Hurricane engine torque

[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.