Physics of Racing - Part 9: Straights
By: Brian Beckman
We found in part 5 of this series, "Introduction to the Racing Line," that a driver can lose a shocking amount of time by taking a bad line in a corner. With a six-foot-wide car on a ten-foot-wide course, one can lose sixteen hundredths by ?blowing? a single right-angle turn. This month, we extend the analysis of the racing line by following our example car down a straight. It is often said that the most critical corner in a course is the one before the longest straight. Let?s find out how critical it is. We calculate how much time it takes to go down a straight as a function of the speed entering the straight. The results, which are given at the end, are not terribly dramatic, but we make several, key improvements in the mathematical model that is under continuing development in this series of articles.The mathematical model for traveling down a straight follows from Newton?s second law: F=ma (eq 1) where F is the force on the car, m is the mass of the car, and a is the acceleration of the car. We want to solve this equation to get time as a function of distance down the straight. Basically, we want a table of numbers so that we can look up the time it takes to go any distance. We can build this table using accountant?s columnar paper, or we can use the modern version of the columnar pad: the electronic spreadsheet program.
To solve equation 1, we first invert it: a=F/m. (eq 2)
Now a, the acceleration, is the rate of change of velocity with time. Rate of change is simply the ratio of a small change in velocity to a small change in time. Let us assume that we have filled in a column of times on our table. The times start with 0 and go up by the same, small amount, say 0.05 sec. Physicists call this small time the integration step. It is standard practice to begin solving an equation with a fixed integration step. There are sometimes good reasons to vary the integration step, but those reasons do not arise in this problem. Let us call the integration step Dt. If we call the time in the i-th row ti, then for every row except the first, Dt = ti - ti-1 = constant. (eq 3)
We label another column velocity, and we?ll call the velocity in the i-th row vi. For every row except the first, equation 2 becomes: 
(eq 4).
We want to fill in velocities as we go down the columns, so we need to solve equation 4 for vi. This will give us a formula for computing vi given vi-1 for every row except the first. In the first row, we put the speed with which we enter the straight, which is an input to the problem. We get: vi = vi-1 + Dt F/m (eq5).
We label another column distance, and we call the distance value in the i-th row xi. Just as acceleration is the rate of change of velocity, so velocity is the rate of change of distance over time. Just as before, then, we may write: 
. (eq 6)
Solved for xi, this is: xi = xi-1 + Dt vi (eq 7)
Equation 7 gives us a formula for calculating the distance for any time given the previous distance and the velocity calculated by equation 5. Physicists would say that we have a scheme for integrating the equations of motion.
A small detail is missing: what is the force, F? Everything to this point is kinematic. The real modeling starts now with formulas for calculating the force. For this, we will draw on all the previous articles in this series. Let?s label another column force, and a few more with drag, rolling resistance, engine torque, engine rpm, wheel rpm, trans gear ratio, drive ratio, wheel torque, and drive force. As you can see, we are going to derive a fairly complete, if not accu
For a complete look at Sempre Ferrari, you may want to check out the rest of the articles from Volume 3, Issue 3 - April/May 1996