Consider a transfer between two circular orbits of similar radius, the only difference being the inclination difference, $\alpha$. What's the minimal $\Delta v$ required to perform this transfer?

Inclination change strategies I have considered so far:

  1. A single burn inclination change. This is simple enough, just the difference between two velocity vectors, which works out to:

$$\Delta v_1(\alpha) = 2\sin(\alpha/2)$$

(measured in unit velocities of the circular orbit)

simple inclination change

  1. However, when $\alpha > 48.9^\circ $, it costs less to accelerate almost up to escape velocity, perform the inclination change at an apoapsis arbitrarily far away, and then burn retrograde back into the target orbit, at a constant cost $2\sqrt{2} -2$ independent of $\alpha$

two inclination change strategies

  1. Like 2), but doing the inclination change at a finite apoapsis, trading a lower acceleration and deceleration cost for a higher inclination change cost at the apoapsis.

$$\Delta v_3(\alpha,A) = 2\left(\sqrt{2-\frac{2}{1 + A}}-1\right) + 2\sin(\alpha/2)\sqrt{\frac{2}{A}-\frac{2}{1 + A}}$$

This only slightly cuts the corner between 1) and 2)

strategy 3

  1. Like 3), but also doing a part of the inclination change, $\beta$, combined with the acceleration and deceleration burns.

$$\Delta v_4(\alpha,A,\beta) = 2\sqrt{\left(\cos(\beta)\sqrt{2-\frac{2}{1 + A}} - 1\right)^2 + \left(\sin(\beta)\sqrt{2-\frac{2}{1 + A}}\right)^2} + 2\sin((\alpha - 2\beta)/2)\sqrt{\frac{2}{A}-\frac{2}{1 + A}}$$

A numerical optimization for $A$ and $\beta$ is drawn in red in the diagram below.

numerical optimization

It's evident that strategies 3) and 4) are slightly more efficient in the region where 2) takes over for 1). Furthermore, 3) as a special case of 4) is never more efficient, so it's always beneficial to split the inclination change between all the burns.

Are there other inclination change strategies that are more efficient for some values of $\alpha$?

Do strategies 3) and 4) have some simple closed form that does not require numerically optimizing their parameters?

Edit: I have been able to find a closed form for 3)

The optimal apoapsis is

$$A(\alpha) = \max\left(1,\frac{\sin(\alpha/2)}{1 - 2\sin(\alpha/2)}\right)$$

Which yields

$$\Delta v_3(\alpha) = 2\left(\sqrt{2-\frac{2}{1 + A(\alpha)}}-1\right) + 2\sin(\alpha/2)\sqrt{\frac{2}{A(\alpha)}-\frac{2}{1 + A(\alpha)}}$$

  • 1
    $\begingroup$ No real information, but if splitting the inclination change across the initial and final burns is an improvement, I'd check the case of multiple midcourse inclination changes just to see if there's any further win to be had there. As long as you're in there plotting things it might also be interesting to look at the time taken by each of these strategies. $\endgroup$ Aug 15 '19 at 15:11
  • 1
    $\begingroup$ I can't find any case where midcourse burns would reduce costs, but they are difficult to handle since they are outside the node alignment. Transfer time is relatively straight forward. 1) is the instantaneous case, and 2) goes towards infinity (and tends towards 3) for realistic cases). Both 3) and 4) are just a single elliptic orbit. 4) is faster than 3) for most of their relevant range. $\endgroup$ Aug 15 '19 at 15:23
  • $\begingroup$ I think I have been able to find a closed form for the optimal apoapsis for 3). Will edit it in. $\endgroup$ Aug 15 '19 at 15:49

I will give my current best shot at this problem, and others should feel free to strengthen the argument with additional mathematics. (Or poke holes!)

You ask two questions, I will answer the first as the second has been partially answered by the update.

Are there other inclination change strategies that are more efficient for some values of $\alpha$?

I say no, that in fact you have found the optimal solution. Any inclination change (or any orbit change for that matter) is simply a change of the angular momentum $\vec{L}$ of the orbit. For a strict inclination change, the magnitude of $\vec{L}$ is constant, only its direction changes.

Now to achieve the inclination change, we can think of any burn as a change in angular momentum integrated over time: $$\Delta\vec{L} = \int{\frac{d\vec{v(t)}}{dt}\times\vec{r(t)}}dt$$ but in the case of impulse burns, it is the sum of discreet components $$\Delta\vec{L} = \sum{\Delta\vec{v(t)}\times\vec{r(t)}}$$ where $t$ need not be time in the second equation, just a parameter to indicate that such a burn occurs at the specified radius.

Your equations are simply explicit versions of the second equation. Then it just remains to see whether any further optimization is possible. Since you have optimized the enlarging, inclination change, and de-larging (is that even a word?) burns, we need only check to see if mid-course burns reduce the total $\Delta v$ expended.

My argument is no. Any such burns, as you mentioned, would be off-axis. Mathematically, they would introduce components to $\vec{L}$ that cannot be removed anywhere else but the exact opposite side of the burn (in the case of 2 mid-course burns) or add further mid-course burns to fix the unwanted components of $\vec{L}$. I am sure someone will find a better way to show this mathematically, but the intuition is that the nodes are being rotated around the orbit and not reduced in magnitude. This is just a waste of $\Delta v$.

For why these burns will always contribute more $\Delta v$ than they remove, I would posit the Oberth effect. Such mid-course burns would occur where the $\Delta v$ has less influence on the radius of the orbit than at the periapsis (which we already burned at!) I mentioned earlier that mid-course burns have unwanted components to them, but they may also have desirable components. I argue that these desirable components (radial, prograde) are better achieved at the initial periapsis burn due to the Oberth effect.

The conclusion is therefore that you have optimized the problem for the 2-body impulse-burn situation. Since any burn must necessarily take finite time, I am sure there are plenty of other optimization parameters that must be taken into account for a finite-time burn. But the gist is the same, a 3 burn inclination change will always be optimal.

I have presented a very hand-wavey argument without too much solid math, but I hope that this lays the framework for someone to hash out the math in a convincing and bulletproof form.

  • 1
    $\begingroup$ That's mostly intuition indeed. For a few exotic cases within these assumed restraints, consider for example a) a five-burn transfer, with an initial apoapsis raising, a periapsis lowering, a second apoapsis raising, a plane change, and then finally the retrograde burn. Such a scheme is not beneficial in planar transfers, but who knows when the inclination is distributed over all those burns? Secondly, b) what about the case where the prograde and retrograde $\beta$ angles are unsymmetrical? My intuition says no in both cases, but that doesn't prove anything. $\endgroup$ Aug 15 '19 at 16:59
  • $\begingroup$ I think a 3-burn solution is likely to be optimal in a lot of cases, but I think there's a distinct possibility something akin to a bi-elliptic transfer could have a window of "optimal-ness". There may be other smooth parameters missed of the list or the use of a local, but non-global minimum. Interesting question. I might have a stab at some point, but I suspect proving anything as optimal would be hard. $\endgroup$
    – ANone
    Aug 20 '19 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.