Hurricane engine torque

[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.

[Revelation]

Everything in this thread makes my brain cry in confusion.

But as far as my 2 cents go, torque feels a little gimped, as someone stated earlier, applying full throttle when near stall speed should send your hurri into one hell of a spin afaik.

[Viper2000]

Originally Posted by b101uk

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”

Welcome to aerospace engineering. Enjoy your stay. You can complain about the conventions we use all you like, but they won't change.

Originally Posted by b101uk

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.

Not really. As I have said, you've got a nice sanity check from the combination of the convention output:input and the use of the word "reduction". If the number and the convention don't line up then this a good cue that you need to check further. Safety is paramount because once you're past V1 you can't just pull over onto the hard shoulder in the event of mechanical failure.

Originally Posted by b101uk

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,

drpm/d? = ? : ?
Answers on a postcard...

Originally Posted by b101uk

HP dose not exist without a quotient of speed

Are you Charlie Sheen?

Originally Posted by b101uk

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

Ah yes, Fast, that well-known holiday destination for propeller blades...

Originally Posted by b101uk

and would need more blades to have sufficient surface area which would yield a heavier more expensive prop with more moving parts

Why must a bigger propeller have more moving parts rather than simply the same number of larger parts?

Originally Posted by b101uk

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.

So in addition to groceries you also sell reduction gear boxes with which to dose those suffering from some kind of peculiar ailment or other.

No tiger blood? But of course that wouldn't be in the grocery department! Silly me!

Actually the main argument for increasing propeller disk area is that it allows you to move more air and thus approach a higher limiting Froude efficiency at any given operating point (TAS, altitude, input power).

Originally Posted by b101uk

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

Torque is not a force...

Originally Posted by b101uk

and coupled with RPM denote power BUT if torque is insufficient something will not turn

This really doesn't make any sense. Torque is a special case of a moment, i.e. force*distance. Force is a vector. If there is zero nett torque then "something" will keep on turning at the same rate as before. If nett torque is negative then "something" will experience an angular acceleration in the opposite direction to that which you first thought of when setting up your coordinate system.

Originally Posted by b101uk

regardless of computed notional power (HP),

There's nothing "notional" about brake horsepower. Hence the need for a whacking great water or electrical dynomometer to measure it.

Originally Posted by b101uk

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

I'm sorry, but this isn't the school playground. However, I suppose your misapprehension that it is would go some way towards explaining your obvious difficulties with basic spelling and punctuation, let alone engineering science. In the adult world, you will find that attempts to disguise your intellectual inadequacy by disparaging that which you fail to understand is not an effective strategy for winning friends or influencing people.

Originally Posted by b101uk

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!

No, as I explained, the torque required to drive a propeller through the air will vary as the square of its rpm; the power required varying as the cube of rpm. I don't need any groceries either.

Originally Posted by b101uk

thus as most engines DON’T have constant engine torque but instead have a curve of torque

No, most engines don't have constant torque simply because that doesn't happen to be how the physics generally works out. In the case of piston engines, torque variation as a function of rpm is caused by changes in induction, scavenging, frictional and heat transfer losses, as well as changes in charge temperature in the case of a supercharged engine.

Originally Posted by b101uk

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!

This is a circular argument.

It is not immediately obvious that an engine must achieve its rated power at a higher rpm then its maximum torque. It is certainly true that this is often the case, but in the end it depends upon the design of the engine and especially upon its physical size.

Generally speaking, the maximum rpm that a piston engine can attain is set by the acceleration loads imposed upon its reciprocating components.

To a first order, torque is independent of rpm because if you think about it, it's just the product of the BMEP and the piston area. The piston area is fixed, and the BMEP is set by the thermodynamics of the cycle.

So you know from the start of the design process that you're most likely to get more power at higher rpm.

Therefore you tend to set up the valve timing with that in mind, and so it's pretty unsurprising that peak power tends to end up close to peak rpm.

However, the engine will generally tend to breathe better at lower rpm because more time is available to fill and empty the cylinders. This means that you'll get a slightly higher MEP at lower rpm.

In the automotive world, people generally install large amounts of excess power in order to achieve rapid acceleration. Since gearboxes are expensive, you'll get better performance per unit cost if the engine's power curve is flat with respect to rpm, even if this costs you peak power.

Therefore you'll tend to tweak the exhaust and induction systems to improve low rpm performance, probably using something like Ricardo Wave if your design organisation doesn't have the necessary technical resources or motivation to produce its own code.

If the engine is intended to be anything like domestic then you'll probably also put quite a bit of effort into giving it a reasonable idle; this can mitigate quite strongly against the use of aggressive valve timing, because the sort of "characterful" refusal to idle smoothly at low rpm which sounds good for a few minutes at a drag strip gets very old very quickly in the real world, especially if you're the one paying for the fuel.

This enforced emphasis upon nice behaviour at low rpm will tend to reduce torque at high rpm because there just won't be enough valve overlap to let the engine breathe properly.

Of course, variable valve timing can solve that problem if you're prepared to suffer the increased cost and complexity...

But in any case, it is the nature of car engines that in the absence of a constant speed drivetrain of some sort they must have excess torque available across their operating range to provide acceleration.

Peak power is an entirely academic quantity in this context, because you can't use it anyway; inevitably it's a transient to be accelerated through shortly before the next gear change.

Aircraft engines are fundamentally different machines. You can quite confidently optimise them for a far smaller rpm range, and with a constant speed propeller you can quite easily maintain constant rpm from takeoff to landing if you so desire.

This means that the design drivers are totally different from the automotive world.

The same sort of argument applies to stationary power or marine engines. A really big industrial diesel engine can easily produce 10^5 bhp. Obviously, such an engine is extremely large, and turns at a relatively low (constant) rpm despite having a perfectly respectable piston speed.

It is quite easy to see how the peak torque and maximum rpm of such a machine might coincide.

So really it's not reasonable to assume that peak torque is naturally and inevitably at some "low" rpm for any given engine.

Thankfully, such an assumption is unnecessary for the reasons which I have already explained.

Originally Posted by b101uk

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

No, I still don't want any groceries to go with your "interesting" views on engineering. But do go on...

Originally Posted by b101uk

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

Writing in capital letters is not an especially effective rhetorical device.

Originally Posted by b101uk

, therefore torque and torque curve is of every relevance so my statement of

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.

Firstly, power has just as much existence as force, distance or time.

Secondly, check your units.

Thirdly, you'll find it much easier to get ahead in life if you learn to spell and punctuate.