Question

Calculate the point of no return for an airport runway of 1.47 mi in length if a jet plane can accelerate at 11.5 ft/s² and decelerate at 6.75 ft/s².The point of no return occurs when the pilot can no longer abort the takeoff without running out of runway. What length of time is available from the start of the motion in which to decide on a course of action?

Answer 1

The jet will reach V^2 = 2aT where a = 11.5 ft/sec^2 and T = take off roll.

When it reaches V^2 at T, it then needs to decelerate V^2 = 2d(L - T); where L = 7920 ft runway length and d = 6.75 ft/sec^2. In which case.. L - T is the amount of remaining runway the plane has to come to a stop.

2aT = 2d(L - T) = 2(dL - dT) so that aT + dT = dL and T = dL/(a + d) = 6.75*7920/(18.25) = 2929.31 ft take off roll and distance of no return as there is insufficient runway left (L - T) to stop after having reached V^2 and T.