The Lambert Problem

[@husseinatetti-balogun1240]

thank you for this i really appreciage and you went into the actual formulas and application of them

[@VercilJuan]

Thank you for this sir

[@goliver1967]

Ottimo, molto chiaro e ben documentato con formule ed esempi 👏👏

[@giuseppescifoni4563]

Grazie amico.

[@emanueleplasmati2564]

Molto ben trattato l'apparato matematico. E il metodo numerico estremamente efficace. Sarebbe molto interessante una narrazione aneddotica della storia di questo problema. Solo una mia personale curiosità. E gli sviluppi nella ricerca degli obiettivi di ingegneria aerospaziale.

[@giuseppescifoni4563]

Certo. Sarebbe molto interessante. Una applicazione molto attuale è la determinazione delle orbite degli asteroidi con potenzialità di impatto sul nostro pianeta...

[@giuseppescifoni4563]

At slide 13 the Venus parameters in the table right up are not correct. The right ones are those at slide 12: Day= 8816 Radius= 0.727 UA Theta= 264.8 deg.

[@foley10x]

I like this video and was using it to validate my calculations and noticed there is an issue between slides 12 and 13. On slide 12 the info for Venus is Day 8816 R.727 Theta 264.8 deg. On slide 13 it is Day 8795 R.724 Theta 231.4 deg. It looks like the data from slide 12 is used throughout, however I calculate the arg pe to be 175, not 165. Can you verify which is correct?

[@giuseppescifoni4563]

Hallo my friend. The Venus parameters at slide 13 are not correct sorry. That must be a type error. Actually the travel time is set to 158 day, also the right date must be 8816=8658+158. The Argument of Periapsis of the transfer orbit is 165 deg instead. Reason I explain here: by getting E1 from formula at slide 10 the inverse cosine has sign ambiguity (+/- 3.006 rad). If you take the positive result (+3.006 rad) and you plug that into the arc-tangent formula below, you find theta_0 (arg_pe)=176 deg. But if you take the negative value (-3.006 rad) then the arg_pe=165 deg. To check that the negative value for E1 is the correct one, you may repeat same procedure for eccentric anomaly E2 and then finding theta_2-theta_0 and finally arg_pe. You see it turns to 165deg in both cases.