Ejection angles for direct transfer, help me understand

[Rainshine]

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

[Zhetaan]

3 hours ago, Rainshine said:

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.

This is correct; assuming that you don't need to start from a high orbit, it's usually better not to do so because the savings from starting in low orbit, with its higher velocity, outweigh most other factors.

3 hours ago, Rainshine said:

But it seems the ejection angle is smaller for the higher orbits, despite the larger burn. Why?

The reason is because you are still in Kerbin's gravity well.  As you climb out of the well, Kerbin's gravity warps your trajectory, so you need to change your starting point so that Kerbin warps your exit beneficially, in order that you leave Kerbin's sphere of influence parallel to Kerbin's prograde vector and have an efficient transfer.  In other words, rather than starting at the 'right' point and having Kerbin warp your trajectory off-course, you deliberately start 'off-course' and let Kerbin correct it for you.

The farther you are from Kerbin when you start, the less it warps your trajectory.  Eventually, you escape Kerbin entirely and begin from a solar orbit, in which case you would just burn prograde to reach outer planets.  You also lose on the advantages of a higher initial velocity, and as I mentioned above, those are normally more valuable.

3 hours ago, Rainshine said:

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)?

It's about the velocity, first and most importantly.  Duna is closer than Dres, so your hyperbolic excess velocity (for practical purposes, that's the velocity that you have at Kerbin's sphere of influence boundary that is over and above escape velocity) is going to be lower for Duna because you're transferring to a lower solar orbit.  It is true that the more powerful your burn is, the less you need to adjust your ejection, but that's because a more powerful burn results in a higher Kerbin-relative velocity, which results in less time spent in Kerbin's sphere, which results in less time for Kerbin to warp the trajectory.  Your angle will affect how far you eventually fly from the sun, but not because the burn is more or less powerful--burn for the same amount of time at different ejection points, and a burn that sends you to Moho from one point will take you past Jool from another.

Let's say that you want to go to Duna; you neither need nor want a burn that, at a different angle, will take you to Jool, because you either want that propellant for use at Duna or else not to take the extra mass.  There is a trade-off:  if we assume that you have enough propellant to reach Duna from half, or nearly half, of possible ejection points from Kerbin (half because the other half will send you in-system rather than out), then for most of those points, you will not only reach Duna but will also fly past it in a classic case of having too much propellant for the task at hand.  If we reduce the propellant for the burn, then the angles that will allow you to reach Duna become constrained because the most inefficient angles at the extremes of the arc will no longer send you quite far enough to get to Duna's orbit.  Eventually, you reach one point at a specific angle that will get you to Duna for minimum propellant cost.  Any other angle would require more propellant, and any less propellant will see you fail to reach Duna no matter where you burn.  That point is the burn point, and that angle is the ejection angle.

This is exactly the same reason why we prefer prograde/retrograde burns to radial ones for transfer manoeuvres.  Your ejection angle tells you the right place to burn so that your eventual exit to the interplanetary transfer orbit is as parallel to the planet's prograde/retrograde vector as possible, such that it makes the burn fuel-efficient.  Burning off-angle is exactly the same as burning off-prograde in the more local sense.  There may be reasons to do it (adjusting arrival times for a flotilla of probes so that you're not capture-burning all of them at the exact same time, for example), but those reasons don't often have anything to do with efficiency.

Edited June 26, 2020 by Zhetaan

[ManEatingApe]

10 hours ago, Rainshine said:

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)?

Your intuition is correct - the faster you burn, the less that Kerbin warps your orbit as you escape, so the closer your angle can be to prograde.

This is why Dres's transfer angle is closer to prograde than Duna's transfer angle. The ejection angle is not about giving you time to make the burn; all maneuver node burns are assumed to be instantaneous. Instead it measures how much Kerbin bends your trajectory.

In real life, escape trajectories are mathematically hyperbolae. (NOTE: Technically also parabolas and in KSP there is a big exception...but more on that in a bit).

The eccentricity of a hyperbola is given by: e = v_p² / (r_p × μ) - 1
Where v_p is your velocity at periapsis after your transfer burn, r_p is the radius at which you made the burn and μ = GM
is the standard gravitational parameter for Kerbin.

The angle of the tangent to a hyperbola at infinity is related to sin^-1(1 / e) and the angle relative to prograde is π/2 + sin^-1(1 / e).
Now we can see that the faster we go, the greater the eccentricity and the closer the angle from prograde becomes to 90 degrees.

For example a parabola (barely escaping, eccentricity = 1) would have a angle to prograde of π/2 + sin^-1(1) = π = 180 degrees
If we could go infinitely fast the angle to prograde would be π/2 + sin^-1(1 / ∞ ) = π/2 + sin^-1(0) = π/2 = 90 degrees

To see this visually here are some graphs of hyperbolae of increasing eccentricity showing the tangent line at infinity.

In KSP however there is a big difference from reality. KSP uses patched conics, where beyond a certain radius the planet's gravity does not affect you at all. This is the planet's sphere of influence (SOI). This makes all orbits instances of the 2-body problem, which is easier to reason about mathematically. It also means that the formula for the tangent that I used earlier is wrong when applied to KSP.

Instead of finding the tangent at infinity we need to find the tangent at the SOI radius r_s (for Kerbin 84,159,286 meters). This can be calculated by:

Semi-major axis is: a = r_p / (1 - e)
Cosine of the eccentric anomaly (E) is: cos(E) = (a - r_s) / (e × a)
Slope of the tangent at radius r_s is: m = cos(E) × sqrt[ (1 - e²) / (1 - cos(E)²) ].
Angle from prograde = 2π - tan^-1(m)

Interestingly, if you examine the source code for Alex Moon's solver, it approximates the tangent using the formula for the true anomaly (θ) at a given radius:

r_s = a × (1 - e²) / (1 + e × cos(θ))
θ = cos^-1[ (a × (1 - e²) - r_s) / (e × r_s) ]

This is probably close enough.

This MIT lectures notes has very nice details on interplanetary transfer orbits. The Wikipedia articles on eccentricity, hyperbolic trajectory, Kepler's equation, true anomaly and eccentric anomaly are also great reading.

What's also fun is that KSP allows you to escape from a planet with an orbit that is mathematically an ellipse, by "chopping" off the top of the ellipse once you go out of SOI. In the Jool system, it's possible to transfer from Tylo to Laythe with such an orbit.
 

Edited June 26, 2020 by ManEatingApe