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?


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:

bi-tangential transfer


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.

  • $\begingroup$ This is all nice for a simplified 2-body problem (1 massive + 1 negligible mass). When you take gravity assists, or orbit-matching (e.g. transferring between two planets) this becomes way more complex. But yeah, for optimal results departure and insertion are pretty much always prograde/retrograde burns, and that results in tangent to the orbit. $\endgroup$
    – SF.
    Commented Jul 3, 2016 at 23:55
  • $\begingroup$ @SF This is just for the 2-body problem, a purely theoretical question. You say "pretty much always", but my question is: it is always the case? :) $\endgroup$ Commented Jul 4, 2016 at 0:23
  • $\begingroup$ At first, I wanted to reply "If you can perform impulsive burns of arbitrary power, optimize for fuel (not time), then yes, always." But then I thought about a specific case: two strongly elliptical orbits that differ only in argument of periapsis. Same eccentricity, same semi-minor, semi-major axis lengths, only with major axis pointing at different azimuth; not far apart. And I'm no longer sure. $\endgroup$
    – SF.
    Commented Jul 4, 2016 at 7:01
  • $\begingroup$ I'm new to this and uncertain of the presumed ground rules. What are the constraints on initial and final eccentricities? Are $\epsilon_1$, $\epsilon_2$ < 1? Any burn schedule of arbitrary magnitude, direction, and duration envelopes is allowed, or only multiple discrete, instantaneous impulses? And the question only applies to co-planar orbit pairs where a single-ellipse bi-tangential transfer is lower $\Delta v$ than any multi-elliptic bi-tangential transfer? There are no massive objects in these orbits, and arbitrarily long transfer time is allowed? $\endgroup$
    – uhoh
    Commented Jul 4, 2016 at 15:59
  • 1
    $\begingroup$ @uhoh 1. All eccentricities are =< 1 2. Burns are assumed to be instantaneous. 3. yes, coplanar by definition, and partially yes, where any transfer has lower $\Delta v$ than a bi-elliptical transfer. 4. The only object with a significant mass is the central point mass, and 5. Arbitrary long transfer times are allowed. $\endgroup$ Commented Jul 4, 2016 at 16:27

2 Answers 2


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.


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.)

Transfer and destination orbits

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 $\sqrt{2/r-1/a}$


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


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

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 $\phi$

Angle between velocity vectors

Will try to add to this soon.

  • $\begingroup$ Thank you for replying directly! I can see your reasoning for how the bi-elliptical transfer I excluded can in a sense be viewed as bi-tangential too. However, bringing in an inclination change makes this overly broad, as I am not sure if the hypothesis is right even for the coplanar case. As I see it, going out of a plane is fundamentally not a tangential manoeuvre. $\endgroup$ Commented Jul 1, 2016 at 17:13
  • $\begingroup$ @Hohmannfan (Googling tangent…) Seems in addition to "just touching", tangent also means going the same direction. So I guess you're right transfers between non coplanar orbits couldn't be bitangential. In the coplanar case, a proof doesn't occur to me. So at this point my belief is just an opinion. $\endgroup$
    – HopDavid
    Commented Jul 2, 2016 at 3:24
  • $\begingroup$ My bounty is about to expire. This is not quite the answer I wanted, but because of your other good answers here, and your interesting work on tether transportation, I think you deserve the 100 points anyway. $\endgroup$ Commented Jul 10, 2016 at 15:27
  • $\begingroup$ @Hohmannfan Thanks, but I certainly don't deserve the bounty. Your question made me aware that a "fact" I had always taken for granted is actually an unproven conjecture (so far as I know). Since you asked it I have been searching for a proof (or disproof it turns out not be true). So far without success. $\endgroup$
    – HopDavid
    Commented Jul 11, 2016 at 3:19
  • $\begingroup$ After a several rounds of bounties, I give up trying to bring attention to the question to get a definitive answer. Case closed. $\endgroup$ Commented Nov 10, 2016 at 9:49

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!

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.

Transfer between equal ellipses using two symmetric impulses.

  • $\begingroup$ At least this is useful as it provides a possible counterexample! $\endgroup$ Commented Sep 14, 2023 at 13:51

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