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Old 08-13-2017, 08:17 AM   #1
dgatsoulis
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Default Elliptical trajectory from 3 points and focus?

In the pic below we have a suborbital trajectory, where one of the foci is set at the origin (0,0) and these are known: A (departure base), B (Apoapsis), C (target base) and the angle between them (from the origin). Is this information enough to determine the shape of the elliptical trajectory needed to get from A to C? What I want to find is the second focus point.



Thanks in advance.
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Old 08-13-2017, 03:05 PM   #2
BrianJ
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Hi,
this seems too easy - I must have gone wrong somewhere......

I think you need to look at it in Polar terms:
Code:
Let distance f1->A = b
Let distance f1->B = d
Let angle φ/2 = β

then from
R(θ) = a(1 – eČ) / (1 – e.cosθ)
where R = distance from focus to point, a = semi-major axis, e = eccentricity,
θ = angle between major axis and f1-to-point line.

we get:
b = a(1 – eČ) / (1 – e.cosβ) 	        → 1/b =  (1 – e.cosβ) / a(1 – eČ)

and  (for apoapsis, where  cosβ = 1)
d = a(1 – eČ) / (1 – e) 		→ 1/d = (1 – e) / a(1 – eČ)

so:
(1/b)  / (1/d) = (1 – e.cosβ) / (1 – e)

re-arrange left hand side:
d/b = (1 – e.cosβ) / (1 – e)

d(1 – e) = b(1 – e.cosβ)

d – d.e = b – b.e.cosβ

d – b = d.e – b.e.cosβ

d – b = e(d – b.cosβ)

e = (d – b) / (d – b.cosβ)	//Now we have e


and from:
R(apoapsis) = (1 + e)a
then

d = (1 + e)a

a = d/(1 + e)			//Now we have a


Let distance from Apoapsis->f2 (B->f2) = m
then
m = a – e.a = a(1 – e)

So, f2 lies along the line f1->B
at a distance of (d – m) from f1
Any help?
Brian
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Old 08-13-2017, 03:24 PM   #3
dgatsoulis
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Thank you very much!
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Old 08-29-2017, 11:03 AM   #4
MontBlanc2012
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In the opening post, dgatsoulis has posed the following problem:

Quote:
In the pic [above] we have a suborbital trajectory, where one of the foci is set at the origin (0,0) and these are known: A (departure base), B (Apoapsis), C (target base) and the angle between them (from the origin). Is this information enough to determine the shape of the elliptical trajectory needed to get from A to C? What I want to find is the second focus point.
This problem is a variant of the classic Lambert's Problem which, according to Wikipedia, can be phrased as:

Quote:
In celestial mechanics Lambert's problem is concerned with the determination of an orbit from two position vectors and the time of flight, solved by Johann Heinrich Lambert. It has important applications in the areas of rendezvous, targeting, guidance, and preliminary orbit determination.
Whereas the classic Lambert's Problem formulation specifies a 'time-of-flight', dgatsoulis' variation specifies an apoapsis radius. This simple change of constraining quantity from time-of-flight to apoapsis radius makes the problem much easier to solve. This is because the time-of-flight constraint requires solving a non-linear equation involving trigonometric functions whereas the apoapsis constraint reduces to solving a quadratic. Sophisticated numerical procedures are needed to accurately, reliably and quickly solve the time-of flight constraint. Meeting the apoapsis constraint reduces to simple arithmetic. Moreover, in dgatsoulis' problem the initial and final radii (at the points 'A' and 'C') are the same. This implied symmetry further simplifies the problem.

In general, there are two Keplerian orbits that solve the dgatsoulis formulation. An illustration is given below:



In the above, which is modeled on dgatsoulis' diagram in his opening post, we have two points - A and C - which are at the same distance R from the coordinate origin. The angular separation of these two points is \theta. For a given apoapsis radius, which in the diagram is at a height R+h, there are two Keplerian arcs that pass through the points A and C with the required apoapsis. One solution places the apoapsis at the point B halfway between the 'short' arc connecting A and C spanning an angle \theta; the other places the apoapsis at the point B' halfway between the 'long' arc connecting A and C and spanning an angle 2\,\pi-\theta. These are precisely two solutions to this problem cast as a quadratic equation which, of course, always has precisely two solution.

Without setting out the proof, the solution to dgatsoulis' problem in terms of the semi-major axis a and orbital eccentricity e can be found as follows:

1. Set A=2\,R

2. Set r_a = R + h

3. Calculate B=\sqrt{A^2 - \ell^2}

4. The 'short' arc solution is, then:
a = r_a\,\frac{B - 2\,r_a}{A + B - 4\,r_a}

e = \frac{A - 2\,r_a}{B - 2\,r_a}

5. And the 'long' arc solution is:
a = r_a\,\frac{-B - 2\,r_a}{A - B - 4\,r_a}

e = \frac{A - 2\,r_a}{-B - 2\,r_a}

Once the shape parameters a and e have been calculated, it is straight forward to calculate the distance between the two foci. This is just 2\,a\,e.

As an example, let's assume that the that points A, C and O form an equilateral triangle such that R=6370 km; \ell=6370 km; \theta = 60^{\circ}. Let's also suppose for sake of argument that we that set h = 1000 km. Then, we calculate the solution to dgatsoulis' problem as follows:

1. Set A=2\times 6370 km

2. Set r_a = 6370 + 1000 = 7370 km

3. Calculate B=\sqrt{A^2 - \ell^2} = 11033.16 km

4. The 'short' arc solution is, then:
a = 4787.13 km

e = 0.539543

5. And the 'long' arc solution is:
a =  6839.27 km

e =  0.0776001

Although the solutions to this problem has been drawn as a pair of ellipses, the solution is also valid for the hyperbolic regime. Whereas for an ellipse, we require that r_a > R; for a hyperbolic motion, we require that r_a<0 for the 'short' arc (where B>0); and r_a<B/2 for the 'long' arc (where B<0).

As an example of two hyperbolae, let's take the same example as in the first diagram - namely where R=1 km; \ell=1 km; \theta = 60^{\circ}. This means that |B|=\sqrt{3} so, for sake of illustration, let's set r_a=-4. In this case, we find that for the 'short' arc:

a = -1.97284

e = 1.02753

and for the 'long' arc:
a = -1.54118

e = 1.59542

The two hyperbolae are plotted in the following graph:



Of course, in the hyperbolic case, specifying a given apoapsis radius is a little fathom since it invariably leads to negative numbers. A more natural way of constraining hyperbolic orbits is to specify the periapsis radius (rather than the apoapsis radius) and it isn't hard to show that in this case for the 'short' arc solution, we have:

a = r_p\,\frac{B + 2\,r_p}{4\,r_p - A + B}

e = \frac{A - 2\,r_p}{B + 2\,r_p}

and for the 'long' arc:
a = r_p\,\frac{-B + 2\,r_p}{4\,r_p - A - B}

e = \frac{A - 2\,r_p}{-B + 2\,r_p}
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