Sync/Rendezvous computation and guidance?

[Artlav]

How can one compute the intersection points of two coplanar orbits?
How to compute closest distances between two vessel in these intersection points at a given time in the future?
How to estimate the best velocity change vector and time to get the orbits and positions synced?

These are the problems i solved in UAP sync orbit autopilot in a crude, brute-force way of propagating the elements of the orbits for the next few periods 1/10th of orbit at a time, looking for nearest distances with local optimising search for them between the tenths, all this for each DV possible within given limit in 1 m/s increments.

It works, and modern computers swallow it whole, but it's bloody inefficient, somewhat uncertain, and is a quarter of the size of the entire UAP system.

So, are there any better solutions to the given problem?
-A way to compute the true/mean anomaly of orbital intersection points analytically?
-Some approximation algorithm for the manoeuvre?

 

[tblaxland]

How can one compute the intersection points of two coplanar orbits?

This should help: http://www.mathlinks.ro/viewtopic.php?p=1209667#1209667. The constants theta_0 and alpha referred can be determined after some algebraic manipulation and the application of the a phase shift trigonometric identity.

How to compute closest distances between two vessel in these intersection points at a given time in the future?

No analytic method AFAIK, but at least if you know the (maximal) two intersection points of the orbit it will reduce the number of computations to determine on which orbit is best to rendezvous.

How to estimate the best velocity change vector and time to get the orbits and positions synced?

I would say use the method above to find which orbit will get you closest (within) limits and then use the Hills-Clohessy-Wiltshire equations to determine the required impulse (this is the technique used by Rendezvous MFD). You can vary the time to intercept to optimise the required impulse. There is an old Apollo era paper on it here: http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19680019369_1968019369.pdf but Google also turns up plenty of others.

It works, and modern computers swallow it whole, but it's bloody inefficient, somewhat uncertain, and is a quarter of the size of the entire UAP system.

Hopefully you can make it more efficient but it will always be a significant portion of you code I think. Like any worthwhile optimisation problem :dry:

 

[Enjo]

Agentgonzo has done some Magicke in Launch MFD regarding some of your questions. Precisely with intersection point and time to intersection, depending on your current orbital velocity. It's some 3D geometry math. I'll elaborate after returning home.
All I can say now is that the intersection is defined by cross product of n1 and n2, where n1 is your orbit's normal and n2 is that of the target.

 

[tblaxland]

All I can say now is that the intersection is defined by cross product of n1 and n2, where n1 is your orbit's normal and n2 is that of the target.

That gives you the node where the two orbit planes intersect but for the coplanar case (the specific case referred to in the OP) these two vectors are parallel and their cross product yields nothing useful about the intersection of the two ellipses.

 

[Enjo]

Ah, I was afraid that you're going to say that while trying to recall what "coplanar" means in English

 

[Pagnatious]

Coplanar orbit intersection

Hi,

I have a similar, if not the same question. Determining if two coplanar orbits intersect, and ideally then calculating time to intersection and from that determining if it results in a rendezvous.

This should help: http://www.mathlinks.ro/viewtopic.php?p=1209667#1209667.

This link doesn't appear to work. Was it's method anything like this? I was thinking using that method to find the intersection of the ellipses, essentially what values of true anomaly for the ships orbit have it crossing the other, then calculating travel time from that and projecting where the planet would be.

Problem is I'm a bit maths rusty and while I think I understand the Newton-Raphson method, there could be a potential 4 intersection points and would it only find one?

It then leaves the question of what if one of the orbits was hyperbolic?

I would greatly appreciate any help on this. It was something I was working on a little while ago but then took a break from and am coming back to now...

 

[tblaxland]

This link doesn't appear to work. Was it's method anything like this?

No, it was somewhat different. IIRC, the method I linked to was essentially solving a set of two equations for the radii of the ellipses as a function of true anomaly. The two equations would yield between zero and two solutions, the maximum possible for two ellipses that share a focus (there was a proof for that in the thread too).

Problem is I'm a bit maths rusty and while I think I understand the Newton-Raphson method, there could be a potential 4 intersection points and would it only find one?

You could expect to find only one for any given initial value. I'm not sure what an optimum strategy for selecting the initial value would be for a general case, but for the more restricted case in orbital dynamics where the two ellipse share a common focus, values of true anomaly 90° apart ought to be sufficient, eg, 0, 90, 180 and 270, or 45, 135, 225 and 315. Actually, two values 180° apart should also be OK but there might be some scenarios where that would not converge, or converge slowly.

It then leaves the question of what if one of the orbits was hyperbolic?

That should still work using the parametric form of a hyperbola.

Having said all that, the original poster was looking for an analytical solution.

 

[mjessick]

Investigating solving the original problem of the elements of intersection for two intersecting coplanar elliptical orbits:

Using the polar equation (radius as a function of true anomaly and fixed elements):
r = p / (1 + ecc * cos( true_anomaly))
write this equation for both orbits, _1 and _2.
At the epoch of intersection, r_1 must equal r_2.

(See Bate, Mueller, White (BWM), section 2.3, classical orbital elements)
define:
L0 = longitude_ascending_node + arg_perigee + true_anomaly
u0 = argument of latitude at epoch = longitude_ascending_node + arg_perigee

u0 is therefore another parameter known and constant for each of the orbits.

Set r_1 = r_2, and substitute:
true_anomaly_1 = L0_1 - u0_1
true_anomaly_2 = L0_2 - u0_2

Since the orbits are coplanar, then at the epoch of intersection L0_1 = L0_2. Let it equal L0. (See BWM figure 2.3-1 for why this must be true)

The r_1 = r_2 equation is then something like:
p_1 + p_1 * ecc_2 * cos(L0 - uo_2) = p_2 + p_2 * ecc_1 * cos(L0 - uo_1)

Here, everything is constant and known except L0 of the intersection. (One equation in one unknown.)

So if you can solve this transcendental equation analytically, then your problem is solved. Note that there should be two solutions. I suspect one can't solve this easily, but I haven't looked at it further. There are lots of identities for cos(A + B) that might be of use, but probably not.

Even if this method doesn't solve the problem directly, it might make it easier to iterate, particularly since you can take analytic derivatives for use in the iteration.

 

[Pagnatious]

The two equations would yield between zero and two solutions, the maximum possible for two ellipses that share a focus.

I didn't realise that, I thought 2 would be most common but I didn't know it was impossible for more. That makes life a bit simpler.

Having said all that, the original poster was looking for an analytical solution.

Ah, of course, that rules out the Newton-Raphson method. Sorry.

 

[Miner1]

mjessick is on the right track to finding the analytic solution. What results is an equation of the form

a*cos(x)+b*sin(x)+c = 0.

There is a very clever way to solve equations in this form, and it makes use of half angle identities. Let y = tan(x/2), then it is easy to show that cos(x) = (1-y^2)/(1+y^2) and sin(x) = 2*y/(1+y^2). Using these identities in our original equation and we get a simple quadratic equation

(c-a)*y^2+2*b*y+(c+a) = 0

The solution of this equation, as is well known, can have two real solutions (corresponding to two points of intersection), one real root (one point of intersection), or two imaginary roots (orbits don't intersect at all). Take the real solutions and solve for the original variable of interest: x = 2*atan. Note that quadrant checks are not necessary.

That's all there is to it. Hope that helps.