Originally Posted by
212man
Surely the TR thrust reduction follows a square law? So assuming a nominal 100% Nr is normal, 90% Nr = 81% TR thrust, 80% Nr = 64% TR thrust etc.
May I kindly ask you for an explanation why it would follow a square law?
Edit: After re-reading I'm afraid I misunderstood your point. The T/R itself follows a square rule but not any different than the main rotor. I misinterpreted this as that behaviour between main and tail followed a square rule. So you can ignore my comment/question
/Edit
It is driven with a fixed gear ratio (something along 1/8) against the main rotor.
Main rotor operating at 80% so will be the T/R. Both will then only create 64% of the respective Lift at 100% if nothing else changes.
@Thomas Coupling: I'm really curious as to why a 5% drop in M/R RPM leads to 20% loss at the back?
Did you get any explanation from the AW guys?
Lift normally follows (Rho/2)*Cl*A*v^2.
Since nothing else changes between front and rear this would require a change of Cl by Factor 4 (e.g. from 1.2 which would be rather high and occur at an AoA 8-10° to 0.3 which would occur at an AoA of ~2-3°). I don't want to say that is impossible but it is at least a bit strange.
The blocking effects of the Fenestron shrouding should rather decrease with reduced volume flow which also contributes to my surprise about these values.