How do you actually calibrate the end speed by just watching a video? Can you please walk me through your calculations.
Middle school physics.
Dist = 1/2 x Acc x (time)²
Dist was known 185m, time was observed 7 sec
(Important Note: Time observation may be fallible with negative degree. Any correction towards higher time decreases acceleration which indicates even lower T/W. If thrust is known, that would mean higher Takeoff weight which is trivial since we are trying to establish a bound safe high limit.)
Acc= 7.5m/s²
Velocity =acc x time
Velocity at end of deck about 200kmph
The deck is angled at 15°,
Vertical component of velocity is sin(15°) = 50 kmph(15m/s)
Horizontal " " " " cos(15°) = 190 kmph(55m/s)
Same with the thrust - Vertical thrust is 25% and Horizontal thrust is 96% of total thurst(26 mgf).
After take off before achieving flight sustaining lift, the jet would rely on its high AoA control characteristics which are unknown, so lets give it a disadvantage of being considered a nominal shape.
It hence follows a ballistic trajectory countered by vertical thrust.
Gravity pulls down with 10m/s², vertical thrust pushes up by about 2m/s² (this holds up until T/W of 0.8) creating a resultant of 8m/s². Trading 100% vertical velocity(15m/s) gives it 2 seconds to increase its horiz velocity over stall speed. The altitude provides extra safety against negative vertical velocity which can be cut by pitching up and increasing vertical thrust.
The horizontal thrust gives it extra 50-60 kmph within those 2 seconds to start sustaining required lift.
(In reality, aero lift will aid the jet within those last 2 seconds as well)
A lot of these numbers can be moved around but only in the same direction(aero lift,over deck wind considered 0, but will certainly be higher), which gives it even more safety. One might note that no aero lift equation is used, this is because the characteritics of the body are considered consistent upto sustained flight and target v has been set to 250 kmph.
