Is the ideal transfer between two arbitrary planar orbits always a bi-tangential ellipse?

Question

I was reading Hollister David's short paper on bi-tangential transfers. The example he uses is a transfer between a planar circular and an elliptical orbit. I am wondering: Is an ideal, that is, the lowest $\Delta v$ cost, transfer between two arbitrary orbits around a single point mass always an ellipse that is tangential to both the orbits? I am excluding the cases where a bi-elliptical transfer is ideal, as the answer is then always a manoeuvre with infinite apoapsis.

I am not entirely comfortable with trusting that as a safe assumption, as although paying a $\Delta v$ expense in another direction than prograde or retrograde is expensive, it is by no means horrible. If this assumption is indeed not true, what criteria must to planar arbitrary orbits satisfy in order to make the ideal transfer between them a bi-tangential ellipse?

Clarification:

A transfer orbit being "tangential" to an orbit means that in the transition between the orbit and the transfer orbit, velocity change is only applied in the prograde or retrograde direction. A "bi-tangential transfer" is when the transfer orbit between a pair of orbits is tangential to both of them. As a consequence, this problem is strictly planar. The "bi-elliptical transfer" I am referring to is when the alternative with the lowest cost is to do a burn at periapsis accelerating to infinity, then performing zero-cost manoeuvres "at infinity" before falling back to the other orbit's periapsis.

Example of a bi-tangential transfer. Both burns are done tangentially:

Bounty:

I had a 100 rep bounty on this question that expired gaining only a partial answer. Because it is a bit unfair to a potential complete answer that another one got the bounty, There is now a 500 rep bounty running.

Answer 1

Thanks for linking to my pdf!

I have always assumed bi-tangential transfers took the least delta V. But your question has made me realize my assumption is conjecture.

My goal is to find a general equation for delta V, integrate it, and then hope the minima of the manifold will correspond to bi-tangential orbits.

Sometimes playing with conics is rewarding. It is a delight when complicated equations reduce to something simple and elegant. But so far I've been frustrated. Poking and prodding at these equations have only caused them swell up like an angry puff fish. I'm sharing my efforts in the hope folks will help me cut a path through this thicket of thorns. I will add to this as I have time.

Units

When using AU (Astronomical Unit) and years, the sun's gravitational parameter GM is easy to describe: $\mu = 4 \pi^2 AU^3/year^2$

Circular orbit velocity is described as $V = \sqrt{\mu / (rAU)}$

For earth's orbit r = 1. Plugging $\mu$ and earth's r into the above, we get earth's velocity is $2 \pi AU/year$ which is reassuring.

Finding speed at arbitrary rendezvous points

Picking an arbitrary point rendezvous $P_1$ sets a quantity $r_1AU$. This quantity $r_1AU$ is distance from $P_1$ to sun. $P_1$ is the rendezvous point where transfer and destination orbits intersect. ($P_0$ will be rendezvous point where transfer and departure orbits intersect.)

Using the vis viva equation we can find speeds of payload and destination at point P.

$V = \sqrt{\mu (2/(r AU) - 1/(a AU))}$

Where aAU is length of ellipse's semi major axis.

Recall with our units $\mu = 4\pi^2AU^3/year^2$. So the vis viva equation becomes:

$V=(2\pi AU/year)* \sqrt{2/r -1/a}$

So the speed of a body in elliptical orbit is earth's speed times $\sqrt{2/r-1/a}$

So...

$V_{payload} = V_{earth} * \sqrt{2/r_1-1/a_2}$

and

$V_{destination} = V_{earth} * \sqrt{2/r_1-1/a_1}$

Flight Path Angles

We have speeds of payload and destination at point $P_1$ but we don't have direction. For that we need to find the difference between the payload and destination flight path angles. I will call that angle $\phi$

Will try to add to this soon.

Answer 2

This is maybe not so much an answer as an observation I made while playing KSP; I don't know or have much interest in the math to look into this more rigorously, but maybe it can be useful as a starting point for further investigation.

I was trying to get from the cyan orbit to the pink orbit on a spacecraft with very limited delta-v and this was the best series of maneuvers I could come up with (and even then my remaining fuel was just barely enough).

The first maneuver (yellow, 54.9 m/s) was made at 60 degrees off tangent (27.5 m/s prograde and 47.5 m/s anti-normal), and so was the second maneuver (purple, 79.8 m/s: 40 m/s retrograde and 70 m/s anti-normal). I determined the position (timing) of the maneuvers experimentally, but they look like they're about halfway between apoapsis and periapsis. I tried playing with other timings and directions but nothing else seemed any better, though fiddling with maneuver nodes in KSP probably isn't exactly the best tool for this kind of optimization.

After seeing this question, I briefly tried to plan the same transfer using a bi-tangential ellipse but couldn't come up with anything that looked as efficient. I would be interested to hear if the strategy I used is provably better!

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I found a diagram of something that looks similar to my strategy from the paper "New Solutions to Impulsive Correction for Argument of Perigee Using Gauss’s Variational Equations" by Gang Zhang, Haiyang Zhang, and Xibin Cao. I think this method comes from "Impulsive Transfer between Elliptical Orbits" by Derek F. Lawden. Interpreting these papers is beyond me, but I hope they're useful to someone with more relevant knowledge.