Horsepower vs. Torque
Which do you care about - horsepower or torque??? This question is raised quite often by enthusiasts who are confused about claims made by various people. The situation is confounded by those who answer the question claiming to be experts, but who aren't, and who give misleading or outright incorrect information. Here is my attempt at a straightforward answer - enough detail for the techies, but simple enough for the non-engineers to understand.
First, we need a clear definition of the terms. This will help the information presented later make more sense.
Horsepower: 1 hp = 33,000 lb-ft /
One horsepower is the ability to lift 33000 pounds one foot in one minute, or 1 pound 33000 feet in one minute.
Torque: Torque = force x moment arm.
(units are lb-ft for Americans)
One pound foot of torque is produced by a one pound force at the end of a one foot lever. (Think of a wrench one foot long - a one pound force at the end of this wrench will produce one pound foot of torque.)
Now lets consider what we're interested in. Most often for an enthusiast, we want our vehicle to either accelerate faster or go faster or both. How does this happen? The vehicle drive system produces a force at the point where the drive wheel/tire contacts the road, and this force must be greater than the resistance to motion. To the extent that the force exerted by the tire onto the road surface exceeds the resistance to motion of the vehicle, the vehicle will accelerate. For the vehicle to remain at a constant speed, the force exerted by the tire onto the road must equal the total resistance to motion of the vehicle. The resistance to motion is the total of all of the friction, rolling resistance, and air resistance that is exerted on the vehicle, and these values of resistance change as the speed of the vehicle changes. If the force generated at the tire contact point is greater than the total resistance of the vehicle, the vehicle will accelerate. The greater the difference between the tire force and the resistance, the faster the vehicle will accelerate. This is described by the equation F = m * a, where F is the net acceleration force (total force minus the total resistance), m is the mass of the vehicle, and a is the rate of acceleration.
So, if we increase the force generated at the point where the drive wheel/tire contacts the road, we will increase the acceleration rate of the vehicle. (This is a fundamental point, so make sure you understand this before you go on.) The wheel/tire is connected to the drive axle, which is connected to the engine through the transmission and final drive system. The force generated by the drive wheel/tire is at a distance from the center of the drive axle (the tire rolling radius, to be specific.) Thus, the force at the tire contact point with the road is the result of a torque generated by the axle (force x distance = torque). Therefore, increasing the torque on the drive axle will cause the vehicle to accelerate more quickly. (It's really quite simple, isn't it?) Therefore, what I really care about in getting my vehicle to accelerate more quickly is more torque at the drive wheels!
Horsepower is just the torque divided by time. I'll explain more later.
There are several ways to accomplish the objective of creating more torque at the drive wheels.
The first is done every time a vehicle is accelerated from a stop. The transmission is engaged in 1st gear. The torque is multiplied by the reduction ratio of the transmission, increasing the torque at the drive wheels. That is why your little pony accelerates faster in first gear than in second. The consequence of doing this is to limit the speed of the vehicle, as the engine may red-line at somewhere between 20 and 60 mph, depending upon the ratios involved and the allowable speed of the engine. The transmission has no effect on the horsepower output of the vehicle, but it does change the torque available at the drive wheels.
A second way to improve acceleration is to put a higher (numerically) ratio final drive in the vehicle. The torque available at the drive wheels is the engine torque multiplied by the transmission ratio multiplied by the final drive ratio. Let's just use some hypothetical numbers to make the math easy. Let's say for an example that the engine has 200 lb-ft of torque all across the usable speed range (a gross oversimplification); let's have a final drive ratio of 3:1, a rolling radius of 1.25' (15" radius, 30" diameter tire); and let's say that the transmission ratios are 3:1 (1st), 2:1 (2nd), 1.5:1 (3rd), 1:1 (4th), and .75:1 (5th). So, the torque available at drive wheels in first gear is 200 (engine torque) x 3 (first gear ratio) x 3 (final drive ratio) = 1800 lb-ft. (The force at the point where the drive wheels contact the road is a total of 1800 lb-ft / 1.25 ft (rolling radius) = 1440 lb.) If we change the final drive ratio to 4:1, the first gear torque will be 200 x 3 (first gear ratio) x 4 (final drive ratio) = 2400 lb-ft. This is a 33% increase in drive wheel torque, and it will result in dramatically improved acceleration. (It will also increase engine speed on the highway, resulting in lower fuel mileage, increased noise & engine wear, and a lowered top speed of the vehicle.) The final drive ratio has no effect on the horsepower of the vehicle, but it does affect the torque available to the drive wheels.
A third way to create more torque
at the drive wheels is to increase the power output of the engine. Increasing
the power output of the engine may be done in several ways, some of which increase
torque and some of which do not, and some of which improve acceleration and
some do not. This may seem confusing. How can all these statements be true?
It has to do with the shape of the torque curve over the engine speed range.
Let's talk more about horsepower. Let's take our example engine that (simplistically)
has 200 lb-ft of torque all across the speed range. Engine horsepower is calculated
hp = torque(lb-ft) x speed (rpm) x 2 x 3.14 (pi) / 33,000 (or torque x speed / 5252)
So, our example engine (with 200 lb-ft. of torque) at 6000 rpm generates 228 horsepower. This same engine operated at 3000 rpm generates only 114 horsepower, even though it still generates 200 lb-ft of torque! If our engine is allowed to run 7000 rpm, the horsepower is increased to 266!
Let's take a second example engine that has 200 lb-ft. of torque at 3000 rpm, but where the torque gradually falls to 125 lb-ft at 6000 rpm. (This is more typical of the performance of a street engine.) This engine will produce a maximum of 143 horsepower at 6000 rpm, although it will produce 114 horsepower at 3000 rpm like the first engine. This engine also benefits from having its red-line increased, as its maximum horsepower will be increased to about 166 horsepower.
From this discussion we learn that the easiest way to increase the power of most street engines is to run them faster, increasing the horsepower, but not increasing the torque - which in many cases would improve the acceleration of the vehicle.
Now this is more confusing - how could it possibly increase acceleration if it doesn't increase torque? - didn't we say it takes more torque to increase acceleration? Remember that when we start the vehicle from a stop in first gear, the torque available at the drive wheels is the engine torque multiplied by the transmission ratio multiplied by the final drive ratio. Let's just use the example engine/drivetrain from above. The torque available at drive wheels in first gear is 200 (engine torque) x 3 (first gear ratio) x 3 (final drive ratio) = 1800 lb-ft. For our simple case of an engine with a flat torque curve where the torque is 200 lb-ft all across the engine speed range, this value doesn't change as the engine accelerates. When the engine reaches red-line, we shift to second gear. Now, the torque available at the drive wheels is 200 x 2 (second gear ratio) x 3 (final drive ratio) = 1200 lb-ft. For a vehicle with these characteristics, increasing horsepower by increasing the red-line (without increasing torque) would improve the acceleration, because during the time the engine is accelerating from 6000 rpm to 7000 rpm, we are still in first gear with a wheel torque of 1800 lb-ft, which means we are accelerating faster than if we were in second gear. However, with our second example engine, at the red-line the torque is 125 lb-ft, so the rear wheel torque has fallen to 125 x 3 x 3 = 1125 lb-ft. In this case, running the engine faster to increase horsepower would not improve acceleration, since 2nd gear at 200 lb-ft results in 1200 lb-ft of drive wheel torque. In the case of the first engine, the fastest acceleration is achieved by running the engine to the red-line, but in the case of the second engine, the fastest acceleration is achieved by shifting to second slightly prior to red-line, so as to maximize drive wheel torque. So, in order to know if increasing power through increasing the top speed of the engine will help your vehicle accelerate, you need to know the shape of the torque curve of your particular engine. If the torque curve has a pronounced peak in the mid-range, raising the red-line may not help acceleration at all, even though it probably raised the horsepower of the engine.
A fourth way to improve acceleration is by increasing the torque output of the engine. Increasing the horsepower of the engine without increasing the speed at which it runs requires increasing the torque output. Increasing engine torque will always improve the acceleration of the vehicle because that increased torque flows through the driveline to where it increases the torque to the wheels. When a claim is made about increased engine horsepower, always look at the graph to see if the power is raised across the speed range, or if the engine is just being run faster. (Increased engine horsepower from 200 to 230 at a speed of 4000 rpm requires a corresponding increase in torque, increasing the engine horsepower from 200 to 230 by running it faster does not.)
So, which is more important - torque or horsepower? Well, both are important, but torque at the drive wheels is what's really important to the performance of your vehicle.
About the author:
Fred Wucherpfennig has a degree in mechanical engineering. He has worked as a licensed professional engineer for over 20 years. He has numerous patents for various inventions, and has extensive experience with designing, building, modifying, and repairing vehicles of many types.