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It might seem logical at first but why is reverse thrust actually more effective at higher airspeeds?

If we assume that engine performance will not change during a 100 knot deceleration on landing roll-out, and engine is kept at constant reverse thrust throughout the whole landing then the engine will produce a constant force to slow the aircraft down. It won't be more effective at higher airspeeds because we are continuously redirecting a constant amount of air at specific velocity. Even if we include an engine performance change on landing roll-out, thrust will probably not change by more than a percent. Am I missing something?

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  • $\begingroup$ FWIW, but perhaps not getting at what you're looking for, if you leave the engines at idle thrust, but just deploy the reverser doors (or whatever you choose to call them) that in itself will cause a significant at amount of drag over their closed position at high speed, especially if they're the old clam-shell doors. That drag will decrease as the airspeed across them decreases. $\endgroup$ – Terry Apr 1 '18 at 21:21
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From "handling The Big Jets" by D.P.Davies.

Reverse thrust is much more effective at high aircraft speeds than low speeds for two reasons. Firstly, the net amount of reverse thrust increases with speed because the acceleration imposed on the (constant) mass flow is greater. This is because the aircraft forward speed is additional when in reverse thrust as opposed to subtractional when in forward thrust. Secondly the power produced is higher at higher speeds because of the increased rate of doing work. In this context it means that the kinetic energy of the aeroplane is being destroyed at a higher rate at higher airspeeds.

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In simple terms when you deploy the reversers (clamshell or other) it is as if you were sticking both your hands out of a cruising car at 200 Km/h, obviously the slower you go the less effective the resistance on your hands becomes, all the way down to 0 Km/H were the resistance is nil. Same with the reversers....

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