Deltav question again

[seaces]

Ok today I was doing some rescue kerbal contracts got some 3 perhaps to rescue in one mission one of them in mun orbit. However this time I pulled out those refreshed deltav maps and summed up the deltav for trip to the mun around 5050. However I am not sure that the correct way of founding deltav from mun to Kerbin is to again sum up all number from mun to kerbin so it will he again 5050 of deltav to and 5050 from mun. Of course I didn't put that deltav on my rescue ship simply becouse before I did trips from mun and experience told me that when I burn from mun orbit prograde at the side of the mun that faces Kerbin the manuver node to bring Kerbin Pe is couple of houndreds only. So clearly I am missing something here.

[SAI Peregrinus]

You're missing that you can aerocapture/aerobrake at Kerbin. So instead of taking 3300 delta-V to get from LKO to the surface it takes 0. Indeed, you don't even have to circularize in LKO, you can just go from the Mun into the atmosphere. So it should be 5950 by WAC's map's values (3300+860+310+580+580+310+0+0).

[seaces]

You're missing that you can aerocapture/aerobrake at Kerbin. So instead of taking 3300 delta-V to get from LKO to the surface it takes 0. Indeed, you don't even have to circularize in LKO, you can just go from the Mun into the atmosphere. So it should be 5950 by WAC's map's values (3300+860+310+580+580+310+0+0).

3300+860+310+580+(580+310+0+0)- this in brackets is return section?

so 580 is to get from surface to Mun orbit next come 310 from orbit to getting Kerbin intercept and end up with some kind of Orbit around Kerbin? So somehow those 860 that I see on map are not anymore needed? And by aerocapture I suppose that is to lower the Pe enough to get into circular orbit around Kerbin if I wish or let it drop to Kerbin?

[Yasmy]

Short answer: Yes.

Returning is pretty much the reverse of going. (This is because gravity is a conservative force.)

Takes 580 to land. Takes 580 to go back to orbit. (Excluding user error and safety issues.)

So think about what you are reversing. The first 310 m/s is circularization at the Mun from a Kerbin to Mun transfer orbit. Thus reversing the 310 just puts you back on a LKO-Mun transfer orbit. Imagine doing the capture burn at the Mun, turning around, and immediately burning 310 m/s in the opposite direction. You would be back on the orbit that brought you to the Mun.

So burning 310 from LMO does not put you in a LKO, but it can get you back close to Kerbin on a Mun-to-Kerbin transfer orbit.

Using Kerbin's atmosphere to slow you down, you can erase most or all of the 860 to turn a Mun-to-Kerbin transfer orbit into a LKO orbit, and most or all of the launch cost as well. This is because friction is not a conservative force. If you didn't want to aerobrake, it would cost 860 m/s to get back into a LKO orbit from your transfer orbit.

Edited June 20, 2015 by Yasmy

[FancyMouse]

3300+860+310+580+(580+310+0+0)- this in brackets is return section?

so 580 is to get from surface to Mun orbit next come 310 from orbit to getting Kerbin intercept and end up with some kind of Orbit around Kerbin? So somehow those 860 that I see on map are not anymore needed? And by aerocapture I suppose that is to lower the Pe enough to get into circular orbit around Kerbin if I wish or let it drop to Kerbin?

Right, except that aerocapture alone cannot reduce you to a circular orbit - you'll need to burn a little bit to let periapsis leave atmosphere to finish circularization. But other than that everything you said is correct, including that aerocapture could slow you down enough so that it directly leads to a landing.

Usually 30km is a good value for landing on Kerbin, and you'll need some higher value for just an aerocapture without landing.