I have a question? e2e = 51.89 (Page 22).

[tom234]

(02-23-2023 12:11 PM)tom234 Wrote:  

(02-23-2023 07:53 AM)EdS2 Wrote:  Just a few notes, no helpful answer from me sorry...

θ angle in orbital plane measured from perigee point; also, true anomaly
subscript 2e equivalent selected position in first orbit

(where equivalent means accounting for the rotation of the earth since launch)

(we can even write the subscripts like so: θ₂ₑ)

So the worked example on p20 of the document (pdf page 22) is

Case A, eastward launch.- For case A, eastward launch, the given launch position is ϕ₁=28.50° N., λ₁=279.45° E.; the selected position is ϕ₂=34.00° N., λ₂=241.00° E.; the number of orbital passes n is 3; and the equivalent selected longitude λ₂ₑ is 309.689° E. The values of the first approximations are
[t(θ₂ₑ) - t(θ₁) ] ≈ 7.693
Δλ₁₋₂ₑ = 32.162
θ₂ₑ = 51.89
t(θ₂ₑ) = 12.702
ψ₁ = 70.468
i = 34.081

From Appendix B, page 18, we have
cos(θ₂ₑ - θ₁) = sinϕ₂ sinϕ₁ + cosϕ₂ cosϕ₁ cosΔλ₁₋₂ₑ

I have done this but still can't get that number.