Hurricane engine torque

[Viper2000]

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).

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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.

[TomcatViP]

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!

[Viper2000]

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 .

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.

[Redroach]

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

[RAF74_Winger]

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

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.

[RAF74_Winger]

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

W.

[Redroach]

Originally Posted by RAF74_Winger

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.

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

[Peril]

What you may be looking for is evidence that the wing, engine mount or rudder has been shaped with this 'counter' torque rotation in mind.

I fly and build RC planes too


I'll see what I can find in my data set on the Hurricane.

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The Vstab of the Hurricane was offset 1.5 deg to the left. That offset should provide some point where the plane rolls to the right. Does it yaw right with throttle off in CoD? If not, why not??

Logical progression here, this offset was intended to counter some of the torque, most likely neutral for cruise speeds?

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See below the tested speeds at which the plane was neutral in trim, under climb power (most torque effect) and at the glide (least effect).

It appears at around 130mph the plane was neutral in 'roll' but did still have a slight side slip. Note here the key is neutral in roll, ie. no aileron input needed to maintain bank angle. Side slip generated in the glide was undoubtedly due to the 1.5deg fixed vstab, thus evidence of this functioning as an intended counter to torque whilst under power. Furthermore at climb speeds above 130mph it was necessary to apply left aileron likely because of the 1.5deg vstab incidence influence was greater than that of the torque under power.

Hope this helps someone make a more accurate sim.

Order of graph:

Aileron Angle
Rudder Angle
Elevator Force
Elevator Angle

[b101uk]

Originally Posted by 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?

yes "output:input" is all well and good but if you look up automotive ratios for e.g. gearbox and rear axels or portal box they are almost exclusively done as per I have stated (input : output) and as most people will be familiar with, what next will we get fractions allowing the use of numbers less than <1 or numbers with decimal placing, in the world of describing things to “normal” people saying ~2.0964:1 instead of 0.477:1 works better because MOST people attribute that number for each ~2.0964 turns of the engine the propeller will turn once which is a reduction, if you give a figure of 0.477:1 it normally signifies to the masses a ratio higher than 1:1 e.g. each part turn of 0.477 of a crank revolution the propeller/prop shaft/wheel will turn once given that last part of a ratio is normally “:1”

Do you not see any perversity of logic in describing a physical system mathematically (reduction gearbox) that acts as a devisor (as dose #:#) by using a multiplier number (* 0.477) rather than a devisor number (/ ~2.0964) that mimics the systems function given most gearbox in this world around us are reduction gearbox that make something turning faster into something that turns slower and has more torque as a logical consequence.

As for “I don't know how accurate the rpm measurement would be, and it doesn't make any difference to the argument, so why worry” lol, because RPM is of every relevance, differential of RPM denotes ratio, HP dose not exist without a quotient of speed which RPM is and the engine speed being reduced via reduction box on a Merlin is not because they need more torque its because as you know you would have to use a smaller diameter propeller to stop the tips going to fast and would need more blades to have sufficient surface area which would yield a heavier more expensive prop with more moving parts which takes proportionally longer to make and blows more air backwards onto the aircraft nose (rather than past it) which is in all less efficient, however obvious torque increase from reducing RPM’s with a reduction box dose enable a big propeller diameter to be used wile keeping tip speed lower.

Originally Posted by Viper2000

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.

Err yes you do need to account for torque fully in every way, the first rule of any mechanical system is there MUST be sufficient force to induce movement, in this case torque is that force and coupled with RPM denote power BUT if torque is insufficient something will not turn regardless of computed notional power (HP), you can forget you fancy pants “CL and CD increase and your force on the blade is proportional to the square of the tip speed …………..etc” guff above and get back to basics to understand torque, you could have 200000000HP but if you lack torque by even 0.0001lbft you are not going to educe movement, if you have a constant torque engine and a fixed gear reduction then err torque will remain constant across the RPM range and HP will be the figure that changes the most HOWEVER if you lack torque at one RPM you will lack it at ALL RPM’s! thus as most engines DON’T have constant engine torque but instead have a curve of torque so there are natural shifts in torque along the rpm range and that peak torque is almost exclusively at a lower RPM than rated power so that means rated power speed has less torque than the peak torque value!

So in the case of the Merlin if you are at 3000rpm and you apply load via increasing prop pitch which exceeds available torque at 3000rpm then rpm’s will fall to the point on the torque curve ware there is sufficient torque, as a natural consequence of something turning slower than it was it consumes less notional power but ALWAYS MUST HAVE SUFFICIENT TORQUE, therefore torque and torque curve is of every relevance so my statement of

Originally Posted by b101uk

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.

is correct with reference to being a larger value than your approximation of ~4808 lb-ft assuming that was near correct @ 3000rpm and a course pitch was applied which dragged the engine down lower than 3000rpm to ware there is more torque available on the torque curve.

You should get it into your head that HP doesn’t really exist its purely notional and in the case of HP is any force that equals 550lb/ft/sec or 33000lb/ft/min equals 1HP, so 1lb @ 550ft/sec or 550lb @ 1ft/sec or 275lb @ 2ft/sec or 33000lb @ 1ft/min or 1lb @ 33000ft/min and so on ALL equal 1HP.