Reading a Delta V Map.

[SpaceSmith]

Okay so I downloaded the Subway style Delta V map of the Kerbol system and I am just curious of how to read it. It is my assumption you just add the numbers on the path from your origin to destination and so on?

So say I wanted to go from Kerbin to land on the Mun and then back to Kerbin again, would it be 4550m/s, +860m/s +210, +640 and then the same back to Kerbin?

But that is only to the Mun and when I come back it never feels like I am using that much Delta V when coming back, I mean I can come back on a quarter tank of Fuel.. so really, how do I find the Delta V cost of returns from say the Mun?

[GoSlash27]

Smith,

The DV map is accurate for continuous burns taking advantage of the Oberth effect, and you've got the idea of how it works. It does not, however, take into account DV saving tactics such as aerobraking, slingshotting, etc. Your return trips to Kerbin will never use as much DV as the map suggests because you're using Kerbin's atmosphere to convert excess DV into heat. If Kerbin had no atmosphere, you'd have to fire engines to soak up the DV and the map would be correct.

Beware using this map for interplanetary trips when you're not doing one continuous burn from launch, tho'. Simple flight plans (escaping Kerbin SoI and then burning for an intercept) will take much more DV, while slingshotting/aerobraking will take much less.

Best,

-Slashy

Edited August 1, 2014 by GoSlash27

[LethalDose]

So say I wanted to go from Kerbin to land on the Mun and then back to Kerbin again, would it be 4550m/s, +860m/s +210, +640 and then the same back to Kerbin?

Well, not the 4550 part, that's to get from surface into orbit. Even without the atmosphere, it wouldn't take as much to make orbit as it would to land, since gravity's on your side going down.

[SpaceSmith]

Okay awesome, so learning how much you need for a return trip is basically a thing of experience then?

[Claw]

Yes, and most dV maps also include markings to indicate that aerobraking is available. So you know you don't have to include that dV in your destination calculations.

They are generally good guides for how much dV it will take unless you go out of your way to do some of those other things mentioned (like slingshots). The other thing to realize is that each dV chart is built off of a set of assumptions. They usually list "craft is in XX meters above the body." (Such as above an atmosphere, or 10km above the highest terrain.) So if you start significantly off from that altitude, the numbers won't match exactly.

Cheers,

~Claw

[Red Iron Crown]

But that is only to the Mun and when I come back it never feels like I am using that much Delta V when coming back, I mean I can come back on a quarter tank of Fuel.. so really, how do I find the Delta V cost of returns from say the Mun?

In addition to the good advice given above, you should note that the second half of a tank gives more dV than the first half. It's not quite as simple as converting units of fuel to m/s of dV.

[Geschosskopf]

Okay awesome, so learning how much you need for a return trip is basically a thing of experience then?

To a certain extent. In my experience, the outward-bound numbers from Kerbin are pretty good (although achieving low orbits costs way less if you can aerobrake). But for coming home, things are different depending on whether you're returning "downhill" or "uphill".

"Downhill" is for interplanetary trips from Duna, Dres, Jool, or Eeloo back to Kerbin, or from Mun/Minmus to Kerbin. These transfers generally only take 30-50% of the delta-V needed to make the same trip in the opposite direction. For instance, transferring from LKO to a low Pe at Mun (not capturing--that's a separate burn) costs about 860m/s just as shown on the chart. But coming back from low Mun orbit to a low Pe at Kerbin only costs about 250m/s (and that's generally all you have to spend due to aerobraking unless you're using Deadly Reentry).

"Uphill" returns are from Moho and Eve. These transfers seem to cost pretty much the same in both directions.

I don't know why it works like this. When I was 1st getting into the Delta-V Map, I asked the same question about return transfers, and was told by more experienced players that the dV map reads the same in both directions. In theory, it takes the same amount of energy to move between the same 2 orbits no matter which way you go. But in KSP it doesn't. Could be a difference between real and KSP physics. I just don't know.

[StuntMax]

Well, not the 4550 part, that's to get from surface into orbit. Even without the atmosphere, it wouldn't take as much to make orbit as it would to land, since gravity's on your side going down.

If it wasn't for the atmosphere, you'd have to use your engines to *fight* the gravity... or else, you'd go litobraking really hard at a freefall-straight-from-the-orbit speed.... try landing on Tylo with less than 3000 dV under the hood

[Red Iron Crown]

If it wasn't for the atmosphere, you'd have to use your engines to *fight* the gravity... or else, you'd go litobraking really hard at a freefall-straight-from-the-orbit speed.... try landing on Tylo with less than 3000 dV under the hood

IME landing on an airless body usually costs a little bit more than ascending from it, due to it being difficult to do a perfect suicide burn. I always end up with a slow descent near the end while I zero out horizontal velocity and make sure there's no obstruction at the landing zone. Ascent is just light it up and go.

[Drew Kerman]

let's not also forget that in case the OP is using FAR and good ascent practices (immediate gravity turn) it only takes 3.4km/s of ÃŽâ€v to reach 70Km orbit

[metaphor]

To a certain extent. In my experience, the outward-bound numbers from Kerbin are pretty good (although achieving low orbits costs way less if you can aerobrake). But for coming home, things are different depending on whether you're returning "downhill" or "uphill".

"Downhill" is for interplanetary trips from Duna, Dres, Jool, or Eeloo back to Kerbin, or from Mun/Minmus to Kerbin. These transfers generally only take 30-50% of the delta-V needed to make the same trip in the opposite direction. For instance, transferring from LKO to a low Pe at Mun (not capturing--that's a separate burn) costs about 860m/s just as shown on the chart. But coming back from low Mun orbit to a low Pe at Kerbin only costs about 250m/s (and that's generally all you have to spend due to aerobraking unless you're using Deadly Reentry).

"Uphill" returns are from Moho and Eve. These transfers seem to cost pretty much the same in both directions.

I don't know why it works like this. When I was 1st getting into the Delta-V Map, I asked the same question about return transfers, and was told by more experienced players that the dV map reads the same in both directions. In theory, it takes the same amount of energy to move between the same 2 orbits no matter which way you go. But in KSP it doesn't. Could be a difference between real and KSP physics. I just don't know.

It should be the same delta-v going both ways, it's just that you're using a different atmosphere for slowing down each time. For example, from LKO to low Duna orbit takes a 1080 m/s burn, followed by aerobraking in Duna's atmosphere to shave 610 m/s off your speed. When you go back, it takes a 610 m/s burn, followed by aerobraking at Kerbin to shave 1080 m/s off your speed. So it seems like you're using less delta-v going back but that's neglecting the contribution of the atmospheres.

Delta-v map

[Red Iron Crown]

It should be the same delta-v going both ways, it's just that you're using a different atmosphere for slowing down each time. For example, from LKO to low Duna orbit takes a 1080 m/s burn, followed by aerobraking in Duna's atmosphere to shave 610 m/s off your speed. When you go back, it takes a 610 m/s burn, followed by aerobraking at Kerbin to shave 1080 m/s off your speed. So it seems like you're using less delta-v going back but that's neglecting the contribution of the atmospheres.

Delta-v map

If I understand it correctly, it's not even that the atmospheres are different, but that the burns that are eliminated or reduced through aerobraking are different magnitudes.

Your dV map is still the best to date, thanks again for making it.

[metaphor]

If I understand it correctly, it's not even that the atmospheres are different, but that the burns that are eliminated or reduced through aerobraking are different magnitudes.

Yes, that's what I meant. Thanks.

[Geschosskopf]

It should be the same delta-v going both ways, it's just that you're using a different atmosphere for slowing down each time. For example, from LKO to low Duna orbit takes a 1080 m/s burn, followed by aerobraking in Duna's atmosphere to shave 610 m/s off your speed. When you go back, it takes a 610 m/s burn, followed by aerobraking at Kerbin to shave 1080 m/s off your speed. So it seems like you're using less delta-v going back but that's neglecting the contribution of the atmospheres.

Um, that's not what I'm talking about. I'm talking about the transfer burn from a given altitude to a given altitude in the same or different SOI. IOW, going uphill raising your AP, vs. going downhill, lowering your Pe. Capturing once you get to the new Ap or Pe is a separate burn (or aerobraking) you do after you do the transfer burn, so I'm not taking that into account at all. I'm just reading the chart in both directions.

Examples:

1. You are in a 100km Kerbin orbit. Raising your Ap to geostationary altitude takes ~650m/s. Now you burn another 430 or so to circularize, which you have to do otherwise you'd still have a 100km Pe and there'd be nothing here to discuss. But for the purposes of this discussion, ignore the circularization cost because I'm only comparing transfer burns. OK, so now I'm in a geostationary orbit. I want to transfer back to a 100km orbit without ever touching Kerbin's atmosphere. If the chart read the same way in both directions, this would cost me another 650, same as it took to get out there. But it doesn't. It actually only takes about 200m/s to lower the Pe, without ever touching the atmosphere. Of course, now you have to circularize with engines but that's a separate thing from the transfer.

2. You are in a 100km Kerbin orbit and want to get to a 10km Munar orbit. This transfer burn takes between 860-900 depending on how well you place your node. After circularizing, you now want to return to a 100km Kerbin orbit. This transfer burn only takes about 220-250m/s, so again the chart doesn't read the same in both directions.

3. You are in a 100km Kerbin orbit and want to get to Duna. You just want an encounter and you'll fine-tune your Duna Pe later on. This will cost you about 1000-1100m/s depending on the day you do it and how accurately you set up and execute it. You do the return trip the same way, just trying to get a Kerbin encounter. If the chart read the same in both directions, this would again cost 1000-1100m/s, but instead it only costs like 500-600. Again, it's cheaper coming downhill than going uphill, and neither atmosphere is involved in either burn.

4. OTOH, you are in a 100km Kerbin orbit and want to get to Eve, again just trying for an encounter you'll tweak later. This will cost you about 1100-1200. Getting back to just a Kerbin encounter will also cost you about 1100-1200. So when coming home uphill, the chart does read the same in both directions for the transfer burns, without involving either atmosphere.

[metaphor]

Um, that's not what I'm talking about. I'm talking about the transfer burn from a given altitude to a given altitude in the same or different SOI. IOW, going uphill raising your AP, vs. going downhill, lowering your Pe. Capturing once you get to the new Ap or Pe is a separate burn (or aerobraking) you do after you do the transfer burn, so I'm not taking that into account at all. I'm just reading the chart in both directions.

Examples:

1. You are in a 100km Kerbin orbit. Raising your Ap to geostationary altitude takes ~650m/s. Now you burn another 430 or so to circularize, which you have to do otherwise you'd still have a 100km Pe and there'd be nothing here to discuss. But for the purposes of this discussion, ignore the circularization cost because I'm only comparing transfer burns. OK, so now I'm in a geostationary orbit. I want to transfer back to a 100km orbit without ever touching Kerbin's atmosphere. If the chart read the same way in both directions, this would cost me another 650, same as it took to get out there. But it doesn't. It actually only takes about 200m/s to lower the Pe, without ever touching the atmosphere. Of course, now you have to circularize with engines but that's a separate thing from the transfer.

2. You are in a 100km Kerbin orbit and want to get to a 10km Munar orbit. This transfer burn takes between 860-900 depending on how well you place your node. After circularizing, you now want to return to a 100km Kerbin orbit. This transfer burn only takes about 220-250m/s, so again the chart doesn't read the same in both directions.

3. You are in a 100km Kerbin orbit and want to get to Duna. You just want an encounter and you'll fine-tune your Duna Pe later on. This will cost you about 1000-1100m/s depending on the day you do it and how accurately you set up and execute it. You do the return trip the same way, just trying to get a Kerbin encounter. If the chart read the same in both directions, this would again cost 1000-1100m/s, but instead it only costs like 500-600. Again, it's cheaper coming downhill than going uphill, and neither atmosphere is involved in either burn.

4. OTOH, you are in a 100km Kerbin orbit and want to get to Eve, again just trying for an encounter you'll tweak later. This will cost you about 1100-1200. Getting back to just a Kerbin encounter will also cost you about 1100-1200. So when coming home uphill, the chart does read the same in both directions for the transfer burns, without involving either atmosphere.

If you're going to ignore the circularization, then yes the burn lower down the gravity well will cost more delta-v than the burn higher up the gravity well in a Hohmann transfer. This doesn't always work with different SOIs though, since you have the Oberth effect which varies with the size of the planet/moon.

But I'm not sure what you mean by the chart not reading the same both ways. It should take the same delta-v to go somewhere as it takes to get back, if you include all the delta-v's including circularization. For example, if you're going to geostationary altitude from a 100 km Kerbin orbit, your first burn would take 650 m/s, then 430 m/s to circularize. On the way back it would take 430 m/s to transfer down to the lower orbit, and 650 m/s to circularize there. It's the same delta-v both ways, but in reverse order.

[Mr Shifty]

2. You are in a 100km Kerbin orbit and want to get to a 10km Munar orbit. This transfer burn takes between 860-900 depending on how well you place your node. After circularizing, you now want to return to a 100km Kerbin orbit. This transfer burn only takes about 220-250m/s, so again the chart doesn't read the same in both directions.

Uh, if you want to get into a 100 km Kerbin orbit at the end, it takes about 850 m/s to circularize. Gravity does not actually violate the conservation of energy.

[Geschosskopf]

But I'm not sure what you mean by the chart not reading the same both ways. It should take the same delta-v to go somewhere as it takes to get back, if you include all the delta-v's including circularization. For example, if you're going to geostationary altitude from a 100 km Kerbin orbit, your first burn would take 650 m/s, then 430 m/s to circularize. On the way back it would take 430 m/s to transfer down to the lower orbit, and 650 m/s to circularize there. It's the same delta-v both ways, but in reverse order.

I agree that it SHOULD take the same amount of delta-V in both directions, but in practice it does not, which is what I've been going on about.

And here's my point about the chart not reading the same in both directions. A person puts his finger on the chart at his starting point, moves his finger along it to his destination, and adds up the numbers along the way. All well and good--that works and matches observations. OK, now he wants to go back so he repeats the process in reverse. He has his finger were he is now and moves it back to Kerbin, adding up the numbers as he goes. So in the case of geostationary....

Outward-bound:

1. 650m/s to raise Ap to geostationary level

2. 430m/s to circularize

TOTAL: 1080m/s

And this will match observations when trying it in the game.

But then:

Homeward-bound

3. 650m/s to lower Pe to LKO

4. ???? to circularize (without aerobraking). Hmmm, this isn't on the chart. But we've heard of this thing called conservation of energy so we'll assume 430m/s to get the same...

TOTAl: 1080m/s

Now, we all know this isn't correct. The downhill transfer (step 3) doesn't take 650m/s. But that's what you get from the chart if you read it backwards. That's what I mean about the chart not reading the same in both directions.

You say above that the downhill transfer actually takes the 430 and then the 650 circularizes, so you end up with the same total. I agree that sounds like how it should work, what with conservation of energy and all that. But it doesn't actually work that way in practice. Step 3 will only cost about 200m/s. Now, if conservation actually worked in KSP, you'd need 1080 = 200 = 880m/s to circularize back in LKO without the atmosphere helping. But it actually only takes in the neighborhood of 500. Don't ask me why, but that's that sort of thing I see, and this applies to any downhill transfer.

Now, we all know that KSP violates all the conservation laws it can all the time. When you EVA a Kerbal, you create mass. The EVA Kerbal has mass yet the mass of the capsule doesn't change. Likewise, when the Kerbal goes back inside, you have destroyed mass. And if this is all done with the ship moving, you violate conservation of momentum. You can change the orbital velocity of a spinning ship simply by pumping fuel from 1 end to the other, thereby creating kinetic energy from nothing. Ships on rails don't use electricity so there you're violating conservation of charge. And when spinning ships shed pieces, the central part doesn't spin any faster, so it violaties conservation of angular momentum. I'm sure there are other examples of broken conservations in KSP. This appears to be just another one.

Edited August 1, 2014 by Geschosskopf

[Mr Shifty]

Start with a circular orbit at 100 km, velocity 2246 m/s.

For KTO: Burn prograde 651 m/s for an Ap of 2863 km and Pe of 100 km.

Coast to Ap: Velocity is now 586 m/s.

Circularize at KSO: Burn prograde 424 m/s for a total of 1010 m/s.

Total burn cost: 1075 m/s.

Going back down is exactly the same thing in reverse. Burn retrograde 424 m/s to lower Pe to 100 km, then 651 m/s retrograde at Pe to circularize there. If you find different, you'd have to supply pictorial proof, because that would be a major error in the game's orbital mechanics system.

[Red Iron Crown]

I'm really curious to see an example of the effect Geschosskopf is describing, too. I haven't paid all that close attention to the symmetry of orbits in KSP, but they should be symmetrical, especially if no SoI boundaries have been crossed. It should be that the sum of apsis changing and circularization are equal in either direction, really curious to see if it's not. (Sadly I am away from my KSP computer for a week or so or I would test this myself).

[LethalDose]

Because people have started talking about "conservation of energy" in a discussion about ÃŽâ€V, I'ma pipe up about a pet peeve of mine.

ÃŽâ€V != change in [specific orbital] energy

And this isn't an issue of units, either, because the two are not linearly proportional either. the relationship is described by the equation:

ÃŽâ€E = V * ÃŽâ€V + (ÃŽâ€V^2)/2

So while you can say energy is conserved, the ÃŽâ€V costs may not be since ÃŽâ€E is dependent on V as well as ÃŽâ€V. This is another statement of the Oberth effect. For the "conservation of energy" to translate into "conservation of ÃŽâ€V", your start and end velocities need to be the same.

I'm not saying this assumption isn't met in this conversation, I'm just saying that statements like

But we've heard of this thing called conservation of energy so we'll assume 430m/s to get the same...

bug me sometimes because those assumptions aren't made explicit, and sometimes the parties aren't even necessarily aware of them.

Okay, I feel better now.

One possible explanation for the phenomenon that Geschosskopf has described (and I think the most like one) is that it's an illusion. While the dV cost may be the same in both directions, the cost in fuel is vastly different, for a bunch of reasons. I agree with Mr. Shifty that in the KSO example provided, the burn to circularize back into LKO is equal to the transfer burn, but after playing the game it's really is hard to believe they're the same. It seems like they should be different, and it's way worse with no dV meter in the stock game (I don't know how you play, but it's really easy to see how this could be unclear).

[LethalDose]

And I FINALLY figured out what bugs me about reading some of the ÃŽâ€V maps! For the target planets, they combine the escape velocities with the transfer velocities! at least this one doesn't.

To transfer from Kerbin to Duna, you need to spend ÃŽâ€Vek (delta V to reach escape velocity from LKO) and ÃŽâ€Vtkd (delta V to initiate the transfer from Kerbin to Duna).

To transfer from Duna to Kerbin, you need to spend ÃŽâ€Ved + ÃŽâ€Vtdk.

The problem is that ÃŽâ€Vdkt != ÃŽâ€Vkdt. In other words, the ÃŽâ€V cost to initiate the transfer from a higher orbit does not equal the cost to initiate the transfer from a lower orbit. This was demonstrated in the LKO/KSO example above (651 m/s to initiate transfer from a low orbit, and 424 m/s to inititate from a higher orbit). To extend the analogy to interplanetary travel, the starting and end states in the LKO/KSO example roughly represent the states of vessels in low orbits at Ve at their respective planets.

The way I read the map above, for the Duna -> Kerbin transfer it either gives either 370 m/s or 370 + 110 m/s.

If my calcs are right, then both of those numbers are wrong.

370 m/s is ÃŽâ€V needed to reach escape velocity (ÃŽâ€Ved) from a 50-ish km orbit only. It's not enough to initiate the transfer.

If 110 m/s is ÃŽâ€Vtkd, then as I pointed out above, ÃŽâ€Vtkd != ÃŽâ€Vtdk.

Now, it's possible that 110 m/s = ÃŽâ€Vtdk + ÃŽâ€Vtkd, but if that's the case, then we shouldn't be aiming for escape + 110 when heading out of Kerbin.

tldr: I think ÃŽâ€V maps should separate escape velocities at low planetary orbits from the additional cost of initiating the interplanetary transfer. Man, that's been bugging me!

[Yasmy]

Geschosskopf, you are reading the charts wrong in reverse. You have to reverse the order of the burns. Delta-v charts work perfectly fine in both directions. A Hohmann transfer is a completely reversible process. It costs the same either way, provided you don't aerobrake.

1) Go from small circular orbit to large circular orbit: Burn V1 raises apoapsis. Burn V2 raises periapsis.

2) Reverse the process: Burn -V2 lowers the periapsis. Burn -V1 lowers the apoapsis.

(And yes, in practice, it does actually work like this. Try it out from LKO to higher orbit and back.)

I.e.: On the reverse trip, the transfer burn reverses the old circularization burn, not the old transfer burn.

Let's look at the ÃŽâ€V cost of a Hohmann transfer:

Now here's the trick. Go ahead and try this at home. It's one line of algebra.

Exchange r1 and r2 in ÃŽâ€V_total and you find ÃŽâ€V(r2, r1) = -ÃŽâ€V(r1, r2).

It makes no difference which one of r1 and r2 is bigger.

If you include the effects of escaping and circularizing around massive bodies at either end, you find the same thing. Escape + transfer going from Kerbin to Planet X is equal to circularization at Kerbin inbound from plaent X. Circularization at Planet X from a Kerbin transfer orbit is equal to escape and transfer from planet X to Kerbin.

Edited August 1, 2014 by Yasmy
linky-link

[Yasmy]

Really, I should just let the above equations stand for themselves. Trust the math.

But I'll just use LethalDose's example of Kerbin to Duna:

LKO to escape: 950 m/s

escape to Duna: 110 m/s

Duna circularization: 370 m/s

Kerbin to Duna in two burns:

ÃŽâ€V1 = 950 m/s + 110 m/s = 1060 m/s for escape + transfer

ÃŽâ€V2 = 370 m/s for circularization

ÃŽâ€V = ÃŽâ€V1 + ÃŽâ€V2 = 1430 m/s

Duna to Kerbin in two burns:

ÃŽâ€V1 = 370 m/s for escape + transfer

ÃŽâ€V2 = 950 m/s + 110 m/s = 1060 m/s for circurlarization

ÃŽâ€V = ÃŽâ€V1 + ÃŽâ€V2 = 1430 m/s

Edited August 1, 2014 by Yasmy

[LordCorwin]

According to the map, intercepting Duna once you've escaped from Kerbin costs 110m/s. Has anyone of you achieved that? Normally it takes me between 1000 - 1200 m/s (10 times higher). Also, it takes me more than the 75 days mentioned.

[LethalDose]

@Yasmy above: Based on what I've posted above, I think there may be problems with your numbers. By my math, 370 m/s is the dV needed to reach escape velocity for Duna (1272 for escape and 902 for orbit, both at 52 km as posted on the map), however, you've listed that value as "escape + transfer". They can't both be right. I think if you burn 370 m/s of dV from a circular 52km orbit around Duna, you will not be in a transfer orbit (even if you do it at the right time).

I think the 110 m/s value represents the sum of the dV cost to start or end the transfer from escape velocity around Kerbin and the dV cost to start or end the transfer from escape velocity around Duna. Someone still needs to actually do the math precisely to see if this is true.

IF this is true, then your two burns would be:

LKO to transfer: 950 + X m/s

Transfer to LDO: 370 + (110 - X) m/s

Where 0 < X < 110

Basically, it should either take ~ 110 m/s total to start and stop the transfer (regardless of the direction) or the numbers on that chart are wrong. I think I'm siding with the former.

According to the map, intercepting Duna once you've escaped from Kerbin costs 110m/s. Has anyone of you achieved that? Normally it takes me between 1000 - 1200 m/s (10 times higher). Also, it takes me more than the 75 days mentioned.

IMO, the exact value of the required dV beyond escape velocity to transfer to Duna from LKO is up for debate, but yes, it's in the ball park of 110 m/s. Its definitely < 200 m/s.

There are reasons for the dV cost and transfer duration to be much, much higher. These include:

• Not transfering during a Kerbin -> Duna transfer window
  
• Waiting until you've left Kerbin's SoI to make your transfer burn
  
• Some other reason you didn't make your transfer burn in LKO
  
• Burning at a suboptimal point in your LKO that screws up your ejection angle from Kerbin
  

Edited August 1, 2014 by LethalDose
clarification and addressing subsequent comment