Delta-V Map for Homeward-Bound Trips?

[Geschosskopf]

I've managed to dig up 2 delta-v maps in this forum that show how much you need to get OUT to a distant body. One of them even shows how much it takes to get back into orbit around that distant body.

What neither of them show, however, is the delta-v needed during the landing itself, nor how much you need to transfer back to Kerbin, orbit, and land there.

So, is there a full delta-v map that works in both directions?

Thanks in advance.

[Johnno]

I've only seen the two maps you're referring to, for return trips I use these two webpages:

http://ksp.olex.biz/

http://alexmoon.github.io/ksp/

[Domfluff]

Most of the delta-v maps I've seen show landing delta-v's - http://forum.kerbalspaceprogram.com/showthread.php/25360-Delta-V-map for example, is shown by the numbers in the circle.

The landing and launch delta-v on an airless world are going to be roughly identical (allowing for in-optimal trajectories and the like)

Return delta-v's are rarely hugely different from the outbound delta-v's, as a rule of thumb I tend to just double it, and build to that specification.

[Sephta]

Actually: everything on these charts is reversible (except the atmospheric starts/reentries) due to the physics used in space flight: you need the same amount of dV change to get there and back. For the rest: simple add the values for your planned journey and add a safety margin of about 1000m/s

If you plan to go to Jool, Eve, Duna or Laythe: you can safe massive amounts of fuel by aerobreaking.

[Geschosskopf]

@ Johnno:

Thanks for the links. I had the Olex one but not the other.

@ Sephta

Thanks for the info. Now please tell me if I'm doing this right.....

I'm looking at this map: http://forum.kerbalspaceprogram.com/showthread.php/25360-Delta-V-map

Suppose I have a Munar lander that's already on its transfer orbit to Munar encounter before its own engine has to do any work. It's job is to land and return to Kerbin. Thus, it needs:

210 to get captured

640 to land

640 to weigh anchor

210 to get back in Munar orbit

860 to shove off for Kerbin

XXX to get into Kerbin orbit and then de-orbit at the chosen target, the atmosphere and parachute slowing the capsule down.

YYY for unforeseen emergencies

-----------

2560 + XXX + YYY total

If YYY is 1000, then the lander needs 3560 + XXX.

Is this correct? And what is XXX?

Thanks again.

[allmappedout]

Well, if you time your burn correctly, XXX could be zero as you can aerobrake in Kerbin and end up landing exactly where to be. Obviously that's fairly difficult.

Also, I think, your 640 to 'weigh anchor' includes getting back into orbit; the 210 estimate is your escape velocity from Mun -> Kerbin. Your 800 would be required to bring you back to your original orbit.

edit: I'm pretty sure about all that, but i'm speaking from experience rather than calculations, so someone who has time to do specific calcs feel free to correct me!

[Colonel_Panic]

@ Johnno:

Thanks for the links. I had the Olex one but not the other.

@ Sephta

Thanks for the info. Now please tell me if I'm doing this right.....

I'm looking at this map: http://forum.kerbalspaceprogram.com/showthread.php/25360-Delta-V-map

Suppose I have a Munar lander that's already on its transfer orbit to Munar encounter before its own engine has to do any work. It's job is to land and return to Kerbin. Thus, it needs:

210 to get captured

640 to land

640 to weigh anchor

210 to get back in Munar orbit

860 to shove off for Kerbin

XXX to get into Kerbin orbit and then de-orbit at the chosen target, the atmosphere and parachute slowing the capsule down.

YYY for unforeseen emergencies

-----------

2560 + XXX + YYY total

If YYY is 1000, then the lander needs 3560 + XXX.

Is this correct? And what is XXX?

Thanks again.

860 is the cost to reach mun from kerbin orbit, and 210 is to circularize from capture. from there, 640 to land (I usually budget 700) and another 640 to return to a circular orbit about the mun (you're not circularizing from capture) Your return cost is not 860 though, it's probably the same as the 210 you used to circularize, since if you burn to munar retrograde low to the surface, the oberth effect can slingshot you back down into kerbin's atmosphere for aerobraking, at which point circularizing or landing is practically free.

Edited June 17, 2013 by Colonel_Panic

[Sephta]

Ok, here's my reading of the chart:

210 to get captured into Mun Orbit

640 to land

640 to take off Mun

210 to leave Mun Orbit to Kerbin

860 to circularize to LKO ( you can save here already with an aerobreak maneuver

XXX to land at Kerbin (would be 4500 for a non atmosphere planet)

YYY to cover minor hickups

-----------

2560 + XXX + YYY total

"If YYY is 1000, then the lander needs 3560 + XXX.

Is this correct? And what is XXX?"

So, aside from the things you do with the dV your calculation was correct. The XXX would be the dV to bring you you into the atmosphere long enough to break there (should be around 50m/s on a 100km LKO)

but keep in mind: all these numbers are estimates and the transfers between the planets are including the plane change, but also a proper injection angle (so you launch directly fom KLO!) if you put your ship on a solar orbit first, you need more fuel.

[capi3101]

Delta-V for landing is the same as it is for a launch provided the target has no atmosphere; makes sense if you think about it. I mean, in a landing, you're going from orbital velocity to zero and in a launch you're going from zero to orbital velocity. In my experience, the numbers on the chart reflect the amount of delta-V required for a launch; on those worlds with atmospheres, the landing delta-V is less than what's listed because you do pick up some free delta-V due to atmospheric drag.

So, your main question is how much delta-V is required "to get into Kerbin orbit and then de-orbit at the chosen target, the atmosphere and parachute slowing the capsule down".

The answer's going to be dependent upon the ejection angle you pick and whether or not you want to use aerobraking, which is why there isn't a hard and fast number available. Most folks just want to land so they set a maneuver node with a Kerbin periapsis around 30k; it does the trick. Setting it around 50k and then being patient will eventually get you to back to a 100k orbit (it usually takes a few passes).

If you can predict your speed at Kerbin periapsis, you'll have a good idea of how much delta-V you're going to need. You'll just need enough to reduce your speed to about ~2300 m/s when you reach Kerbin periapsis. Once you're in orbit, a final de-orbit burn isn't going to take much; at 2200 m/s, you'll already be in a sub-orbital path. In any case, XXX shouldn't be more than 860 m/s, since that's how much you needed to get to the Mun in the first place.

[Geschosskopf]

Thanks to all for the help with getting home from Mun. I think I kinda get that . Now let's take talk to getting to and from the other planets....

The chart shows going to other planets as a 2-step thing with Kerbol in the middle. I take it the 950 between Kerbin and Kerbol is what you need in all cases just to escape Kerbin, then the number between Kerbol and the target planet is the extra needed to get there. Am I correct so far?

But I imagine the step of escaping from the planet and heading back to Kerbin differs depending on whether you're going "uphill" from an inner planet or "downhill" from an outer planet. Is that a correct assumption?

From what I gather from the inputs above, on the way home from a moon, you ignore the number in the middle of the line. Is it the same way for the outer planets? For instance, Duna shows a total of 1160 between it and Kerbin, which you need to get out there. But you'd just need 370 to escape Duna get home headed home. But I imagine you'd have to pay the 1680 between Moho and Kerbin coming back if you went there. Is that correct?

[bsalis]

Return delta-v's are rarely hugely different from the outbound delta-v's, as a rule of thumb I tend to just double it, and build to that specification.

I don't think it is that rare. The difference can be very significant. For example, landing on Gilly from Eve orbit is about 1900 m/s. Getting back? .. you can get back to *Kerbin* on EVA!

Delta-V for landing is the same as it is for a launch provided the target has no atmosphere;

Assuming the descent is done the same as the ascent. MechJeb may do this but if attempted manually will mostly end with undesirable results.

[tomf]

The return trip always (if you pick the optimum transfer point) takes the same delta-v as the outgoing trip. The differences you see (e.g. returning from mun or gilly) is due to aerobraking in the parent bodies atmosphere.

For a full mission to Duna (numbers taken from the current chart on the kerbin wiki page http://wiki.kerbalspaceprogram.com/w/images/7/73/KerbinDeltaVMap.png)

Takeoff ~4550

Kerbin escape 950

Duna intercept 110

Duna capture 370

Duna landing 1380

Duna Takeoff ~1700

Duna escape 370

Kerbin intercept 110

Kerbin capture 950

Kerbin landing ? very small

As you can see the chart is symetrical.

However the numbers in red are orbit changes for which you can use aerobraking, you should be able to save about 90% of the given delta v, which does mean that it ends up being masively cheaper to get back from Duna than it was to get there.

Also not that the chart has split out the number to escape Kerbin and to intercept Duna. However the numbers have in fact been calculated on the basis that you will do a single burn from a circular orbit of the departure body.

[Geschosskopf]

For a full mission to Duna (numbers taken from the current chart on the kerbin wiki page http://wiki.kerbalspaceprogram.com/w/images/7/73/KerbinDeltaVMap.png)

Takeoff ~4550

Kerbin escape 950

Duna intercept 110

Duna capture 370

Duna landing 1380

Duna Takeoff ~1700

Duna escape 370

Kerbin intercept 110

Kerbin capture 950

Kerbin landing ? very small

As you can see the chart is symmetrical.

However the numbers in red are orbit changes for which you can use aerobraking, you should be able to save about 90% of the given delta v, which does mean that it ends up being masively cheaper to get back from Duna than it was to get there.

First off, thanks for the link to the new map. I hadn't been able to find that one. Where's the link to it within the Wiki? I keep turning up the old chart.

Next, thanks for the info. But I'd like to understand how you got the "~1700" for Duna takeoff. I see the "1380" on the chart, which I get landing totally on engines, not using the atmosphere. But that's the only number the chart has, so how do you get the 1700 to take off again? Do you have to calculate that somehow, is it from experience, or is there some rule of thumb you can apply to the landing number to take the atmosphere into account when taking off?

Thanks.

[capi3101]

The ~1700 accounts the amount of delta-V lost due to atmospheric drag. Take Kerbin for example; you know you need ~4500 delta-V to make orbit. That's going from zero to orbital velocity, of course. If Kerbin didn't have an atmosphere, by definition you should be going 4,500 m/s when you make orbit. Because there is atmosphere, though, you make it up there only going about 2,300 m/s or so. Duna also has the same atmosphere, thus more delta-V is required to compensate for it.

Unless I'm mistaken, the 1,380 is an average value for both launch and landing (i.e. it takes ~2700 delta-V total to go from orbit to the surface and then back to orbit; if ~1700 of that is needed for takeoff, about ~1000 will be needed for landing - assuming you don't use chutes). If you plan for that, you oughta be able to make it there and back without any issues.

[tomf]

It looks like you are correct, which surprises me a little (that the chart is done that way - not that you are right )

If Duna had no atmosphere it would take 1009 each way.

[Geschosskopf]

Ah, so to make sense of the landing/take-off numbers on the chart for planets with atmospheres, you need to know the orbital velocities at the altitudes the chart lists for low orbits. Then you can compare that to the chart value to determine the effect of the atmosphere. Without this piece of info, it appears that atmospheres are traps for the unwary noobs like me. "Hey, I'll build an Apollo-style 2-stage lander for Duna and the top stage only needs 1380m/s to dock with the mothership in LDO." And then I name the newest Duna crater after the crew .

I take it the speed for Duna is 1009. I'll have to dig up or calculate the others I guess, unless somebody can save me the work .