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The question How to intuitively explain that reaching geostationary orbits consume more Delta V than escape orbits? includes the graphic below, which illustrates that when transferring between circular co-planar orbits where the final orbit is higher than the initial orbit, it is sometimes more efficient to use two elliptical transfers.

For example going from r=1 to say r=20, it requires less delta-v to use one Hohmann transfer from r=1 to r=100 then a second from r=100 to r=20, than a single Hohmann transfer from r=1 to r=20.

I've never heard of a tri-elliptic transfer, so I'm assuming it would not be more efficient in terms of delta-v than a bi-elliptic transfer.

Question: If two ellipses are better than one, why aren't three ellipses better than two? How can one intuitively explain that tri-elliptic transfers are not more efficient than bi-elliptic transfers in terms of delta-v?

bi-elliptic transfer efficiency

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    $\begingroup$ I find it difficult to visualise such tri elliptic orbit. Would you add a picture to illustrate what's in your mind? $\endgroup$ – Diego Sánchez May 22 '18 at 6:09
  • $\begingroup$ @DiegoSánchez I've given some thought to this and I'll add some illustrations tomorrow. $\endgroup$ – uhoh May 22 '18 at 18:30
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We'll only consider the simplified circular-circular transfer because it makes our lives easier. In this case the same considerations apply as with the bi-elliptic transfer:

  • We only want to burn at apsis or periapsis - any other point will waste dV changing the argument of periapsis. This means we will always be raising or lowering the opposite Ap/Pe.
  • We also only want to burn exactly prograde or retrograde - any other direction will waste dV changing the eccentricity without changing the energy of the orbit.
  • We'll burn every time we reach Ap or Pe - in reality we may want to wait several phasing orbits between burns, but we can ignore this. If we burn at Pe, miss the next Ap, then burn again at Pe, we could have just done one larger burn. So we burn on alternating Ap and Pe.

Taking some current orbit O - with apoapsis a and periapsis p - and our circular target orbit T, we can see that there are three possible relative 'configurations', followed by the final orbit:

  1. Both a and p below T
  2. a above (or at) T and p below T
  3. a above T and p above (or at) T
  4. Final orbit - a and p at T

With a single burn, we can either remain in the same configuration or change to a new configuration that shares either a or p with our current configuration. Note: the 'or at' means we can share a or p with the final orbit.

However, we want to change the configuration every time we burn - if we don't change the configuration, we will essentially be doing a rough spiral and therefore lose out on Oberth-efficiency as discussed in this answer. So we always want 'changes' - if we find ourselves with a particular combination of a and p values that mean we can't change to a new configuration, we know the previous burn can't have been the most efficient.

Because we burn on alternating a/p, it's impossible to return to a configuration we have just left, so we only want to use 'forward-changes'

Therefore starting at 1 we can see:

  • From 1, we only share p with 2, so we have to move to 2.
  • From 2, we have two possibilities:
    • If a is at T, we share a with 4 so can move directly there.
    • If a is above T, we share a with 3 so have to move there.
  • From 3, we have to move to 4.

The sequence 1-2-4 is the special-case of our old friend the Hohmann Transfer and the sequence 1-2-3-4 is the Bi-elliptic Transfer. Since we never defined any actual altitudes for the orbits, they can be arbitrarily selected and we can choose which of the two options is more efficient.

Et voilà! - we can logically reduce a sequence of burns to the key 'forward-configuration-changing' ones because any other burns are either not possible or - at best - as efficient as a single burn.

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  • $\begingroup$ I'm confident my reasoning is sound here, but obviously if I haven't articulated it well enough, it won't count as intuitive! $\endgroup$ – Jack Aug 6 '18 at 21:38
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    $\begingroup$ I'm going to sit myself down with paper and pencil today and work through this. I've often wondered if there could be some symbolic operators similar to QM's raising and lowering operators for Keplerian orbits. It might work more like optical ray tracing matrix multiplication where each burn, and each a-to-p or p-to-a delay is one in a long string of matrices. $\endgroup$ – uhoh Aug 7 '18 at 0:48
  • $\begingroup$ I'm having trouble with the "above or at" parts. For example, if at 2, a is above T, then I don't see how I can get to 4 in one step. I'll need two burns: one at a to raise p to T, and another at p to lower a to T. Similarly, if at 3, both a and p are above T, I'm going to need two burns to get to 4: one at a to lower p to T, and another at p to lower a to T. I wonder if there's ambiguity in the wording that wouldn't trouble me if I knew more about transfer orbits. $\endgroup$ – Wayne Conrad Aug 7 '18 at 12:37
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    $\begingroup$ @WayneConrad You're right, the 'or at' wording is a little confusing, I'll edit some to make it clearer what I mean $\endgroup$ – Jack Aug 7 '18 at 12:50
  • $\begingroup$ Much better, and I like how you explain that you can only move from a one state to another when they have a or p in common. I think there may still be something ambiguous in the wording. For state 3, "a above T and p above (or at) T," should this perhaps read "a above T and p at T?" Otherwise you could have both a and p above T, and now you need two burns to get to state 4. $\endgroup$ – Wayne Conrad Aug 8 '18 at 0:13
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Let me try at "Why, and when 2 ellipses are good", and since I can't picture a scenario where 3 could be remotely good for any similar reasons, I think this should suffice. Also, I'll describe raising orbits, as it's easier to "feel" - but the situation is always reversible so lowering orbits works the same.


When performing orbital maneuvers against a single body, there are two situations where you are the lord over your trajectory; spend little delta-V, modify your orbit by a lot.

  • Situations where you use Oberth effect - zipping past the central body at insane speeds; that $^2$ in your kinetic energy equation squares the value of your fuel there. You are limited in what you can do, because your kinetic energy is directed: you can cheaply raise/lower apoapsis, or escape (/capture). Other maneuvers become quite expensive, but that one is very important.
  • Situations where you are barely subject to gravity and inertia at all. You're far, far away, moving slowly. By performing small, lazy maneuvers you can change your trajectory to pretty much anything legal. An enormous circular orbit, over the outskirts of the system? Why not? Make it polar? Cheapo. Turn from prograde orbit to retrograde? Sure, it's like a dozen m/s cost! Since everything is so lazy and slow, it will take years, or decades, until you actually complete one of the orbits you achieved. But hey, you did achieve these fancy orbits and it cost peanuts!

As you see, down, low, and high up are your places of power. It's all these damned middlings that are a hellish limbo hard to enter and hard to escape. Not enough speed for the $^2$ of Oberth to truly kick in like a legendary turbo, still enough speed that it costs an arm and a leg in fuel to change anything meaningfully.


Hohmann transfer is a straightforward thing. Raise Ap, circularize. Nothing weird, nothing fancy, works completely fine for small changes, and when you go very far, to the distant outskirts, it still works fine, as you're going through your "two places of power". But it starts ailing for the middlings. You raise your Ap just fine, and then what? And arm and a leg in fuel to circularize; no Oberth to help; the entire 3km/s of GEO speed out of your fuel tanks at no discounts!

That's when you start wishing for a discount - and use both discounts you can have.

As you start off with the Hohmann transfer orbit, you first apply your Oberth discount at raising apoapsis, to go far, far above, on the cheap. Then you go there and using your wild liberties to change orbits as you desire, you make the Pe align with the orbit you want. And yeah, you still need to pay the circularization cost, and you've already spent extra fuel to access your 'discount maneuvers'. But hey! Now you're zipping in from the outside! And that means you're now subject to another Oberth discount! Not as large as before, because the orbit is higher and slower, but hey, you're moving faster than necessary and need to slow down, instead of accelerating from standstill where your Oberth benefits were zero. That's where we're getting our savings!


And as you see, "two happy places", you start off at one, reach the other, then go to the destination. Two transfers, two ellipses. Might be different if you were starting at one middling and going to a different middling... except then you'd be near enough that plain Hohmann would come in cheaper.

...unless you're doing tricks like orbital plane change. But that's a different tale.

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