Firstly, I came across the following words in 'From the Mun' Training.
“You'll find that you have the lowest Pe when your ejection vector is parallel to Mun's orbital track but heading the opposite direction. (If the Kerbin Pe disappears under Kerbin's surface, drag on the retrograde handle to reduce the burn until you can see the Pe again. This will keep your reentry survivable.) You've now created an ejection maneuver which, despite being prograde in munar orbit, gives you a retrograde kick in terms of your eventual Kerbin orbit! You can use the same principle to travel to other planets from Kerbln, although there you'll need to be mindful of "transfer windows" (a term for the most efficient times to travel between planets).”
It means what blizzy78 made in his reply here: http://forum.kerbalspaceprogram.com/index.php?/topic/59299-ejection-angles/
According to what the Training says, this kind of 'parallel' can achieve the lowest Pe (and seems to be optimal efficient way).
But that quite confusing me! I wonder WHY the method could be the best one.
In other words, I have severals following questions about it:
1, What is the relationship between the lowest Pe and the optimal efficiency? Why is the-lowest-Pe way equal to the optimal way?
2, What is exactly the reason why there should be such a 'parallel' to achieve the most efficient way? What does the 'parallel' really mean?
3, I have learnt about Hohmann transfer and this website:http://ksp.olex.biz/ but each of them demonstrates the transfer of bodies orbiting the same parent. Does they have something to do with this case? And how to say that?
Or we can just have a discussion. This question has confused me for couple of days..... thanks to everyone helping me out.