## How Tires Support a Car

You may have wondered how a car tire with 30 pounds per square inch (psi) of pressure can support a car. This is an interesting question, and it is related to several other issues, such as how much force it takes to push a tire down the road and why tires get hot when you drive (and how this can lead to problems).

The next time you get in your car, take a close look at the tires. You will notice that they are not really round. There is a flat spot on the bottom where the tire meets the road. This flat spot is called the **contact patch**, as illustrated here.

If you were looking up at a car through a glass road, you could measure the size of the contact patch. You could also make a pretty good estimate of the weight of your car, if you measured the area of the contact patches of each tire, added them together and then multiplied the sum by the tire pressure.

Since there is a certain amount of pressure per square inch in the tire, say 30 psi, then you need quite a few square inches of contact patch to carry the weight of the car. If you add more weight or decrease the pressure, then you need even more square inches of contact patch, so the flat spot gets bigger.

You can see that the underinflated/overloaded tire is less round than the properly inflated, properly loaded tire. When the tire is spinning, the contact patch must move around the tire to stay in contact with the road. At the spot where the tire meets the road, the rubber is bent out. It takes force to bend that tire, and the more it has to bend, the more force it takes. The tire is not perfectly elastic, so when it returns to its original shape, it does not return all of the force that it took to bend it. Some of that force is converted to heat in the tire by the friction and work of bending all of the rubber and steel in the tire. Since an underinflated or overloaded tire needs to bend more, it takes more force to push it down the road, so it generates more heat.

Tire manufacturers sometimes publish a **coefficient of rolling friction** (CRF) for their tires. You can use this number to calculate how much force it takes to push a tire down the road. The CRF has nothing to do with how much traction the tire has; it is used to calculate the amount of drag or rolling resistance caused by the tires. The CRF is just like any other coefficient of friction: The force required to overcome the friction is equal to the CRF multiplied by the weight on the tire. This table lists typical CRFs for several different types of wheels.

Tire Type |
Coefficient of Rolling Friction |

Low rolling resistance car tire | 0.006 - 0.01 |

Ordinary car tire | 0.015 |

Truck tire | 0.006 - 0.01 |

Train wheel | 0.001 |

Let's figure out how much force a typical car might use to push its tires down the road. Let's say our car weighs 4,000 pounds (1814.369 kg), and the tires have a CRF of 0.015. The force is equal to 4,000 x 0.015, which equals 60 pounds (27.215 kg). Now let's figure out how much power that is. If you've read the HowStuffWorks article How Force, Torque, Power and Energy Work, you know that power is equal to force times speed. So the amount of power used by the tires depends on how fast the car is going. At 75 mph (120.7 kph), the tires are using 12 horsepower, and at 55 mph (88.513 kph) they use 8.8 horsepower. All of that power is turning into heat. Most of it goes into the tires, but some of it goes into the road (the road actually bends a little when the car drives over it).

From these calculations you can see that the three things that affect how much force it takes to push the tire down the road (and therefore how much heat builds up in the tires) are the weight on the tires, the speed you drive and the CRF (which increases if pressure is decreased).

If you drive on softer surfaces, such as sand, more of the heat goes into the ground, and less goes into the tires, but the CRF goes way up.

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