Originally Posted by
Machinbird
Henra, did you forget the vertical component of velocity? They add vectorially. And as the velocity vector is largely pointed down, drag (which is substantial) helps support much of the weight of the aircraft.
No, I did not, at least not intentionally (may be I was a bit unclear?!)
I said horizontal component was 60kts. Vertical was ~60kts as well.
But even if you combine both you end up with ~85 kts along the trajectory.
And these 60kts and 85kts is even TAS not IAS, so when entering it into a L/D diagram, you would have to use IAS which is roughly TAS divided by 1,8 at altitude, i.e. ~35kts horizontally or ~50 along the trajectory, increasing to 60/85kts at S/L.
That's not really much.
You would have to enter the L/D diagram of the A330 at an Alpha of ~45° with 85 kts and see if the combined Lift and drag higher than the 2100 kN (+ thrust of the engines).
That would mean ~1500kN along the trajectory, if no thrust of the engines available (idle, stalled) and ~ 1800kN at Cruise Thrust along the 45° path.
The projected surface at 45° being roughly 520 sqm, at 44m/s you get a resisting force of 620 kN at Cd = 1.
This means the Fuselage + wing would have to have a drag coefficient of ~ 2,5 - 3.
That's definitely too much.
I would expect something around 1.
At Cd = 1 it would mean ~80m/s => ~160kts along the trajectory. =>
~110kts horizontally + 110 kts vertically
On top of all that you would additionally have to shed the initial kinetic energy of it flying @470kts.