November 11, 2014 by Concentric

[arkie87]

Another way of wording my question/problem/confusion:

A thought experiment. Which satallite will take less deltaV to get to Jool?

(a) A satallite in LKO

( A satallite in the same orbit as Kerbin, but on the opposite side of Kerbol (thus, same orbit as Kerbin, but not in Kerbin SOI)

I maintain answer has to be B since A first must escape Kerbin SOI. If I am wrong, why?

Please explain in detail, not just quoting Wikipedia articles on Oberth Effect etc...

[arkie87]

Okay. Bringing up the Hop's Blog link I edited in earlier...

Basically, with the spacecraft on the opposite side, you're producing the entirety of your kinetic energy by thrusting, many times your escape requirement (2750m/s rather than 950m/s). However, close in, in Kerbin orbit, you can take advantage of Kerbin's v_inf and your orbital speed, needing only to increase your escape by (in this case) 965m/s over the 950m/s to increase your specific orbital energy by the same amount. What you're doing is taking advantage of the squaring: (u+v)² = (u² + v²) + 2uv. The bolded term is what's making the difference. With your existing escape velocity u, you don't need nearly as high a v to get the same energy if you add before squaring rather than after - which means doing the burn at the same time as you do the escape burn. Additionally, as you leave Kerbin's SoI faster, you lose less of your orbital velocity to Kerbin's gravity, so you make more use of your escape burn.

The Hop's Blog link is talking about Earth-Mars relationship, specifically figuring out the required difference over your escape burn to perform the transfer if you perform them simultaneously. It's much, much smaller than the burn that would be needed if performed after you barely escaped, precisely because of this root-of-sum-of-squares - the critical difference being when you sum and when you square, which is entirely about when you burn.

What you're talking about there is the two-burn method. A quick look seems to say that that's more effective in some situations, but not necessarily all the time. Additionally, you'd have to get your rocket up to whatever starting point in the first place, which may cost you more overall.

Yes, thank you.

My mistake was saying it only takes 1 km/s to escape kerbin SOI from LKO.

In fact, it takes more, but you already have orbital velocity (2.3 km/s), so therein lies the problem.

Thus, in my post above, ( is the correct answer as long as the spaceship doesnt start in LKO but rather on the surface of Kerbin.

Otherwise, if it is in LKO, by burning 2 km/s, it can escape Kerbin SOI with (4.3^2-3.3^2)^0.5 = 2.76 km/s (which, incidentally, is more than it burned).

Thanks again.

Edited November 11, 2014 by arkie87

[Concentric]

Another way of wording my question/problem/confusion:

A thought experiment. Which satallite will take less deltaV to get to Jool?

(a) A satallite in LKO

( A satallite in the same orbit as Kerbin, but on the opposite side of Kerbol (thus, same orbit as Kerbin, but not in Kerbin SOI)

I maintain answer has to be B since A first must escape Kerbin SOI. If I am wrong, why?

Please explain in detail, not just quoting Wikipedia articles on Oberth Effect etc...

You are technically correct. The worst kind of correct, because to get to where it is, ( has already expended escape delta-v.

So we have our satellites, (a) and (. Lets call Kerbin's orbital velocity v_K, the speed of the Hohmann perihelion from Kerbin to Jool v_H. Thus, satellite (, which benefits not at all from any help, must increase its velocity by v_H - v_K.

Close to Kerbin, the Hohmann transfer orbit looks like a hyperbola. Its v_inf = v_H - v_K, as before. (a) is in a circular orbit, with velocity v_orb, and v_esc the escape velocity at that altitude. To just escape, (a) needs to increase velocity by v_esc - v_orb. However, we are performing the transfer also, so (a) needs to increase velocity by v_hyp - v_orb, where v_hyp² = v_esc² + v_inf².

So, the delta-v expenditure for (a) to get to Jool encounter from LKO is v_hyp - v_orb. While this is (possibly*) larger than v_inf, what ( spends to get from Kerbin-height orbit to Jool, if you factor in the cost of escape that ( has already paid (you didn't build it there, after all), you get v_hyp - v_esc, considerably smaller than v_inf. About a third of the size, in this case.

Edit: Where the Oberth Effect comes in is as stated by Slashy below: it's about the fuel consumption. At higher speeds you're getting more kinetic energy out of your fuel - for one thing, you can get a hold of the kinetic energy of the fuel itself as well as its chemical energy, and for another, a greater proportion is converted to kinetic energy rather than waste heat or light. Delta-v scales with the logarithm of the fuel mass ratio, so we can talk about delta-v requirements, instead of energy level differences. The effect on specific orbital energy is the same whether you escape first and burn out in Kerbol orbit or do the burns at the same time, it's just that you use less fuel the second way as your engines are working more efficiently (Oberth), so the delta-v is lower.

Edit*: v_inf² = (v_hyp - v_esc)(v_hyp + v_esc). So, v_inf is the geometric mean of v_hyp - v_esc, v_hyp + v_esc, while v_hyp is the arithmetic mean, slightly larger. If v_orb is larger than this difference, (a) actually expends less fuel from LKO than ( does from its position.

Edited November 11, 2014 by Concentric
Small correction - (a) can do less than (b)

[GoSlash27]

The oberth effect is:

Let's say it takes 1 km/s to escape a planets SOI from low orbit. If you burn 1 km/s in the right direction, you will end up with 0 m/s velocity. If, instead, you burn 1.1 km/s, you will have more than 100 m/s velocity left over since gravitational wells require a certain amount of energy to escape them rather than deltaV. In fact, you will have (1100^2-1000^2)^(1/2) = ~ 458 m/s left over (an additional 358 m/s). THIS is the Oberth effect. That additional velocity can be used towards raising apoapsis towards Duna etc... In addition, if you were to burn 10 km/s instead of 1.1 km/s, you would escape SOI with 9,950 m/s and as the burn approaches infinity, the dV required to escape SOI approaches zero... However, you will notice that the velocity you have upon escaping SOI is never larger than the initial burn (obviously) and the velocity difference is never larger than the amount of dV needed to escape SOI i.e. 1 km/s....

No it's not. You *think* you understand the Oberth effect, but you don't. I think that's the source of your confusion.

Kinetic energy (which is what gets you places) is not measured in m/sec, it's measured in Joules.

Kinetic energy is (M*V^2)/2.

If you add 100 M/sec when you're going 2.3 Km/ sec, Your gain in kinetic energy is a whole heckuva lot more than if you add it when you're doing zero.

Say you have a 1 T mass at rest. Kinetic energy zero. You burn for 100M/sec. Your final kinetic energy is 5 megajoules, so that's what you've added. Now say you do the same at an initial velocity of 2.3KM/sec. Starting Ek 2.645 GJ. Add a hundred M/sec and your new Ek is 2.880 GJ. Now you've added 235 MJ... with the same burn.

It's because a rocket expends part of it's chemical energy spitting exhaust out the back. The faster you're going, the more of your chemical energy gets converted into kinetic energy instead of fire.

Given that this is what's *actually* going on rather than what you've assumed, your follow-up question is based on a false assumption and therefore immaterial.

Best,

-Slashy

Edited November 11, 2014 by GoSlash27

[boxman]

I am sure it would be easier for newbies, but what fun would it be if most of the challenge is gone??

And you dont need to know any math at all. I just learned from trial and practice and after a few visit to a planet I ended up just memorizing the phase angle.

I think the easist way for a newbie to learn to do this manually is by doing it from a solar orbit and by trying different angles while moving the node around the orbit until you get a good encounter. It does not really take many trips before you just know where the node should be placed.

[Kamuchi]

What you're talking about there is the two-burn method. A quick look seems to say that that's more effective in some situations, but not necessarily all the time. Additionally, you'd have to get your rocket up to whatever starting point in the first place, which may cost you more overall.

True to that, but if you would use an orbiter stage, then in theory it would benefit most of the time and if launching from a low grav moon like Minmus, that stage could be tiny compared to a Kerbin launch while having maximum slingshot speed.

Just escaping Kerbin`s orbit takes 800dV :/

Ofcourse timing is another issue

[arkie87]

You are technically correct. The worst kind of correct, because to get to where it is, ( has already expended escape delta-v.

So we have our satellites, (a) and (. Lets call Kerbin's orbital velocity v_K, the speed of the Hohmann perihelion from Kerbin to Jool v_H. Thus, satellite (, which benefits not at all from any help, must increase its velocity by v_H - v_K.

Close to Kerbin, the Hohmann transfer orbit looks like a hyperbola. Its v_inf = v_H - v_K, as before. (a) is in a circular orbit, with velocity v_orb, and v_esc the escape velocity at that altitude. To just escape, (a) needs to increase velocity by v_esc - v_orb. However, we are performing the transfer also, so (a) needs to increase velocity by v_hyp - v_orb, where v_hyp² = v_esc² + v_inf².

So, the delta-v expenditure for (a) to get to Jool encounter from LKO is v_hyp - v_orb. While this is larger than v_inf, what ( spends to get from Kerbin-height orbit to Jool, if you factor in the cost of escape that ( has already paid (you didn't build it there, after all), you get v_hyp - v_esc, considerably smaller than v_inf. About a third of the size, in this case.

Edit: Where the Oberth Effect comes in is as stated by Slashy below: it's about the fuel consumption. At higher speeds you're getting more kinetic energy out of your fuel (and not wasting it as heat and light). Delta-v scales with the logarithm of the fuel mass ratio, so we can talk about delta-v requirements, instead of energy level differences. The effect on specific orbital energy is the same whether you escape first and burn out in Kerbol orbit or do the burns at the same time, it's just that you use less fuel the second way as your engines are working more efficiently (Oberth), so the delta-v is lower.

I meant the question as posed. It was a thought experiment, so for the question's sake, satellite ( originated where it was, and we do not need to factor in how much energy/dV it took to get there.

Not sure what the rest of the post was adding though...

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No it's not. You *think* you understand the Oberth effect, but you don't. I think that's the source of your confusion.

Kinetic energy (which is what gets you places) is not measured in m/sec, it's measured in Joules.

Kinetic energy is (M*V^2)/2.

If you add 100 M/sec when you're going 2.3 Km/ sec, Your gain in kinetic energy is a whole heckuva lot more than if you add it when you're doing zero.

Say you have a 1 T mass at rest. Kinetic energy zero. You burn for 100M/sec. Your final kinetic energy is 5 megajoules, so that's what you've added. Now say you do the same at an initial velocity of 2.3KM/sec. Starting Ek 2.645 GJ. Add a hundred M/sec and your new Ek is 2.880 GJ. Now you've added 235 MJ... with the same burn.

It's because a rocket expends part of it's chemical energy spitting exhaust out the back. The faster you're going, the more of your chemical energy gets converted into kinetic energy instead of fire.

Given that this is what's *actually* going on rather than what you've assumed, your follow-up question is based on a false assumption and therefore immaterial.

Best,

-Slashy

Please explain to me how what i said about the oberth effect is incorrect. In fact, I am saying exactly what you are saying, but for some reason you think i am wrong. We are approaching the Oberth effect from two different (phase :-P ) angles....

[GoSlash27]

Another way of wording my question/problem/confusion:

A thought experiment. Which satallite will take less deltaV to get to Jool?

(a) A satallite in LKO

( A satallite in the same orbit as Kerbin, but on the opposite side of Kerbol (thus, same orbit as Kerbin, but not in Kerbin SOI)

I maintain answer has to be B since A first must escape Kerbin SOI. If I am wrong, why?

Please explain in detail, not just quoting Wikipedia articles on Oberth Effect etc...

Shockingly, it's 1.1Km/sec cheaper to make the trip to Jool from the bottom of the gravity well at Kerbin than it is to do it from the opposite side of Kerbol. The reason is the extra kinetic energy you have at the time of burn from whipping around Kerbin, which makes your conversion of DV into Ek more efficient.

[GoSlash27]

Please explain to me how what i said about the oberth effect is incorrect. In fact, I am saying exactly what you are saying, but for some reason you think i am wrong. We are approaching the Oberth effect from two different (phase :-P ) angles....

What you're saying is *completely* different from what I'm saying, which is why I get it and you don't.

Transfers are kinetic energy. They're not velocity.

Your explanation of the Oberth effect didn't mention kinetic energy at all, and that's how it works.

Best,

-Slashy

Edited November 11, 2014 by GoSlash27

[Concentric]

I meant the question as posed. It was a thought experiment, so for the question's sake, satellite ( originated where it was, and we do not need to factor in how much energy/dV it took to get there.

Not sure what the rest of the post was adding though...

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Please explain to me how what i said about the oberth effect is incorrect. In fact, I am saying exactly what you are saying, but for some reason you think i am wrong. We are approaching the Oberth effect from two different (phase :-P ) angles....

In the thought experiment, ( is there ex nihilo, but this is misleading when making the comparison with (a). Certainly it's easier to move to an orbit when already orbiting the same thing and meeting that orbit at a single point than it is to go from elsewhere. Just as it's easier to go to Mun from Kerbin high orbit than it is from Kerbin's surface. That doesn't mean that going to HKO and from there to Mun is in any way a good option compared to going to LKO and then to Mun when launching from the surface. That's why I said "You are technically correct - the worst kind of correct", because while it is true, it is misleading. A better comparison is between ( and ©, a probe just barely escaping from Kerbin, at its periapse at LKO height, also where it is ex-nihilo. If left alone, © will move into the same orbit as (, just a different position.

From the escaping situation, much less fuel (and thus delta-v) is needed to get the same kinetic energy than in the orbiting situation, as you are travelling more quickly. That kinetic energy is the amount needed to change the Kerbin-height orbit into the transfer orbit, and is so constant between ( and ©. So, © needs less fuel to do the same job, i.e. less delta-v. But if left to escape first, © would need the same delta-v as (.

Edit: Correction coming made, see here. In short, (a) has a lower requirement than ( after all.

Edited November 11, 2014 by Concentric

[arkie87]

What you're saying is *completely* different from what I'm saying, which is why I get it and you don't.

Transfers are kinetic energy. They're not velocity.

Your explanation of the Oberth effect didn't mention kinetic energy at all, and that's how it works.

Best,

-Slashy

My explanation used the kinetic energy formula to calculate velocity after leaving gravity well... was that not clear? or do you just know the term "kinetic energy" but not know what it means or how to calculate it???

[arkie87]

In the thought experiment, ( is there ex nihilo, but this is misleading when making the comparison with (a). Certainly it's easier to move to an orbit when already orbiting the same thing and meeting that orbit at a single point than it is to go from elsewhere. Just as it's easier to go to Mun from Kerbin high orbit than it is from Kerbin's surface. That doesn't mean that going to HKO and from there to Mun is in any way a good option compared to going to LKO and then to Mun when launching from the surface. That's why I said "You are technically correct - the worst kind of correct", because while it is true, it is misleading. A better comparison is between ( and ©, a probe just barely escaping from Kerbin, at its periapse at LKO height, also where it is ex-nihilo. If left alone, © will move into the same orbit as (, just a different position.

From the escaping situation, much less fuel (and thus delta-v) is needed to get the same kinetic energy than in the orbiting situation, as you are travelling more quickly. That kinetic energy is the amount needed to change the Kerbin-height orbit into the transfer orbit, and is so constant between ( and ©. So, © needs less fuel to do the same job, i.e. less delta-v. But if left to escape first, © would need the same delta-v as (.

I understand. I wasnt trying to be "technically correct" (i think its stupid when people try to be technically correct) and i understand why you think its misleading. I wasnt suggesting its better to do ( and then go to Jool.

My question arose from a thought experiment i.e. it doesnt make sense that to get from (a) to jool takes less energy than (, since (a) is inside two gravitational wells (the same as ('s but also in Kerbin's) and should have to expend more energy to get out of those wells. Do you understand where i am coming from?

However, it seems like GoSlash27 thinks we are both wrong, and that actually, it takes less dV/energy even from Kerbin's surface. I'd like to see the calculations on that....

[arkie87]

Shockingly, it's 1.1Km/sec cheaper to make the trip to Jool from the bottom of the gravity well at Kerbin than it is to do it from the opposite side of Kerbol. The reason is the extra kinetic energy you have at the time of burn from whipping around Kerbin, which makes your conversion of DV into Ek more efficient.

I'd like to see the math, please since i think this answer goes against intuition (as you implied with "shockingly").

[Concentric]

Edit*: v_inf² = (v_hyp - v_esc)(v_hyp + v_esc). So, v_inf is the geometric mean of v_hyp - v_esc, v_hyp + v_esc, while v_hyp is the arithmetic mean, slightly larger. If v_orb is larger than this difference, (a) actually expends less fuel from LKO than ( does from its position.

Noticed a mistake in my earlier post - it is possible that (a) has a lower delta-v requirement than ( even from the positions in the thought experiment - in fact it is so in this case. I forgot to factor in subtracting the v_orb from v_hyp when comparing. With sufficiently large v_orb - which isn't too much, as the difference between the means here is at most v_esc over the square root of 2 ^{[Corollary 1]} - (a)'s delta-v requirement drops below v_inf.

In this case, v_esc/sqrt(2) = 2430 m/s to 3 significant figures, and v_orb = 2480 m/s to 3 significant figures (obtained by subtracting LKO-escape delta-v from escape velocity, as I couldn't find an LKO orbital velocity figure), so (a) has a lower delta-v requirement than (. In fact, as this is independent of destination, the fact that (a) requires less delta-v than ( is also independent of destination.

Edited November 11, 2014 by Concentric

[Superfluous J]

Hey thanks for pimping my video

One thing important to note, though:

They are fragile in the sense that if you click the node, it will reset, at least according to 5thHorsemen. Regardless, i think everyone agrees a built in tool would be ideal.

Clicking the node won't ruin it. DRAGGING it is what ruins it. You can click on it and modify the grade/normal/radial tuggies, and you can click the +/- orbit buttons, but if you drag it one pixel left or right it'll snap to the current orbit.

[arkie87]

Noticed a mistake in my earlier post - it is possible that (a) has a lower delta-v requirement than ( even from the positions in the thought experiment - in fact it is so in this case. I forgot to factor in subtracting the v_orb from v_hyp when comparing. With sufficiently large v_orb - which isn't too much, as the difference between the means here is at most v_esc over the square root of 2 ^{[Corollary 1]} - (a)'s delta-v requirement drops below v_inf.

In this case, v_esc/sqrt(2) = 2430 m/s to 3 significant figures, and v_orb = 2480 m/s to 3 significant figures, so (a) has a lower delta-v requirement than (. In fact, as this is independent of destination, the fact that (a) requires less delta-v than ( is also independent of destination.

interesting... going to check the math, but GoSlash27 claimed 1.1 km/s not 50 m/s....

Also, thanks for having a mature, intellectual conversation with me, rather than trying to "outnerd" me :-)

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Hey thanks for pimping my video

One thing important to note, though:

Clicking the node won't ruin it. DRAGGING it is what ruins it. You can click on it and modify the grade/normal/radial tuggies, and you can click the +/- orbit buttons, but if you drag it one pixel left or right it'll snap to the current orbit.

Oops. Didn't mean to put words in the Horse's mouth :-P

[GoSlash27]

I'd like to see the math, please since i think this answer goes against intuition (as you implied with "shockingly").

I've already shown you the math. 1/2M*(V2)^2- 1/2M*(V1)^2.

Look... all you have to do is answer your thought experiment in practice. You have a suitable simulator right here. Put a ship in LKO and another in Kerbin's orbit away from Kerbol and try it yourself.

It will tell you the same thing I'm telling you and I've already told you why it is so.

If you stop thinking of transfers in terms of "DV" and start thinking of them in terms of "DEk" it will all make sense.

Velocity <> kinetic energy.

Best,

-Slashy

Edited November 11, 2014 by GoSlash27

[Concentric]

interesting... going to check the math, but GoSlash27 claimed 1.1 km/s not 50 m/s....

Also, thanks for having a mature, intellectual conversation with me, rather than trying to "outnerd" me :-)

That 50m/s is a lower bound on the improvement, not an exact figure. It's just that it's independent of v_inf and v_hyp, so it doesn't really matter where you're going. Jool, Duna, Moho... Given a specific destination, particularly further away, the difference increases.

[arkie87]

I've already shown you the math. 1/2M*(V2)^2- 1/2M*(V1)^2.

Look... all you have to do is answer your thought experiment in practice. You have a suitable simulator right here. Put a ship in LKO and another in Kerbin's orbit away from Kerbol and try it yourself.

It will tell you the same thing I'm telling you and I've already told you why it is so.

If you stop thinking of transfers in terms of "DV" and start thinking of them in terms of "DEk" it will all make sense.

Velocity <> kinetic energy.

Best,

-Slashy

I am well aware of this. You seem to not be reading what i am writing, and instead, just attacking my perceived lack of understanding.

EDIT: I am also less interested in what the correct answer is, and more interested in the "why", so telling me to check it for myself using KSP since, after all KSP is a orbit simulator is not a satisfactory answer for me. Also, when explaining why, i appreciate detailed responses that explain/hone in on the misunderstanding/cause of confusion/error, rather than telling me vaguely i dont understand the oberth effect or the difference between kinetic energy and dV.

Edited November 11, 2014 by arkie87

[arkie87]

That 50m/s is a lower bound on the improvement, not an exact figure. It's just that it's independent of v_inf and v_hyp, so it doesn't really matter where you're going. Jool, Duna, Moho... Given a specific destination, particularly further away, the difference increases.

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 (?

[GoSlash27]

Also, thanks for having a mature, intellectual conversation with me, rather than trying to "outnerd" me :-)

*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