Engine torque output

AMCKen said:

That page says "mpg (imp.): 23.5-28.3 / 17.3-20.8 / 18.2-21.9 / 22.7"

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I know it does, that is with a warm engine and in a time before the oil shock when people didn't question such figures. Mine was 7 years old and when you started it the exhaust was not only black smoke but stank of petrol.

pbehn said:

I know it does, that is with a warm engine and in a time before the oil shock when people didn't question such figures. Mine was 7 years old and when you started it the exhaust was not only black smoke but stank of petrol.

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Carb problems, sounds like to me. Running way too rich.

Peak torque for aircraft engines is like peak torque for marine engines - not that relevant since the propellor can't absorb it all as soon as the rpm drops away (variable pitch excepted). So you only ever see the horsepower/propellor curve for marine engines vs. rpm, never torque.
On a dynamometer it would be possible to map it, but only a VP propellor would let you use it.
Perhaps the manifold pressure (and detonation onset) becomes the limiting factor once rpm falls and a VP propellor is in use - that might explain why peak torque is seldom ever seen....... any engineers out there that can comment on this?

AMCKen said:

It seems all aeroengines have listed output in horsepower, but I have yet to see what their output is in foot-pounds of torque (or other applicable units). One can calculate the output from the rpm and horsepower, but that only applies to the torque at the same rpm as the horsepower (2100hp@2700 = 4085ftlbs@2700). As we know from automobile engines, the peak torque is usually higher than the torque at hp peak, and at a lower rpm. Anyone have info? Thanks.

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For piston engines, normally you do not see torque readings; but when you go to turbine aero engines you do have torque limits for the output shaft. Most turbine/jet aircraft have readings in torque to measure the power available.

AMCKen said:

Horsepower is calculated from torque and rpm, which is what a dynamometer actually measures. Obviously it will take maximum boost and fuel.

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A large slow running engine will put out considerable torque, but may not be especially powerful. The reason torque gets attention is that in automotive applications, it is often useful to have an engine that has a nearly constant level of torque across a large range of speeds for reasons of drivability. An example of an engine that was poor in this respect was the BRM V16. Partly because of the centrifugal supercharger (which was uncommon in automotive applications), the torque (and to an even greater extent, the power output) at low engine speeds was very poor. Consequently, the driver had to do a lot of gear shifting to get good performance. The Auto Union V16, in contrast, had torque characteristics that allowed it to be raced with a minimum of gear shifting. In aviation applications, engines will usually operate over a much smaller range of speeds than those designed for automotive use and can be optimized for those conditions.

Rubenova said:

Not much out there. Check out
Aircraft Engine Dynamometer Testing | Allison Competition Engines, Inc

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Thanks. As you see, they do measure torque.
The problem with that picture is that is shows the torque at the same rpm as the horsepower. So it probably still isn't the peak for that engine.

Michael Hope said:

For piston engines, normally you do not see torque readings; but when you go to turbine aero engines you do have torque limits for the output shaft. Most turbine/jet aircraft have readings in torque to measure the power available.

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And many large piston engines have torquemeters installed too.

swampyankee said:

During the time I worked in the aircraft industry (Sikorsky and what was then called Hamilton Standard), torque rarely came up in conversation unless we were talking about gearboxes and shafting. Outside of sizing shafts and gears, torque has really very little importance to the people designing the main and tail rotors or propellers.

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And yet the power capacity of many gearboxes is rated in footpounds.

AMCKen said:

And yet the power capacity of many gearboxes is rated in footpounds.

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Yes, because the limit of the shafts and cogs in a gear box is a function of the feet pounds and diameters of shafts and surface area of cogs, often the most difficult condition for a gear box is at low speed and low RPM for example in a car towing a caravan or going up hill crunching gears between mis matched engine and road speed..

rinkol said:

A large slow running engine will put out considerable torque, but may not be especially powerful. The reason torque gets attention is that in automotive applications, it is often useful to have an engine that has a nearly constant level of torque across a large range of speeds for reasons of drivability. An example of an engine that was poor in this respect was the BRM V16. Partly because of the centrifugal supercharger (which was uncommon in automotive applications), the torque (and to an even greater extent, the power output) at low engine speeds was very poor. Consequently, the driver had to do a lot of gear shifting to get good performance. The Auto Union V16, in contrast, had torque characteristics that allowed it to be raced with a minimum of gear shifting. .

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For many years that was "the game" in motorsport, a broader flatter power curve or band that was driveable verses more power in a narrower band requiring more gear changes, with more gears in the box requiring a faster changing, more reliable gear box and stronger clutch. I can still remember the days when in F1 they used to mention the number of gear changes required to win at Monaco with a manual "box" something like 4000. There are also other aspects to it, for example on motorcycles with the same HP, a single cylinder 4 stroke has more traction out of a corner than a twin, and a twin has more traction than a four stroke 4, they all have more traction than two stroke twins and fours. This is due to the interaction between the engine and the tyres. In the dry this is academic because everyone pretty much has as much as they can use, on a wet circuit it is another matter, I had a few embarrassing moments in open practice with vintage BSAs and Norton singles. The world of big aviation engines is completely different as far as I have read simply because of forced aspiration, carburettor options and C/S props.

AMCKen said:

And yet the power capacity of many gearboxes is rated in footpounds.

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Power capacity isn't measured in foot-pounds; it no more a measure of power than is kilograms. As mentioned above, by pbehn, gears and shafts are sized by torque. This is one of the reason that the V-1710 in the P-39 and P-63 had its propeller reduction gearbox mounted in the nose, as a higher rpm shaft will be smaller for the same power than will a low rpm shaft. Do note that I said that the people who designed gearboxes and shafting cared about torque. None of the aero people did.

swampyankee said:

Power capacity isn't measured in foot-pounds; it no more a measure of power than is kilograms. As mentioned above, by pbehn, gears and shafts are sized by torque. This is one of the reason that the V-1710 in the P-39 and P-63 had its propeller reduction gearbox mounted in the nose, as a higher rpm shaft will be smaller for the same power than will a low rpm shaft. Do note that I said that the people who designed gearboxes and shafting cared about torque. None of the aero people did.

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It is a concept that is sometimes hard to grasp. Just looking on the "net" a Triumph Bonneville 790 has 44 ft/Lbs of torque at 3,500RPM and 62 BHP, I can easily generate 44 ft/Lbs of torque with a torque wrench or just a big piece of wood, but that isn't at 3,500 RPM.

torque is the instantaneous force acting on the output shaft.

The output shaft can be the end of the crankshaft or the shaft coming out of any manner of gearbox.

In an unsupercharged engine the max torque should be at the rpm with maximum volumetric efficiency. The cylinders fill the most completely for the most fuel burn per cubic in or sq in of piston area which gives the most pressure on the crown of the piston which in turn gives the most force on the connecting rod to turn the crankshaft. At a lower rpm there may be inefficiencies in the intake system that do not allow maximum filling of the cylinder. At higher rpm there may also be inefficiencies that limit cylinder filling.
However a 10% reduction in cylinder filling at 20% higher rpm means the engine will make more power per minute or second even if the shaft isn't being twisted quite as hard (torque) each revolution. Car/motorcycle engine power falls off when the cylinder pressure (BMEP=force on the crankshaft/torque)) is falling faster than than the increase in rpm (more little power pulses) can make up for it.

In a supercharged aircraft engine things get a lot more complicated. Cylinder filling is no longer done at near the same temperatures. Depending on the supercharger and gear ratio max boost may only be obtainable at near max rpm. Max boost should give the best cylinder filling (most fuel/air per piston stroke) but hotter air is less dense so manifold pressure alone doesn't tell the story. Gated throttles introduce pumping losses.

To pick an example of the first some American radials with two speed superchargers could hold take-off power to about 2500ft using full rpm. Much less rpm at sea level would give less power not only because of the lower rpm but because the torque would be lower because of the lower manifold pressure because the cylinders are getting less fuel/air at the lower rpm.

The Merlin III is the opposite example. The same rpm and manifold pressure at sea level for take-off gives about 150hp less than at 16,500ft.

Were does the power go at sea level or more importantly for this discussion , what would be the torque readings at sea level and at 16,500ft?
The throttle is part shut at sea level so there are pumping losses, but the engine is not getting the same amount of fuel and air to cylinders at sea level as it is at 16,500ft. It is making less torque at the same rpm. But once you have a constant speed prop, nobody really cares except the guys designing the reduction gears and the prop mechanism.

Airplane engines, as mentioned by others, operate over a much narrower range than car or motorcycle engines. The max torque rpm and max power rpm may be a lot closer or actually the same in some cases.

AMCKen said:

It seems all aeroengines have listed output in horsepower, but I have yet to see what their output is in foot-pounds of torque (or other applicable units). One can calculate the output from the rpm and horsepower, but that only applies to the torque at the same rpm as the horsepower (2100hp@2700 = 4085ftlbs@2700). As we know from automobile engines, the peak torque is usually higher than the torque at hp peak, and at a lower rpm. Anyone have info? Thanks.

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Aero engines do not have to operate over a wide range of speeds as do automotive/motorsport application engines. Therefore they are (usually) designed such that peak rpm with peak boost does not exceed the point where breathing & friction begin to reduce the gain in speed (which is why torque goes down in the first place).

As others have mentioned, "torque" is of very little use to the airframe designer from an aerodynamics perspective (i.e performance) because unlike a car, the propelllor is not mated directly to the propulsive medium (ie. tarmac). The plane designer cares about the speed and power available at the prop, to determine available thrust. Torque will only be used for mechanical design considerations, like the strength of the shafts and so on.

As an example the DB 605 A engine, has its power begin top drop off about 3000rpm at 1.3 ata manifold pressure. But at the actual maximum specified running speeds, power does not show any sign of dropping off over the whole operational range. (hence we stop increasing rpm before the torque curve "hump" is reached).

Car engines, do not have this luxury, and have to run over many very sub-optimal speeds to allow a normal person to drive it sensibly without having a very complicated gearbox (not sure if you have ever driven a car with a CVT... its really weird, and they never seem to have caught on). Hence in a car power curve you`ll see the full effect of the torque drop-off.

I`ve taken the liberty of taking another test which went to higher rpm and doing an EXCEL sheet to show you. Its from a graph which was a bit rough, and so its not exact (extrapolating, in these units they ought to cross at 5252rpm, which they dont quite, but I think its pretty close).

Engine Torque is directly related to horsepower, using the constant 5252 (min-hp)/(rev-(lb-ft)) multiplied by the measured torque, Lbf-Ft and speed, rev/min. This constant is the definition of horsepower, 33,000 Lb-Ft/Min divided by two Pi (i.e., 6.28) per revolution.
The "curves" we are all use to seeing for automobile engines have the shapes they do because they are usually naturally aspirated. As engine rpm increases the specific air consumption (pounds of air to produce one horsepower for an hour) changes because of the internal shapes of the air passages, valve timing and choking within the passages.
Most aircraft engines are supercharged and are not limited by these factors until reaching very high flow rates. These engines are BMEP limited by the fuel quality and pressure in charging the cylinder, and can operate up to the detonation limit for the engine. They in fact can achieve such high induction pressures that at well below the maximum RPM they can reach the BMEP limit.
I am attaching a power curve for a V-1710 run on a calibrated dynamometer a few years ago. You can see the direct relationship of torque to horsepower and that the curves have the same shape. Note also the values on this chart are the horsepower and torque developed at the crankshaft. When the usual 2:1 reduction gear is installed the numeric values of "engine" Torque double, while the RPM is halved, though the Horsepower remains the same.

I'm sorry, but I don't understand the initial question !

For a wheeled land vehicle, the wheels are fixedly connected to the engine shaft and therefore for a given vehicle speed, i.e. for a given engine rpm, it is useful to know what power is available at that speed. So, it is important to know the shape of the engine's power curve to assess its ability to accelerate (or climb slopes).

A good approximation of the general shape of an engine power curve is given by 2 points: the maximum power and corresponding rpm, and maximum torque with its rpm.

In a propeller plane, propeller is constantly « slipping » in the air : if you open full throttle, the engine runs at its maximum power rpm (or about), regardless of plane speed. So, general shape of engine power curve is useless - as is dual data max torque + its rpm !

This is the performance curve of the E-185 in the early Bonanza. You will note there are two power curves. One is the power that the prop can accommodate, the other is the max the engine can put out.

To a limited extent, different props will absorb different amounts of power.
This means that for a given rpm, the maximum torque the engine can put out is known however the actual torque depends on the prop.

I expect that any airframe designer will be very interested in the maximum torque. How else are they meant to design the engine mount and adjacent structure?

HP = (Torque * rpm) / 5252 where: hp = horsepower (550 ft-lbs/s); Torque = ft.-lbs. or lb.-ft,, whichever you like.

At 5,252 rpm, torque = HP. Anything below 5,252 rpm and Torque is higher. Anything above 5,252 rpm and HP is higher. Simple mathematical relationship.

AMCKen said:

It seems all aeroengines have listed output in horsepower, but I have yet to see what their output is in foot-pounds of torque (or other applicable units). One can calculate the output from the rpm and horsepower, but that only applies to the torque at the same rpm as the horsepower (2100hp@2700 = 4085ftlbs@2700). As we know from automobile engines, the peak torque is usually higher than the torque at hp peak, and at a lower rpm. Anyone have info? Thanks.

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HP = (torque * rpm) / 5252. At 5,252 rpm, torque and horsepower are equal.

That's horsepower as in 550 ft-lbs /sec horsepower.

GregP said:

HP = (torque * rpm) / 5252. At5,252 rpm, torque and horsepower are equal.

That's horsepower as in 550 ft-lbs /sec horsepower.

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HP = (Torque * RPM)/5252. At 5,252 RPM Torque and HP values are the same.

QUESTION: For Automobile engines the peak torque is usually higher than the torque at HP peak, and at a lower RPM. How is it for Aircraft Engines?

ANSWER: Aircraft engines have to deliver Torque to the propeller. It takes power to do this.

Below is a Torque and BHP versus RPM curve for an Allison V-1710-73(F4R) engine, having 8.8:1 supercharger gears and rated at 3,000 RPM to deliver 1560 BHP using 60 inHgA (14.8 psi Boost) manifold pressure. The torque data was obtained 5-11-2001 on the ACE Allisons calibrated dynamometer, and the horsepower values calculated per the above relationship.
See V-1710-73 Chart

The above chart covers the full range of useful power from the engine. Of significance, is the linearity of the shown data. For comparison, similar curves for the Chevrolet LS-1, a V-8, are shown below. This chart clearly shows the intersection at 5,252 rpm where torque and horsepower have the same values as well as the "linear" range from 1,000 to just over 4,000 rpm.
See LS-1 Chart

This linear range is where the cylinders are being fully filled on each intake stroke, i.e., Volumetric Efficiency is not changing as rpm increases, for both naturally aspirated and supercharged engines. When supercharging, intake manifolds and/or the valves have reached a flow rate where they can no longer deliver a full charge to the cylinder torque begins to drop off, as does the produced power. This is very apparent on the Torque/Power curves for an engine capable of reaching 5,252 rpm, or above.

Large aircraft engines, due to their relatively long strokes and heavy pistons are not capable of operating at speeds in the area of 5,252 rpm. As such, aircraft engine power curves do not show the peaking and subsequent drop-off. Instead, they operate in the very linear range where, for the most part, horsepower is a direct function of rpm.

Note that supercharged aircraft engines are rated, at each available rpm, by identifying the manifold pressure at the point of incipient detonation, as established by the fuel and mixture strength being used. This then describes their "available power" curve, which is always going to be in the "linear" range shown on the Torque/Horsepower versus RPM curve.

Another interesting aspect of the above formula is the origin of the 5252 factor. It is simply the definition of horsepower, 33,000 lb-ft/minute, divided by 2(pi). The 2(pi) comes from the conversion of RPM into Radians.

Dan Whitney

Orangevale, CA 2-4-2023