STS-119

Awesome job Discovery, I was actually at the launch when it happened coolest thing I have ever seen.
 
Small question:
during the deorbit burn, they weren't in retrograde attitude as we use in orbiter, there was a small angle between the velocity vector and the engines vector.... why is that? to make small corections to the path they will go until landing?
 
Small question:
during the deorbit burn, they weren't in retrograde attitude as we use in orbiter, there was a small angle between the velocity vector and the engines vector.... why is that? to make small corections to the path they will go until landing?

By burning out of plane they will burn excess prop. This is done to improve the vehicle's center of gravity (cg) for landing, then after the deorbit burn there is a dump of prop from the forward RCS to finalize the cg.
 
By burning out of plane they will burn excess prop. This is done to improve the vehicle's center of gravity (cg) for landing, then after the deorbit burn there is a dump of prop from the forward RCS to finalize the cg.

But how does the burning out of plane affects the path they will travel? Does it change (slightly) the inclination or any other orbit parameter? Is that change taken into account?
 
But how does the burning out of plane affects the path they will travel? Does it change (slightly) the inclination or any other orbit parameter? Is that change taken into account?
You didn't properly read David413's post. The out of plane angle is to burn off excess OMS propellant as the total delta V of the burn is higher for an Out-Of-Plane(OOP) de-orbit burn than a pure retrograde de-orbit burn.

With other words, it takes more OMS propellant to de-orbit OOP than purefly retrograde. This is done to control where the Center of Gravity(CG) is for entry. The CG needs to be as far aft as possible for better vehicle control during entry and landing.
 
Furthermore, I suspect the out-of-plane burn had little to almost zero impact on the orbital plane inclination. I believe the deorbit burn also didn't take place at the descending node (or essentially any node)!
 
But how does the burning out of plane affects the path they will travel? Does it change (slightly) the inclination or any other orbit parameter? Is that change taken into account?

Burning out of plane will change both the inclination and the RAAN of the orbit ever so slightly, but this change is neglegible, and the initial ground track will be altered greatly by aerodynamic forces during the S-turns.
 
Burning out of plane will change both the inclination and the RAAN of the orbit ever so slightly, but this change is neglegible, and the initial ground track will be altered greatly by aerodynamic forces during the S-turns.

Exactly. The shuttle can change it's orbit plane by burning only by about 1°, but each roll reversal happens after the Delta azimuth (difference between orbit plane and direction to runway) is 10.5°.
 
Ok, I get it now... I thought it was something like that, but I wanted to ask the experts Thanks!
 
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