Orbiter-Forum Elliptical trajectory from 3 points and focus?
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 08-13-2017, 08:17 AM #1 dgatsoulis Orbinaut 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.
 08-13-2017, 03:05 PM #2 BrianJ thing 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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 08-13-2017, 03:24 PM #3 dgatsoulis Orbinaut Thank you very much!

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