Hurricane engine torque

[b101uk]

[QUOTE=Viper2000;270501] The reduction gear of the Merlin III is 0.477:1, so the prop rpm is about 1431.QUOTE]

the reduction gear is ~2.0964:1 to give an output of ~1431rpm from an input of 3000rpm.

0.477:1 would give an output ~6289.3rpm from an input of 3000rpm

You seam to have a habit of fudging your figures I.e. showing the correct result but the wrong equation ware RPM’s is concerned.

Also the torque at 3000rpm will be somewhat less than the torque available at peak torque rpm which will be at a lower rpm than rated power, hence why rpm’s fall lower than 3000rpm back towards peek torque rpm when a course pitch is selected.

[Viper2000]

3000*0.477 = 1431

Rocket science it ain't...

The convention is that the gear ratio is output:input. The maintenance of this convention obviates the need to say "reduction" or "step-up"; but doing so provides an additional check. The same sort of logic applies to the Pressure Ratio of a gas turbine compressor (such that if you want a nice number >1 when considering turbine performance, you'd call it Expansion Ratio instead).

Now, since the above answer is exact, you may be wondering why I said "about" 1431 rpm. Well, there are several reasons. Firstly, I haven't counted the teeth so I don't know if 0.477 is exact or whether it's an approximation. Secondly this whole business is somewhat approximate anyway; I don't know how accurate the rpm measurement would be, and it doesn't make any difference to the argument, so why worry?

You also don't need a torque curve to explain the fact that rpm falls when pitch is coarsened.

Blade alpha increases, therefore blade CL and CD increase. The power required to drive the prop is larger than the power supplied (since input power hasn't changed, and the system was in equilibrium before).

However, the force on the blade is proportional to the square of the tip speed; thus the power required is proportional to the cube of the tip speed.

At constant engine torque (i.e. roughly constant BMEP) the engine power varies directly with rpm.

Therefore as the prop slows down its power demand falls much faster than the engine power output and so a new equilibrium rpm is reached.

No torque curve required.

[jf1981]

Originally Posted by Viper2000

Most of the aeroplanes I've flown in the sim will actually sort themselves out in cruise. However, you have to be flying at the correct speed, boost, rpm and altitude, otherwise it's not going to work.

Filling in the numbers I get roughly 4808 lb-ft. The maximum weight of the aeroplane is rather less than 7000 lb. Go figure...

Hi,

I double checked and found roughly your figures. I further estimate the plane would fall (roll) on the side by 10 to 15 degree in a second if you would'nt compensate.

How comes the counter torque effect is lowered with hight speed (when not compensated, it would roll fast at low sped, but roll less at high speed) ?

Regards
JF

[Sternjaeger II]

Originally Posted by jf1981

Hi,

I double checked and found roughly your figures. I further estimate the plane would fall (roll) on the side by 10 to 15 degree in a second if you would'nt compensate.

How comes the counter torque effect is lowered with hight speed (when not compensated, it would roll fast at low sped, but roll less at high speed) ?

Regards
JF

because at higher speeds, when the wing lift is more pronounced (efficient), torque is partially compensated by your wing lift.