Changing direction causes ΔV (change in velocity), often more than a change in speed. Compare the velocity vectors below. When going the same direction, the difference is 1 km/s. When at right angles the difference is 5 km/s. We know this from driving in traffic. Two cars going almost the same speed hit each other. If they’re in the same lane going the same direction, it’s a mild bump. If one car runs a red light and T-bones a car in cross traffic, the impact is serious:

This is the strength of a Hohmann transfer orbit. Velocity vectors are pointing the same direction at departure as well as destination. No direction change is needed, only a speed change:

Note the Hohmann transfer path moves 180 degrees about the sun:

A Hohmann transfer assumes the departure and destination orbits are co-planar. But what if the destination orbit is inclined?

**Orbit Planes and Spherical Trigonometry**

A plane passing through a sphere’s center cuts the sphere along a great circle. A group of planes all sharing a common point can be represented as great circles on a sphere. Since every orbit about the sun is a conic section having the sun as a focus, each orbital plane shares the sun as a common point. Representing the orbital planes as great circles is convenient. There are already a lot of theorems in spherical trigonometry which gives us a suite of tools for looking at angles between orbital planes.

The shortest path (or geodesic) along a spherical surface between two points is an arc of a great circle. If we set the sphere’s radius to be 1, the arc length is also the angular separation in radians.

A familar group of great circles are the longitude lines on a globe. The equator is the only great circle among the latitude lines. All the longitude lines are great circles passing through the poles.

Let’s use the equatorial great circle to represent the departure plane. Recall the Hohmann transfer moves 180 degrees about the center. In this illustration, latitude and longitude for departure and destination is (0º, 0º) and (7º, 180º). The only great circle connecting these points is a polar orbit nearly 90º from the departure and destination planes! Big plane changes at departure and destination destroys the virtue of a Hohmann orbit.

I’ve also tried to demonstrate this in this video:

The big delta V needed for large plane changes makes the ridge in a porkchop plot:

Porkchop plots are drawn by doing iterations of various Lambert Space Triangles. Lambert iterations give polar transfer orbits when departure and destination longitudes differ by 180º.

Does this mean Hohmann transfers are no good if the destination orbit’s inclined? No, the big plane changes can be avoided with a mid course plane change. Here is a broken plane transfer where a plane change burn is done at the ascending node:

Does this mean Hohmann transfers are no good if the destination orbit’s inclined? No, the big plane changes can be avoided with a mid course plane change. Here is a broken plane transfer where a plane change burn is done at the ascending node:

The line where the destination and departure planes intersect form the ascending and descending nodes. Starting in the departure plane and doing a plane change at the node avoids the two major plane changes. The departure and destination planes differ by an angle called i, for inclination.

Changing a vector by an angle i takes dv of v * 2 * sin(i/2).

The Vis Viva Equation tells us v = sqrt(μ(2/r - 1/a)). So v ranges from sqrt (μ((1-e)/(a(1+e)))) at aphelion to sqrt (μ((1+e)/(a(1-e)))) at perihelion. Let's look at a Ceres transfer orbit. An ellipse with a 1.88 a.u. semi major axis and eccentricity .47 will have speeds ranging from 36 km/s (at perihelion) to 13 km/s (at apohelion). Inclination's about 10.6 degrees. So plane change ranges from 36 km/s * 2 * sin(10º/2) to 13 km/s * 2 * sin(10º/2) or from 6.7 to 2.4 km/s. Is the a 2.4 km/s plane change at aphelion the best we can do? No, it's possible to have less plane change expense.

Launch is at the perihelion of an outbound Hohmann orbit. If the launch coincides with a node, the entire plane change can be done during earth departure or at arrival. Then the delta V entails a speed change as well as a direction change. Doing a single plane change/speed change burn saves delta V as shown by this diagram:

Law of cosines tells us for a triangle a, b, c, a

^{2}+ b^{2}- 2ab cos(i) = c^{2}. In this case, i is the angle between a and b and c is the delta V needed from the combined plane change and speed change.
At aphelion, a combined speed change/plane change only costs .76 km/s more than the speed change alone.

When launching deep in earth’s gravity well, we enjoy an Oberth benefit. Ceres' gravity well lends a little Oberth benefit at the destination. If the line of nodes coincides with transfer orbit's line of apsides, plane change can cost as little as .52 km/s extra.

This indicates as much plane change as possible should be made at departure and arrival. What sort of plane changes should we make to minimize the angle of the midcourse plane change?

The fattest part of an orange slice is right in the middle:

The angular separation at launch has to be some part of the orange slice. To minimize the angle between transfer plane and destination plane, the angular separation at launch should be in the middle. Having the transfer plane intersect the destination plane 90º from launch minimizes plane change angle.

An object on an elliptical path moves slower as it moves further from the sun, so doing plane changes further out are cheaper. The 90º from launch is a minimum. There will be a larger plane change angle 100 degrees from launch, but velocity will be slower. Also plane change lessens as flight path angle increases. I hope to talk about this more when I have time.

But for now I believe this shows that the Lambert iterations greatly exaggerates plane change expense for a Hohmann path where departure and destination points are 180º degrees apart. Most of that plane change expense can be eliminated by choosing a good place to do a midcourse plane change.

A PDF on Broken Plane Maneuvers Fernando Abilleria of NASA Jet Propulsion Laboratory

This indicates as much plane change as possible should be made at departure and arrival. What sort of plane changes should we make to minimize the angle of the midcourse plane change?

The fattest part of an orange slice is right in the middle:

The angular separation at launch has to be some part of the orange slice. To minimize the angle between transfer plane and destination plane, the angular separation at launch should be in the middle. Having the transfer plane intersect the destination plane 90º from launch minimizes plane change angle.

An object on an elliptical path moves slower as it moves further from the sun, so doing plane changes further out are cheaper. The 90º from launch is a minimum. There will be a larger plane change angle 100 degrees from launch, but velocity will be slower. Also plane change lessens as flight path angle increases. I hope to talk about this more when I have time.

But for now I believe this shows that the Lambert iterations greatly exaggerates plane change expense for a Hohmann path where departure and destination points are 180º degrees apart. Most of that plane change expense can be eliminated by choosing a good place to do a midcourse plane change.

A PDF on Broken Plane Maneuvers Fernando Abilleria of NASA Jet Propulsion Laboratory

## 3 comments:

I very nearly understood quite a lot of this from the spherical trig I learned over 40 years ago, from a book by a man called Nobbs. I still have it. The lecturer used to march in and say "Right lads, get your Nobbs out!"

Googled Nobbs and spherical trig and the book seems to be out of print :(. I've heard several mathematicians lament that spherical trig is fading out of college curricula.

A shame.Spherical trig has some pleasing geometry and seems quite useful. I would think it would be a useful part of aerospace curricula.

Do you mean "Trigonometry" by C. G. Nobbs? Alas, does seem out of print. Bookfinder dot Com only has four used copies, all about $30 and all from bookstores in the UK. Pity.

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