All about Horsepower and Torque --
-- and power and leverage and gear ratios and...
By Greg Heumann
|For M5 Drivers (and others)|
I've seen this topic written up many times and none of the explanations satisfied me. I've also seen many more casual attempts on a variety of internet boards - and some of them are way wrong. So I'm not going to just tell you how it is. I am going to help you understand where these terms come from, and how they are related, so you will know. To some extent, I wrote this article for me, because I knew once it was done I would understand this stuff better than I did.
It is long, but it builds (hopefully) on very simple concepts. At no point does it require any more than middle school math. You may "get it" even better if you follow me through some of the simple computations. Or, you can just assume that they're all correct. Hell, they might be!
I designed this to be read end-to-end. It starts with general concepts and concludes with very specific application of these concepts, and covers not only HP and ft-lbs, but force, acceleration, gear ratio choices and more. If you make it through, you should truly understand the relationship between horsepower and torque.
First of all, let me define some terms - torque, work and power:
Torque is a measure of rotational force. It is typically expressed in ft-lb, Dyne-cm, kg-meter or newton-meter. If you put a one foot long wrench on a lug nut and press down on the free end with 50 pounds of force, you are exerting exactly 50 ft-lbs of torque on the nut, whether it is moving or not.
Work is what happens when you exert a force through a distance. You can lean on that lug wrench with 50 pounds of force until you are drenched in sweat, but if the lug nut doesn't turn, by definition you have not done any work. If, on the other hand, you succeed in rotating the nut, you have.
|Cute little illustrations should go here some day.|
Power is the ability to do work over time. Let's say it takes 100 ft-lbs of torque to turn the nut, and that it takes a constant 100 ft-lbs to keep it turning. (Maybe we're rolling the car up a slope.) If you turn the nut through one revolution in 10 seconds, you have done the work in a certain amount of time, so we can express your strength, or the ability to do work over time in terms of Power. Power can be expressed in Watts, Horsepower, calories per hour, ergs per second or many other measures.
Let's say your strength is equal to 1 horsepower. If I can turn the nut the same 360 degrees in 5 seconds, I would have 2 horsepower, because I would have done the same work in half the time. (Hey, I'm writing this - I get to be the strong guy!).
To review, we exerted a force on the wrench, which resulted in a torque on the nut. Because we both turned the nut the same amount, we did the same amount of work. But I have more power than you, because I did the same work in less time.
Now, we're getting somewhere!
Horsepower is related to torque by RPM. The formula is: Horsepower = (RPM * Torque)/5252. It is RPM (revolutions per minute) that adds the dimensions of work and time to torque in order to reach a horsepower figure. A side-effect of this formula is that when RPM=5252, the horsepower number = the torque number. But concluding, as some do, that this means "horsepower = torque" is foolish - it is like comparing apples and oranges. One is an expression of power and one is an expression of something far simpler - rotational force.
Hopefully you now understand the proper, legal, scientific definitions of torque, work, and power. That's good, because we need them as a common ground to move forward.
Now, we need to look at two more vital concepts before we can truly understand how torque and horsepower affect our car's performance:
Acceleration and Leverage.
|Did you know that 1 Horsepower = 746 Watts?|
|Acceleration is the rate of change of speed. Acceleration is expressed in terms like "ft/sec2", which is also read as "feet per second per second". An object accelerating at 1 ft/sec2 is going ft/sec after 1 second, and, because it is accelerating, is going 2 ft/sec after 2 seconds, etc. Thus "feet per second per second."|
|Around cars, acceleration is also frequently expressed in terms of "G's". G stands for "Gravity" and 1 G is the rate at which an object in the earths gravitational field changes speed when dropped. That rate just happens to be 32.17 ft/sec2.||Do the math: with a constant acceleration of 1 G, we should be travelling at 128.68 feet per second at the end of 4 seconds. That's about 88mph. 0-88mph in 4 seconds flat! That would be pretty good, huh? But not as fast a top fuel dragster, whose acceleration is around 5G's!!!|
|Acceleration is the result of a force acting on a mass. One of the Physics' many famous laws is F=MA: Force = Mass * Acceleration. Another way of looking at this relationship is that A = F/M, which in turn makes it easy to see that increasing F (torque) or decreasing M (weight) both result in better acceleration.
Our cars are ultimately accelerated by the tires exerting a force on the road. That is a linear force, parallel to the road surface. But we get that force by turning the axle connected to the wheels that hold the tires, so let's carry on from here in terms of torque, which is nothing more than rotational force.
We know what our car weighs, and we know the torque our engine is supposed to make, so we can figure out how fast the car should accelerate, right? Wrong!!
The torque at the engine is not the same as the force at the wheels! Why? Because we have transmissions, differentials, and tires which "multiply" the engine's torque through the concept of leverage.
|Mass and weight aren't in fact the same, but as long as we're on earth, and for the purposes of this discussion, we can reasonably think of them as equivalent.|
Remember the wrench in the example above? 50 pounds of force on the end of a 1 foot long wrench gave us 50 ft-lbs of torque on the nut. If we extend the wrench to 2 feet, your wimpy 50 pounds of force now results in 100 ft-lbs of torque on the nut. The wrench (lever) multiplies the force in direct relation to its length resulting in greater torque. Note that the wrench end now has to travel twice as far to complete a revolution, however. The torque was twice as great, but the distance through which we had to move the wrench with our 50 pound force was also twice as great. There is no free lunch.
This kind of leverage is familiar to most of us. It can be achieved with levers, but also with hydraulic systems (think about the ratio of the master cylinder diameter to the brake cylinder diameters in your car) or gears. Gears are merely levers that multiply rotational force, or torque.
Why do we need leverage anyway? Here is an example that may be helpful. I've exaggerated it intentionally to help demonstrate the underlying points, but the principles are all sound. Suppose we connected the M5's engine through a clutch to the drive shaft (no transmission) which in turn powered the rear wheels through a 1:1 gear ratio right angle drive unit. The center of our wheel is almost exactly 1 foot from the ground, so a torque at the rear axle of 282 ft-lbs in fact exerts a linear force of 282 pounds at the tire patch. (282 ft-lbs is the M5's torque at 2000 RPM, a number I picked out of a hat assuming we want to start somewhere and accelerate to red line.)
We know that A=F/M, and we know M, and we just figured out F! So we can calculate acceleration of this particular gear ratio: 282 pounds force / 4000 pounds weight = 0.0705G. Less than 1/10 of a G? That sucks!!!! We won't go anywhere!
And it is really worse than that. Our tires are 6.7 feet in circumference, so they want to roll 6.7 ft*2000 revs/min = 13,400 feet in one minute at this RPM - a speed of, (get this) 152 M.P.H.!!! At 2000 RPM! At our red line of 6500 RPM at 1:1, we would have a theoretical speed of 495 m.p.h. I don't know how much torque we'd need at 6500 RPM to keep us going 495mph, but I know it is a boatload more than we have.
No, no - it is even worse than that! To launch the car from 0 to the a point where we can let the clutch hook-up -- say, at 1000 RPM, we're going to have to slip the clutch from 0 to 76 m.p.h. And at something less than 0.0705G, that's going to take a long, long time. We just don't have enough clutch for this. So - 1:1 ain't going to work. We need more leverage Pure and simple.
Clearly, to get more torque
|And here is where it gets really interesting. If we are in control of the leverage ratios (a function of tranny, rear end and tire diameter), then we can choose any amount of rear wheel force, and hence acceleration we want, (and here is the big catch of the day) for that gear.
So let's go to the other end of crazy for a moment. Let's say we want 5G of peak acceleration in 1st gear from our M5, so we can play top fuel dragster. Really - if we can get enough traction, we can make our M5 accelerate at 5G!!! We simply need to know how much force that would take. Piece of cake. F=MA - so 4000 pounds * 5G = 20,000 pounds of force at the tire patch. How do we get 20,000 pounds of force from our peak engine torque of 367 ft/lbs? Leverage! We simply need a gear ratio that multiplies engine torque by 54.5 - voila 20,000 pounds of force - 5G! Wahooo!
|Of course, we're smart now. We know there is no free lunch. Just what is the effect of this "leverage trade off"?
Well, the engine must turn 54.5x faster than the wheels with this gear ratio. At 6500PM red line, then, the wheels will be turning 298.165 times per minute. At 6.7 feet per revolution of the tires, our maximum speed in 1st gear is about 2000 ft/min, or about 22.75 m.p.h. But 5G acceleration says we should be going 160.85 feet per second after 1 second - that's about 110mph!! I don't think we're going to get to use 1st very long.
Even at, say, 2G, or 1.8, we know we're going to hit red line in a real hurry. The sooner we "run out" of 1st gear, the sooner we have to shift to 2nd, a new ratio that gets the engine back into an acceptable RPM range while we continue to accelerate. Each successive gear has to be a higher (numerically lower) gear ratio, or we'll just over-rev the engine when we shift into it, since we're already moving at the red line speed of the previous gear. And that, in turn, means that each successive gear delivers less torque multiplication, and hence less force with which to accelerate the car. That's not all.
The faster we let the engine accelerate relative to the car, the more losses we have from accelerating the engine's own rotating masses. And we only have room for 6 gears in the transmission, and shifting takes time, and fuel economy is a factor, and noise, and cost, and, and, and, etc. Clearly, the choice of gear ratios is a study in trade-offs. There is an optimum set of gear ratios for a single purpose, but it may not be optimal for another.
|The effective multiplier of the stock M5 is (4.227 * 3.15)/1.066 =12.493 - this is the 1st gear ratio * the rear end ratio divided by the rolling radius of the rear tire (which determines the "moment arm", like the length of the wrench above, of the tire - and determines how axle torque becomes force at the tire patch.). This gives us 37mph at red line, and, at peak engine torque of 367 ft/lbs, then, we get 4585 pounds of force at the rear tires. Once again, using A=F/M, we should have a peak acceleration of 1.146 G. We don't actually get this. There are losses all over, including spending some of the torque to accelerate the engine's mass (particularly the flywheel), the rotating mass of all 4 wheels (takes more energy to accelerate a rotating mass than a linear one) and "drive line loss" which is losses to friction, heat, noise and vibration in the clutch, transmission, drive shaft, differential, universal or CV joints (if any) and wheel bearings. Oh yeah, and tire friction and slip.|
|OK. Now you're lost and confused again. You thought you'd be able to take all this new found knowledge and make some sense out of it, but you still have no idea about what you really want to know - at what RPM should you shift?|
|Force at the rear wheels, not horsepower, determines how fast our car accelerates at any moment. Because of the multiplier effect, that force is almost always greater in the lower gear all the way to red line. There are some cars (the M5 is not one of them) whose engines have a characteristic of making far less torque at the upper end of the rev range. But unless that torque falls off so much that the post-shift engine torque * multiplier of next highest gear results in more force at the rear wheels, you're still better to stay in the lower gear.
In the M5, engine torque begins to fall off above 5500 RPM. However the force at the rear wheels in 1st at red line is still greater than it will be at any RPM in 2nd gear -- and the same is true all the way up through 5th. The following Excel Sheet has the numbers for the M5 and compares the predicted performance of the stock 3.15 differential to the Dinan 3.45 rear end. If you know Excel you can easily adapt this chart to your own numbers and needs. You can download it here.
I hope you've found this helpful. I hope it made you think. It sure as hell made me think.
Horsepower IS the factor that determines whether you win, because it is an expression of your cars ability to do work over time. However force at the rear wheels at any moment is a function of torque and gear ratios, making it much easier to think about and calculate with. Ultimately they are related, but not the same. One isn't better than the other. They're just different!
What truly matters from a performance perspective is how well the torque and power characteristics of the engine are leveraged through the careful selection of transmission and rear end gear ratios. That selection must by definition be a compromise between many opposing requirements - so the stock gear ratios in your car may or may not be the absolute best for your purposes. But they're pretty damn good for mine.
|Sometimes you just need to hear something many different ways to help it sink in.There are other articles on the web that cover the subject. Here are a few I think are well written:||http://www.vettenet.org/torquehp.html|
|I also recommend this book, which has all the formulas you'll ever need to do basic performance analysis. The math is all simple - no College Calculus required.||Auto Math Handbook -- by John Lawlor (HPBooks -1020)|