- #1
bobster4562
- 7
- 0
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 thoughts are - resolve instantaneous velocity V into its X and Y components (say Y is North-South pole and X is perpendiculer to it).
Lift being perp. to V means Lx and Vx would be in the same direction, but Vy and Ly would be in the opposide direction. I could then do
Acc x = gx + Lx - Vx [-Vx is basically subtracting the drag force)
Acc y = gy - Ly - Vy
gx, gy being the gravitational acceleration.
Am I on the right lines here ?
Cheers !
Bob
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 thoughts are - resolve instantaneous velocity V into its X and Y components (say Y is North-South pole and X is perpendiculer to it).
Lift being perp. to V means Lx and Vx would be in the same direction, but Vy and Ly would be in the opposide direction. I could then do
Acc x = gx + Lx - Vx [-Vx is basically subtracting the drag force)
Acc y = gy - Ly - Vy
gx, gy being the gravitational acceleration.
Am I on the right lines here ?
Cheers !
Bob