WW2 engines... the issue of... torque

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Ioshic

Airman
11
18
Feb 17, 2014
Hello everyone,
there is a question which I've long wanted to ask to the people who are more knowledgeable than myself,
but I've been mostly refrained from asking for fear of appearing a kind of "ignorant" in such an easy matter.
Please excuse me if my question is really stupid or very easy to reply.

But , anyway, here I am.
We don't know in real life, so I won't have to be embarassed everytime we could meet again!

I'm not a mech. engineer or similar; in fact I studied political sciences and international relations in my University times, so
my basic knowledge of engines and physics is very primitive. Amateurish we may say.

I love planes and old cars, and I basically know how en engine works, but not in detail unfortunately.
I know the difference in cars between HPs and NMs. A 2.0 liter turbo-charged engine with tons of NMs
has a totally different feeling compared to a 2.0 liter normally-aspirated one.

What I wanted to ask is: why are WW2 engines rated and discussed always by their Horsepower and never by their Torque and Nm values?
Is it because , with constant speed propellers, the engines were almost always "rotating" at the same speed, and hence, torque values
are of secondary importance in a plane?
(I tried to read about in in Wikipedia, but I admit I haven't understood much...)

And did they have an impact anyway in each different engines? On take-off? (I remember reading about the "torque effect" during take-offs)
In other flight maneuvers?


Or, am I right in understanding that Torque is the same as the "Manifold Pressure" on those engines?
If so, why it's never written as Newton-Meters?
 
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Hello everyone,
there is a question which I've long wanted to ask to the people who are more knowledgeable than myself,
but I've been mostly refrained from asking for fear of appearing a kind of "ignorant" in such an easy matter.
Please excuse me if my question is really stupid or very easy to reply.

But , anyway, here I am.
We don't know in real life, so I won't have to be embarassed everytime we could meet again!

I'm not a mech. engineer or similar; in fact I studied political sciences and international relations in my University times, so
my basic knowledge of engines and physics is very primitive. Amateurish we may say.

I love planes and old cars, and I basically know how en engine works, but not in detail unfortunately.
I know the difference in cars between HPs and NMs. A 2.0 liter turbo-charged engine with tons of NMs
has a totally different feeling compared to a 2.0 liter normally-aspirated one.

What I wanted to ask is: why are WW2 engines rated and discussed always by their Horsepower and never by their Torque and Nm values?
Is it because , with constant speed propellers, the engines were almost always "rotating" at the same speed, and hence, torque values
are of secondary importance in a plane?

And did they have an impact anyway in each different engines? On take-off? (I remember reading about the "torque effect" during take-offs)
In other flight maneuvers?

Torque is the turning force (Drei Moment in German). Horse power is basically the torque x the RPM (revolutions per minute) I have never flown a plane (but I have ridden a Moto Guzzi) as the propellor turns in one direction the plane (or Moto Guzzi) wants to turn the other way. All WW2 piston engined A/C had massive torque. And streamlining for max performance made torque reaction worse.
 
Next time your facing a tractor trailer at an intersection, watch what happens when the light turns green. If he's loaded and on the throttle, the cab of the truck will lean to one side as he accelerates. The truck is twisting opposite to the rotation of the crankshaft. Now imagine the same thing in a plane. The plane will try to roll opposite the direction of the prop.

Manifold pressure is not torque, it's boost provided by the supercharger. More boost will result in more torque, however.

Aircraft engines are measured in HP because its a measurement of work over time.
 
Horsepower = (torque * rpm) / 5252. The HP is 550 foot-pounds / second HP and torque is in foot-pounds.

Conversdely, Torque = (5252 * HP)/rpm.

In the USA, we usually quote HP, but it is simple to convert. These formulas are for Horsepower (550 foot-pounds / second, and torque in foot-pounds). It is easy to convert to N-m ... 1 foot-pound = 1.355816 N-m (or M-n as you prefer). 1 N-m = 0.7375633 foot-pounds.

From the formula, horsepower equals torque at 5,252 rpm. Below that rpm, torque is greater than HP and above it, HP is greater than torque.

I think WWII engines are rated in HP due simply to tradition. James Watt defined horsepwoer, and almost everything else grew from there. In WWII, they had simply clung to HP as the yardstick. Diesels are usually lower rpm engines and produce a LOT of torque. Many are rated by their torque. Gasoline engines are usually higher rpm engines and are usually rated by HP. But converting is simple. Assuming you use a spreadhseet of some sort, make up a conversion sheet and then it is as simple as entering one value or the other.

By way of example, let's say you have a Honda S-2000 engine. It is rated at 237 HP at 8,300 rpm. Then (237 * 5252) / 8300 gives 149.97 foot-pounds of torque. 149.97 ft-lbs * 1.355816 = 203.33 N-m. Since the rpm is above 5252, you'd expect the torque (in foot-pounds) to be less than HP.

Metric horsepower is very close to SAE horsepower. Metric horsepower is usually quoted on cv or PS. One 550 ft-lbs/s HP = 1.013869359 cv or PS.

So a 1,600 HP engine in WWII would produce 1,622.2 cv or PS. So, it's close, with SAE HP being a slightly smaller number. A 1,600 cv engine produces 1,578.1 HP. Close enough to use either rating, really. It won't make a large difference either way. The difference is only about 1.3% and that is usually less than the uncertainty in the HP or cv rating.

As stated above, the aircraft tends to roll in the opposite direction to the rotation of the crankshaft. Most engineers tried to alleviate this by putting a gear between the crankshaft and the propeller shaft so the prop turns opposite to the crankshaft. That assumes a geared engine, not direct drive. That reduces the net torque seen by the airframe by a bit, but it is still there. You have to be easy on acceleration when slow in a high HP aircraft.

You can torque-roll the aircraft into the ground if you snap the throttle open when just airborne. The torque overcomes the control authority afforrded by the rudder and ailerons and the aircraft can roll against the control input completely over. If you are low and slow, it is usually fatal. You can survive it up high by throttling back and recovering from whatever attitude you find yourself in.

There comes a point in lower HP aircraft when the torque is insufficient to be much of an issue. A 160 HP Cessna 172 has very little trouble with torque. A 2,000 HP F8F Bearcat will torque roll right over if you slam the throttle open from idle on the ground.

Hope that helps a bit. Welcome to the forum! There really aren't any questions that anyone should be embarassed to ask. All of us started out rather bereft of aviation knowledge and we had to learn somewhere, too. We have guys in here who were and maybe still are practicing aeronautical engineers and guys who have never flown and are just getting interested.

Again, welcome.

So ... if you want to talk torque, go ahead. Most of the posters in here can convert it if they choose to do so.
 
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The reason engineers cite horsepower is that horsepower is determined by torque and RPM. This is important because, for example, if an engine has the same torque at 2000 and 4000 RPM, the higher RPM will have 50% greater horsepower. Thus, if by a reduction gear, the 4000 RPM crankshaft speed is reduced to 2000 RPM at the prop, the torque will be twice that of the 2000 RPM engine setting.

A variable speed prop plays into this by varying the engine speed depending on the prop pitch or bite. It's analogous to shifting gears in a car. For cruise you want a high gear and low RPMs. For power you want a low gear and high RPMs.

The "torque" reaction in a plane is more a function of power than engine torque I believe. You won't notice it as much at low RPM and high pitch as at high RPM and low pitch.
 
The "torque" reaction is exactly that, a reaction to torque. Torque is a force exerted over a distance, and that is what causes the aircraft to move with the torque force.

Horsepower is sort of an aritficial, made-up unit. Dynamometers measure torque and then it is converted into horsepower. There is no direct measurement of horsepower as far as I know.

Power is measured in many other units besides horsepower, depending on location and application. Watts (W) and kilowatts (kW) are in common use worldwide. Less common measures include British thermal units per hour (Btu/hr) and foot-pound force per minute (ft-lbf/min). Since any or all of these could be valid performance measures, technically it's more correct to talk about an engine's power than its horsepower. However, in practice most people in the United States use the two words synonymously.

Standard mechanical horsepower is defined as about 33,000 ft-lbf/min, or 745.7 watts. However, there are many different official and unofficial definitions of horsepower. Some of these definitions refer to different ways of measuring power for a specific application. For example:

•"Brake horsepower" generally means that the power was measured on a type of dynamometer called a "prony brake"
(sometimes incorrectly called a "pony brake"--prony refers to the inventor, Gaspard de Prony)

•"Drawbar horsepower," used in railroad applications

•"Air horsepower," used in fan calculations.

Some types of horsepower define power at different magnitudes. Listing them in terms of relationship to the standard horsepower, we see:

HP.jpg



How did we end up with the "horse" in horsepower? The term was coined by James Watt (1736-1819), the British inventor best known for his improved steam engines, who used the term to relate steam engine performance to that of horses. At the time horses were the primary energy source for applications ranging from pumping water from mines and turning grinding mill wheels to pulling carts and loads. Although sources differ on exactly how Watt arrived at the number, it's generally thought that in 1782, he noted how quickly a brewery horse could turn a mill wheel of a certain radius, estimated the amount of force the horse needed to exert to turn the wheel, did the math, and came up with a value of 32,400 ft-lbf/min, later rounded to 33,000 ft-lbf/min.

Comparing the power output of a steam engine to an equivalent number of horses was an easy way for prospective engine purchasers to compare power ratings, so the term stuck. It seems likely the horse in question was either a Suffolk punch or a shire horse, but nobody ever quoted "shire horsepower" or "Suffolk punch horsepower." Given the year, it is very unlikely James Watt measured horsepower with a Clydesdale.

I know, too much information ... cheers.
 
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Manifold pressure is not a mesure of torque. Purely theoretically, (manifold) pressure is a force divided with area, torque is force multiplied with arm length.
 
Spitfire pilots used to Merlin engined versions could get into trouble when they flew a Griffon engined version as the engines rotated in opposite directions and therefore the torque reaction was opposite. I believe it led to several crashes where experienced pilots applied the wrong rudder and the reaction as the plane swung the wrong way led them to apply more and more rudder in the wrong direction.

I drive an automatic but 30 plus years of driving manuals means I still sometimes press the non existent clutch pedal :lol: ingrained automatic reflexes can be hard to unlearn.
 
Getting back to the original question it may have to do that most aircraft engines of a given era and type operated at rather similar rpm. There were few high rpm 'buzz' bombs and the few large displacement slow turning engines were generally the older, sliding into obsolescence, ones.
If you have two 1000 hp engines that operate within a few hundred rpm of each other they are going to have similar torque. Torque is also related to displacement. Not directly but the more displacement you have the more torque you have, in general, compression and manifold pressure do affect torque ( so do some other things but not to large extents). Up until 1940/41 most countries sued similar fuel so manifold pressure was was also similar. A 27 liter 1000hp engine at 3000 rpm may have less torque than a 30 liter 1000hp engine at 2700rpm but difference is going to minor.

And then you have propeller "slippage". Props, at their best, transmit about 80% of the power to the air, unlike tires. Torque, in a car or motorcycle, affects acceleration. HP affects speed. A high torque but low revving engine gives good acceleration off the line (or allows you to pull trailers/weights) but has limited speed. A 'buzz bomb' engine is going to show it's power only after the vehicle is moving. (assuming normal gearing).

Aircraft in the 20s and 30s used fixed pitch props ( or light aircraft today) and the prop "slippage" at low speed ( taxi/take-off) was enormous.

The also used different reduction gears on the engines to match props and aircraft. Higher ratios (slower but bigger props) for larger, slow moving planes and low ratios for fighters. The constant speed props cut down on the need for a large variety of reduction gears.

The Napier Dagger was one of the few "buzz bomb" engines, 1000hp at 4200 rpm from 17 liters (low torque) but it's .308:1 reduction gear resulted in a prop speed of about 1300rpm. Prop speeds were governed by diameter and tip speeds. Merlins used either .477 or .42 gears for prop rpms of 1430 or 1260.

You know just looking at the displacement and RPM that the Dagger is a low torque engine but the engine maker supplies the reduction gear.
 
The problem with the aircraft rolling in the opposite direction to the prop rotation was part of the logic behind Blohm Voss's asymmetric aircraft designs. (I've just been posting some kit photos of two asymmetric B&Vs over at http://www.ww2aircraft.net/forum/your-completed-kits/blohm-und-voss-strangeness-40074.html).

Placing the weight of the cockpit off to one side of the propellor counteracted the tendency to roll, as well as offering a better all-round view for the pilot - see the glasshouse canopy on my BV 141 pics (first set of pics). The BV P194 (second set of pics) was not only aysmmetrical in design but had a turbojet under the crew compartment which delivered slight upward thrust in level flight, I presume to add further oomph to counteract the torque from the big propellor and engine. It's a shame the 194 never got built - it would have been fascinating to see how (if?) the thrust and torque from the piston engine and the turbojet could have been coordinated.
 
Horsepower is sort of an aritficial, made-up unit. Dynamometers measure torque and then it is converted into horsepower. There is no direct measurement of horsepower as far as I know.

Power is measured in many other units besides horsepower, depending on location and application. Watts (W) and kilowatts (kW) are in common use worldwide. Less common measures include British thermal units per hour (Btu/hr) and foot-pound force per minute (ft-lbf/min). Since any or all of these could be valid performance measures, technically it's more correct to talk about an engine's power than its horsepower. However, in practice most people in the United States use the two words synonymously.

Torque and (horse) power are different. Torque is a force measurement and horsepower is, well, power, i.e., the ability do work. For instance, a simple electric motor can develop a high torque at zero RPM. It can push hard but can't do any work. Once it starts rotating the torque diminishes and horsepower builds.
Torque is analogous to volts and power is analogous to watts in electrical terms (which probable doesn't help).
 
Spitfire pilots used to Merlin engined versions could get into trouble when they flew a Griffon engined version as the engines rotated in opposite directions and therefore the torque reaction was opposite.
By way of illustration, here's a quote from Spitfire Command, by Group Captain Bobby Oxspring:
The first priority was to get the pilots fully at home with the Spit XIV which differed considerably from the previous Marks of Spitfires in that it was powered by a new Rolls Royce engine, the Griffon Mark IV. To absorb the 2,035 horse power, the Griffon required a five bladed Rotol propeller which rotated clockwise in the opposite direction to the Merlin. This considerable increase in power coupled with excessive torque needed the most gentle handling on take-off. No more than two-thirds throttle was required to get the aircraft off the ground.
In converting to the Spit XIV, pilots who appreciated the radical differences from the Mark IX coped quite easily, but those who relaxed or were careless caused some hair-raising incidents. One pilot in particular, Warrant Officer Red Bloomer, who had recently been shot down in France and returned to 91 Squadron after evading capture, was so anxious to get airborne again that he took off in an XIV without adequate briefing. The aircraft leapt off with a gigantic swing which with full rudder he could not hold, then headed at right angles from the runway towards the hangars. After his wheels missed the top of the buildings by inches, it took those of us who witnessed his hairy exhibition a considerable time to get our toes unknotted.
 
Horsepower and torque are not "different" ... they are related by a simple formula, given above. As long as you have the rpm and either torque or horsepower, you can get the other one easily. They have different units, true, and so measure different quantities, but they are directly interrelated to each other ... in a running piston engine anyway. Away from a piston engine, they are not related.

You can produce a torque wihout any rotation at all, so you can torque with zero power. All of my stuff above was intended solely for running pistion engine calculations, not as as a Physics definition.

In Physics, torque is not related to power unless work is being done somehow in a circular motion. Anyone who has used a torque wrench knows you can apply torque to a stationary bolt. Since it isn't moving once toqued no "work" is done, but you are definitely pulling the toque wrench with a force. After awhile, it can get tiring, especilly if you are applying 200 foot-pounds with a 4-foot torque wrench.
 
Like many things, there are advantages and dis-advantages to many types of engines/motors. Electric motors can develop maximum torque at stall, one reason they don't need transmissions/gear boxes, however most electric motors also have max current draw at stall (like a dead short) and don't last long stalled without going up in a cloud of smoke. Doesn't do battery packs much good either.
Piston Steam engines also develop max torque at stall. Very handy for rail locomotives (or even steam cars) as no reduction gear is needed. External combustion engines tend to be both heavy and fuel hogs though :)

For airplanes torque swings propellers and most of the time the engines came with the appropriate reduction gear to change the available engine torque into the needed torque to swing the propeller. Since most of the calculations in aircraft performance include time (speed or climb) HP figures (which include time) make things easier that starting with a torque figure and having to add time into the equation.

If you want torque the formula is

HP = Torque x RPM / 5252

or rearranged

TORQUE = HP x 5252 ÷ RPM

Or Merlin making 1030hp at 3000 rpm is making 1803 ftlbs of torque.
 
Manifold pressure is not really a pressure. It is actually a measure of vacuum.

Let's say you are at sea level on a standard day. The atmospheric pressure is therefore about 29.92 inches of Mercury. If you are sitting in a Cessna 182 with the engine stopped, your manifold pressure gauge will read about 29.92 inches of manifold pressure. You can't read it to that resolution, but it will be very close to 29.9 or the instrument is wrong.

What moves the air in the engine's induction system when it is running? There is no fan. The only thing that moves air in the intake manifold is a piston moving downward when the intake valve is open, perhaps more than 1 piston at a time depending on the engine and crankshaft angles. If your engine has a carburetor, then there is a plate that opens and closes off the intake manifold. It can't close it off completely since even when closed it has to pass enough air for the engine to idle.

If you are standing still with the engine at idle power (still at sea level), the carburetor throttle plate is closed and the engine is sucking air is around the edges of the throttle plate, but not enough. The pressure you see on the gauge will be around 14 inches or so. The poor engine is sucking all it can, but the plate is blocking the air flow and we get a partial vacuum (29.92 - 14 or whatever). That gives about 15.92 inches of vacuum in the intake under these conditions. You could measaure it with a vacuum gauge. Some older cars had a vacuum gauge in the dash. Ask your grandfather about vacuum leaks in his old cars.

If you are standing still with the engine at full power (still at sea level), the plate is open and the manifold pressure will read close to 29.9 inches of MAP (manifold absolute pressure). I say "close to 29.9" because the plate has some area and the air filter has some resistance and you won't quite see 29.9 … more likely 28.9 or so. Once the plane is moving, you tend to regain that lost inch or so due to ram effect of air being forced into the intake due to forward speed.

Manifold pressure, then, is NOT an indication of power. The gauge shows 29.9 when the engine is not running and about 28.9 at full power on the ground sitting still.

In order to get power from manifold pressure you also have to know the rpm and something about the engine. You won't see any boost pressure above ambient atmospheric pressure unless you are at considerable speed or your engine is boosted with either a supercharger or a turbocharger. If you have a turbo that generates, say, 6.5 pounds of boost, then at full throttle you will see about 28.9 + 6.5, or about 35.4 pounds per square inch on the MAP gauge, again assuming sea level.

Up to about 10,000 feet, the pressure drops off at about a pound per 1,000 feet so, if you are at an airport that is a 4,000 feet MSL, your MAP reading in the turbocharged airplane at full throttle as you take off would be about 35.4 – 4, or 31.4 inches on the MAP gauge. Anything else is an error and something is wrong.

That brings up another point. Days aren't always standard. We have days of high and low pressure. But, if you are a pilot, then you should know the elevation MSL of the airport close to the tower. If they don't have an operating ATIS, you simply set your altimeter to the airport elevation and read the pressure in the Kolzman window. Suppose you are at sea level and set the altimeter to zero, and you read 28.9.

Then your MAP gauge should read about 28.9 before you start the engine. If you are still in the turbocharged aircraft above that makes 6.5 inches of boost, your gauge will read about 28.9 + 6.5 – 1 at full throttle, or about 34.4 inches of MAP.

You CAN get power from MAP using the engine handbook. It assumes the engine is running at some specified rpm and from that, it is possible to extract power. I don't usually try to calculate it, I usually use the Pilot's Operating Handbook (POH) for the aircraft. In it you will find the percent power for various rpm and MAP values (or fuel flow values for injected engines). Rather than go to extreme lengths, 99.99% of all pilots just set their power (usually to 55 - 75% at cruise) by the POH.

Here in the USA we tend to learn power mangement around the traffic pattern by setting rpm in fixed-pitch aircraft. Over in Europe they yend to set the power using the manifold presure gauge in the same plane. It works out either way, and mostly a matter of habit. The reason you learn that is in the case that your airspeed indicator fails, you can land safely using power settings alone without fear of stalling at low altitude.
 
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One more thing about manifold pressure. For various reasons, various countries had their own traditional units of manifold pressure. The British used pounds of boost (boost above ambient atmospheric pressure). The USA used inches of Mercury absolute. The Soviets and Japanese used either millimeters of Mercury or water. The Germans used technical atmospheres, or ata. This had resulted in a lot of confusion.

So, here's a quick and dirty chart for conversion that is short enough to post and still be readable, and has a useful range of values. I believe I have the formulas correct, but I welcome any discussion.

Boost_Quick.jpg
 
For those of you who may be interested and don't already know, the conversions are as follows:

1. Convert inches Hg to psi boost: psi boost = ((14.696/29.92)* inches Hg) - 14.696

2. Convert inches Hg to ata: ata = (3.386 * inches Hg) / 98.0665

3. Convert inches Hg to Bar: Bar = (3.386 * inches Hg) / 10

4. Convert inches Hg to mm HG: mm HG = (inches Hg – 29.92) * (760 / 29.92)

With a little algebra you can easily go back and forth. Naturally: Hg = Mercury, mm = millimeters. The 3.386 * inches Hg factor gives you kPa. Bar is just kPa / 10 and ata is just kPa / 98.0665.

Again, inches Mercury is a US unit; ata is German; psi boost is British; mm Hg is Russian and Japanese (they also used mm H2O).

1 inch of Hg at 0°C = 13.5951002 inches of water at 4°C; and 1 inch of Hg at 0°C = 345.3250843 mm of water at 4°C.
 
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