Engines are measured for both power and torque. I once read in a magazine that horsepower sells a car, but torque moves it. This is because many people don't understand the difference between torque and horsepower, and how they affect a car's performance.

          First, let's look at a car accelerating constantly. Using the equation     d = 1/2at², we can see how far a car accelerating at .5-G's (4.9-m/s²) goes.

                                                d = 1/2at²

                                                d = 1/2(4.9)(1²)

                                                d = 1/2(4.9)

                                                d = 2.45-m

          In the first second, the car will only get 2.45-m, or about 8 feet, which isn't too far. Let's see after two seconds:

                                                d = 1/2(4.9)(2²)

                                                d = 1/2(4.9)(4)

                                                d = 9.8-m

          This is quite a bit farther: 9.8-m or 32-feet. Because the time is squared in the equation, the car travels quite a bit farther with each second. This shows why a good start is necessary to win a race. If you spin the tires, you'll lose valuable time, and your opponent will gain increasingly more on you.

          Up until now, we have been using values we just pulled out of the sky. Can your car really accelerate at 1-G, or even .5-G's? Now we'll see how the engine's torque and horsepower determine this.

          Our car's engine can produce 448-Nm, or 330-ft-lb, of torque at the crankshaft. If it had a first gear ratio of 3.06, and a final drive ratio of 3.07, we would multiply the torque by the ratios to find that in first gear we have about 4210-Nm or 3100-ft-lb of torque! To find the actual force it can send to the wheel, we need to divide by the wheel radius, .33-m or 1.08-ft. Thus, it can send 12760-N or 2870-lb of force to the rear wheels.

          At rest, our car has an even weight distribution. This means 7130-N of weight is on the rear tires. If it is fitted with 1-G tires, we know from the previous section that anything past 7130-N of force will spin the tires. If we stomp the gas and send all 12760-N to the tires, they will spin excessively. Thus, at the start, we can send up to 7130-N. This will give an acceleration of .5-G's.Using what you learned in the section on weight transfer, you can find the amount of weight on the rear tires right after the car gets going at .5-G's.

                                       L_r = (1-.5)(14260) + (7130)(.51)/(2.54)

                                       L_r = 7130 + 1430

                                       L_r = 8560-N

          After the car gets going, you can give a little more gas and increase the torque to 8560-N on your 1-G tires. This gives a new acceleration of:

                                       G = 8560 / 14260

                                       .60-G's

          The funny thing is that once you're accelerating at .60-G's, more weight will transfer rear, allowing more throttle. Eventually, enough weight will shift back so you can use the full capability of your tires for acceleration.

          Now that we know how torque accelerates our car, where does horsepower fit in? Horsepower is a unit of power, work over time. This power is necessary to keep the net force on the car in the positive direction. While driving, several forces act against your motion, including air resistance, and friction. First we'll examine air resistance.

          Air resistance, or drag, is the most important force that opposes the car's motion, and uses up horsepower. To find air resistance at a certain velocity, you can use the following equation:

                                                F = 1/2 C_dApv²

          Where C_d is the coefficient of drag (.30 in our car), A is the frontal area of the car (1.84-sq meters), p is the density of air, and v is the car's velocity. For our calculations, we will use the density of air as 1.29-g/l.

          Now we can calculate the force of air resistance on our car at various speeds. I will work out the first calculation, 6.7-m/s or 15-mph.

                                                F = 1/2(.30)(1.84)(1.29)(6.7²)

                                                F = 16-N

Using this method, we can send the force at various speeds:

      As you can see, every 30-mph more increases the air resistance by about 4 times. This is what the horsepower is used to overcome, because at 150-mph, you'll need to use 1612.5-N of force just to overcome air resistance.

     To find just how much horsepower is expended in overcoming this resistance can be found by multiplying the air resistance by speed, and then dividing by 745 to get horsepower. With the English system, you need to convert to ft/s, then multiply through and divide by 550 to get HP. Using this conversion we get the following values:

      Again, the horsepower usage increases incredibly over short speed intervals. Going from 90-mph to 120-mph requires more than twice as much horsepower. Finding the rolling resistance of friction is much more difficult and we cannot go over it in this web page. However, because a late-model corvette has a top speed of 150-mph, you have to assume that at 150-mph about 95-HP is used to overcome the friction. This uses up all of our 240-HP.