Consider a very short interval of time, far less than a second. Call it dt (d stands for "delta," a Greek letter mathematicians use as shorthand for "tiny increment"). In time dt, let us say we go forward a distance dx and sideways a distance ds. The forward component of the velocity of the car is approximately v = dx / dt. At the beginning of the time interval dt, the car has no sideways velocity. At the end, it has sideways velocity ds / dt. In the time dt, the car has thus had a change in sideways velocity of ds / dt. Acceleration is, precisely, the change in velocity over a certain time, divided by the time; just as velocity is the change in position over a certain time, divided by the time. Thus, the sideways acceleration is

How is ds related to r, the radius of the circle? If we go forward by a fraction f of the radius of the circle, we must go sideways by exactly the same fraction of dx to stay on the circle. This means that ds = f dx. The fraction f is, however, nothing but dx / r. By this reasoning, we get the relation:

We can substitute this expression for ds into the expression for a, and remembering that v = dx / dt, we get the final result

This equation simply says quantitatively what we wrote before: that the acceleration (and the force) needed to keep to a circular line increases with the velocity and increases as the radius gets smaller.

What was not appreciated before we went through this derivation is that the necessary acceleration increases as the square of the velocity. This means that the centripetal force your tyres must give you for you to make it through a sweeper is very sensitive to your speed. If you go just a little bit too fast, you might as well go much too fast - you're not going to make it. The following table shows the maximum speed that can be achieved in turns of various radii for various sideways accelerations. This table shows the value of the expression

which is the solution of a = v2 / r for v, the velocity. The conversion factor 15/22 converts v from feet per second to miles per hour, and 32.1 converts a from gees to feet per second squared. We covered these conversion factors in part 3 of this series.

Table 1. Speed (Miles Per Hour)

Acceleration	Radius (Feet)
(Gees)	50.00	100.0	150.0	200.0	500.0
0.25	13.66	19.31	23.66	27.32	43.19
0.50	19.31	27.32	33.45	38.63	61.08
0.75	23.66	33.45	40.97	47.31	74.81
1.00	27.32	38.63	47.31	54.63	86.38
1.25	30.54	43.19	52.90	61.08	96.57
1.50	33.45	47.31	57.94	66.91	105.79
1.75	36.13	51.10	62.59	72.27	114.27
2.00	38.63	54.63	66.91	77.26	122.16

For autocrossing, the columns for 50 and 100 feet and the row for 1.00G are most germane. The table tells us that to achieve 1.00G sideways acceleration in a corner of 50 foot radius (this kind of corner is all too common in autocross), a driver must not go faster than 27.32 miles per hour. To go 30 mph, 1.25G is required, which is probably not within the capability of an autocross tyre at this speed. There is not much subjective difference between 27 and 30 mph, but the objective difference is usually between making a controlled run and spinning badly.