1310 bhp at +12 psi, 3000 rpm at 9000'.

R.M.1.S. rating.

Alternatively, on 87 octane, you'd get 1000 bhp at +6¼ psi, 3000 rpm at 15500' (which is preserved in the R.M.2.S rating).

Harvey-Bailey, A. H. 1995 The Merlin in Perspective - the combat years 4th edition. Derby: Rolls-Royce Heritage Trust.

AFAIK these are static ratings (ie without intake ram effect; therefore the FTHs are somewhat lower than would be achieved during a level speed run).

///

4800 lb-ft of torque into rather less than 7000 lbm of aeroplane means that the torque at the prop shaft would be able to oppose the entire weight of the aeroplane with a moment arm a bit more than 6" long.

To put it another way, a hurricane has a wingspan of 40'; so the semi-span is 20'.

The torque is therefore equivalent to a 240 lb mass sat on one wing tip.

There is also propwash and p-factor to consider.

I am not sure those value can be applied to the Hurri or the Spitfire.

Does it include compressor neg torque (some said that it was pumping out 25% of available Pow) and the prop gear box losses ?

Modern turbin are quoted in Shaft HP, value that does not include gearbox etc...

The 240lb mass does not shock me. We can compare that to the force generated by the portion of the wing fitted with the aileron let says 1/3 of the wing area -> 7000lb/3 = 2300lb = 10 time more ;-)

~S!

It's the rated bhp, i.e. the nett horsepower measured on the dyno.

If you go back through the original data, then you find that bhp is at the end of a long chain of subtractions. IIRC it goes like this:

IHP>SHP>BHP

However the details are in this book, my copy of which has vanished (presumably "borrowed") by somebody :evil:.

It's the definitive work on the Merlin engine, and was originally published in 1941. You can actually build a pretty accurate model of the Merlin with the equations provided (and indeed, since they are general, you can also input data for the Griffon and get good results too).

Note that bhp doesn't include the exhaust thrust, which is very roughly 1/10th of the bhp in lbf.

For this reason ehp was invented in the turboprop era, though often the conversions used were somewhat approximate and primarily used for brochure purposes rather than performance calculation.

and what I don't get, in this case, is that the devs (the real devs) shouldn't have been aware of this and tried to counter it. Maybe some asymmetrical dihedral in the wings, or trim tabs pre-set (mechanically) to some value.
I did fly model airplanes for some time and the standard solution for counteracting engine torque there was to mount the engine slightly off-center in an angle. I had an easymode-to-fly standard motorplane trainer and bought a very strong engine for it, waaaay overmotorizing the plane. Nevetheless, the building instructions recommended mounting the engine about 3° off-center and that worked quite well, actually.
And right now, I can't see why that shouldn't be possible for real airplanes.
That said, theory aside, there's got to be some way to counter that engine torque, inherently. And there's got to have been some dev at Hawker to realize this.
Therefore, I still consider current engine torque too strong :)

Probably because you can't counter it over the whole of the speed range for the aircraft - any 'fix' would have to be an aerodynamic one; so a compromise was reached such that the effects were minimised in the normal operating range.

BTW real 'devs' are called engineers.

W.

Oh, and thanks for the link Viper. Thanks to the wonders of Amazon, I should soon have a copy.

W.

[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.

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.

A engine running at 1431 rpm, making propel go 3000 rpm, seems quite odd to me.
More like 3000/0.477 = 6289 rpm in engine. Thats quite high exept for formula1 cars.

yes, that's what I'm 'campaigning' for. Obviously, the aircraft development team (engineers are the lowly folks that actually calculate with numbers and stick prototypes together ;) ) of the Bf-109 took the problem into account and trimmed the plane level in a throttle range of 50-55% (constant prop rpm), as modelled in Il2-1946.
But, as already said, I just can't find a throttle/PP/maybe mixture - combination to go more or less straight ahead (while allowing rudder trim to cancel sideslip, of course) on the hurricane in CoD. It seems to get better at full open throttle and not too coarse PP, but I would project the 'crossover point' at about 140-150% throttle at least...
If I've been just too noobish and somebody *has* found a certain cruise setting, torque-wise, please enlighten me! After all, nobody wants to do a lengthy cruise and then engage the enemy with cramps in his right hand :rolleyes: