Bottle Fatigue, "
If we assume that the engines are producing constant reverse thrust force throughout the manouvre ( yes I accept your second contention ) you have a constant negative acceleration."
The observation (also stated in most manuals) was that you
don't have a contant negative acceleration! Hence, you can't assume a constant reverse thrust! I know that the reversers are still effective, and I have reversed an aircraft using the reversers myself - the point is that they are
not as effective as at speed.
norodnik: Bucket reversers are slightly more effective than high bypass cold stream reversers, but that isn't why you don't see 737s using power-back. The higher wear mentioned earlier is due to dust and sand ingestion kicked up by the reverser flow, and tail mounted engines with their intakes over, or close to, the wings are more protected from this than wing mounted engines. Conversely tail mounted engines can have vibration troubles (i.e. cause cracking) with the empenage at high reverse.
simfly: There is a big difference between drag and reverse thrust. We are not considering the effect of drag devices here, such as speed brakes or spoilers. Reverse thrust isn't proportional to the square of the speed.
Mad (Flt) Scientist: Some engines limit the thrust output in reverse, but I don't know of any that schedule reverse thrust with speed.
BIK_116.80: Dynamic pressure formulae don't relate to thrust. Thrust, (or reverse thrust, for that matter)
doesn't "push[es] against the force returned by ambient air ", in the same way that jet blast deflectors raised behind military aircraft on carriers don't increase take-off thrust, and a Saturn V rocket exhaust doesn't "push against the Earth" on take-off. All of these engines are
reaction engines, and operate according to Newton's third Law of Motion - i.e. throw something out the back, and you will be thrown forward a bit.
sprucegoose, Gotta love D.P. Davies!
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.
The Ram effect is considering increased thrust from increased mass flow through the engine, while he is talking about momentum change - which I think is a bit of a red herring, as he is ignoring the attendant momentum drag, but anyway...
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.
While true, this statement is also a bit of a red herring, in that we aren't looking for efficiency or the rate of destroying kinetic energy, but decceleration.
Power = Work done, divided by time.
Work Done = Force (in this case reverse thrust) divided by distance.
If the reverse thrust were constant, then the amount of force produced would be contant for each metre that the aircraft travels down the runway during the landing roll.
The Work Done by the reversers would be constant over each of those metres, but at the start of the landing run (i.e. at high speed) you are covering more of those metres in each second - so the engines are providing more power then at the end of the landing run (i.e. at low speed). As the fuel is burnt according to time, the fuel used over each metre is less at the start of the landing roll, than at the end (which is one of the reasons why jet aircraft are more efficient at high speed.)
However (before you get all excited
) Jet engines are (more or less) "constant force" engines, rather than "constant power" engines, and the acceleration is still related to the force, not the power. The rate of decceleration provided by the reversers (number of braking 'G') would still be constant throughout the landing roll,
if the reverse thrust remains constant.
In short, if the statement "
The reverse thrusters are more effective at high speed" is true,
and "
more effective" means a higher rate of decceleration,
then
the thrust provided must decrease with speed.
