OK. Here's the deal, in moderately plain english.
In order to apply these measurements to automobiles and their performance (whether you're speaking of torque, horsepower, newton meters, watts, or any other terms), you need to address the three variables of force, work and time.
Awhile back, a gentleman by the name of Watt (the same gent who did all that neat stuff with steam engines) made some observations, and concluded that the average horse of the time could lift a 550 pound weight one foot in one second, thereby performing work at the rate of 550 foot pounds per second, or 33,000 foot pounds per minute, for an eight hour shift, more or less. He then published those observations, and stated that 33,000 foot pounds per minute of work was equivalent to the power of one horse, or, one horsepower.
Everybody else said OK. :-)
For purposes of this discussion, we need to measure units of force from rotating objects such as crankshafts, so we'll use terms which define a *twisting* force, such as foot pounds of torque. A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum.
Now, it's important to understand that nobody on the planet ever actually measures horsepower from a running engine. What we actually measure (on a dynomometer) is torque, expressed in foot pounds (in the U.S.), and then we *calculate* actual horsepower by converting the twisting force of torque into the work units of horsepower.
Visualize that one pound weight we mentioned, one foot from the fulcrum on its weightless bar. If we rotate that weight for one full revolution against a one pound resistance, we have moved it a total of 6.2832 feet (Pi * a two foot circle), and, incidently, we have done 6.2832 foot pounds of work.
OK. Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement:
Torque * RPM Horsepower = ------------ 5252This is not a debatable item. It's the way it's done. Period.
First of all, from a driver's perspective, torque, to use the vernacular, RULES :-). Any given car, in any given gear, will accelerate at a rate that *exactly* matches its torque curve (allowing for increased air and rolling resistance as speeds climb). Another way of saying this is that a car will accelerate hardest at its torque peak in any given gear, and will not accelerate as hard below that peak, or above it. Torque is the only thing that a driver feels, and horsepower is just sort of an esoteric measurement in that context. 300 foot pounds of torque will accelerate you just as hard at 2000 rpm as it would if you were making that torque at 4000 rpm in the same gear, yet, per the formula, the horsepower would be *double* at 4000 rpm. Therefore, horsepower isn't particularly meaningful from a driver's perspective, and the two numbers only get friendly at 5252 rpm, where horsepower and torque always come out the same.
In contrast to a torque curve (and the matching pushback into your seat), horsepower rises rapidly with rpm, especially when torque values are also climbing. Horsepower will continue to climb, however, until well past the torque peak, and will continue to rise as engine speed climbs, until the torque curve really begins to plummet, faster than engine rpm is rising. However, as I said, horsepower has nothing to do with what a driver *feels*.
You don't believe all this?
Fine. Take your non turbo car (turbo lag muddles the results) to its torque peak in first gear, and punch it. Notice the belt in the back? Now take it to the power peak, and punch it. Notice that the belt in the back is a bit weaker? Fine. Can we go on, now? :-)
Because (to quote a friend), "It is better to make torque at high rpm than at low rpm, because you can take advantage of *gearing*.
For an extreme example of this, I'll leave carland for a moment, and describe a waterwheel I got to watch awhile ago. This was a pretty massive wheel (built a couple of hundred years ago), rotating lazily on a shaft which was connected to the works inside a flour mill. Working some things out from what the people in the mill said, I was able to determine that the wheel typically generated about 2600(!) foot pounds of torque. I had clocked its speed, and determined that it was rotating at about 12 rpm. If we hooked that wheel to, say, the drivewheels of a car, that car would go from zero to twelve rpm in a flash, and the waterwheel would hardly notice :-).
On the other hand, twelve rpm of the drivewheels is around one mph for the average car, and, in order to go faster, we'd need to gear it up. To get to 60 mph would require gearing the wheel up enough so that it would be effectively making a little over 43 foot pounds of torque at the output, which is not only a relatively small amount, it's less than what the average car would need in order to actually get to 60. Applying the conversion formula gives us the facts on this. Twelve times twenty six hundred, over five thousand two hundred fifty two gives us:
Oops. Now we see the rest of the story. While it's clearly true that the water wheel can exert a *bunch* of force, its *power* (ability to do work over time) is severely limited.
A very good example would be to compare the current LT1 Corvette with the last of the L98 Vettes, built in 1991. Figures as follows:
Engine Peak HP @ RPM Peak Torque @ RPM ------ ------------- ----------------- L98 250 @ 4000 340 @ 3200 LT1 300 @ 5000 340 @ 3600The cars are geared identically, and car weights are within a few pounds, so it's a good comparison.
First, each car will push you back in the seat (the fun factor) with the same authority - at least at or near peak torque in each gear. One will tend to *feel* about as fast as the other to the driver, but the LT1 will actually be significantly faster than the L98, even though it won't pull any harder. If we mess about with the formula, we can begin to discover exactly *why* the LT1 is faster. Here's another slice at that formula:
Horsepower * 5252 Torque = ----------------- RPMIf we plug some numbers in, we can see that the L98 is making 328 foot pounds of torque at its power peak (250 hp @ 4000), and we can infer that it cannot be making any more than 263 pound feet of torque at 5000 rpm, or it would be making more than 250 hp at that engine speed, and would be so rated. In actuality, the L98 is probably making no more than around 210 pound feet or so at 5000 rpm, and anybody who owns one would shift it at around 46-4700 rpm, because more torque is available at the drive wheels in the next gear at that point.
On the other hand, the LT1 is fairly happy making 315 pound feet at 5000 rpm, and is happy right up to its mid 5s redline.
So, in a drag race, the cars would launch more or less together. The L98 might have a slight advantage due to its peak torque occuring a little earlier in the rev range, but that is debatable, since the LT1 has a wider, flatter curve (again pretty much by definition, looking at the figures). From somewhere in the mid range and up, however, the LT1 would begin to pull away. Where the L98 has to shift to second (and throw away torque multiplication for speed), the LT1 still has around another 1000 rpm to go in first, and thus begins to widen its lead, more and more as the speeds climb. As long as the revs are high, the LT1, by definition, has an advantage.
Another example would be the LT1 against the ZR-1. Same deal, only in reverse. The ZR-1 actually pulls a little harder than the LT1, although its torque advantage is softened somewhat by its extra weight. The real advantage, however, is that the ZR-1 has another 1500 rpm in hand at the point where the LT1 has to shift.
There are numerous examples of this phenomenon. The Integra GS-R, for instance, is faster than the garden variety Integra, not because it pulls particularly harder (it doesn't), but because it pulls *longer*. It doesn't feel particularly faster, but it is.
A final example of this requires your imagination. Figure that we can tweak an LT1 engine so that it still makes peak torque of 340 foot pounds at 3600 rpm, but, instead of the curve dropping off to 315 pound feet at 5000, we extend the torque curve so much that it doesn't fall off to 315 pound feet until 15000 rpm. OK, so we'd need to have virtually all the moving parts made out of unobtanium :-), and some sort of turbocharging on demand that would make enough high-rpm boost to keep the curve from falling, but hey, bear with me.
If you raced a stock LT1 with this car, they would launch together, but, somewhere around the 60 foot point, the stocker would begin to fade, and would have to grab second gear shortly thereafter. Not long after that, you'd see in your mirror that the stocker has grabbed third, and not too long after that, it would get fourth, but you'd wouldn't be able to see that due to the distance between you as you crossed the line, *still in first gear*, and pulling like crazy.
I've got a computer simulation that models an LT1 Vette in a quarter mile pass, and it predicts a 13.38 second ET, at 104.5 mph. That's pretty close (actually a tiny bit conservative) to what a stock LT1 can do at 100% air density at a high traction drag strip, being powershifted. However, our modified car, while belting the driver in the back no harder than the stocker (at peak torque) does an 11.96, at 135.1 mph, all in first gear, of course. It doesn't pull any harder, but it sure as hell pulls longer :-). It's also making *900* hp, at 15,000 rpm.
Of course, folks who are knowledgeable about drag racing are now openly snickering, because they've read the preceeding paragraph, and it occurs to them that any self respecting car that can get to 135 mph in a quarter mile will just naturally be doing this in less than ten seconds. Of course that's true, but I remind these same folks that any self-respecting engine that propels a Vette into the nines is also making a whole bunch more than 340 foot pounds of torque.
That does bring up another point, though. Essentially, a more "real" Corvette running 135 mph in a quarter mile (maybe a mega big block) might be making 700-800 foot pounds of torque, and thus it would pull a whole bunch harder than my paper tiger would. It would need slicks and other modifications in order to turn that torque into forward motion, but it would also get from here to way over there a bunch quicker.
On the other hand, as long as we're making quarter mile passes with fantasy engines, if we put a 10.35:1 final-drive gear (3.45 is stock) in our fantasy LT1, with slicks and other chassis mods, we'd be in the nines just as easily as the big block would, and thus save face :-). The mechanical advantage of such a nonsensical rear gear would allow our combination to pull just as hard as the big block, plus we'd get to do all that gear banging and such that real racers do, and finish in fourth gear, as God intends. :-)
The only modification to the preceeding paragraph would be the polar moments of inertia (flywheel effect) argument brought about by such a stiff rear gear, and that argument is outside of the scope of this already massive document. Another time, maybe, if you can stand it :-).
Finally, operating at the power peak means you are doing the absolute best you can at any given car speed, measuring torque at the drive wheels. I know I said that acceleration follows the torque curve in any given gear, but if you factor in gearing vs car speed, the power peak is *it*. An example, yet again, of the LT1 Vette will illustrate this. If you take it up to its torque peak (3600 rpm) in a gear, it will generate some level of torque (340 foot pounds times whatever overall gearing) at the drive wheels, which is the best it will do in that gear (meaning, that's where it is pulling hardest in that gear).
However, if you re-gear the car so it is operating at the power peak (5000 rpm) *at the same car speed*, it will deliver more torque to the drive wheels, because you'll need to gear it up by nearly 39% (5000/3600), while engine torque has only dropped by a little over 7% (315/340). You'll net a 29% gain in drive wheel torque at the power peak vs the torque peak, at a given car speed.
Any other rpm (other than the power peak) at a given car speed will net you a lower torque value at the drive wheels. This would be true of any car on the planet, so, theoretical "best" top speed will always occur when a given vehicle is operating at its power peak.
Thanks for your time.
But this is not all on the subject. There are other views as well, continue reading...
Over the past decade, I have encountered any number of articles, on the Internet, that endeavor to explain torque and power. One of those articles, authored by a fellow named Bruce Augenstein, has appeared on dozens of independent Web sites, and seems to have had a strong influence on the popular understanding of this subject. Regardless of his intentions, his article has promoted several fallacious ideas, along with a dubious overall understanding of the subject. Before we look at what he wrote, it will be helpful for us to begin by identifying specific criteria that are useful in assessing the merit of his (or any other) effort to explain this subject:
Before looking at what he wrote, we should also take a quick review of the essential facts. If you turn the crank of a well to lift a 1-lb bucket at a steady rate of 1 ft/min, you are doing work at the rate of 1 ft-lb/min. This is simply the product of the force and the velocity, and turning it around, force is equal to power divided by the (non-zero) velocity. If you substitute that expression for force into the familiar equation that relates force, mass and acceleration, you get:
acceleration = power / (mass x velocity)
If you multiply the product of torque and rotational speed by twice pi, you effectively translate that product to the equivalent product of force and linear velocity, i.e., you calculate the power.
acceleration = engine_torque x 2 x pi x engine_speed / (velocity x mass)
Hence, given the vehicular velocity that is applicable to some point in time, the acceleration that you get, for a given amount of engine torque and a given mass, depends on the engine speed. Of course, if the ratio of engine speed to vehicle speed is given, as it effectively is while the gear ratio is held constant, acceleration will then vary according to the engine torque. Here we go:
“First of all, from a driver's perspective, torque, to use the vernacular, RULES :-). Any given car, in any given gear, will accelerate at a rate that *exactly* matches its torque curve … Torque is the only thing that a driver feels, and horsepower is just sort of an esoteric measurement in that context. 300 foot pounds of torque will accelerate you just as hard at 2000 rpm as it would … at 4000 rpm in the same gear, yet … the horsepower would be *double* at 4000 rpm. Therefore, horsepower isn't particularly meaningful from a driver's perspective, and the two numbers only get friendly at 5252 rpm, where horsepower and torque always come out the same.
In contrast to a torque curve (and the matching pushback into your seat), horsepower rises rapidly with rpm. ... However, as I said, horsepower has nothing to do with what a driver *feels*.
You don't believe all this? Fine. Take your non turbo car (turbo lag muddles the results) to its torque peak in first gear, and punch it. Notice the belt in the back? Now take it to the power peak, and punch it. Notice that the belt in the back is a bit weaker? Fine. Can we go on, now? :-)”
It is true that 300 lb-ft of torque will yield the same acceleration at 2000 rpm as it will at 4000 rpm in the same gear, and the power will of course be double at 4000 rpm. However, the force that the driver feels at any instant is proportional to the driver’s acceleration, which is the same as the vehicular acceleration, and because vehicular acceleration is always proportional to power, it is obvious that the force felt by the driver at any instant is proportional to power. Yet, he asserts that “torque is the only thing that the driver feels”, that “horsepower is just sort of an esoteric measurement”, and that “horsepower has nothing to do with what a driver *feels*”. No honest, unbiased assessment of what he wrote could deny that the bulk and gist of it is simply nonsense. The only place where what he was thinking is uncertain, is where he says that torque and horsepower “get friendly at 5252 rpm”. It’s anyone’s guess what he was thinking when he wrote that, but the number 5252 is merely an artifact of using English units of measure for torque and power.
He argues that power is meaningless since according to him, the acceleration that you get for a given amount of engine torque is the same no matter the engine speed. That is what he encourages the reader to infer from the fact that, in a given gear, the acceleration that you get for a specific amount of engine torque does not vary. That is simply a ruse. The pertinent facts are that wheel torque at a given wheel speed depends equally on engine torque and engine speed, and that the acceleration associated with a given amount of engine torque always depends on the engine speed. These highly pertinent facts can be understood via the fact that power is the same at both locations (ignoring friction), and via the fact that power is essentially equal to the product of torque and rotational speed.
That excerpt came from his section titled, “The Case For Torque”, which took us right to the heart of the problem with his understanding of this subject. Near the beginning of his article, the section titled “Force, Work and Time”, offers this explanation of power:
“If you have a one pound weight bolted to the floor, and try to lift it with one pound of force (or 10, or 50 pounds), you will have applied force and exerted energy, but no work will have been done. If you unbolt the weight, and apply a force sufficient to lift the weight one foot, then one foot pound of work will have been done. If that event takes a minute to accomplish, then you will be doing work at the rate of one foot pound per minute.”
This explanation of power muddles the connection between power and acceleration, because it doesn’t reveal the meaning of instantaneous power, as distinct from that of average power, which has no simple relationship to instantaneous acceleration. Additionally, energy isn’t spent unless work is performed, and while a minimum force is required to overcome the force of gravity and move the weight at all, it makes no sense to talk of the force sufficient to move an object a specific distance. These misunderstandings clearly reveal a lack of basic knowledge, but they are innocent, and do not suggest any dubious agenda. However, smack dab in the middle of his theoretical explanation of power, he spiced things up a bit:
“Now, it's important to understand that nobody on the planet ever actually measures horsepower from a running engine. What we actually measure (on a dynomometer) is torque, expressed in foot pounds (in the U.S.), and then we *calculate* actual horsepower by converting the twisting force of torque into the work units of horsepower.”
Does this mean that power cannot be measured except by first measuring torque? Why else would this be “important to understand”, or worth mentioning? The notion, that power is somehow less real than torque, is easily identifiable as a theme of the article. Yet, it has no meaning or interpretation that can be confirmed experimentally, and as far as the orthodoxy and methodology of empirical science is concerned, notions of that sort are meaningless. This criticism would be no less valid even if it were true that power is only ever deduced by measuring torque and rotational speed. Of course, with inertial dynamometers, you can deduce power from the drum’s angular acceleration and its inertial moment, without measuring torque. For that matter, if you were to apply an engine to the task of lifting an elevator car and you inserted a continuously variable transmission between them to allow you to stabilize the speed of both the engine and the elevator car at any desired engine speed, you would then have a brake dynamometer of sorts. You would deduce power by multiplying the elevator’s steady velocity by its weight (minus the counter-weight), and as with brake dynamometers in general, that measurement will be unaffected by the engine’s inertial resistance to acceleration.
Next came the section I discussed first, and then a section titled, “The Case For Horsepower”:
“OK. If torque is so all-fired important, why do we care about horsepower?
Because (to quote a friend), "It is better to make torque at high rpm than at low rpm, because you can take advantage of *gearing*.
For an extreme example of this, I'll … describe a waterwheel I got to watch awhile ago. This was a pretty massive wheel …, rotating lazily on a shaft which was connected to the works inside a flour mill. … the wheel typically generated about 2600(!) foot pounds of torque. …it was rotating at about 12 rpm. If we hooked that wheel to, say, the drivewheels of a car, that car would go from zero to twelve rpm in a flash, and the waterwheel would hardly notice :-). … twelve rpm of the drivewheels is around one mph for the average car, and, in order to go faster, we'd need to gear it up. To get to 60 mph would require gearing the wheel up enough so that it would be effectively making a little over 43 foot pounds of torque at the output, which is not only a relatively small amount, it's less than what the average car would need in order to actually get to 60. Applying the conversion formula gives us the facts on this. Twelve times twenty six hundred, over five thousand two hundred fifty two gives us: 6 HP.
Oops. Now we see the rest of the story. While it's clearly true that the water wheel can exert a *bunch* of force, its *power* (ability to do work over time) is severely limited. ”
Even though there are no errors per se in this, I still find it annoying. He started by saying that we care about horsepower because making torque at higher rpm means that you can take advantage of gearing. The gist of the anecdote is that even though the torque of the waterwheel itself is substantial, if gearing is applied to increase the output speed, the torque is reduced accordingly. He didn’t say anything about why that happens. He produced a value for the output torque, without any explanation of how it was calculated. He calculated the power, but he did not mention that power, being the same at the output as it is at the input, explains why the output torque must decrease in order to compensate for the increase in output speed. At the end, the point of this anecdote seemed to be to reiterate the fact that torque by itself doesn’t determine the capacity to perform work over time. The formula that you use to calculate power from torque and rotational speed tells you that, and although that is certainly relevant, that fact by itself doesn’t shed much light on the connection between power and acceleration.
Next came the long section titled, “At the Dragstrip”, the essence of which is:
“… some examples of how horsepower makes a major difference in how fast a car can accelerate, in spite of what torque on your backside tells you :-). A very good example would be to compare the current LT1 Corvette with the last of the L98 Vettes, built in 1991. … The cars are geared identically …. First, each car will push you back in the seat … with the same authority - at least at or near peak torque in each gear. One will tend to *feel* about as fast as the other to the driver, but the LT1 will … be significantly faster than the L98, even though it won't pull any harder. …. Where the L98 has to shift to second (and throw away torque multiplication for speed), the LT1 still has around another 1000 rpm to go in first, and thus begins to widen its lead ...
Another example would be the LT1 against the ZR-1. Same deal, only in reverse. The ZR-1 actually pulls a little harder than the LT1... The real advantage, however, is that the ZR-1 has another 1500 rpm in hand at the point where the LT1 has to shift….There are numerous examples of this phenomenon. The Integra GS-R, for instance, is faster than the garden variety Integra, not because it pulls particularly harder (it doesn't), but because it pulls *longer*...”
In this section, he argues that greater power can yet be advantageous because it allows the driver to wait longer before shifting to the next gear. He repeatedly asserts that greater power allows you to pull “longer”, but not “harder”. He consistently applied the constraint that the two cars that he was comparing in order to illustrate what he was trying to say, share identical transmissions and identical overall gear ratios. That constraint obscures the pertinent fact that the car with greater peak power may well exhibit greater peak wheel torque in each individual gear, even if its peak engine torque is less than that of the other vehicle. This section, which accounts for nearly half of the article, mistakenly assumes that the significance of power can be understood and explained by considering only the peak power. It then compounds that mistake by giving a ridiculous, bogus explanation of the advantage of greater peak power.
Toward the end of the article, in the section titled, “At The Bonneville Salt Flats”, he talked about the fact that the power peak is the best engine speed (regardless of the vehicle speed):
“I know I said that acceleration follows the torque curve in any given gear, but if you factor in gearing vs car speed, the power peak is *it*. An example, yet again, of the LT1 Vette will illustrate this. If you take it up to its torque peak (3600 rpm) in a gear … However, if you re-gear the car so it is operating at the power peak (5000 rpm) *at the same car speed*, it will deliver more torque to the drive wheels, because you'll need to gear it up by nearly 39% (5000/3600), while engine torque has only dropped by a little over 7% (315/340). You'll net a 29% gain in drive wheel torque at the power peak vs the torque peak, at a given car speed. ”
Note first that it isn’t generally necessary to re-gear a car in order to select an engine speed at or near the power peak in lieu of the torque peak. Low vehicle speed, where 1st gear is the gear that offers the greatest power, is the exception of course.
That was the closest that he ever got to saying that acceleration is proportional to power. It is in the vicinity of the target, but because it deals specifically with the power peak and does not say plainly that acceleration is proportional to power, it doesn’t hit the bulls-eye. Without a clear understanding of the fact that acceleration is always proportional to power, there is no understanding of why optimal shifting consists of always selecting the gear that yields the greatest power. Note also that even though his calculation correctly implies that wheel torque is proportional to the product of engine torque and engine speed (and thus to power), he never plainly said so, and he never said anything about why it is true.
For his summary, he chose to repeat his perspective on why power is relevant:
“The Only Thing You Really Need to Know
Repeat after me. "It is better to make torque at high rpm than at low rpm, because you can take advantage of *gearing*." :-)”
This is not particularly conducive to an insightful understanding of why power and engine speed matter. Engine torque reveals the amount of work performed over any specific interval of crankshaft rotation, whereas acceleration at any time is proportional to the rate at which work is being performed, which rate depends as much on engine speed as it depends on engine torque.
The facts that are pertinent to a proper understanding of this subject are conspicuously missing from Augenstein’s article, having been replaced by bogus ideas. Instead of saying plainly that acceleration is proportional to power, he defiantly asserted that only engine torque has anything to do with what a driver feels. He made audacious claims about the measurability of power, and his explanations of the significance of power, were bogus. There is very little in his article that qualifies as a usefully correct explanation of anything, and most of what he espouses is bogus.
And one more...
Plato: Dude, nice toga. Say, I’ve just been reading up on torque and power. Torque, it seems, is the rotational equivalent of force in straight-line motion.
Socrates: To fully appreciate what that means and get us off on a solid footing, let’s start with a quick look at the familiar equation: F = M x A. This equation tells us that whenever an object’s present velocity is changing, the acceleration is given by the ratio of force to mass. Manifestly, the greater an object’s mass, the greater the force needed to yield a specified amount of acceleration. Mass also determines how much kinetic energy an object contains when it is moving at a given velocity. In rotational motion, that familiar equation is replaced by a similar equation: Torque = Moment-of-Inertia x Angular Acceleration. The moment of inertia determines how much torque is needed to yield a given amount of angular acceleration, as well as how much kinetic energy an object contains when spinning at a given angular velocity. When an ice skater in a spin brings his or her arms in closer to the torso, there is no loss of kinetic energy, and the observed increase in angular velocity reveals that the moment of inertia has been made smaller.
Plato: To find the torque associated with a straight-line force, you multiply the force by the length of the lever arm, which is always measured along a line of direction that is square to the direction of the force. Whenever I tighten a bolt, if I vary the length of the lever arm, the force that I sense in my hand and arm will change, yet the torque doesn’t change unless the amount of friction in the threads changes. Whenever I think about overcoming friction, I think about work and power.
Socrates: To find the work associated with any steady force, you multiply the force by the distance covered. Whereas work is cumulative over time and distance, power is the measure of how quickly work is being performed, instantaneously in time. Work and energy are truly the same concept, so the measure of how quickly work is being performed, is also the measure of how quickly energy is being spent. If you turn the crank of a well to lift a 1-lb bucket at a steady rate of 1 ft/min, you are doing work at the steady rate of 1 ft-lb/min, which is simply the product of the force and the velocity. If the radius of the spool is 1 ft, then the torque applied to the crank by the 1-lb bucket, will be 1 lb-ft. In each complete rotation of the crank, the bucket will move a distance equal to the radius of the spool multiplied by twice pi. It follows that the work associated with a specific amount of torque, for one complete rotation, may be found by multiplying the torque by twice pi. The calculation of power from torque and rotational speed is similar. The following chart summarizes:
force x distance
torque x number of rotations x 2 x pi
force x velocity
torque x angular velocity x 2 x pi
The expression for power in rotational motion reveals that you always multiply by the same constant value (2 x pi) to calculate power from the product of torque and angular velocity. Note, though, that this assumes that angular velocity is measured in complete rotations per unit of time. You could just as easily measure angular velocity using a smaller angular distance, such that you would have to multiply that smaller angular distance by twice pi in order to yield one complete rotation. If you measured angular velocity using that smaller angular distance (which is known as a “radian”), the expression for power would be simply the product of torque and angular velocity, i.e., you would not multiply by twice pi to calculate power. Hence, the business of multiplying by twice pi is equivalent in effect to converting from one unit of measure to another, and it is correct to say that power is simply equal to the product of torque and angular velocity.
Plato: If the bucket is raised at steady velocity, its kinetic energy will be steady. Only its potential energy will be changing, and the power will be simply the static weight of the bucket multiplied by the steady velocity. It is easy enough to measure instantaneous velocity when the velocity is steady, but in real-world scenarios, doesn’t it get more complicated?
Socrates: A common approach to measuring the power of an engine is to use a regulated brake to hold the engine steady at the desired speed. You have to measure the force that resists the pull of the engine on the brake, so that you can deduce the engine torque from that force and from the lever arm, which you also have to figure out. And, of course, you have to measure the engine speed. Dynamometers of this sort are known as “brake dynamometers”. Conceptually, you could implement a brake dynamometer of sorts by applying the engine to the task of lifting an elevator car, using a continuously variable transmission in the coupling. The CVT would allow you to stabilize the speed of both the engine and the elevator car at any desired engine speed. To deduce power, you would multiply the elevator’s steady velocity by its weight, and as with brake dynamometers in general, those measurements would be unaffected by the engine’s inertial moment. The other common approach is to hitch the engine to a massive drum that spins freely. As long as the increase in the kinetic energy of the drum is the only energy sink, the power will be given by the instantaneous rate of increase of the drum’s kinetic energy, which can be deduced from the drum’s moment of inertia and its instantaneous angular acceleration. Dynamometers of this type are known as “inertial dynamometers”. The angular acceleration can be measured with the help of an accelerometer, or deduced from closely spaced measurements of time and angular distance. The drum’s moment of inertia can be measured separately, or calculated from its dimensions and the density of its substance. The increase in the kinetic energy of the engine itself is an energy sink. The measurements are influenced to a degree by the engine’s moment of inertia, and they reveal, to a degree, the ability of the engine to quickly increase its work output. As such, measurements taken on an inertial dynamometer give a more realistic picture of an engine’s actual performance on the road. For purposes of ordinary performance tuning on a test bench, that sort of accuracy isn’t particularly beneficial, whereas the ability to keep the engine running steadily for extended periods can be beneficial.
Plato: I read somewhere that to measure power, you measure torque and then you deduce power from torque. That supposedly demonstrates that power is just an abstraction of torque.
Socrates: Clearly, there are various ways to measure power independently of torque. Moreover, the notion, that power is less real than torque, has no meaning or interpretation that is capable of being confirmed experimentally. As far as the orthodoxy and methodology of empirical science is concerned, notions of that sort are meaningless.
Plato: I should be able to measure the power of my mare, by measuring how quickly she is able to lift a large bucket of water from my well. If I adjust the amount of water such that the velocity is steady, the actual force will be equal to the weight of the bucket. That way, I won’t have to measure the actual strain in the rope, and of course, it will be easier to measure the velocity.
Socrates: In the future, a fellow by the name of James Watt will determine that his horse is able to perform work at an ongoing, instantaneous rate of 33,000 foot-pounds of work per minute. If you measure torque in lb-ft and rotational speed in rpm, and you want to express the power in hp, you can use the conversion factor: 1 hp = 33,000 ft-lb/min. The value of twice pi is 6.283, and that divided by 33,000 is about 1/5252. So, as long as torque is measured in lb-ft, rotational speed is measured in rpm, and you want to express the power in hp, you can take a short cut and divide the product of torque and rotational speed by 5252.
Plato: Does that mean that torque and power are equivalent at 5252 rpm?
Socrates: Nope. Torque and power are distinct properties, with each being analytically related to acceleration in its own special way. The value 5252 is merely an artifact of the English system of measure, and that value is not the least bit special if another system of measure is used. In most of the world, torque is expressed in Newton-meters (N-m), and power is expressed in Watts or kilowatts (kW), which we use for electrical power. The engine speed where torque in lb-ft and power in hp coincidentally take on the same numerical value, happens to fall within the operating range of most engines, so on dynamometer plots, it is convenient to use a single number scale for both torque in lb-ft and power in hp. When that is done, the two curves will cross at 5252 rpm.
Plato: But, it seems that torque should determine acceleration, so why does power matter?
Socrates: Power matters because at any point in time, acceleration is proportional to the rate at which the engine is performing work. Engine torque tells you how much work is performed over any specific interval of crankshaft rotation, but does not tell you how quickly the work is being performed. It is of course possible to deduce acceleration from the engine torque using other information such as the overall gear ratio and wheel diameter, but that doesn’t change the pertinent and useful fact that at any point in time, acceleration is proportional to power. Recall that power is equal to the product of force and velocity. If you turn that around, it says that force is equal to power divided by the (non-zero) velocity. If you substitute that expression for force into the familiar equation that relates force, mass, and acceleration, you get this:
acceleration = power / (mass x velocity) =>
acceleration = engine_torque x 2 x pi x engine_speed / (velocity x mass)
Hence, given the vehicular velocity that is applicable to some point in time, the acceleration that you get, for a given amount of engine torque and a given mass, depends on the engine speed. Of course, if the ratio of engine speed to vehicle speed is given, as it effectively is while the gear ratio is held constant, acceleration will then vary according to the engine torque. (Note that if you plug a set of values into that equation to calculate acceleration, in order to get proper units of measure for acceleration, you need to use lbf instead of lb for the force component of the torque. 1 lbf is the force of gravity on 1 lb of mass: 1 lbf = 1 lb x 32.2 ft/s^2 = 32.2 ft-lb/s^2 = 4.45 N.)
Plato: But, if acceleration is proportional to power, why does acceleration track with the engine torque curve as the engine speed and the vehicle speed increase in a given gear?
Socrates: The perception of a contradiction, between the fact that wheel torque tracks with the engine torque while the gear ratio remains fixed, and the fact that acceleration is proportional to power, is a false perception. The equations reveal that the proportionality between acceleration and power is different at different vehicle speeds. The acceleration that you get for a given amount of power decreases as the vehicle speed increases, yet, at any point in time, acceleration is proportional to power, and depends as much on engine speed as on engine torque.
Plato: What does this mean from the perspective of gear selection strategy?
Socrates: Whenever you change gears, as long as you are quick to avoid any significant loss in vehicle speed during the up-shift, the proportionality between power and acceleration will be steady across the up-shift. Hence, in order for acceleration to be steady across the up-shift, power must be steady across the up-shift, which means that the engine torque must increase to compensate for the drop in engine speed. If the throttle is held open so that actual power follows the engine’s power curve, the engine speed must transition between two equal-power points on opposite sides of the power peak. Note that shifting such that power will be steady across the up-shift, and shifting such that you are always using the gear that yields the greatest power, are two different ways to describe the same optimal strategy.
Plato: What would happen if the engine torque were to be held steady across the up-shift, i.e., you kept the engine speed within the flat region of the engine torque curve?
Socrates: The acceleration would drop abruptly at the up-shift, matching the drop in engine speed. Let’s look at it another way, and let’s take a quick side trip that may help to put the significance of power into better perspective. In an electrical transformer, any increase in voltage between the primary and the secondary windings, must be accompanied by a compensating decrease in current. Power is equal to the product of voltage and current, and as the saying goes, “power in is power out”. That saying applies as well to the physics of mechanical motion. Except for the energy losses due to friction, the product of torque and rotational speed will be the same at the wheel as it is at the engine, and as it is anywhere else that you measure it along the drive train. You want the wheel torque to be steady across the up-shift, and since the wheel speed will also be steady at the up-shift, the product of torque and rotational speed will be steady at the up-shift, not only at the wheel, but at the engine as well. That, of course, means that the engine torque must increase to compensate for the drop in engine speed.
Plato: Okay, but given two vehicles that are identical except for the engines, the one with the greater peak engine torque will still exhibit greater peak acceleration in each gear, right?
Socrates: If the vehicle with greater peak power is allowed to use a different final drive ratio, then by shifting its engine torque peak to lower vehicle speed, the corresponding wheel torque will increase. Thus, the vehicle with greater peak power may exhibit greater peak acceleration in each gear, even if its peak engine torque is less than that of the other vehicle.
Plato: Well, there are still certain benefits to emphasizing torque in lieu of power, aren’t there?
Socrates: Certain effects, such as improved acceleration from a full stop and less frequent shifting, are the result of a comparatively flat, uniform spread of engine output, starting at comparatively low engine speed. It makes perfect sense to attribute such effects to a de-emphasis on peak power. However, logically speaking, torque and power are not opposites, and it does not follow from the fact that you have de-emphasized peak power, that you have emphasized torque. Of course, if there exists some other justification for the practice of equating the engine’s low-speed performance to torque, that will also constitute justification for equating a de-emphasis on peak power to an emphasis on torque, never mind that torque and power are not opposites. At the wheel, the affinity between low rotational speed and torque is quite genuine, owing to the fact that the transmission is used to exchange rotational speed for torque. But this effect does not apply to the engine. The practice, of equating engine performance at low and moderate engine speed exclusively to torque, seems to derive essentially from the fact that the peak engine torque occurs at a lower engine speed than does the peak power. This seems a weak justification when you consider that the peak engine torque reveals the engine performance accurately at only a single engine speed. That engine speed is often above the midpoint of the engine’s operating range, and no matter how low the actual engine speed, the actual performance depends partly on the engine speed, and is fully revealed by the actual power.
Plato: What else?
Socrates: Many people seem to believe that the full explanation, for why longer stroke generally means improved low-end performance, is simply that by increasing the effective lever arm (the crank throw distance is one-half of the stroke distance), you increase the torque. For whatever reason, they don’t realize that if it were that simple, the improvement in engine torque would be uniform over the operating range, which would not explain why the performance improvement is specific to low engine speed. They have somehow gotten the idea that any change, that directly improves engine torque, will automatically favor lower engine speed. Clearly, it isn’t that simple. If you increase the stroke while keeping the volume displacement constant, the piston surface area will decrease, which will nullify the effect of the increased lever arm, since the force depends on the surface area of the piston face. Engine torque corresponds to the amount of energy spent over any specific interval of crankshaft rotation, and that amount of energy depends on the amount of oxygen used. It follows that the variation in engine torque with engine speed reveals the variation in the amount of air captured per individual intake stroke. Cylinder shape interacts with the duration of the intake stroke to influence the amount of air that is captured on the intake stroke. When the cylinder is made long and skinny, the effect is to increase the amount of air captured at low engine speed, and to decrease the amount of air captured at high engine speed. Note also that if the relationship between stroke and torque were as direct as the naïve explanation suggests, you could get free energy just by making the cylinder long and skinny.
Plato: I need to go get measured for a new toga, but before I run along, I’d like to know what you think about the various claims that engine torque is the true indicator of engine performance.
Socrates: Those sorts of claims have to be interpreted to mean that you are always supposed to get the same acceleration for a given amount of engine torque, no matter the engine speed at which that much engine torque is delivered. There simply is no other meaningful, tangible interpretation of those claims. Yet, as we have already seen, wheel torque depends just as much on engine speed as it does on engine torque. Anyone who is not convinced of that, need only discover for themselves that at any of the various vehicle speeds where the transmission will permit you to choose between two equal-torque points on opposite sides of the torque peak, the acceleration will be dramatically greater in the lower of the two gears. It is logically dubious to infer, from the fact that the peak power does a poor job of revealing the engine’s performance at low and moderate engine speeds, that torque is the true indicator of engine performance.
OK Bruce, if you are still out there. It's your turn.