Recent content by bobster4562

Thanks very much, I'll see if I can come up with a simplified expression based on the information in there !

The last NASA article is very interesting - they do not use a constant lift but a constant-ish L/D ratio. Thanks very much !

For ballistic re-entry case, I am assuming Cl does not change that much (we are talking a few minutes or so of re-entry flight time I guess). I appreciate that there is incredible amounts of drag and hence Cd must be a function of velocity or derived from a table or something.

Cheers, would also agree that this approach ie - L = 1/2 x rho x v^2 /(2 x Beta_L) D = 1/2 x rho x v^2/(2 x Beta_D) [and D acts opposite to V] Beta_D = m/(A x Cd) where m = mass, A = aerodynamic area and Cd = drag coefficient Beta_D = m/(A x Cl) Cl = lift coefficient would be...

Thanks, I appreciate your point. At the moment I am considering y to be Earth's N-S axis and x to be perpendicular to it, so all my forces would be resolved in this coordinate system. Assuming my body is descending in the +X, -Y direction (ie to the bottom right corner of your screen)...

My apologies, I was a bit careless when writing the equations down - By L I mean the acceleration due to lift force where L = 1/2 x rho x v^2 /(2 x Beta_L) and V I'm implying the acceleration due to drag where D = 1/2 x rho x v^2/(2 x Beta_D) [and D acts opposite to V] Beta_D...

Hey guys, I understand that 'ballistic re-entry' is usually used in the context of drag only with 0 lift, but if I were to look into the effects of lift how would I go about resolving the forces acting on a body at each instant to determine the instantaneous body acceleration ? My initial...