Often overlooked by most drivers, the tires are the only connection between your massive car and the road. In fact, the contact patch is extremely small when compared with the size of the car. Those four small squares transmit your input into the car's action.

    A tires grip can be measured by finding the force necessary to push it sideways. If a 225-N tire takes 330-N to push sideways, you can find it's adhesive limit. Simply divide the force by the weight on the tire. In this case it comes out to 1.47. We can express this as G's due to the idea that 1G is the weight on the tire. Thus, it takes 1.47-G's of acceleration to make that tire slip. However, when actually mounted on a car, this value is reduced because of inflation, temperature, and speed. All these factors greatly reduce the adhesive limit of a tire. Most sports cars have an adhesive limit in the high .80's. High-end sports cars can have adhesive limits in the high .90's and even over 1.00-G. Generally, better rated tires have a higher adhesive limit. It is also important to note that this is the rating in perfect conditions. If the road is wet, the value is reduced by up to 30%. Snow reduces the adhesive limit to between .15 and .2 for most tires. This is why it is so difficult to drive on snow.

          So, how does this relate to real driving? When traveling in a straight line, it is fairly easy. If your tires have an adhesive limit of .90, then any acceleration over .90-G, or braking over -.90-G will cause the tires to slip. When accelerating, this means spinning the tires, and wasting valuable time. When braking, this can mean skidding and possibly locking the wheels, causing a very difficult to control car. However, when you are not traveling in a straight line, it becomes more complicated. You have to find the sideways acceleration on the tire. To do this, you will need to know the radius of the turn, and the velocity of the car. You can then use the following formula:

                                                a = v²/r

Where a is the acceleration, v is the velocity, and r is the turn radius. For example, in a 15-m radius turn, at 6-m/s, you will experience the following lateral acceleration:

                                                a = 6²/15

                                                a = 36/15

                                                a = 2.4-m/s

The lateral acceleration on the tire will be 2.4-m/s or about .25-G's (2.4/9.8). The following chart shows some common values. The two value under each radius is the velocity in m/s and mph:

    Thus, if you were in a 30-m radius turn at 12.2-m/s or 27.3-mph, your tires would undergo .50-G's or 4.9-m/s² of lateral acceleration. As you can see in the chart, just a few mph faster can mean the difference between a low G and a high-G turn. In a 15-m turn, less than 14-mph separates .25-G's from 1-G.

          In the above example, your tires should have plenty of adhesion left for braking or acceleration. However, just how much is left? For this we turn to the "Traction Circle". 

    It is a circle centered on an x-y axis graph. The radius of the circle is 1 unit. 1 unit is equal to the adhesion limit of a tire. Thus, on the traction circle, .45-G of acceleration would be .5 graph units, since it is half the tire's limit. The y-axis represents acceleration and braking. The x-axis represents turning. Anyone who has basic geometry can find if the tire will slip or not using the traction circle. basically, if you have graph paper, you can just pinpoint the spot. For example, if you are using 1-G rated tires, and are accelerating at .5-G and are in a 15-m turn at around 20-mph (.5-G of lateral acceleration, see above) your tires should hold. To prove this mathematically, we can use the Pythagorean Theorem. Draw a right triangle with the horizontal leg representing lateral (turning) acceleration, and the vertical leg representing acceleration/braking force. Using the Pythagorean theorem we can find the hypotenuse:

                                       R² = x² + y²

                                       R² = .5² + .5²

                                       R² = .25+.25

                                       R² = .5

                                       r = .71-units

          Thus, your tire would be using up 71% of its traction. In our 1-G tire we used, the actual force on the tire would be .71-G's

          If you try to accelerate too much, or turn at too high of a speed, the tires will skid. As another example, we will use 1-G tires again. The driver enters a 30-m turn at 38-mph (1-G), and realizes he's going a bit fast so he brakes at .50-G's. Will his tires hold?

                                       R² = 1² + .5²

                                       R² = 1 + .25

                                       R² = 1.25

                                       R = 1.12-units

          His tires are at 112% of their limit, so they will lose their traction. Since he is braking, his car will most likely oversteer because the weight is shifted forward.

          In real situations, a traction circle may be less of a perfect circle, and flattened at one or more axes. For example, a RR car tends to oversteer, so the lower end of the circle may be only .8 units from the center, meaning it can't use as much brakes as our "ideal" car in the above examples.