9
$\begingroup$

If the orbits are co-planar (lie in the same plane), the required minimal $\Delta v$ is calculated by the well-known Hohmann transfers. But what is the case if they aren't co-planar?

How many orbital corrections are needed in this case, and in which direction?

$\endgroup$
  • 1
    $\begingroup$ When you say "non-parallel", do you mean not in the same plane? $\endgroup$ – Ryan Caskey Jan 3 '17 at 19:08
  • 1
    $\begingroup$ Can we get clarification on "parallel" orbits? Do you mean when the starting orbit and destination orbit has the same inclination? $\endgroup$ – IT Bear Jan 3 '17 at 19:12
  • $\begingroup$ @RyanCaskey Yes. For example, there was a plan to move the Hubble Space Telescope to the ISS, which could make possible its continuous service. (But this question is not specific to this suggestion.) $\endgroup$ – peterh - Reinstate Monica Jan 3 '17 at 20:07
  • $\begingroup$ @ITBear Exactly. $\endgroup$ – peterh - Reinstate Monica Jan 3 '17 at 20:08
  • $\begingroup$ Are you asking about the cost of changing inclination when the orbits are otherwise very similar? $\endgroup$ – Blake Walsh Jan 4 '17 at 16:35
4
$\begingroup$

A single plane change maneuver is required to make the orbits coplanar. Plane change maneuvers are expensive and are most easily applied where the original and desired orbital planes intersect. The burn is more efficient the closer it is to the apogee of the orbit -- so it is often better to do your in-plane adjustments first. You can get the direction of the burn by normalizing the cross product of the velocity vector and radial vector at that point.

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ To be as efficient as possible, you would probably want to combine your Hohmann transfer second burn with your plane change burn. In order for this to be possible, you would need to try to synchronise your orbits so that you could start your Hohmann transfer at either the ascending or descending nodes of the intersection of the two orbits. Your velocity will be lower at the second burn point, thus making the plane change not as expensive as it could be. $\endgroup$ – Ryan Caskey Jan 3 '17 at 21:48
3
$\begingroup$

Another answer started to hit at the right answer, but I'm not entirely sure that it's completely correct from start to finish:

By combining the Hohmann apogee burn with the plane transfer you gain an additional efficiency from the vector addition of the velocity changes.

I say the majority, because I believe when you do all the math the optimal solution performs a fraction of the inclination change at perigee as well.

I believe this is right in that a plane change is done both at apogee and at perigee. However, I believe that it throws around some erroneous assumptions about why and how much.

For a Hohmann transfer, your first burn is at perigee. Think about 2 separate planes for a second. If the 2nd burn is to be done at apogee, then that alone determines the angle that the perigee burn must have. This doesn't have anything to do with optimization at this point. If we are doing a perigee + apogee burn to complete the entire transfer, then we have necessarily set the angles that each will require. They are determined by geometry.

Here's some academic stuff:

http://www.kdm.p.lodz.pl/articles/2011/V15_2_07KAM_SOL_2_2011.pdf

And from that, allow me to restate a proper name for it - "Hohmann orbit transfer with split–plane change"

You can consider alternatives, but you'll back yourself into a corner regarding optimization if it looks anything like a Hohmann transfer. If you burned within your own plane at perigee and waited to do a plane change, then you'll need to cover more distance in a shorter time, and it doesn't make sense. If you separated the plane change and the Hohmann burns, then you're trading the hypotenuse of a triangle for the sum of its sides.

Bi-Elliptic is still a possibility for energy minimization, but I do not know how to categorize what that combined (plane change plus other parameter changes) maneuver would look like, and obviously, it would only be relevant for a subset of all possible transfers.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Hohmann transfer can be an orbit raising or orbit lowering maneuver - so your first burn could be at apogee or perigee. $\endgroup$ – CoAstroGeek Jan 4 '17 at 3:47
  • $\begingroup$ True, I wrote it as if I were speaking of going from a lower orbit to a higher orbit. But it's reversible the other way. $\endgroup$ – AlanSE Jan 4 '17 at 3:59
  • $\begingroup$ "If the 2nd burn is to be done at apogee, then that alone determines the angle that the perigee burn must have." This isn't correct. The plane change is entirely separable from the hohmann transfer. Taking the orbit raising case where the burns are at the line of nodes, you could do all of the inclination change at the perigee burn, or all of it at the apogee, all of it half a rev, later or any combination.thereof. But there is an optimum combination of how much of the inclination change you do at perigee vs apogee. $\endgroup$ – CoAstroGeek Jan 4 '17 at 4:28
  • 1
    $\begingroup$ A plane change will always be more efficient the closer you are to apogee. $\endgroup$ – Erik Jan 4 '17 at 4:47
  • $\begingroup$ > The plane change is entirely separable from the hohmann transfer. I think you should elaborate on that if you feel strongly about it, because we seem to have at least some failure of communication. Perhaps we are not even really talking about the same problem. $\endgroup$ – AlanSE Jan 4 '17 at 13:53
2
$\begingroup$

I'm going to assume you mean non-coplanar rather then parallel.

First - a Hohmann transfer isn't always the most efficient transfer between coplanar orbits. When the ratio between the initial & final semi major axis is large enough, (> approximately 12) a Bi-Elliptic Transfer may be more efficient.

$\Delta v$ for orbit inclination change is given by $\Delta v_i=2v \sin \frac{\Delta i}{2}$, where $v$ is the orbital velocity. So, for the simple case of a transfer between circular orbits, you'll want to do you the majority of the inclination change at the apogee of the Hohmann transfer. By combining the Hohmann apogee burn with the plane transfer you gain an additional efficiency from the vector addition of the velocity changes.

I say the majority, because I believe when you do all the math the optimal solution performs a fraction of the inclination change at perigee as well. I'll see if I can find the reference for that.

Here's one reference I was thinking of for the general solution: On non-coplanar Hohmann Transfer using angles as parameters

The reference from AlanSE looks good as well, from a quick look: On The Generalized Hohmann Transfer with Plane Change Using Energy Concepts Part I

| improve this answer | |
$\endgroup$

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.