Please Help! About the ejection angle

[OME]

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.

 

[Boris-Barboris]

1. Lowest and precisely as needed Pe means you won't need additional burns to correct you trajectory in the middle of the transfer -> can mean fuel efficiency.
    It is only optimal when you define optimal as "get lowest Pe". Task of getting optimal ejection when optimal is defined like "get this particular Kerbin Pe from this particular Mun orbit while wasting minimum fuel" is more interesting.

2. It's a rule of thumb, and a convenient one, but it is not optimal.

3. No: different problem. Neither there is any relation to Hohmann transfer.

Most of optimisation tools you'll find for KSP handle planet-planet hops, because they are the ones that usually matter. People rarely care about spare 10m\s when playing with Kerbin moons. I'm not even sure that KSP TOT cares.

I'd recommend to just play the game the way intuition tells you to, it's more fun anyways. Word 'optimal' is mentioned far too often for my taste on this forum.

[FullMetalMachinist]

1 hour ago, OME said:

1, What is the relationship between the lowest Pe and the optimal efficiency? Why is the-lowest-Pe way equal to the optimal way?

That relationship only exists because the objective of that tutorial is to lower your Kerbin PE. It would be better to think about it more like "a burn of a certain m/s at this spot in my Mun orbit has a bigger effect on my Kerbin PE, and that same burn has less effect on my Kerbin PE on that other spot." 

The 'most efficient' spot to burn is all relative based on what you want to do after you eject from the Mun. It's all about using the same size of burn at different times in your Mun orbit to effect your Kerbin orbit in a specific way. 

1 hour ago, OME said:

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?

You've basically already shown what is meant by 'parallel' in that screenshot. 

The reason that it most efficient (for what you're trying to do) has to do with relative velocities. The Mun has an orbital velocity of 542.5 m/s. Let's say that your ship is in Mun orbit of 200 m/s. That means that on the 'far' side, when your ship is traveling in the same direction as the Mun's orbit, you add that, and your ship's velocity around Kerbin is 742.5 m/s. When it's on the 'near' side and traveling opposite (retrograde) of the Mun's orbit , you subtract, and your ship's velocity around Kerbin is 342.5 m/s. 

In the case of returning to Kerbin, the near side is more efficient because to lower your PE to the atmosphere, you need to burn retrograde. And on the near side 200m/s of that burn is given to you for free. It's the same rough principle that gravity assists work on. 

1 hour ago, OME said:

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?

I'm not really sure what you're asking here. But that site is mainly for finding the transfer windows, which is simply the timing of when to leave Kerbin to have the least expensive dV flight to another planet. 

[Waxing_Kibbous]

This video may help, although the technique described is best for Eve and Duna- around 4:30 I believe the concept of the ejection angle being parallel is mentioned:

https://www.youtube.com/watch?v=7tQOZl1d_m8

For Kerbin's moons though I usually plop a node about where it is in the OP image, pull prograde until the Pe approaches Kerbin, adjust the position of the node on the orbit using the ring until I get the most efficient position, which is the position where the Pe is at it's lowest. As you move the node along the orbit you will see the Pe sink then rise again, you want the node at the position where it is lowest. If that position happens to be inside of Kerbin that's bad, a good height when returning from the moons is a Pe around 25,000m. Lastly adjust the prograde burn accordingly.  Doing this will show you visually what the tutorial says.

https://www.youtube.com/watch?v=u1SSHWM_phU

this video discusses the idea of slowing down/speeding up to meet objects in higher or lower orbits.

[Nicias]

Basically your ejection burn does two things: 

1) Gets you out of Mun's SOI, since the SOI's are finite, you exit with some surplus velocity.

2) Changes your orbit around Kerbin, using the surplus velocity. I think about it like you start in the same orbit as Mun. You want this to be retrograde, so you should make your surplus velocity is retrograde, so (anti-)parallel to your "orbit" (really the Mun's orbit)

[Plusck]

1) You've already seen in the first orbiting tutorials that the best place to raise/lower Ap is at Pe, and the best place to raise/lower Pe is at Ap.

When you leave the Mun's SOI, your surplus speed (called V_∞ or V_inf) has the effect of a retrograde burn from a circular, Mun-altitude orbit. Therefore you want that velocity change to be perfectly retrograde. That means leaving the SOI on a line parallel to its direction of movement.

So, if you imagine you were following the Mun's orbit (without the Mun being there), and had the absolute minimum amount of fuel on board to get home, you would have no option but to burn directly retrograde. If you burned at any other angle, your Pe would necessarily be higher.
Therefore, watching your Kerbin Pe as you drag the node around indicates whether you have that retrograde ejection angle perfect, or not.

2) Actually, it isn't quite the most efficient if you make it perfectly parallel at the time of the burn. It takes time to leave the Mun's SOI, by which time the Mun will have travelled along its orbit. You should find that the most perfectly efficient return trajectory is therefore going to be rotated slightly (anti-clockwise, from the point of view shown in your pic).

The faster the ejection speed, the closer to perfectly parallel it should be (if you leave at the Mun's orbital velocity, you are essentially coming to a dead stop in space and won't move from spot you're in right now as the Mun moves away from you - and in this case you would want the angle to be perfectly parallel - but on leaving the Mun's SOI you will just fall straight down into Kerbin).

3) When you leave the Mun to go home, you're optimising the angle of your V_inf velocity to reduce Pe around the major body, Kerbin.

When you go interplanetary, you want to optimise the angle of your V_inf velocity to raise Ap or reduce Pe around the major body, the sun, to get to a different level orbit around the sun. It's exactly the same principle: for maximum efficiency, you change orbit by changing velocity purely prograde or retrograde, which means leaving along a line parallel to your starting orbit.

Edited October 25, 2016 by Plusck

[Reactordrone]

7 hours ago, OME said:

 

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?
 

 

Think of it as if the Mun isn't there. If your ship was just orbiting where the Mun is and you wanted to drop your periapsis down to Kerbin you'd have to burn retrograde to do so. Making your ejection burn from the Mun give you that retrograde heading also takes advantage of the Oberth effect. Returning from low Mun orbit uses about 270m/s, if you were free flying at the same altitude as the Mun returning to Kerbin would use about 370m/s and if you burn to just barely escape from Mun orbit and then do a separate burn to get back to Kerbin  you'd use over 500m/s.