How to calculate DeltaV to get to orbit or to get from orbit to land

[MH60AV8R]

Hey guys,

I'm trying to figure out how to calculate the Delta V required to get from Kerbin surface to orbit (any alt) and from Munar orbit to Munar surface. Basically I'm trying to get my numbers to come out similar to the Delta V map numbers. I know the Hohmann Transfer equations and Delta V equations that I need to use. I understand that the Kerbin to orbit calculation might be a little convoluted due to the atmosphere so if I have to hit the "I Believe" button, I will.

Basically I am using this example:

The Delta V map shows from a 14km munar orbit it requires 640 Delta V to land.

Im using a Mu value of 6.5138398*10¹⁰ m³/s² and a Munar radius of 200 km. r₁=214000 m and r₂=200000 m. If I plug the values into the Hohmann equations I only get between -9.4 and -9.7 m³/s² Delta V depending on whether I use delta v₁ or delta v₂ equation respectively.

Where am I going wrong?

Edited July 9, 2013 by MH60AV8R

[purpletarget]

Your hohmann numbers are probably correct, but...that assumes that your already doing orbital velocity at 0m altitude, which for the Mun is about 570m/s. The surface of the Mun however trundles along at a leisurely 9ish m/s...and good thing too, otherwise jumping over a trashcan would put a poor Kerbal into orbit instantly!

So, launching from the Mun requires a bit of Delta-V to get to altitude, but mostly on vacuum worlds, you need to gain (or kill) the orbital velocity needed for your desired orbit.

[K^2]

You are sitting on the surface. At equator that puts you at sidereal rotational speed. On the Mun, that's 9.04 m/s. Periapsis velocity for Hohmann is sqrt((r2/r1)GM/a) = sqrt((214,000m/200,000m)6.5138398x10¹⁰/(207,000m)) = 580.26. This means your first burn is going to be 571.22m/s. (I just subtract off the velocity you already have, the 9.04m/s. Assuming Eastward launch, of course.) The apoapsis velocity, using the same formula, will be 542.30m/s. You need 551.71m/s to circularize, so that's another 9.41m/s burn at apoapsis.

The total delta-V required to reach orbit, assuming Eastward launch from equator is, therefore, 580.63m/s. Realistically, it will be slightly higher, depending on how high your TWR is and how well you perform the gravity turn. For ascent in vacuum, the higher the TWR the better.

For a rough estimate you can also take escape velocity, which is 807.08m/s and divide that by 1.41 to get 572.40m/s. It's always going to be an underestimate, but it gives you the absolute minimum delta-V to break away from the surface.

Finally, the atmosphere. The simple formula to estimate additional delta-V to leave atmosphere is 4gH/v_t. Here, g is surface gravity, H is scale height of the atmosphere, and v_t is the surface terminal velocity. For Kerbin that works out to 1,832m/s. This assumes TWR of roughly 2 of your rocket, which optimizes fuel consumption on ascent.

[MH60AV8R]

Anyone else care to weigh in? This is REALLY bugging me because math doesn't lie...and don't say to download a Mod because I like the vanilla version of things.

[K^2]

What's wrong with equations I gave you? They are absolutely precise.

[MH60AV8R]

Sorry, I posted that last post and didn't see you had responded.

K^2, I like the math you presented. Where did you get the equation you used for the first Periapsis equation?

The equations I was using were Delta V1=sqrt(Mu/r1)*(sqrt(2*r2/r1+r2)-1) and Delta V2=sqrt(Mu/r2)*(1-sqrt(2*r1/r1+r2)). The equation you used looks like a derivation of the one I'm using. What are the definitions of 'M' and 'G'? And why did you use the average of the radii?

[MH60AV8R]

Ok, figured out the G and the M. I also found the equation you used. You worked the problem backwards from what I was asking though. Lets say I am parked in orbit, what Delta V does it take to get from that orbit to the surface?

[K^2]

Yes, these are the same equations, but for actual velocity, rather than change. Works out the same once you take everything into account.

Without atmosphere, perfect landing is the exact reversal of perfect takeoff. So ideally, delta-V is going to be the same. However, "perfect landing" requires a suicide burn. So you perform a de-orbiting burn to put yourself on transfer orbit that just touches the surface, and then you burn to halt right before hitting the surface. Of course, if you mess up the suicide burn, splat. Hence the name. So in practice, you burn earlier which means you'll need more fuel to fight gravity as you do final adjustments. How much extra fuel you use will depend on how much of margin you want for errors. I would recommend about 20% in reserve.

Then, of course, there are alternative landing techniques if you want to land at a precise location. These tend to require a lot more fuel, but it all depends on exactly what you are planning to do.

[Kosmo-not]

Of course, if you mess up the suicide burn, splat.

If your periapsis just touches the surface, you can easily avoid going splat.

[K^2]

Not once you started the deceleration burn. That drops apoapsis, causing your trajectory to cut through the planet/moon. So if it's just a matter of, "Oh, crap. Missed my window." Then yes, a small correction burn will get you to miss the surface and you can go for another try. But if you only realized that you are not going to make it half way through your burn, you're done.

[Arrowstar]

Please note that the calculations for atmospheric drag loss are, in general, highly path (that is, trajectory) dependent. How much fuel you expend on your way to orbit depends on how you fly. Any mathematical relationship you see for atmospheric drag loss should be regarded as a loose approximation until you get in game and can verify manually.

[K^2]

Very much so. Yet 4gH/v_t formula goes very nicely with all of the empirical results available on "best" ascent scenarios. While a loose approximation, it's a pretty good tool for estimating how much you are going to save by change of elevation, for example, which will change v_t value.

[MH60AV8R]

You are sitting on the surface. At equator that puts you at sidereal rotational speed. On the Mun, that's 9.04 m/s. Periapsis velocity for Hohmann is sqrt((r2/r1)GM/a) = sqrt((214,000m/200,000m)6.5138398x10¹⁰/(207,000m)) = 580.26. This means your first burn is going to be 571.22m/s. (I just subtract off the velocity you already have, the 9.04m/s. Assuming Eastward launch, of course.) The apoapsis velocity, using the same formula, will be 542.30m/s. You need 551.71m/s to circularize, so that's another 9.41m/s burn at apoapsis.

The total delta-V required to reach orbit, assuming Eastward launch from equator is, therefore, 580.63m/s. Realistically, it will be slightly higher, depending on how high your TWR is and how well you perform the gravity turn. For ascent in vacuum, the higher the TWR the better.

For a rough estimate you can also take escape velocity, which is 807.08m/s and divide that by 1.41 to get 572.40m/s. It's always going to be an underestimate, but it gives you the absolute minimum delta-V to break away from the surface.

Finally, the atmosphere. The simple formula to estimate additional delta-V to leave atmosphere is 4gH/v_t. Here, g is surface gravity, H is scale height of the atmosphere, and v_t is the surface terminal velocity. For Kerbin that works out to 1,832m/s. This assumes TWR of roughly 2 of your rocket, which optimizes fuel consumption on ascent.

So I'm having a little trouble understanding where the following numbers came from: "You need 551.71m/s to circularize" and "The total delta-V required to reach orbit, assuming Eastward launch from equator is, therefore, 580.63m/s" I got the rest of the numbers figured out but I can't figure out where the 551.71 comes from. Also, the 580.63 is very close to what was calculated at 580.26 m/s, why are these different?

[K^2]

551.71 m/s is the orbital velocity at 14km altitude. You can use the formula I posted earlier with r₁ = r₂.

As for the difference, I suspect rounding errors. They are more than sufficient to introduce a discrepancy of 1 part per 1,000.

[MH60AV8R]

I really appreciate your help K^2. It's been awhile since I took an orbital mechanics class. Can you explain why we calculate both the apoapsis and periapsis velocities?

I figured out where you got the 580.63 number from. You took the difference of the orbital velocity and apoapsis velocity which was 9.41 and added it to the periapsis velocity to get the total required. So, from my understanding, we calculated we needed 571.22 m/s of velocity to get to 214km altitude and 9.41 Delta V to circularize?

[K^2]

So, from my understanding, we calculated we needed 571.22 m/s of velocity to get to 214km altitude and 9.41 Delta V to circularize?

Correct. The actual altitude will read as 14km, of course, since the game will subtract the radius of the Mun.

Can you explain why we calculate both the apoapsis and periapsis velocities?

Because the ship will slow down as it races from periapsis to apoapsis. (Or vice versa, speed up on the way down.) This is just consequence of conservation of energy.

[tavert]

Not once you started the deceleration burn. That drops apoapsis, causing your trajectory to cut through the planet/moon. So if it's just a matter of, "Oh, crap. Missed my window." Then yes, a small correction burn will get you to miss the surface and you can go for another try. But if you only realized that you are not going to make it half way through your burn, you're done.

You don't have to point retrograde during the whole braking burn. I thought that would be most efficient too, until Nao demonstrated otherwise in some other thread... It turns out to be better to pitch up and keep your vspeed at 0 during the braking burn, though the difference is only noticeable if your TWR is in the 1-1.5ish range.

[K^2]

That is not true. Burning retrograde optimizes fuel consumption. It's simple calculus. Something wasn't done right in these experiments.

[tavert]

Let's do the calculus then. Simple-ish, nonlinear so we'll need to pick some example cases and integrate numerically. Perhaps a new thread on numerical trajectory optimization is in order?

I convinced myself with a quick little Euler-integrating spreadsheet, but it's not hard to do something more sophisticated. If you allow your altitude to drop, you also gain horizontal speed which you have to counteract. By my numbers it looks worse than the steering losses you incur in a constant-altitude landing.

[Mr Shifty]

Perhaps a new thread on numerical trajectory optimization is in order?

Please post a link here if you do this. I would love to see the math.

[Arrowstar]

I would love to participate as well!

[Mr Shifty]

Here's a test case to start with:

Start from a 10 km x 0 km, equatorial, eastbound orbit over the Mun.

The landing burn starts at 1 km altitude; here are some parameters for the orbit at that point:

(These speeds are orbital, not surface speeds.)

Vertical speed: 8.4 m/s

Horizontal speed: 574.7 m/s

Total speed: 574.8 m/s

Time to periapsis: 228.7 seconds

Contest: Total m/s required using a constant thrust retrograde burn vs. constant altitude burn followed by a soft landing burn.

Edited July 8, 2013 by Mr Shifty

[tavert]

Posing this as a challenge would be a good way of experimentally confirming the math here, and might get Nao's attention - he's much better at flying the constant-altitude landings than I am. But maybe Minmus would be a better target, since there we could require landing on one of the lakebeds to rule out any differences due to landing altitude. The results will be most interesting for a fixed craft with low TWR, which for Minmus would mean a tiny engine.

[Mr Shifty]

Posing this as a challenge would be a good way of experimentally confirming the math here . . .The results will be most interesting for a fixed craft with low TWR, which for Minmus would mean a tiny engine.

Or a very large craft:)

Sounds like a good idea; I'll have to work on setting up the challenge when I get a chance. Probably a persistence file with a fixed, low TWR craft in a low circular orbit around Minmas.

I'd still like to see the math though.

[tavert]

Let me dig up some papers I was looking at last month...

How much of http://link.springer.com/article/10.1007%2Fs12555-011-0618-0 (click "look inside") can you guys see if you're not coming from an academic domain? I was going to start from those dynamic equations on page 2, but do a simpler discretization than the pseudospectral stuff they used, and not impose some of the constraints they had that are unnecessary for crazy KSP landings. Start with equatorial for simplicity, so phi = alpha = v_phi = 0.

I was planning on using http://openopt.org for formulation and interface to optimization solvers, though any simple ODE integrator can tell us the answer for the restricted cases of retrograde burn vs constant-altitude burn.