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Is this data consistent with a parabola flown with the horizontal axis of the aircraft parallel to the velocity vector?

For any projectile in freefall the only acceleration is in the direction of the gravitational field and so the horizontal component of velocity must remain constant. If this is the case then the initial horizontal velocity must be the same as the velocity at the peak of the parabola. If the horizontal axis is parallel to the velocity vector then the angle of attack can be calculated from basic trigonometry:

 Hookes Law

 Hookes Law

This is considerably steeper than the stated 47°. There are several possible reasons for the discrepancy:

Assume that the velocity figures are correct. Calculate the maximum increase in altitude and the total length of the ground track over the period of one parabola (excluding the two hypergravity stages).

First it is important to make the units consistent, so:

 Hookes Law

The initial airspeed, the ground speed and the initial rate of climb form a right angle triangle and so:

 Hookes Law

The maximum altitude will occur when the vertical velocity (rate of climb) is zero. The altitude that this occurs at is given by the equation:

 Hookes Law

This is consistent with the stated altitude increase of 1000 m. The time to achieve this change in altitude is given by:

 Hookes Law

This is considerably longer than the stated ten seconds of microgravity. However taking this as the correct figure the entire parabola will last for twice this duration i.e. 29.4 s. During this time the ground speed of the aircraft remains constant and so the ground track covered is:

 Hookes Law

The aim of the initial hypergravity stage is to gain sufficient velocity and altitude so as to produce a lengthy period of microgravity. Why do the pilots not fly at a higher altitude and rely only on the downside edge of the parabola? This would remove the need for the first hypergravity stage.

One benefit of flying both sides of the parabola is that the aircraft begins and ends the complete parabolic sequence at the same altitude and at the same velocity and so the end of one parabola puts the aircraft into a condition ready to start the next. Flying just the down side of the parabola would involve the aircraft in a long climb back to the starting altitude. Flying both sides of the parabola saves both time and fuel.

Another, and more dramatic, point comes out of the drop tower questions. Nearly ten seconds of microgravity on a drop tower can be achieved by climbing up and then dropping down a 110 m tower or just dropping down a 440 m tower. A straight drop requires four times the change in altitude. So to achieve 20 seconds of microgravity by flying one side of the parabola would require four times the altitude change compared to flying both sides. The final rate of decent would also be twice as large and so the pull out phase of hypergravity would also be longer. In addition the aircraft would end the microgravity phase at a far steeper dive angle. The total altitude change over the entire manoeuvre would be closer ten kilometres rather than the two and a half kilometres currently experienced. The A300 Zero G may be classified as a test aircraft but it is still essentially a modified passenger aircraft, not a jet fighter.

Alternately why do the pilots not increase the initial airspeed (810 km/h) so as to gain more vertical velocity and so fly longer parabolas?

All aircraft have strictly defined operational parameters, exceeding these parameters in anything but an emergency is irresponsible and endangers the aircraft, together with the passengers and crew. 810 km/h is the maximum permitted airspeed of the A300 Zero G.

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