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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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