# [Orbital Mechanics] Efficient Plane Change I

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Noticed you have more Keostationary satellites than you need and want to convert some into polar orbits to extend the network coverage? Having launched a probe mindlessly firing eastward into space just to find out that you actually need a polar orbit? Burning normal does help you change the inclination of an orbit, but if you have some experiences in orbital maneuvers, you probably already know that it is one of the most fuel-consuming one, and you probably also know, either from your own experience or from other players, that it is a fuel eater because you are burning perpendicular to your velocity, which requires a lot more delta-v than other common maneuvers. Are there any other ways than just burning directly to achieve the same goal with a smaller delta-v (fuel) requirement?

In this thread we will cover a simple type of such maneuver: a plane change of a circular orbit by 90 degrees into another circular orbit of the same radius. It is possible to get the idea by doing a number of experiments, but a bit of math is certainly needed to have a complete picture. Nevertheless, if the math is too much for you, a simple answer to the question above is:

If we limit ourselves to

(1) 90 deg changes,

(2) circular initial and final orbits, and

(3) same radius for both orbits,

then the delta-v requirement is always less if we raise the other part of the orbit and burn normal at the apoapsis, and finish by circularizing the orbit when you return to the periapsis.

Qualitatively speaking, this maneuver exploits the fact that an acceleration perpendicular to the direction of motion is the most efficient when the speed is the smallest, and it turns out that while some delta-v is needed to raise/lower the apoapsis, the amount you save for the plane change at lower speeds is greater and outweighs what you spend to change the eccentricity. So at this point it is perfect to just have this in mind and go ahead with your missions; but if you want to know more, here comes the math.

Before writing down equations I would like to emphasize that everything that follows is a result purely from physics/math and has nothing to do with rocket designs (efficiency of engine, TWR optimization and so on). One important separation between the theory and practice, though, is that the changes in velocities are assumed to be instantaneous so you might find deviations if the thrust of the engine you are using is very small.

Let's first consider the "direct burn" way: Say the initial (eastward) orbit is circular and equatorial marked in green, and the orbital velocity (the prograde vector) is the black arrow labeled v1. The final orbit, our goal, is also circular but polar marked in orange, then right after you finish the plane change burn, your velocity, labeled as v2, should be to the north (or the south but it doesn't really matter here) in the +z-direction. For our simplification that the orbits have the same radius we have v1 = v2 = v. The change in velocity, delta-v, is then the subtraction of v1 from v2, and can be drawn by connecting the tips of the vectors, as shown in red.

The amount of delta-v is its magnitude, easily found by the Pythagoras' Theorem:  One interesting thing to note here is that the delta-v vector does not point entirely to the north, but inclined to the west; it is because to complete the maneuver, you have to not only obtain a northward velocity, but also cancel all of your eastward motion. It is the combination of the northward and the westward components that makes the required delta-v inclined.

We are basically done with the direct-burn case! The delta-v you need is the square root of 2 times, or about 140% (yes, 140...) of, your initial (and also final) orbital speed.

Now, let's look at the more complex, alternative way: This scheme involves 3 burns in total, as marked by red dots in the figure. The first one is burning prograde to raise the other side of the orbit to a certain apoapsis altitude. The second one is performing the actual plane change maneuver at the apoapsis. The last one, taking place at the same location where the first maneuver was, is burning retrograde to recover a circular orbit. There are more velocities we need to take care of here. Let's take a closer look with a "top view" from the north: It is time to bring up the prime equation we need to deal with orbital velocities, the vis-viva equation: This equation is the direct result of the conservation of energy and momentum, although we will not go into the details here. Once we fix the gravitational parameter Ã‚Âµ, which is just a constant related to the parent body, and the periapsis a, the equation tells us the speed v at whatever position r we are interested in.

The first maneuver (we call it maneuver I) takes place at the leftmost position in the figure. Let's say your orbital velocity in the green, circular orbit is v1I, and after burning prograde, the velocity is v2I. Using the vis-viva equation we get Remember that after the prograde burn, the orbit is no longer the green one -- it becomes the blue ellipse. v2I is a part of this elliptical orbit, and the corresponding semi-major axis a is half the sum of r1 and r2. Therefore, the speed right after the prograde burn is The delta-v of the first maneuver is Maneuver II takes place at the apoapsis of the elliptical orbit, in the right hand side of the figure. The vis-viva equation again gives This maneuver ends up with a orbit perpendicular to the screen, with a velocity pointing out of the screen in the +z-direction. In the figure we use the arrow head/tail symbols: where a cross represents a vector pointing into the screen and a dot represents a vector out of the screen. This maneuver is a plane change, so the delta-v is the square of two times the speed: The final step is circularization of the orbit when we reach the periapsis. Since the only difference between the two elliptical orbits before and after the plane change burn is the inclination, the situation is exactly the same except that we are burning retrograde this time: The total delta-v is then the sum of all for the three maneuvers: which is not something very good to look at, but to know how much delta-v has been saved, we need to subtracted the delta-v for a direct burn from the total of the three maneuvers (recalling that in the direct-burn case, the orbital speed is equal to v1I): This is an expression that depends on r1 and r2, the distances to the periapsis and the apoapsis. If the difference turns out to be negative, then we can indeed save some delta-v via the 3-maneuver scheme. So when is it negative? The difference is negative if the expression in the square brackets is: The two parameters r1 and r2 are a bit too much to handle, though we can transform it into a simpler one by dividing by r1 and letting and get Rearranging by a bit we get a quadratic inequality in s: Solving this for s gives: Recall that s is the ratio between the periapsis and the apoapsis distances, so the negative solution here is meaningless and is discarded. We are left with s > 1, or, in other words, We got a very simple solution here: as long as r2 is bigger than r1, the delta-v requirement for the 3-step maneuver is smaller; the delta-v needed to bring up (and down) the apoapsis is worth it -- you will always (90-degree plane change between circular orbits of the equal size) save some delta-v by doing so.

It is also very useful to know, after all these calculations, that how much delta-v we can actually save. If we express the total delta-v budget in terms of a fraction, f, of that for the direct plane change: we can work out that which is not very illuminating to look at as well. But let's ask a question: "How much can we save at most?" It turns out that if we push out the apoapsis extremely far away, that is, as s grows indefinitely large, f approaches the number 0.586, which means that the total delta-v can only get as small as ~58.6% of that in the direct-burn case (or we can say we save ~41.4%). If you want to cut, say, 50% of the delta-v requirement, there is no way you can get there. This is a very interesting result; you cannot cut your budget indefinitely by pushing out your apoapsis as far as you want.

But we knew this already, didn't we? Let's pause a bit to recover ourselves from the tedious math. We knew that there is a minimum amount of delta-v we need to deal with -- to bring the apoapsis up and down! In the case where the apoapsis is indefinitely far away, we are just barely escaping the parent object's gravity, and at that very-far-away place, our velocity reaches zero exactly. The plane change actually takes no delta-v at all to complete; the budget will entirely be spent for the prograde burn to "ellipticize" the orbit, and after a very long period of time when we finally come back, the retrograde burn to circularize the orbit. The speed we need to achieve to reach the very-far-away apoapsis is, of course, the escape velocity ve: The total budget then comes from the two burns (one prograde, one retrograde), which require an identical amount of delta-v: This is exactly what we got above!

Knowing that we are indeed getting consistent results, we may as well look at some "experiments". Using Kerbin as the central body, some tests were carried out to verify our calculations with the help of Jebediah (in fact he was issuing commands to MechJeb to ensure that the burns were accurately performed). The craft was first HyperEdited to a circular, equatorial orbit at an altitude of 100 km. It turns out that FL-T800 + FL-T100 has just enough fuel for an LV-T45 engine to achieve (3288 m/s) a delta-v of 3177 m/s (we also have a couple of solar panels to survive long eclipses in highly eccentric orbits) in the direct-burn case. Next, the three-step maneuver was carried out with an apoapsis at 200 km. This is repeated for 300, 500, 1000, 2000, 5000 and 10000 km. Finally another run was done the apoapsis at 80000 km, near the edge of Kerbin's SOI. Some example screenshots are shown below and f was calculated for each ratio s and plotted along with the red curve from the theoretical point of view. (please open this in a new tab) The plot: As you may have noticed, the delta-v requirements for the first and the third maneuvers were not exactly the same. This comes from the imperfectness of the burns and pointing even with MechJeb, although the error is practically negligible. The 3-step maneuver is evidently advantageous, from the point of view of delta-v. One sacrifice of this method is the time to complete the objective -- the orbital period of the elliptical orbit. Nevertheless, as seen from the plot, you don't actually have to go for a highly eccentric orbit; a value of 10 for s already gives an f less than 65%; in fact, it doesn't make a lot of difference (in the amount of delta-v saved) if we go from s = 40 to over a hundred.

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Hello! I don't really quite understand the math but if delta-v required for the change will always be lower than a "direct burn", then would it also apply to the situation where I change my apsides before raising one apsis for the normal burn?

For example whenever I go to Minmus:

1. Launch and circularize into a equatorial orbit
2. Change apsides so they are the same with the orbital nodes to Minmus, usually where apoapsis is on the side where Minmus is currently going to
3. Raise apoapsis
4. Inclination change
5. Fiddle with maneuvers to get an encounter I've seen Scott Manley's video to Minmus and he doesn't bother with the plane change at Kerbin, but I can't get an encounter if I try it myself, just aiming at the inclination nodes.

Is my step 2 unnecessary?

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• Change apsides so they are the same with the orbital nodes to Minmus, usually where apoapsis is on the side where Minmus is currently going to

Not sure what this step is supposed to mean. In step 1 you mentioned going into a circular orbit, so why would it matter where your periapsis and apoapsis are with respect to Minmus?

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Not sure what this step is supposed to mean. In step 1 you mentioned going into a circular orbit, so why would it matter where your periapsis and apoapsis are with respect to Minmus?

Yes, sorry bad wording there I used circular loosely. I should have said eccentric because I rarely bother circularizing them, as long as it's off the atmosphere I'm good! Then I move the apsides so they are "on top" of the inclination nodes (Ap/Pe is on AN/DN), so after I've raised my apoapsis, I can plane change at the node too. I don't really know if that matters but I thought that it's more efficient to burn on or near the node marker itself. I'm thinking that I pushed my apoapsis closer to Minmus, and used up lower delta-v by changing on the inclination node which is at my apoapsis too.

It works for me but I'm wondering if I'm just complicating things up and what I'm doing is actually inefficient.

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Yes, sorry bad wording there I used circular loosely. I should have said eccentric because I rarely bother circularizing them, as long as it's off the atmosphere I'm good! Then I move the apsides so they are "on top" of the inclination nodes (Ap/Pe is on AN/DN), so after I've raised my apoapsis, I can plane change at the node too. I don't really know if that matters but I thought that it's more efficient to burn on or near the node marker itself. I'm thinking that I pushed my apoapsis closer to Minmus, and used up lower delta-v by changing on the inclination node which is at my apoapsis too.

It works for me but I'm wondering if I'm just complicating things up and what I'm doing is actually inefficient.

My understanding is that after the plane change at Ap you come down to the periapsis to push Pe out to Minmus' orbit? The efficiency of the plane change burn depends on how far out the AN/DN is but is better than one at low-Kerbin orbit, although Minmus would have displaced some distance by the time you go back to the Pe so you may have to delay your burn. Burning at points other than the AN/DN would actually change the positions of AN/DN; you can still minimize the relative inclination but if you want it zero you have to burn at AN/DN. I usually do my plane change at AN/DN on the way to Minmus like what Scott Manley did in his tutorial.

Looking solely at the plane change, doing it at apoapsis should be optimal, but considering things like the displacement of Minmus it gets a little bit complex (for example, whether it is still beneficial given now we have a delayed burn being non-optimal as we can't burn at Pe because Minmus has moved) and practically we might as well just do a plane change on-the-way. This is quite interesting as a general plane change problem and it definitely deserves to be a sequel to this thread if I have the time ##### Share on other sites

This is backed up by an exert from an artical I recently read here:http://www.braeunig.us/space/orbmech.htm (more maths and orbital mechanics information if you're interested)

Plane changes are very expensive in terms of the required change in velocity and resulting propellant consumption. To minimize this, we should change the plane at a point where the velocity of the satellite is a minimum: at apogee for an elliptical orbit. In some cases, it may even be cheaper to boost the satellite into a higher orbit, change the orbit plane at apogee, and return the satellite to its original orbit.

Nice tutorial OP, we'll all be expert fuel misers in no time!

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• 1 year later...

At the risk of necro-ing this thread, I was doing some research on the same topic and wanted to share my results. Wmheric covers the case of a 90-degree plane change, but what if you have some other plane change angle you need to perform?

I did a bunch of Mathematica calculations to come up with a more exact answer. For a circular (or nearly circular) orbit, the distilled version is pretty simple:

-- For plane changes of less than about 45 degrees, you're best off to just do them directly.

-- For plane changes from about 45 to 60 degrees, you want to first raise your apoapsis to about double the periapsis* (and make sure to orient your burn so the apoapsis lines up with the plane change node!), then do the plane change, then lower your apoapsis again.**

-- For plane changes of more than about 60 degrees, you want to raise your apoapsis as much as possible (to the edge of the SOI), then do your plane change, then lower your apoapsis again.

* Double relative to the center of the planet, not to its surface. So if you're orbiting Kerbin with its 600 km radius at an altitude of 100km above the ground, you are actually 700km above the center and would therefore need to increase that to 1400 km from the center, which is to say 800km above the surface.

** This is an approximation. The actual optimum is to use a higher and higher apoapsis as your required plane change angle increases from about 45 to about 60 degrees, but you will be darn tootin' close to optimal if you just use a 2x increase as a rule of thumb.

This Mathematica plot shows the required delta-V of doing the plane change directly vs. using the bi-elliptic approach. The blue line is the delta-V required by a direct burn (as a fraction of your current V). So for example, if you want to make a 180-degree plane change (i.e. stopping and reversing direction), it will require 2.0x your current velocity, which makes intuitive sense. The orange lines show the total delta-V of the bi-elliptic approach. The steeper orange lines are for smaller boosts to your apoapsis, while the totally horizontal orange line is for an infinite boost to your apoapsis.

A couple interesting things jump out from the chart. The first is that if done optimally, no plane change should ever require more delta-V than about 85% of your current speed. Why? If you increase your apoapsis to infinity (or thereabouts), your orbital velocity falls asymptotically to zero, so it costs you nothing extra to do any plane change you like. All of your delta-V is spent on the apoapsis maneuvers and a vanishingly small amount is required for the actual plane change.

The second interesting thing is that there's a surprisingly small range where the optimal amount of apoapsis boost varies. Below (about) 45 degrees, you are best off to just do a direct plane change, and above (about) 60 degrees, you just want the largest apoapsis boost you can get. And between those two numbers, simply raising your apoapsis to double your current orbital radius will get you most of the possible benefit. (BTW the curve for doubling your apoapsis is the 2nd steepest orange line. The steepest is for a 1.5x increase.)

(Remember when we talk about apoapsis heights here, it's relative to the center of the planet, not to the surface. So be sure to add in the planet's radius when calculating your burn altitudes.)

That's pretty much it for circular orbits. The case for how to handle highly eccentric (elliptical) orbits is a bit more complex... I will try to come back to address it later if I have time to get through all the math.

Edited by Yakky
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• 6 years later...

I can smell the grave wax from here. This has been such a help and saved one of my launches after I inadvertently launched 80 instead of 90, and I was really stingy about fuel.

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