Why does it matter if I transfer to another planet directly from Kerbin orbit?

[Bane1998]

So given my frustration with dealing with maneuver nodes, I've been wondering why all the guides for going to Duna or anywhere talk about building your node directly from Kerbin orbit. It seems to me, that it ought to be the same delta-V planning the route directly from Kerbin, as it would doing an escape from Kerbin, then doing the transfer after the escape.

My thought process is, no matter what, you have to escape Kerbin. In fact, it doesn't even matter which direction you escape it in. You burn just prograde until you just barely get an escape. If you only burn exactly that far you resulting orbit should be almost identical to Kerbin's orbit around the Sun. From there you plan your transfer, and do another burn using the more convenient solar-level maneuver nodes.

My brain tells me this SHOULD be equivalent. There will be some Delta-V negligible losses if I overshoot my Kerbin escape maybe, or escape via Kerbin's retrograde when I need a planetary prograde burn, but the first step of going anywhere is to escape Kerbin, so why not just do that first?

One could argue time perhaps. If you work out the phase and such ahead of time, you wouldn't even have to sit around in space waiting for it, but most guides have you sitting in Kerbin orbit anyway. Why does it matter if my Kerbals are sitting in a convenient Solar orbit nearly identical to Kerbin orbit, or sitting in a Kerbin orbit when I start time warping to get the phase?

Then again, I figure I must be wrong, and would be awesome to have someone explain why I am, because I'm not hotting the Delta-V targets on this map at all:

http://wiki.kerbalspaceprogram.com/w/images/7/73/KerbinDeltaVMap.png

My Kerbin escape is around 1000 Delta-V, which is close to the map, but my Duna intercept is another 1000 Delta-V, and the map says that's 100.

This is hard for me to test, as I get way too frustrated trying to intercept a planet directly from Kerbin to actually see if there's a difference.

Anyway, thanks ahead of time!

Edited January 2, 2014 by Bane1998
Seems to be answered

[Dave Kerbin]

The reason is the Oberth effect.

[Woopert]

I still need to do more research on the Oberth effect, thanks for saying that. I have known about it and all but still need to learn how it works.

I usually do transfers from Kerbin orbit anyways because, well, that's how I learned to play KSP.

[Eric S]

Dave is correct. To give a bit more of an explanation, it basically comes down to this. Consider the baseline being how you do this. Velocity X is just enough velocity for you to coast to the edge of the SoI. If you increase this velocity slightly, say X+Y, then because of the way the energy is conserved, when you reach the SoI, you haven't lost all of your X speed this time, so you're moving faster than Y when you reach the SoI.

You can imagine this "free" velocity coming from one of two places. First, the actual "correct" explanation is that the amount of kinetic energy an object has is 1/2*(mass of object)*(velocity squared). Escaping the gravity well costs a fixed amount of energy. So, if it takes x velocity to escape and we add one extra m/s, we have 1/2*mass*((x+1)squared) energy. We extract the energy of x, and this leaves us with 1/2*mass*(2x+1) energy, which as long as x is larger than 0, means that we have more velocity than the "extra" amount we put in. The larger x is (meaning the gravity well is stronger), the more "free" velocity you gained.

If looking at it from a mathmatical/physics point of view is too much, then think of it this way. Think of x as escape velocity, the amount of acceleration backwards the gravity well is going to produce on the craft while the craft moves to the SoI boundary, assuming it hits zero velocity right at the SoI boundary. Now, if you move faster, gravity has less time to slow you down, so you get slowed down by a total velocity of less than x, which means that in addition to keeping the extra velocity, you get to keep some of the base x velocity as well.

The implications of the Oberth effect are all over the place in this game, and knowing about it and its implications will help a lot. For example, this also explains why a higher orbit is actually less efficient for achieving a planetary transfer from, because a higher orbit will have a smaller x value.

Edited January 1, 2014 by Eric S

[1greywind]

"it doesn't even matter which direction you escape it in." - You are not correct on this point.

Your velocity around Kerbol is summ of Kerbin orbital velocity and your orbital velocity around Kerbin. Vectors summ is dependent on vectors orientations. So kerbolocentric orbits that you achive then you just "burn prograde" will be diffirent if you starting points are diffirent. Even if starting points are on same perfectly circular Kerbinocentic orbit and burn dV is the same.

dV map gives your values calculated for instant burns, that will place you on Hohmann transfer orbit. If you first escape Kerbin SOI in some random direciton using value from dV map, your will not be on correct transfer orbit. So you will need additional dV to change your orbit into correct one. Oberth effect is not an issue there.

Edited January 1, 2014 by 1greywind

[Bane1998]

Dave is correct. To give a bit more of an explanation, it basically comes down to this. Consider the baseline being how you do this. Velocity X is just enough velocity for you to coast to the edge of the SoI. If you increase this velocity slightly, say X+Y, then because of the way the energy is conserved, when you reach the SoI, you haven't lost all of your X speed this time, so you're moving faster than Y when you reach the SoI.

You can imagine this "free" velocity coming from one of two places. First, the actual "correct" explanation is that the amount of kinetic energy an object has is 1/2*(mass of object)*(velocity squared). Escaping the gravity well costs a fixed amount of energy. So, if it takes x velocity to escape and we add one extra m/s, we have 1/2*mass*((x+1)squared) energy. We extract the energy of x, and this leaves us with 1/2*mass*(2x+1) energy, which as long as x is larger than 0, means that we have more velocity than the "extra" amount we put in. The larger x is (meaning the gravity well is stronger), the more "free" velocity you gained.

For the record, I don't disagree with the effect existing, I just can't quite grok it intuitively. Which for me needs I try to somehow find that intuition. I guess my problem in understanding the Oberth effect is it's based on 'speed'

From Wikipedia:

In astronautics, the Oberth effect is where the use of a rocket engine when travelling at high speed generates more useful energy than one at low speed.

What I can't follow is... isn't speed relative? I have a speed relative to Kerbin, Kerbin has a speed relative to the Sun. Why I don't get the effect is escaping Kerbin just zeroes out my Kerbin speed, it doesn't zero out my Sun speed which is much higher, almost to the point that it almost seems the Kerbin speed is a minor point when summing it all up. Yet somehow it seems it's not. That's the voodoo part of it, to me. You could choose an arbitrary reference frame and say that I'm going 'fast' in that reference frame, and will gain a magic Oberth effect from that unknown reference frame.

If looking at it from a mathmatical/physics point of view is too much, then think of it this way. Think of x as escape velocity, the amount of acceleration backwards the gravity well is going to produce on the craft while the craft moves to the SoI boundary, assuming it hits zero velocity right at the SoI boundary. Now, if you move faster, gravity has less time to slow you down, so you get slowed down by a total velocity of less than x, which means that in addition to keeping the extra velocity, you get to keep some of the base x velocity as well.

This explanation makes sense to me. The faster I escape Kerbin, the less time Kerbin has dragging me back towards it, so the less velocity I lose on my way out. But, is that the Oberth effect? The effect seems to have nothing to do with gravity technically, so gaining an understanding of it in this way seems like I might be understanding it incorrectly, still. And that worries me some.

The implications of the Oberth effect are all over the place in this game, and knowing about it and its implications will help a lot. For example, this also explains why a higher orbit is actually less efficient for achieving a planetary transfer from, because a higher orbit will have a smaller x value.

Does this mean the dV map I linked in the OP is not quite correct? Going to Kerbin from Duna is more expensive than going to Duna from Kerbin? Or is this back to the magic voodoo gains from exiting the Duna system at high 'Duna speed' somehow effecting my solar speed?

[Bane1998]

"it doesn't even matter which direction you escape it in." - You are not correct on this point.

Your velocity around Kerbol is summ of Kerbin orbital velocity and your orbital velocity around Kerbin. Vectors summ is dependent on vectors orientations. So kerbolocentric orbits that you achive then you just "burn prograde" will be diffirent if you starting points are diffirent. Even if starting points are on same perfectly circular Kerbinocentic orbit and burn dV is the same.

dV map gives your values calculated for instant burns, that will place you on Hohmann transfer orbit. If you first escape Kerbin SOI in some random direciton using value from dV map, your will not be on correct transfer orbit. So you will need additional dV to change your orbit into correct one. Oberth effect is not an issue there.

I think I might disagree with this in at least principle. If my goal (as it is before trying to understand the Oberth effect better) is to escape Kerbin, and join Kerbin in an orbit around the Sun as close to Kerbin's orbit as I can, then I don't think it matters which way I escape, if I can perfect my burn so that I stop the very instant I achieve escape velocity. Or at least, not practically. I suppose in theory I have to have a different trajectory than the Kerbin system by some infinitesimally small amount, but the direction I escape will be not even worth the bother of choosing the 'correct' direction. You can see this in KSP if you plan a burn just up to escape and not a single dV further, you basically have the same orbit as Kerbin, and it doesn't matter which way you chose to exit. At least up to as good as you are timing your burn to stop the moment you have escape trajectory.

This of course is only if your goal is to first just escape Kerbin, and then do a transfer to another planetary orbit as completely separate operations.

[Mobjack]

Here is the analogy I like to use. It uses some math but it is rather simple.

Let's assume you throw a ball straight up at 10 m/s. It will go about 5 meters high.

Now if you throw a ball straight up at 20 m/s, it will go about 20 meters in height. Assuming gravity is 10 m/s^2, for the first second, it will go 15 meters in the air. After one second, the ball will be traveling 10 m/s which will get an additional 5 meters.

So if you apply 20 m/s dV all at once it will make you go 20 meters in the sky. If you apply 10 m/s dV initially, wait a second, then apply an additional 10 m/s, it will only go 10 meters despite using a total of 20 m/s dV.

When you barely go outside Kerbin's SOI then burn again to reach Duna, it is like throwing the ball up, waiting for it to stop, then throwing it again. Instead of using two burns, if you did it all at once near Kerbin it will be more efficient.

Another way of looking at it is if you initially have your apoapsis at the edge of Kerbin's SOI leaving the system at 10 m/s. Burning an additional 100 m/s dV close to Kerbin could make you leave the SOI traveling an additional 200 m/s. Test it out yourself and see.

[Eric S]

What I can't follow is... isn't speed relative?

Yes, all velocity is relative to something (and gravity assists depend on this fact), and I don't have a precise answer to this, though I suspect it's because if nothing else, you'd need to pay attention to the energy in all the gravity wells you'd have to traverse to get from where you are to your point of reference. I never made it that far in physics in college before getting distracted by how much money my pre-graduate programming skills could earn me in the "real world."

This explanation makes sense to me. The faster I escape Kerbin, the less time Kerbin has dragging me back towards it, so the less velocity I lose on my way out. But, is that the Oberth effect? The effect seems to have nothing to do with gravity technically, so gaining an understanding of it in this way seems like I might be understanding it incorrectly, still. And that worries me some.

Here's the thing. If you ignore gravity and other external effects, the Oberth effect is just navel gazing. It's neat that speeding up from 20 m/s to 30 m/s gives you more energy than speeding up from 10 m/s to 20 m/s, but since it took the same amount of delta-v (the very definition of delta-v) and delta-v isn't affected by your velocity (assuming you're not in an atmosphere), that's all it is. The Oberth effect doesn't affect how far you can travel or how fast until those external forces come into play, and gravity is the most common external effect that we deal with. Atmospheric resistance is probably the only other one we deal with in this game as far as navigation is concerned.

So while the Oberth effect itself doesn't require the presence of gravity, it's kind of useless without it (which isn't the same as not existing without an external force). The Oberth effect is about energy, and gravity wells are about kinetic and potential energy (at any point in an orbit, the sum of your kinetic and potential energy should be the same value). The Oberth effect is about how much delta-v it takes to increase/decrease your kinetic energy, which in turn affects your total energy. However, that's drifting back towards the "correct" explanation of the Oberth affect that I described previously.

Does this mean the dV map I linked in the OP is not quite correct? Going to Kerbin from Duna is more expensive than going to Duna from Kerbin? Or is this back to the magic voodoo gains from exiting the Duna system at high 'Duna speed' somehow effecting my solar speed?

Not exactly, in both cases, the planet you're departing from is a more significant gravity well than the solar gravity well at the point you're departing from, so for purposes of escaping that gravity well, the solar gravity well isn't the important factor. In fact, because of the patched conics physics we work in, the solar gravity well doesn't even exist within Kerbin or Duna SoI (relative to the planet, that is), all that maters is the velocity vector you have relative to the Sun when you exit the planetary SoI and enter the Solar SoI. If you expend delta-v to accelerate inside the Solar SoI, then this would be a factor.

Edited January 2, 2014 by Eric S

[Superfluous J]

Here is the analogy I like to use. It uses some math but it is rather simple.

Thank you. I've been playing this game about half a year now and I finally understand the Olberth Effect

And actually, now that I think about it, it's the same idea behind the Suicide Burn only in reverse (and with more of a chance of exploding)

[LarryWallwart]

I think I might disagree with this in at least principle. If my goal (as it is before trying to understand the Oberth effect better) is to escape Kerbin, and join Kerbin in an orbit around the Sun as close to Kerbin's orbit as I can, then I don't think it matters which way I escape, if I can perfect my burn so that I stop the very instant I achieve escape velocity. Or at least, not practically. I suppose in theory I have to have a different trajectory than the Kerbin system by some infinitesimally small amount, but the direction I escape will be not even worth the bother of choosing the 'correct' direction. You can see this in KSP if you plan a burn just up to escape and not a single dV further, you basically have the same orbit as Kerbin, and it doesn't matter which way you chose to exit. At least up to as good as you are timing your burn to stop the moment you have escape trajectory.

This of course is only if your goal is to first just escape Kerbin, and then do a transfer to another planetary orbit as completely separate operations.

I believe once you crossed that plane of SOI, you would continue going farther away. I believe if you wanted to stay just inches outside the SOI, you would have to do a retro maneuver after you left SOI. So in essence you are wasting a LOT of thrust if you go the wrong way.

[Yasmy]

I think I might disagree with this in at least principle. If my goal (as it is before trying to understand the Oberth effect better) is to escape Kerbin, and join Kerbin in an orbit around the Sun as close to Kerbin's orbit as I can, then I don't think it matters which way I escape, if I can perfect my burn so that I stop the very instant I achieve escape velocity. Or at least, not practically. I suppose in theory I have to have a different trajectory than the Kerbin system by some infinitesimally small amount, but the direction I escape will be not even worth the bother of choosing the 'correct' direction. You can see this in KSP if you plan a burn just up to escape and not a single dV further, you basically have the same orbit as Kerbin, and it doesn't matter which way you chose to exit. At least up to as good as you are timing your burn to stop the moment you have escape trajectory.

This of course is only if your goal is to first just escape Kerbin, and then do a transfer to another planetary orbit as completely separate operations.

You are on the right track, Bane1998. You are correct in theory, but slightly incorrect in the KSP model.

It is still better to take advantage of the Oberth effect in terms of delta-V savings,

by doing your interplanetary burn near Kerbin such that your ejection angle is prograde or retrograde.

But if you burn in any direction just enough to escape Kerbin's SOI, you will almost, but not exactly, be on Kerbin's orbit around Kerbol.

Let's do the math. As usual, all we need is the vis-viva equation: v² = mu (2/r - 1/a).

Let r_p = your periapsis around Kerbin (altitude + Kerbin radius)

Let r_s = the SOI radius

Then the semi-major axis is a = (r_p + r_s)/2

Thus the velocity at the SOI of an orbit that just touches the SOI is v = sqrt( mu (2/r_s - 2/(r_s + r_p))).

Let's put in some numbers for Kerbin:

[table=width: 500]

[tr]

[td]r_p = 0 km[/td]

[td]v = 0 m/s[/td]

[/tr]

[tr]

[td]r_p = 100 km altitude + 600 km Kerbin radius[/td]

[td]v = 26.3 m/s[/td]

[/tr]

[tr]

[td]r_p = 200 km altitude + 600 km Kerbin radius[/td]

[td]v = 28.1 m/s[/td]

[/tr]

[/table]

So if you went straight up from the surface (but westward enough to cancel Kerbin's surface velocity)

you would have a periapsis of zero, and you would be moving at zero relative to Kerbin at the SOI.

However, if you start from orbit, and send your apoapsis exactly to the SOI,

you will have some small velocity relative to Kerbin, perpendicular to the semi-major axis.

Since your velocity relative to Kerbin is perpendicular to the semi-major axis,

it is actually better to exit the SOI either at Kerbin noon or midnight,

if your plan is to exit the Kerbin SOI, and then do your interplanetary burn.

But we're talking about at most about 60 m/s difference for exiting on opposite sides of Kerbin.

Eve injection will cost between approximately 60 and 120 m/s (ignoring plane change) based

on where you exit the SOI. I'd call that significant. On the other hand, when I go to Jool, I go in style,

and 60 m/s is entirely negligible.

As one final complication, if you really care about the nitty-gritty details, you will almost always

exit the SOI at a different radius around Kerbol than Kerbin's radius (by at most +/- Kerbin's SOI).

This has an additional effect on the interplanetary burn delta-v. Instead of using Kerbin's

orbital radius in the transfer calculation, use the spacecraft's Kerbol orbital radius.

Edited January 3, 2014 by Yasmy

[LtKraftKrackers]

you know, i was preparing a really long paragraph of how i see this effect and how it makes sense to me. but i think its best if i just tell you this.

Trust on the guys above me. Trust the people who have earned this knowledge from scratch, evolved it, and refined it. Sometimes you just have to let go.

[Camacha]

Reading this the effect should be relatively small when only reaching a low parking orbit before burning to Duna, right? As opposed to burning to the edge of the SOI and then burning.

Trust on the guys above me. Trust the people who have earned this knowledge from scratch, evolved it, and refined it. Sometimes you just have to let go.

What others do is their own choice, but for me finding that intuitive understanding is the point of playing this game. Knowing and reading about things is easily enough, truly understanding them and even getting a feel for them is the hard part.

[FenrirWolf]

I try to think of the oberth effect like this...

The ideal point to raise your apoapsis is at periapsis, and vice versa. Conversely, a burn performed right at apoapsis will have no effect on the apoapsis itself, and the same holds true for the periapsis.

In other words, energy spent raising your periapsis does not contribute to raising your apoapsis. So higher parking orbits mean more delta-v was used to make the orbit in the first place (more was spent on raising periapsis). So with a lower orbit and spending as delta-v as possible near periapsis, the most energy possible goes into speeding up the rocket to raise apoapsis and eventually escape Kerbin's SoI

Naturally there will always be some losses in that regard since we cant impart all of our delta-v instantaneously at the proper point, but we have various ways to minimize the loss like spreading a long burn across multiple passes and so on.

Edited January 3, 2014 by FenrirWolf