Daniel W. Koon

St. Lawrence University Physics Department Phys/ENVS 107:

July, 2005

There is a lot you can learn about your car from some readily available data. To start with, if you can measure how long it takes for your car to slow down by itself on a level highway, you can determine the frictional force acting on it.

Let T = the time required for your car to slow down by 10 miles per hour = 10mph = 4.47m/s.

Then the acceleration, a, is given by

Notice that we put the units that we will assume that you will be using for T in brackets after the “T”. This is the convention we will use throughout. Some of the equations on these pages will work only if you follow the convention.

This acceleration is not a particularly useful quantity, so let’s calculate the force that this represents. A moving car comes to rest only if there is a net force acting on it. If there were no frictional forces acting on your car, it could travel in a straight line at a constant speed all day without needing any fuel. You only need to give it gas to either increase your speed, climb a hill, or overcome friction. The fact that you keep giving it gas even when you are on level ground traveling at a constant speed suggests that the bulk of the gas you supply the car goes into fighting friction, whether wind resistance, road friction, or friction inside the engine or car. Now, to calculate the force, we need to know the weight of your car. You can find this information in the car’s registration. If it is registered in the US, it will be given in pounds. The net frictional force acting on your car is given by

For a 3000lb Honda Accord that takes 12s to drop from say 60mph to 50mph, this represents a resistant force of 115lb. If you take a bathroom scale and use it to push that car in a parking lot, that’s about how much force the scale will read as the car starts to budge.

By the way, when I take the car out of gear, it takes about twice as long to slow down from 60mph to 50mph. This tells me that at least half of that 115lb is coming from inside the car, but there’s no way to power my wheels without having the car in gear, so I’m stuck with that.

As you may have guessed, the more friction the car needs to overcome, the worse its gas mileage. Let’s say that my car gets 35 miles to the gallon. I’ll write that as X=35mpg. That means that the amount of work I have to supply (from the gas tank) to travel X miles is equal to the energy content of a gallon of gasoline, approximately 130 megajoules (130MJ). Work is equal to force times distance, so

(The terms on the extreme right-hand side are just unit conversion factors. Notice that because the top and the bottom of each of these two fractions is identical, multiplying by a unit conversion factor is equivalent to multiplying by 1, which is equivalent to folding your arms and doing nothing.)

So this suggests that a 35mpg Honda Accord will be fighting a 514 pound frictional force when it's in motion. Why doesn’t this agree with the 115lb figure above? Because we made one crucial incorrect assumption: we assumed that all of the energy of the gasoline went into fighting the frictional force of the road. This would represent a “tank-to-wheel efficiency”, as it is called, of 100%. More typical values for an internal combustion engine are closer to 20-25%.

So, by taking a simple measurement, and looking up a few others, you can determine how inefficient your automobile is at converting gasoline energy into mechanical energy. By the way, hydrogen-powered fuel cells can be about twice as efficient at converting fuel into mechanical energy as internal-combustion engines, and would thus use half as much gasoline.

There’s another important number that comes up when talking about car performance, namely horsepower. This is a unit of power, the rate at which energy is consumed. We can use the data from above to calculate the minimum amount of horsepower the engine must put out in order to allow you to continue to cruise down the level highway. You’ll want extra horsepower on occasion when climbing hills or passing the car in front of you.

As you may have guessed from the preceding calculation, there are two types of relevant horsepower. There is the horsepower as measured by the rate of consumption of gasoline, and then there is the horsepower that actually gets to the vehicle’s wheels. Let’s start with the horsepower at the wheels:

The rate of energy consumption is equal to force times speed. We will assume that force, F, is given in pounds and speed, V, is given in mph:

The Honda, traveling at 60mph, and experiencing 115lb of force, needs about 18hp at the wheels.

Now, let’s consider the rate at which your vehicle is guzzling gas. You can calculate the number of gallons per hour consumed by the car by chasing the units in the following expression:

But, since each gallon is worth 130MJ and each hour is 60×60s,

That’s 83hp for a 35mpg car traveling at 60mph. But it overestimates the actual output of the engine, because it assumes that the gas’ energy is completely converted to work, which we know is nonsense.

Now, both these numbers are less than the values you are accustomed to seeing for the specs of actual cars. Why? Well, first of all, real car engines can go much faster than 60mph. Shouldn’t usually, but need to occasionally. Second, actual engines need more “pickup” climbing steep highways. The values we just calculated are average values, values for the horsepower that the car generally uses, not the peak value. And who would want to buy a car with only 18 horses under its hood anyway?