Just a quick check of logic

[seaces]

I was digging around wiki KSP and found this one: http://wiki.kerbalspaceprogram.com/wiki/File:KerbinDeltaVMap.png

So if I am correct for example to get dV to Moho from Kerbin I just add up all the number up to Moho and I get dV for Kerbin-Moho and if I want to get back from there I would add up again those all numbers from Moho to Kerbin?

Edited May 5, 2015 by seaces

[Randazzo]

Basically, yes. The LKO Delta V is different now, so starting from space:

LKO - LMO (50k)(worst case): 950 + 2520 + 2410 = 5880

LMO - Surface: ~870 (Always give yourself some margin for error)

Moho Surface - LKO (worst case): 870 + 2410 + 2520 + 950 = 6750

Total dV: 5880 + 870 + 6750 = 13500 (from LKO)

This is "worst case" based on the map. If you use the Launch Window Planner and give it your in-game date accurately, it can help you find out ahead of time how much it may actually cost. The windows are not indentical.

I suppose it goes without saying (yet here I am saying it) that if you send a ship with 14000 dV but the lander only has 1000 dV, your kerbals are staying on Moho.

Edited May 5, 2015 by Randazzo

[Gaarst]

Yes that's right but remember that atmospheric delta-v predictions are outdated on this map since 1.0 and that these delta-v are usually for optimal manoeuvres (especially for interplanetary trips), always have more delta-v than you actually need.

EDIT: ninja'd

[seaces]

Good. Just one more question to Mr. Randazzo. In your example you said: LKO - LMO (50k)(worst case): 950 + 2520 + 2410 = 5880. I have looked at the map and I see that your mentioned 2520 stands above the red line leading to Moho above 760dV(which is btw never mentioned in your post). Why did you ignore the number 760 and just used teh 2520?

[Randazzo]

Good. Just one more question to Mr. Randazzo. In your example you said: LKO - LMO (50k)(worst case): 950 + 2520 + 2410 = 5880. I have looked at the map and I see that your mentioned 2520 stands above the red line leading to Moho above 760dV(which is btw never mentioned in your post). Why did you ignore the number 760 and just used teh 2520?

If I have read the map key correctly the black number above the line is the highest possible delta V amount that could be required at that node.

Edit: Nope, that's incorrect. So you would have to add 760 each way for absolute worst case scenario dV. Good catch.

[Gaarst]

I always though that the number above the delta-v requirements meant the delta-v required to align orbital planes

Guess I learnt something today

[Randazzo]

I always though that the number above the delta-v requirements meant the delta-v required to align orbital planes

This is correct. It is the maximum dV required for plane change.

[seaces]

So it would be like this: LKO - LMO (50k)(worst case): 950 + 2520+760 + 2410 = 5880 ?

[Randazzo]

So it would be like this: LKO - LMO (50k)(worst case): 950 + 2520+760 + 2410 = 5880 ?

LKO-LMO: 950+2520+760+2410 = 6640

LMO-Surface- ~870

Surface-LKO: 870 +760+2410+2520+950 = 7510

Total: 6640 + 870 + 7510 = 15020

[Gaarst]

@seaces

Yes and no.

That would be best case scenario for good launch window, but bad orbit orientation (ie: wrong plane, radial correction needed): 950 + 2520 + 760 + 2410 = 5880 m/s

Best case scenario would be: 950 + 760 + 2410 = 4120 m/s

Worst case scenario would be launching at the opposite of the optimal launch window, then transfer delta-v can go up to several km/s

[seaces]

Well, thank you very much! That would be it.