I'm trying to wrap my head around ejection angles for transfers. The phase angle has to do with where you're headed, and knowing where it will be when you travel from where you are to there -- essentially that it will be there to catch the 'ball' of your spacecraft. That makes some decent sense to me.
The angle to exit your system (Kerbin, for my purposes) is a bit trickier to my head. The idea is to make the velocity you already have as a result of the system's orbit around the sun work for you and harness as much of it as you can. Burning a parallel prograde vector from 90 will send you to a higher orbit with nice efficiency, burning parallel to retrograde will send you in-system. The more d/v you burn in that direction, the further you'll go. Of course, you can't burn instantaneously, so you need to start a bit further back maybe. But that's not actually the way the ejection angles work. For instance, I've been plotting to transfer a probe out to Dres in the next window. I've got a comm sat in high (75 Mm) Kerbin orbit, I put some nodes on it to get an idea of the intercept for my probe. The dv from such a high orbit is greater than that for a low orbit, presumably because you don't have quite the orbital velocity from such a high altitude. But it seems the ejection angle is smaller for the higher orbits, despite the larger burn. Why? And why is the angle greater to reach closer orbits (i.e. Duna ejection angle is 150, Dres is 123 from the base orbit, per https://ksp.olex.biz/?) Is it that the closer your ejection angle is to 90, the (roughly) more powerful your burn is, so by burning at a wider angle, you're not going as far outward (perpendicular to the sun)?
I've been reading
https://ksp.olex.biz/
https://alexmoon.github.io/ksp/
but I want to understand why the numbers come out the way they do