Discussion on engine "torque" misnomer

[kenamond]

Maybe this belongs in a different forum. Moderators feel free to move it.

I have to admit that I've been utterly confused by engine torque talk and how it relates to power since I was in college as a mechanical engineering student. Nobody explained it to me properly, and it is my suspicion that many describe it incorrectly all over the place (I found 4 references online in the past week, and all had it wrong).

I'm thinking I'm correct in what I'm about to say, but I tend to find out just how wrong I am by pounding my righteous chest on these forums. All I want out of this thread is a discussion of engine torque and a consensus in the end of what it's all about.

I'll start by stating some facts about work, energy, and power. I say they are facts, because they are straight out of the physics world. I've chosen my own wording just to keep it in context for this discussion.

Work and Energy
Work is the amount of energy that is expended in moving something against its will. By this, I mean that you have to apply a force to something and move it in the direction of that force. It is the amount of energy that is required to do something.

Example #1: Lifting a sack of potatoes from the floor to the back of a pickup truck. You have to apply a force away from the ground *and* move the sack of potatoes a few feet away from the ground.

Example #2: Compression/expansion of a gas. Assume you have a piston and cylinder full of gas. You have to apply force to the piston to decrease the volume of the gas in the cylinder. The physics geeks out there call this "PdV work", as you have to apply a pressure P to a fluid and change its volume some amount dV. Keep in mind that the gas can apply a pressure to its surroundings and increase its volume - but that's also PdV work. Obviously, we see the former case of work during the compression stroke of a 4-stroke engine. We smash the gas in the cylinder into an ever-decreasing volume by pushing on a piston. We also see the latter case of work during the power stroke. Gas at high pressure pushes the piston down which results in an increase of the volume of the gas. The compression stroke requires the engine to do work on the gas (compress the air/fuel mix) while the power stroke requires the combustion gas to do work on the engine (move the piston).

The units of work are the same as the units of energy. They are force*distance such as ft-lb or N-m. I'll get into this more later, but just because this is also the same units as torque, IT'S NOT THE SAME THING! That's where I was misled.

If you do the math for PdV work, you have pressure*volume. Pressure has units of force/area which is force/(length*length) or force/(length^2) like lb/in^2 or psi. Volume has units of length*length*length or in^3 or cubic inches. So if you multiply pressure and volume, you still get units of force*length.

Power
Power is the rate at which work is done. It is a work-rate. The units of power break down into force*distance/time like ft-lb/s or N-m/s. Notice that it is work per unit time or work/time. It states how quickly you can do work.

Lifting a sack of potatoes is another perfect example:

I can lift a whole 60-lb sack of potatoes in about 1 second from the ground to the bed of my pickup truck. The work is force*distance, so for a 60lb sack and a 3 foot lift, that works out to 180 ft-lb of work required to complete that task. The power is work/time. I did it in 1 second, so that's 180ft-lb/s of power.

Now, my 2 year old son can lift the sack of potatoes as well, but he'd probably do it two potatoes at a time with the help of a stool. He'd eventually do 180ft-lb of work to get the whole sack in the back of my truck, too. So we'd both do the exact same amount of work. But it may take him 60 seconds to do so. So my son is only generating 3ft-lb/s of power.

Pulleys, leverage, and hydraulics are all examples of this. The same amount of work is done in each case, but it takes longer. A 10:1 lever can lift 10x as much, but it does so 10x slower. The work is the same, but the power required is 1/10th as much.

Horsepower is a specific unit of power just like a foot is a unit of length. The definition of a horsepower is 33,000 ft-lb/min. A 1hp motor can lift 33,000 pounds up a height of one foot in one minute.

Work not Torque
Torque in the traditional sense is a twisting force. It is measured in units of force*length just as work and energy are, but that's where the similarities end.

Work requires you to apply a force and move something in the direction of that force. You can apply 100ft-lb of torque to a bolt without the bolt budging. That works out to exactly zero work done by you on the bolt. A ton of bricks sitting on the ground does no work either; even though it's pushing on the ground with 2000 pounds, it's not sinking, so no work is being done. If you apply 100ft-lb of torque to a bolt using a 4-foot breaker bar and rotate the bolt 90* in the process, you did work: you applied force to something (the handle of the breaker bar) and moved it (rotated it 90*). However, the work you did on that bolt was not 100ft-lb!!

The actual work done is the force applied to the breaker bar times the distance the breaker bar moved. If you were applying 100ft-lb of torque to a 4-foot breaker bar, then you were pushing on the bar with 25lb (25lb * 4feet = 100ft-lb of torque). The distance the handle of the bar moved to make that 90* turn is 1/4 of the circumference of a 8-foot diameter circle. The math works out to 6.28 feet. Taking the force of 25lb and the distance of 6.28 feet, that works out to 157 ft-lb of work...NOT 100ft-lb. Torque and work are not the same.

Work and Power and Engines
So how does this all fit together?

What everyone calls "torque" for engine performance is actually the work that is produced by the engine in one revolution. Since we're dealing with 4-stroke engines which only fire every other revolution, you could think of this as the average work done per revolution - half the work done in two revolutions or 1/20th the work done in 20 revolutions.

If we think about what goes on in the engine, there's a lot of work that the engine has to use to drive itself, and this work never makes it to the flywheel or the wheels:

- The compression stroke requires the engine to use some of its energy to compress the air/fuel mixture (and do PdV work on the air/fuel gas). It uses the work produced by other cylinders during their power strokes and the kinetic energy (think inertia) of all of the rotating and reciprocating mass in the engine to make it through the compression strokes of the different cylinders. The energy of the battery is used to get it all going in the first place when the motor is first started.

- Also, it takes energy to suck the intake charge into the cylinders and to expel the exhaust gasses during the exhaust stroke.

- And it takes energy to drive the turbine/compressor wheels. The turbocharger is actually a turbine engine in and of itself that is doing all 4 strokes at the same time. The turbine is driven by high-pressure exhaust gasses which cause it to turn (like the power stroke). This turning drives the compressor at the same time (compression stroke). Since the air is continuously feeding through both wheels, the intake and exhaust strokes are happening all the time. But the high pressure exhaust gasses which generate power in the turbocharger are produced during the exhaust stroke of the engine - and that is more work that the engine uses to make itself function and which doesn't make it to the flywheel or wheels. The piston has to work harder to push the exhaust out of the cylinder, because of the increased energy demand of the turbine.

- Then you have all of the friction and viscous losses, losses of energy via heat conduction into the coolant (from the block) and intercooler (from the hot compressor output), and vibrations (sound) which all leave the system and do not contribute to turning the flywheel or wheels.

The laws of thermodynamics state that energy is neither created nor destroyed. The only energy source in our engine is the combustion of the fuel-air mixture, and the trick is to get as much of that energy to the wheels.

Unless something has gone really wrong, even after all of these losses, we end up with work making it to the flywheel (analagous to brake horsepower) and to the wheels (analagous to wheel horsepower), although there are even more energy losses between the flywheel and road.

We can measure the power and/or energy (I refuse to call it torque) output of an engine at either the flywheel or at the wheels, and this is strictly controlled by where you actually measure it. You CANNOT back out power at the flywheel from power at the wheels, but I'm not going to go there in this discussion. If you bolt the engine up to an engine dyno, you'll get flywheel numbers. If you use a rolling road dyno, you'll get wheel numbers.

Regardless, the "torque" numbers are the amount of energy that the engine produced that actually made it to the measuring point (flywheel or wheels) PER REVOLUTION. In order for this to be useful, we need to know what RPM that "torque" value was measured. There are two reasons for this: 1) you need to know RPM to compute power and 2) engines produce different "torque" at different RPMs.

As for 1) above - since power is just the rate at which work is done, and the "torque" numbers are the work per revolution, we can compute power from (work per rev)*(rev per minute). There are some conversion factors to get it to horsepower, but that's the basic rule. As for the conversion factors, I already stated that 1hp=33000ft-lb/min. Also, 1rev=2*pi radians. A revolution is a dimension. It can be converted to a dimensionless value by converting to "radians". So putting all of this together to get power (in hp) in terms of rpm (in rev/min) and torque (in ft-lb), we get P=T*rpm/5252. But keep in mind that it's just taking the energy output from one revolution (that infamous "torque" number) and using the rpm to get the energy production rate of the engine.

As for 2) above - An engine running at 1rpm is NOT going to produce any work. It will stall. Our turbocharged motors aren't going to produce much work per rev at 1000rpm either. If you look at a dyno sheet, the most important number on it is probably peak "torque". That's the point in the rpm spectrum where the engine is producing the most energy per cycle. It's the engine's "sweet spot". In our engines, it's where everything is working best together. Turbo is at full boost, timing advance is good, intake and exhaust can handle the flow, cam timing is good for the current flow, etc. If you go a bit higher or lower, the energy production per rev drops off a bit; the engine is not quite as effective at producing energy per cycle. However, if your "torque" doesn't drop off very much past the peak "torque" value (at higher rpm than the peak "torque"), you can still get higher power at higher rpm. Since power is "torque"*rpm and the torque is holding in there as the rpm go higher, the power can continue to go up. Keep in mind, though, that the engine is doing a poorer job of generating energy.

Summary
What folks call torque is actually the energy produced per rev of a motor. It has nothing to do with twisting force other than the units are the same. High torque numbers mean that your engine is producing more energy per rev than low torque numbers. Since engines run differently at different rpm, the torque curves for an engine change with rpm. The torque curve is a bunch of snapshots of engine energy production capability at various engine speeds. Engines typically have a peak torque at a specific rpm, and this is where the engine is producing the most energy per rev than any other point in the curve.

Power is the rate of energy production of the engine. It is important, because it indicates how quickly energy can be produced from an engine. Lifting a 60lb sack of potatoes 3 feet is a fixed amount of work. Doing it in 1 second is better than doing it in 60 seconds. That's the difference between work (torque) and work rate (power).

Power in an engine is just torque*rpm, so it also varies as rpm changes (torque changes with rpm and (obviously) rpm changes with rpm). Peak power usually occurs at a higher rpm than peak torque, because rpm increases faster than torque drops off (usually), so the product of torque and rpm continues to go up a little while after the torque peak.
Nowhere am I saying power is not important. Let's say you have a motor with a peak torque of 300ft-lb at 5k rpm. It drops off after that, but still holds at 200ft-lb at 17k rpm if your engine could rev that high. The power at 5k rpm is 300*5000/5252=286hp. The power at 17k rpm is 200*17000/5252=647hp!! The trick is reaching 17k rpm (think about a 2.0 liter V12 motor with low reciprocating mass and HEFTY rods). In this case, you're pumping out 2.26 times as much energy per second at 17k rpm than you are at 5k rpm. That'll get you there faster! A lot of power comes from trying to keep the torque numbers good into higher rpm ranges and even in raising the redline.

So that's my piece. Whoever decided to call it "torque" messed up in my opinion. That word was already being used for something completely different, and all it's done is confuse people.

 

[D_Eclipse9916]

Not to be rude, but when are you guys gonna learn that it is all just a marketing scheme, jesus christ. Horsepower is technically a strange way to measure, but it was came up by a guy in the late 1900s to compare the power of horseless buggies to one another. A MARKETING point.

 

[kenamond]

Not to be rude, but when are you guys gonna learn that it is all just a marketing scheme, jesus christ. Horsepower is technically a strange way to measure, but it was came up by a guy in the late 1900s to compare the power of horseless buggies to one another. A MARKETING point.

Although this wasn't the type of post I was hoping for, I don't agree. Power and work are both extremely meaningful. Sure, the particular unit of horsepower was invented so that James Watt could give horse-and-buggy-types a means of comparison to what they were familiar with (horse power), but that has nothing to do with the fact that power is a real quantity that is very relevant to engine performance. If you'd prefer to have your engine performance quoted in Watts or BTU, feel free. Nobody invented them just as a "marketing scheme." The "torque" and power curves you get from a dyno run are valuable data for evaluating engine performance. If they are more impressive, car companies will point out that they are more impressive. If you know what they mean, you can judge for yourself how impressive they are.

 

[Rice Over Wheat]

IMO, torque is perfectly appropriate as a term relative to automobile engines. See here: http://science.howstuffworks.com/fpte4.htm

 

[kenamond]

IMO, torque is perfectly appropriate as a term relative to automobile engines. See here: http://science.howstuffworks.com/fpte4.htm

And that's one of the four sites I saw over the past week which are incorrect w.r.t. their connection to dyno torque values. The torque they're talking about is the true torque (like a torque wrench). It is, however, unrelated to horsepower (well, it's related but in a very different and much less direct way). The "torque" that you see quoted for engine performance is NOT the type of torque they talke about on that website. If you try to make it be the same, you'll just confuse yourself.

Twisting force times rpm gives you complete garbage.

A 2.3L stroker kit for the 4G63 motor increases "torque" by increasing the distance the combustion gasses can push the piston down the cylinder (work=force*distance, and the distance is larger in a stroker motor). They also have a larger moment-arm at the crank for larger real torque, but that's the related-but-in-a-very-different-way mechanism for real torque affecting "torque" or work.

Torque numbers on a dyno sheet are not twisting force torque; they tell how much work the motor does per rev. That site describes torque as a twisting force and then goes on to talk about dyno "torque" numbers even though the torque they explained has nothing to do with the dyno torque numbers.

 

[gob4sho89]

Wow, my head kinda hurts. I was thinking the same thing when I learned about torque and work and power and all of that good stuff in physics. I just let it go though, as I pick and choose my battles. I am now looking up all kinds of info on the origins of hp and touque. I will post if I find anything intresting... Good thread IMO

 

[Rice Over Wheat]

Sounds like you're reinventing the wheel here and most of this is just semantics.

 

[kenamond]

Sounds like you're reinventing the wheel here and most of this is just semantics.

Calling work done per cycle "torque" is misleading and in my opinion a misnomer, because torque is a twisting force which has nothing to do with work. I think that's a bit different than semantics. Nowhere in that "howstuffworks" set of articles do they explain the "torque" numbers in the simulated dyno, and that leads readers to believe that it is the twisting torque they describe in the earlier section. Their description of energy is way too limited, too, but that's beside the point.

I'm just trying to clarify dyno torque and power for myself and hopefully for others. Maybe I'm the only one in the world confused by dyno torque. Since I've not yet found any reference that properly describes dyno torque, I've not seen this wheel I'm reinventing.

I'd be grateful for a reference that correctly describes it.

 

[ThingyNess]

What he's trying to say (and he's right in doing so) is that engine torque is irrelevant when it comes down to it, at least as far as how quickly your vehicle accelerates.

What matters is your average horsepower throughout the powerband you're using.

Part of the reason people used to get all in a tizzy about torque being so important versus horsepower was the typical comparison of, say, an old carbed 305 camaro with 140 official factory horsepower, versus a tuned 1.6L honda civic Si making 140hp. Even given the same weight, the 305-powered vehicle will win every time.

Why? Not because it has more low-end torque, but because its Average horsepower over its useable powerband is much greater. If you look at a dyno graph of a detuned, low HP 305, it gets a lot of peak torque early, and as you move into the higher RPM range, the torque drops like a stone. With a 5L V8 making 140HP, it wasn't uncommon back in the day for the engine to make very closer to peak horsepower (not torque) over more than 40% of its usable powerband. For example, the 305 might make 140hp officially, at like 3800rpm. However, at 4400RPM, it's still making 130HP or so, and ditto at 3000ish. What this means is, when you shift at 4400RPM, if your revs drop down to, say, 3000, you're still making close to your peak horsepower number.

With the cammed, tuned, 140HP honda as an example, it might make its 140HP at 7000rpm. However, its redline, and even fuel cutoff is around 7500. Even shifting the poor thing at redline, because the torque curve in the honda is (relatively, in comparison to the 305) flat, your RPMS will drop down closer to 4500 or so, in which case you're only making closer to 90-95HP.

The average horsepower in the Honda's case is considerably lower, which is why it loses in the race. Its the peaky horsepower curve that kills it. But the fact that it might only have 100lb/ft of peak torque versus the 260ish of the 305 is *not* the reason it loses.

 

[kenamond]

What he's trying to say (and he's right in doing so) is that engine torque is irrelevant when it comes down to it, at least as far as how quickly your vehicle accelerates.

What matters is your average horsepower throughout the powerband you're using.

Part of the reason people used to get all in a tizzy about torque being so important versus horsepower was the typical comparison of, say, an old carbed 305 camaro with 140 official factory horsepower, versus a tuned 1.6L honda civic Si making 140hp. Even given the same weight, the 305-powered vehicle will win every time.

Why? Not because it has more low-end torque, but because its Average horsepower over its useable powerband is much greater. If you look at a dyno graph of a detuned, low HP 305, it gets a lot of peak torque early, and as you move into the higher RPM range, the torque drops like a stone. With a 5L V8 making 140HP, it wasn't uncommon back in the day for the engine to make very closer to peak horsepower (not torque) over more than 40% of its usable powerband. For example, the 305 might make 140hp officially, at like 3800rpm. However, at 4400RPM, it's still making 130HP or so, and ditto at 3000ish. What this means is, when you shift at 4400RPM, if your revs drop down to, say, 3000, you're still making close to your peak horsepower number.

With the cammed, tuned, 140HP honda as an example, it might make its 140HP at 7000rpm. However, its redline, and even fuel cutoff is around 7500. Even shifting the poor thing at redline, because the torque curve in the honda is (relatively, in comparison to the 305) flat, your RPMS will drop down closer to 4500 or so, in which case you're only making closer to 90-95HP.

The average horsepower in the Honda's case is considerably lower, which is why it loses in the race. Its the peaky horsepower curve that kills it. But the fact that it might only have 100lb/ft of peak torque versus the 260ish of the 305 is *not* the reason it loses.

Thanks for the reply I agree with everything except the first paragraph. Engine torque over the rpm band and horsepower over the rpm band tell you exactly the same thing. A torque curve is all you need or a horsepower curve is all you need; you can compute one from the other independent of the type of engine that was tested. That means (to me) that torque and horsepower matter the same. I agree that the four numbers a manufacturer gives you (W max. torque at X rpm and Y max. power at Z rpm) don't tell you much about how quick the car will be. A single torque value without accompanying rpm tells you squat. You need to know one of the curves. One exception would be a CVT car where you can hold the revs constant at the peak power point as you accelerate.

I finally found an old college textbook, and it confirms the "torque=work per revolution" notion, though it's only mentioned once in passing.

 

[jking29]

They way I have always understood it is that horsepower is a function of torque. Torque is the amount of work an engine can produce, and horespower is a measurement of how fast the engine can do it. The standard conversion rate to figure horespower from torque is 5252. That is why when you see a dyno chart, the torque and horsepower curves always cross at 5252rpm.

 

[kenamond]

They way I have always understood it is that horsepower is a function of torque. Torque is the amount of work an engine can produce...

Torque is the work an engine can produce in one revolution at a specific rpm. Wish I had always understood it.

 

[Low Impedance]

I think the problem is the internet itself! While what your talking about makes considerably more sense when it comes to actually engineering a combustion engine or an electric motor, there still is a standard twisting torque occuring, which can be measured. The reason i think that this method, while similiar but misleading, is so commonly used, is that when the average joe with some ideas about physics looks at the rotating assembly, they see the rods connected to the crankshaft and picture the textbook example of a wrench on a bolt. Usually, we tend to apply the easiest principles of science in situtations that look like it might make sense. Kinda like using the quotient principle in dervitives. It works for derivitives, but when applied to integrals, its complete difference, but that tends to be the method that is first used. then you have the whole fundamental theorem of calculus crap...

...anyway back on track. In the sytem, both forms of torque exist. But if you want a better example of why the theorem you describe "works" better (no pun intended) look at a rotary engine. There is no distinct up and down acting like you have on the piston engine. It has a rotating action with the combustion chamber decreaseing and increasing as the rotor face goes through the housing. While there is still is a twisting torque (rotor spins around the electric shaft) you still have the energy being produced over an area of the combustion chamber.

Yeah so this is getting hard to understand in terms of what im saying...mostly because im rethinking what im typing as i type.

Actually if you think about it. Determining the torque produce on the crankshaft while it seems straight forward enough DOESNT satify all the needs of the torque produced. The rotation of the crankshaft itself stores energy (due to its rotating mass) and can itself release energy prodducing torque.

The term flywheel actually relates to a object that stores energy through the use of rotation. That is why a heavier flywheel can actually produce more torque, as it can store more energy in its rotating mass. (and why my rx7 still has the factory flywheel of almost 30 lbs...). Ever notice that you lose alot of rpms and boost between shifts with a light weight flywheel in comparison to a heavier one? This is why.

 

[deletrius]

really though, how can there be any work done per rev? you yourself said that work=distance*force, and in one revolution you have the piston moving down the cylinder and back up to its starting position, in effect the distance from start to end would be zero. IIRC from my Mechanics of Materials and Thermodynamics II classes we had some problems similar to this and never did you deal with work, just torques and forces. Literally, like in your first post, you have a force on the crankshaft that is offset in such a way that it creates a torque on the crankshaft, a twisting force.

 

[Low Impedance]

i dont think its work PER rev as much as it is work OVER a rev. Meaning that instead of thing of it interms of number of revolutions, have far does the spin object travel in the period of one revoltuion.

damn it, looks like im signing up for the internal combustion class next year!

 

[luvmygst]

I think the problem is the internet itself! While what your talking about makes considerably more sense when it comes to actually engineering a combustion engine or an electric motor, there still is a standard twisting torque occuring, which can be measured. The reason i think that this method, while similiar but misleading, is so commonly used, is that when the average joe with some ideas about physics looks at the rotating assembly, they see the rods connected to the crankshaft and picture the textbook example of a wrench on a bolt. Usually, we tend to apply the easiest principles of science in situtations that look like it might make sense. Kinda like using the quotient principle in dervitives. It works for derivitives, but when applied to integrals, its complete difference, but that tends to be the method that is first used. then you have the whole fundamental theorem of calculus crap...

...anyway back on track. In the sytem, both forms of torque exist. But if you want a better example of why the theorem you describe "works" better (no pun intended) look at a rotary engine. There is no distinct up and down acting like you have on the piston engine. It has a rotating action with the combustion chamber decreaseing and increasing as the rotor face goes through the housing. While there is still is a twisting torque (rotor spins around the electric shaft) you still have the energy being produced over an area of the combustion chamber.

Yeah so this is getting hard to understand in terms of what im saying...mostly because im rethinking what im typing as i type.

Actually if you think about it. Determining the torque produce on the crankshaft while it seems straight forward enough DOESNT satify all the needs of the torque produced. The rotation of the crankshaft itself stores energy (due to its rotating mass) and can itself release energy prodducing torque.

The term flywheel actually relates to a object that stores energy through the use of rotation. That is why a heavier flywheel can actually produce more torque, as it can store more energy in its rotating mass. (and why my rx7 still has the factory flywheel of almost 30 lbs...). Ever notice that you lose alot of rpms and boost between shifts with a light weight flywheel in comparison to a heavier one? This is why.

A heavier flywheel cannot produce more torque it can only store more energy as torque. It spins down slower than a light flywheel because of this. But on the flip side when you shift gears you are wasting energy putting it back into the heavy flywheel. The purpose of a light flywheel is to eliminate having to put more work into turning it while accelerating, the extra power goes to the wheels.

This makes my head hurt.

 

[donmagicjuan]

Torque, as it pertains to automobile engines, is exactly what the name implies. If you remember from your physics classes, torque = moment of inertia X angular acceleration. In the example of a car on a dyno, the torque applied to the drum is actually what's determined by measuring the angular acceleration of the drum. From there, the ratio of the angular velocity of the drum to the speed of the engine (in rad/s) is used to figure the torque of the engine for each point on the torque vs. RPM graph. This is necessary because the drivetrain acts as a rotational transformer that steps up/down the torque from the engine based on the overall gear ratio. This is also why you will get (close to) the same numbers on a dyno run regardless of gear selection. Even though the acceleration of the drum will be greater for a lower gear, the aforementioned ratio of velocities will exactly cancel out the effect. Another useful equation is the one that says power = torque X angular velocity. This is why power and torque are related by engine speed, and the simple linear function HP = torque (ft-lb) X RPM / 5252 is applied to the torque value from the dyno to obtain engine horsepower at a given engine speed. Interestingly enough, when you rearrange the power-torque equation, you get torque = power / angular velocity or, more simply put, torque (X 2*pi) = work per revolution. Test this with your example of 100 ft-lbs applied to a bolt for 1/4 revolution to see for yourself!

Hope that helps.

 

[Low Impedance]

A heavier flywheel cannot produce more torque it can only store more energy as torque. It spins down slower than a light flywheel because of this. But on the flip side when you shift gears you are wasting energy putting it back into the heavy flywheel. The purpose of a light flywheel is to eliminate having to put more work into turning it while accelerating, the extra power goes to the wheels.

This makes my head hurt.

i think produce is not the correct word. Was the best i could come up with at the time though. I think that "sustain" torque might be a better choice.

 

[donmagicjuan]

An ideal flywheel (rotating mass) can be modeled as a "rotational capacitor" that supplies torque as an output in response to a change in angular velocity, just as an electrical capacitor will supply current in response to a change in voltage across its terminals. In this sense, a flywheel releases stored energy in a manner that resists a change in angular velocity. Therefore lighter flywheel = less drivetrain loss during acceleration, heavier flywheel = better launches.

 

[Low Impedance]

actually my school is working on a flywheel device for electrical storage. no idea how it works.

 

[Defiant]

Not to be rude, but when are you guys gonna learn that it is all just a marketing scheme, jesus christ. Horsepower is technically a strange way to measure, but it was came up by a guy in the late 1900s to compare the power of horseless buggies to one another. A MARKETING point.

Well, James Watt came up with the definition of horsepower to determine the power of steam engines in 1782, but what's two-hundred years between us. His name is where the electrical term of "watts" comes from as well (746 per hp).

And, he invented the centrifugal governor to regulate the speed of steam engines, using a flyball mechanism, which is where the term "balls to the wall" comes from.

Due to the mathematics involved in calculating horsepower and torque, both curves will always cross at 5250 RPM.

 

[kenamond]

Torque, as it pertains to automobile engines, is exactly what the name implies. If you remember from your physics classes, torque = moment of inertia X angular acceleration. In the example of a car on a dyno, the torque applied to the drum is actually what's determined by measuring the angular acceleration of the drum. From there, the ratio of the angular velocity of the drum to the speed of the engine (in rad/s) is used to figure the torque of the engine for each point on the torque vs. RPM graph. This is necessary because the drivetrain acts as a rotational transformer that steps up/down the torque from the engine based on the overall gear ratio. This is also why you will get (close to) the same numbers on a dyno run regardless of gear selection. Even though the acceleration of the drum will be greater for a lower gear, the aforementioned ratio of velocities will exactly cancel out the effect. Another useful equation is the one that says power = torque X angular velocity. This is why power and torque are related by engine speed, and the simple linear function HP = torque (ft-lb) X RPM / 5252 is applied to the torque value from the dyno to obtain engine horsepower at a given engine speed. Interestingly enough, when you rearrange the power-torque equation, you get torque = power / angular velocity or, more simply put, torque (X 2*pi) = work per revolution. Test this with your example of 100 ft-lbs applied to a bolt for 1/4 revolution to see for yourself!

Hope that helps.

UGGGHHH!!! I just typed in a frikkin novel, and my session timed out and I lost it when I went to post. Good news is that I'll shorten it this time and now actually have my mind around what I'm trying to say.

Thanks Don! Glad you chimed in. You always show me the light eventually.

I didn't understand what you wanted me to test w.r.t. the 100 ft-lbs on a nut example. Please elaborate.

Moving on...

First, it's actually "sum of torques = moment of inertia X angular acceleration" which is analagous to "sum of forces = mass X acceleration".

For a rolling road dyno (I don't know too much about them, so correct me if I misspeak) - the car tire applies a traction force to the drum. The drum accelerates. The dyno may be equipped with a brake for controlling the acceleration, doing steady state tests, etc. They measure the acceleration of the drum and know its mass moment of inertia, so they can calculate the acceleration torque. The force on the brake is measured (which they can translate into a torque using the moment arm of wherever the brake is acting on the drum and moment arm to the load cell used to measure the braking force). And then there's the torque applied to the drum by the car which is trying to accelerate. The sum of the three torques (tire, acceleration of drum, brake) must be zero, and they measured two, so they can calculate the torque due to the tire trying to spin the drum. They could back out a traction force on the surface of the drum using the radius of the drum if they wanted to. Anyway, you know the traction force applied to the surface of the drum or the torque applied to the drum by the tire - same information.

An analogy using linear forces would be a tractor pull type of dyno. The mass of the sled is analagous to the mass moment of inertia of the drum. The friction of the sled on the ground is analagous to the brake. The linear acceleration of the system as the tractor tromps the gas is analagous to the angular acceleration of the drum. The force exerted by the tractor on the sled hitch would be analagous to the torque produced by the car on the rolling road dyno (or would be the same as the traction force applied to the drum).

You can translate the rotational variant of the rolling dyno into forces rather than torques.

However, you're still left with forces, not work. You still haven't done anything about the distance any force was applied or the work done by the engine over the period of one revolution (to keep things in radians). If you're doing a steady state dyno run, the drum isn't accelerating, so the acceleration torque is zero, but that motor is certainly doing work!

In the tractor pull dyno, you measure the velocity of the tractor/sled system at some instant, and you measure all of the other "dyno" parameters. You also measure the rpm of the tractor at that instant. From the velocity and rpm, you can calculate the distance the tractor moved during one revolution of the engine. That distance multiplied by the force on the tractor/sled hitch gives you the work (torque) on the dyno chart at that rpm.

For the rolling road dyno, you measure the angular velocity of the drum and the engine rpm along with all of the other stuff at one instant. You compute the angle the drum rotated in one revolution of the motor. You then compute the arc length traveled by the tire on the drum over that angle of rotation to get the distance. Then you multiply the distance and the computed tire traction force on the drum to get the dyno "torque" number for that rpm. If the tire is not slipping on the drum (big if), you can use the gear ratios and tire diameter to compute the drum rotation angle from the rpm, but I'd rather not assume the tire's not slipping.

So I'm not convinced that the dynamics definition of torque equates to work done during one revolution of the engine.

You've had to bark at me for a few posts in the past before I saw the light (that MBC thread was great!), so I'm hoping you'll lay more smack down on me until we get to the answer.

Thanks again!

 

[donmagicjuan]

I'd love to type up a longwinded response for you right now, Mack, but I have to turn in for the night. I'll be back tomorrow morning to edit this post with more thoughts. I think right now the confusion is just a matter of differing perspectives complicated by some extra intermediate steps in your example computation.

Food for thought: treating torque as work per revolution is the same as treating force as work per unit distance. A bit unorthodox, but it can be useful in some analyses.

 

[MyÜberFastGSX]

Good stuff man.

 

[donmagicjuan]

Alright, I'll try to do a better job of explaining what I was saying earlier. As for the example with the torque applied to the bolt, I was just confirming your statement about torque being equal to work done per revolution through a derivation based on the power-torque relationship. You calculated work done on the bolt to be 157 ft-lbs, which works out to be 628 ft-lbs per revolution. The torque applied during the process is just the value of work per revolution divided by 2*pi. The reason torque and energy have the same units lies behind the fact that angular displacements (radians) are necessarily treated as dimensionless quantities in calculations. You can convince yourself that rotational work = torque X angular displacement, just as translational work was equal to force X distance.

However, you're still left with forces, not work. You still haven't done anything about the distance any force was applied or the work done by the engine over the period of one revolution (to keep things in radians).

Torque IS the work done per revolution of the engine, just as force is the work done per unit distance. The problem I think you're having is that this definition of torque requires a nonzero angular velocity. It breaks down when angular velocity (and thus displacement) is zero, as in the case of a torque applied to a stationary bolt. For these cases the definition becomes undefined, because you're left with a zero term in a denominator. Because the dyno and engine can safely be assumed to be continuously rotating with nonzero velocity, this definition is perfectly accurate.

If you're doing a steady state dyno run, the drum isn't accelerating, so the acceleration torque is zero, but that motor is certainly doing work!

In this case, the net torque applied to the drum is zero, but this is simply because the torque applied by the wheels is equal and opposite to the moment exerted by the force of the brake acting at its radial distance from the axis of rotation. Computing the torque applied by the wheels is then just a matter of equating it to the moment of the brake.

For the rolling road dyno, you measure the angular velocity of the drum and the engine rpm along with all of the other stuff at one instant. You compute the angle the drum rotated in one revolution of the motor. You then compute the arc length traveled by the tire on the drum over that angle of rotation to get the distance. Then you multiply the distance and the computed tire traction force on the drum to get the dyno "torque" number for that rpm. If the tire is not slipping on the drum (big if), you can use the gear ratios and tire diameter to compute the drum rotation angle from the rpm, but I'd rather not assume the tire's not slipping.

This will get you the exact same results that I explained earlier, but you've added some extra intermediate steps for arriving at the torque of the engine. All the stuff about computing arc lengths can be eliminated if you agree that the angular acceleration multiplied by the drum's polar moment of inertia gives you the same answer for torque applied to the drum (differrent from torque applied at the wheels). Torque at the wheels could be computed by utilizing the ratio of drum to tire radii, but this is yet another unnecessary step. Because torque at the wheels depends on what gear you're in, what we really care about is torque applied to the input shaft of the transmission (torque at the flywheel). The most logical way to compute this is to use the ratio of drum to the input shaft gear radii, but this is obviously not practical. A better approach is to use the ratio of the two rotational speeds, which will give you the same number. What I'm suggesting, if it isn't clear, is that the ratio of 2*pi radians of rotation of the engine (one revolution) to the angular displacement of the drum for one of these revolutions, as used in your above example, is exactly the same as the ratio of the engine RPM to the drum RPM.

I think you are on the same page as me when it comes to computing power from the obtained torque value, so I won't discuss that any further, but I hope this clarifies my original post. In this case I don't think you're wrong about any of the concepts-- it just seems like you're overcomplicating things.