Reading a Delta V Map.

[LordCorwin]

I think I see what you say. I thought I had to reduce speed to match Duna's velocity and so I always escaped Kerbin retrograde. When it is the opposite, I have to increase energy to be able to reach Duna's orbit, so I must escape Kerbin's orbit prograde.

What a waste of fuel!!

Edited August 2, 2014 by LordCorwin

[LethalDose]

I think I see what you say. I thought I had to reduce speed to match Duna's velocity and so I always escaped Kerbin retrograde. When it is the opposite, I have to increase energy to be able to reach Duna's orbit, so I must escape Kerbin's orbit prograde.

What a waste of fuel!!

Well, now you know, and knowing is half the hassle.

Seriously, though, there are some super handy IP transfer planning tools like Olex's http://ksp.olex.biz/. That one gives you the skinny on ejection angles.

[metaphor]

That delta-v chart is a little bit wrong in that regard. It doesn't take 370 m/s to get from a Duna intercept to low Duna orbit.

This delta-v chart is more accurate there (right click and open image in new tab if you can't read the numbers):

From low Kerbin orbit (70 km), you burn 950+130 m/s or 1080 m/s to get into a Kerbin-Duna Hohmann transfer orbit. From that orbit, when you get to Duna you burn 250+360 m/s or 610 m/s retrograde to get into a low Duna orbit (60 km). From Duna low orbit, if you burn 360 m/s you'll barely get out of its SOI. Instead, you'll need to burn 610 m/s to get into a Duna-Kerbin transfer orbit. Then when you get to Kerbin you burn 1080 m/s retrograde to get into a low Kerbin orbit.

Since Duna has a slightly eccentric orbit, instead of always being 610 m/s the Duna burn can vary between about 550 and 670 m/s, but it's never as low as 370 m/s. Likewise, the Kerbin burn can vary between 1050 m/s and 1110 m/s. Here's another map with ranges of values:

It looks more complicated than the subway map, but it's not that hard to read. If you just sum the values along the vertical lines up to a certain transfer, that's what you need to burn at Kerbin. The sum along the horizontal lines is what you need to burn at the other planet. (The diagonal lines count as vertical.)

I think I might just make another one that just shows the planets instead of the moons and other things so that it's easier to use.

[Yasmy]

@LethalDose

1) If the delta-v map that we used is inaccurate, well, sure, my numbers will be wrong. I didn't vet the dV map.

But I am not wrong that circularization at planet B on an interplanetary trajectory from planet A (capture + circularization, if you prefer) is the same delta-v as escape + transfer from B to A. Just play the maneuver backwards in your head. (Gravity is symmetric under time reversal.)

Looks like metaphor gave us a better map. It appears he also supports my statement that for the reverse trip, just a) read the numbers in the opposite direction, and regard the outgoing transfer burn as part of your incomming circularization burn.

2) It is not true that the transfer cost from Kerbin to Duna is the same as Duna to Kerbin.

Why speculate? I gave the formulae for a Hohmann transfer. You don't even have to plug in numbers.

Clearly dv1 is not equal to dv2. Which is the same as saying swapping r1 and r2 in dv1 gives a different transfer burn.

Edited August 2, 2014 by Yasmy

[LethalDose]

@LethalDose

1) If the delta-v map that we used is inaccurate, well, sure, my numbers will be wrong. I didn't vet the dV map.

But I am not wrong that circularization at planet B on an interplanetary trajectory from planet A (capture + circularization, if you prefer) is the same delta-v as escape + transfer from B to A. Just play the maneuver backwards in your head. (Gravity is symmetric under time reversal.)

Looks like metaphor gave us a better map. It appears he also supports my statement that for the reverse trip, just a) read the numbers in the opposite direction, and regard the outgoing transfer burn as part of your incomming circularization burn.

Yes, I agree with all of this. That's why I there was a problem with your numbers. The math and the concept are fine.

As for the numbers on Metaphor's map above, they are substantially different from what is reported on the subway style map we've been discussing. It appears to take 130 + 250 = 380 m/s to make the transfer from Kerbin to Duna. I'm guessing the 250 dV cost is the burn to go from the transfer orbit to get captured in Duna. This is specifically the value that I thought was missing from the subway chart (what I referred to as ÃŽâ€Vtdk above), and it does not appear to be part of the 110 reported on the subway map.

So that subway map is going to underestimate the required dV for a Duna round trip by 500 m/s... Assuming Metaphor's numbers are correct, that's pretty substantial...

2) It is not true that the transfer cost from Kerbin to Duna is the same as Duna to Kerbin.

Why speculate? I gave the formulae for a Hohmann transfer. You don't even have to plug in numbers.

Clearly dv1 is not equal to dv2. Which is the same as saying swapping r1 and r2 in dv1 gives a different transfer burn.

It depends on what you mean by "transfer cost", but I think we're conceptually on the same page.

Edited August 2, 2014 by LethalDose

[GoSlash27]

The maps shown here do not work the same way backwards because they assume a single burn from LKO to use the Oberth effect.

This one works in either direction:

props to mhoram.

-Slashy

[LethalDose]

@Slashy:

First, The values on that map you posted are wrong: If you burn from LKO, it doesn't take 1050 - 1100 more m/s beyond escape velocity to reach Duna, nor does it take an additional 800 m/s to reach Eve.

Second, The values on the previous maps do work in both directions because the Oberth effect works in both directions.

EDIT: I read the very short thread that was posted in, and those are dV recommendations for beginners. It's not how most dV maps are written, it's misleading to re-post it here without the massive caveat that those flight paths are intentionally suboptimal to facilitate learning, and the only reason it would work in both ways is because it's basically telling players to vastly over-engineer their vessels.

I could just say "A round trip to every planet requires 25 km/s of dV" and it would work in both ways. That doesn't mean the statement is useful.

Edited August 2, 2014 by LethalDose

[GoSlash27]

LD<

No, the numbers are correct for the assumptions it's built for. It's not assuming that you're burning from LKO; That's the point. If you want to burn from Kerbin to Jool and you're doing it after you leave their SOI, it'll cost 3,000 M/sec. Same deal in reverse.

And the oberth effect is velocity- dependent. It would only cost the same both ways if your orbital velocities were the same during the burn, but they never are.

Best,

-Slashy

[LethalDose]

LD<

No, the numbers are correct for the assumptions it's built for. It's not assuming that you're burning from LKO; That's the point. If you want to burn from Kerbin to Jool and you're doing it after you leave their SOI, it'll cost 3,000 M/sec. Same deal in reverse.

Yes, and when you posted it you should have noted it's assumptions in your post. This thread is about teaching new players to read dV maps, and posting a dV map with such vastly different dV values is only going to confuse players if you're not making those assumptions clear.

And the oberth effect is velocity- dependent. It would only cost the same both ways if your orbital velocities were the same during the burn, but they never are.

Best,

-Slashy

Yes, the Oberth effect is velocity dependent. Your orbital velocity is greatest when you are at periapsis. You should be aiming your insertion trajectory so that your Pe is at your target circularization altitude so you are captured directly into low orbit. This is equivalent to making your escape burn in an equivalent low orbit. Both burns I have just described are taking advantage of the Oberth effect.

If your return transfer trajectory is basically the same orbit as your original transfer trajectory, then your speed at periapsis around the return body will be the same as it was at the end of your original ejection burn. Obviously, the orbital speed at your low orbit is going to be the same before the ejection and burn and after the capture. Since the start and end velocities are the same (simply transposed) the difference between those velocities (aka the dV) are the same.

The dV costs are the same, so long as the return trajectory is the same as the approach trajectory, as I stated above. They won't be exact (human error), but they'll be so close that it won't matter.

I can't make it any simpler. In this case, the Oberth effect has as much to do with the altitude at which a burn is made as it does with the velocity at which it is made (in orbital mechanics, these are linked). It's accounted for in both places.

I'm guessing what I've posted simply won't convince you, though, so I'm not going to try any further. I'll leave it to other members of the community either convince you, or to demonstrate that I'm wrong.

[LordCorwin]

Well, now you know, and knowing is half the hassle.

Seriously, though, there are some super handy IP transfer planning tools like Olex's http://ksp.olex.biz/. That one gives you the skinny on ejection angles.

Thanks for the link, I've tested it twice today and has been very helpful!

[K77LostNSpace]

Great information so far from everyone who posted. Learning quite about about this Dv chart.

Question:

My situation is landing on the Mun. First I have really bad pilot skills and my decent profile is horrid.

Reviewing the Dv chart Metaphor posted ( http://i.imgur.com/UUU8yCk.png ) it says 580Dv from a 10k orbit to the Mun.

So I made the assumption of adding 580 + 580 = 1160Dv minimum required for my landing stage to get on the surface and return to 10k orbit and meet up with the transfer stage to return home.

The lander has a bit over 1160Dv (don't have the true number with me at the moment, but it was over the estimated Dv listed)

I find I was unable to get the lander back into a 10k orbit, or any orbit for that matter.

Current landing profile: When trying to land I drop the orbit so it over shoots my landing spot, and slowly burn till I'm just over my landing spot then zero out my velocity, upright and lower at a slow pace until touchdown.

Does anyone have a good decent profile to the Mun I should try to follow when landing to conserve fuel, or should i just stack on more tanks and boosters for greater Dv for my lander stage?

[indroth]

lower at a slow pace until touchdown.

Ideally, you would let yourself drop and then burn at full throttle at the right time to come to a stop on the ground. An approximation I find useful is if you are at 0 velocity at altitude h over the surface (that's radar altitude), start your burn at altitude h / TWR. You will actually reach 0 m/s before reaching the ground though, or sometimes you are moving so fast that you can't throttle up in time and crash. So take some extra delta-v and throttle up a little early.

[mhoram]

Does anyone have a good decent profile to the Mun I should try to follow when landing to conserve fuel, or should i just stack on more tanks and boosters for greater Dv for my lander stage?

You can find a good desctiption about efficient landing here:

http://forum.kerbalspaceprogram.com/threads/39812-Landing-and-Takeoff-Delta-V-vs-TWR-and-specific-impulse