Elliptical trajectory from 3 points and focus?

dgatsoulis

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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.
 

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 – e²) / (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 – e²) / (1 – e.cosβ) 	        → 1/b =  (1 – e.cosβ) / a(1 – e²)

and  (for apoapsis, where  cosβ = 1)
d = a(1 – e²) / (1 – e) 		→ 1/d = (1 – e) / a(1 – e²)

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)

[B]So, f2 lies along the line f1->B
at a distance of (d – m) from f1[/B]

Any help?
Brian
 

MontBlanc2012

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In the opening post, dgatsoulis has posed the following problem:

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:

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 - [MATH]A[/MATH] and [MATH]C[/MATH] - which are at the same distance [MATH]R[/MATH] from the coordinate origin. The angular separation of these two points is [MATH]\theta[/MATH]. For a given apoapsis radius, which in the diagram is at a height [MATH]R+h[/MATH], there are two Keplerian arcs that pass through the points [MATH]A[/MATH] and [MATH]C[/MATH] with the required apoapsis. One solution places the apoapsis at the point [MATH]B[/MATH] halfway between the 'short' arc connecting [MATH]A[/MATH] and [MATH]C[/MATH] spanning an angle [MATH]\theta[/MATH]; the other places the apoapsis at the point [MATH]B'[/MATH] halfway between the 'long' arc connecting [MATH]A[/MATH] and [MATH]C[/MATH] and spanning an angle [MATH]2\,\pi-\theta[/MATH]. 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 [MATH]a[/MATH] and orbital eccentricity [MATH]e[/MATH] can be found as follows:

1. Set [MATH]A=2\,R[/MATH]
2. Set [MATH]r_a = R + h[/MATH]
3. Calculate [MATH]B=\sqrt{A^2 - \ell^2}[/MATH]
4. The 'short' arc solution is, then:
[MATH] a = r_a\,\frac{B - 2\,r_a}{A + B - 4\,r_a} [/MATH]
[MATH] e = \frac{A - 2\,r_a}{B - 2\,r_a} [/MATH]
5. And the 'long' arc solution is:
[MATH] a = r_a\,\frac{-B - 2\,r_a}{A - B - 4\,r_a} [/MATH]
[MATH] e = \frac{A - 2\,r_a}{-B - 2\,r_a} [/MATH]
Once the shape parameters [MATH]a[/MATH] and [MATH]e[/MATH] have been calculated, it is straight forward to calculate the distance between the two foci. This is just [MATH]2\,a\,e[/MATH].

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

1. Set [MATH]A=2\times 6370[/MATH] km

2. Set [MATH]r_a = 6370 + 1000 = 7370[/MATH] km

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

4. The 'short' arc solution is, then:
[MATH] a = 4787.13 [/MATH] km

[MATH] e = 0.539543 [/MATH]
5. And the 'long' arc solution is:
[MATH] a = 6839.27 [/MATH] km

[MATH] e = 0.0776001 [/MATH]
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 [MATH]r_a > R[/MATH]; for a hyperbolic motion, we require that [MATH]r_a<0[/MATH] for the 'short' arc (where [MATH]B>0[/MATH]); and [MATH]r_a<B/2[/MATH] for the 'long' arc (where [MATH]B<0[/MATH]).

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

[MATH] a = -1.97284 [/MATH]
[MATH] e = 1.02753 [/MATH]
and for the 'long' arc:
[MATH] a = -1.54118 [/MATH]
[MATH] e = 1.59542 [/MATH]
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:

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

MontBlanc2012

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Just as a bit of an afterthought re the above post:

The above post sets out a simple method for calculating the semi-major axis and orbital eccentricity given the start-point and end-point of a trajectory and one other piece of information - e.g., either the apoapsis radius or periapsis radius of the trajectory.

The assumption behind these equations is that the radius of the trajectory start-point and the end-point are the same. But how do these equations get modified when we relax that assumption and assume different orbital radii for the start- and end-points?

After thinking about this for a while, I eventually came up with the following:

The theory

Let's suppose that the orbital radius of the start-point is [MATH]r_i[/MATH] and the orbital radius of the end-point in [MATH]r_f[/MATH]. Let's also suppose that the the end-point is at an angle $\theta$ away from the start-point when the angle is measured in a standard counter-clockwise direction. Now, suppose that the one additional piece of information that we are given is the semi-major axis, [MATH]a[/MATH], rather than the periapsis radius or the apoapsis radius.

Then let:
[MATH]A = r_i + r_f[/MATH][MATH] \Delta h = r_i - r_f [/MATH][MATH] B = 2\,\sqrt{r_i\,r_f}\,\cos\theta/2 [/MATH][MATH] \ell = \sqrt{A^2 - B^2}[/MATH]
Next, solve the quadratic equation for [MATH]X[/MATH]:

[MATH]A - B\,X = 2\,a\,\left(1 - X^2 \right)[/MATH]
Label the two solutions [MATH]X_1[/MATH] and [MATH]X_2[/MATH]. These two values correspond to the two possible trajectories that pass through the start- and end-points with semi-major axis [MATH]a[/MATH]. We'll call the trajectory associated with the value [MATH]X_1[/MATH] "Trajectory 1"; and the trajectory associated with the value [MATH]X_2[/MATH] "Trajectory 2".

With this piece of nomenclature in mind, the orbital eccentricity of Trajectory 1 is:

[MATH]e_1 = \sqrt{X_2^2 + \frac{\Delta h^2}{\ell^2}\,\left(1 - X_2^2\right)}[/MATH]
and the orbital eccentricity of Trajectory 2 is:

[MATH]e_2 = \sqrt{X_1^2 + \frac{\Delta h^2}{\ell^2}\,\left(1 - X_1^2\right)}[/MATH]
(Note: the cross-over of the subscripts '1' and '2' is intentional!)

In addition, the time-of-flight of the passage from the start-point to the end-point (in the standard counter-clockwise direction) along Trajectory 1 is given by:

[MATH]TOF_1 = 2\,\sqrt{\frac{a^3}{\mu}}\,\left(\arccos X_1 + X_2\,\sqrt{1-X_1^2}\right)[/MATH]
and along Trajectory 2 is given by:

[MATH] TOF_2 = 2\,\sqrt{\frac{a^3}{\mu}}\,\left(\arccos X_2 + X_1\,\sqrt{1-X_2^2}\right) [/MATH]
where [MATH]\mu[/MATH] is the gravitational constant 'GM' for the central gravitating body.

(Note: again, the cross-over of the subscripts '1' and '2' is intentional!)


Of course, these equations can be re-expressed in terms of [MATH]r_a[/MATH] and [MATH]r_p[/MATH] if that is that one piece of additional information that one has to work with - but the resulting equations are most easily expressed in terms of [MATH]a[/MATH] as above.

These time-of-flight equations are closely related to Kepler's equation. Indeed, if in the first of these time-of-flight expressions, we set [MATH]\mathbf{E} = \cos X_1[/MATH] and [MATH]\mathbf{e} = -X_2[/MATH] we get:

[MATH]TOF_1 = 2\,\sqrt{\frac{a^3}{\mu}}\,\left(\mathbf{E} - \mathbf{e}\,\sin \mathbf{E} \right)[/MATH]
which, all of a sudden begins to show a marked resemblance to Kepler's equation. Of course, it isn't Kepler's equation - but, rather unsurprisingly, it shows a strong family resemblance.


A worked example

As a simple worked example of the above, let's take an object starting from low Earth with [MATH]r_i = 6671[/MATH] km transferring to a higher orbit [MATH]r_f = 7000[/MATH] km and is [MATH]\theta=45^\circ[/MATH] away from start-point. Let's suppose that we know that the transfer orbit has a semi-major axis of [MATH]a = 8000[/MATH] km.

Then, we can immediately write:

[MATH] A = r_i + r_f = 6671 + 7000 = 13671 [/MATH][MATH] \Delta h = r_i - r_f = -329 [/MATH][MATH] B = 2\,\sqrt{r_i\,r_f}\,\cos\theta/2 = 9664.06 [/MATH][MATH] \ell = \sqrt{A^2 - B^2} = 5240.49[/MATH]
So, we need to solve the quadratic for [MATH]X[/MATH]:

[MATH] 13671 - 9664.06\,X =2\times 8000\,\left(1 - X^2\right) [/MATH]
The two solutions of this quadratic are [MATH]X_1 = -0.154287[/MATH] and [MATH]X_2 = 0.943455[/MATH]. OK, so now that we have calculated the two solutions of the quadratic, we can immediately write down the orbital eccentricity for each of the two valid trajectories:

[MATH]e_1 = \sqrt{0.943455^2 + \frac{\Delta h^2}{\ell^2}\,\left(1 - 0.943455^2\right)} = 0.943685[/MATH]
and

[MATH]e_2 = \sqrt{(-0.154287)^2 + \frac{\Delta h^2}{\ell^2}\,\left(1 - (-0.154287)^2\right)} = 0.166289[/MATH]
So, the orbital eccentricity of Trajectory 1 is 0.943685; and the orbital eccentricity of Trajectory 2 is 0.166289.

Finally, we can calculate the time-of-flight. Again, we just plug values into the given equations with [MATH]\mu = 3.986004418\times 10^{5}[/MATH]:

[MATH]TOF_1 = 2\,\sqrt{\frac{a^3}{\mu}}\,\left(\arccos X_1 + X_2\,\sqrt{1-X_1^2}\right) = 6024.6 [/MATH]
and along Trajectory 2 is given by:

[MATH]TOF_2 = 2\,\sqrt{\frac{a^3}{\mu}}\,\left(\arccos X_2 + X_1\,\sqrt{1-X_2^2}\right) = 649.97[/MATH]
So for Trajectory 1, with eccentricity 0.943685, the time-of-flight between the start and end-points is 6024.6 seconds; and for Trajectory 2, with eccentricity 0.166289, the time-of-flight is 650.0 seconds. The first trajectory corresponds to a very high looping 'lob; whereas the second trajectory is a much 'flatter' and faster throw from the start-point to the end-point.
 
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