"Slowin, Fastout!" or, Advanced Analysis of the Racing Line by Brian Beckman PhD
You may remember way back in
Introduction to the Racing Line that we did some
simple calculations by hand to show that the classic racing line through a 90degree righthander is better than the either the line that
hugs the inside or the line that hugs the outside of the corner. 'Better' means 'has lowest time.' The 'classic racing line' was, under the
assumptions of that article, the widest possible inscribed line.
In this and the next instalment of The Physics of Racing, we raise the bar. Not only do we calculate the times for all lines
through a corner, but we show a new kind of analysis for the exit, accounting for simultaneously accelerating and unwinding the
steering wheel after the apex. This kind of analysis requires us to search for the lowest time because we cannot calculate it
directly. We apply the approximation of
the traction circle to stay within the capabilities of the
car. We also model a more complex segment than in
Introduction to the Racing Line,
including an allimportant exit chute where we take advantage of improved cornerexit speed. This style of analysis applies directly to
computer simulation that we now have in progress in other continuing threads of The Physics of Racing.
The whole point of this analysis is to back up the old mantra: "slowin, fastout." We will find that the quickest way through the
whole segment does not include the fastest line around the corner. Rather, we get the lowest overall time by cornering more slowly so
we can get back on the gas earlier. It's always tempting to corner a little faster, but it frequently does not pay off in the context
of the rest of the track.
This analysis is sufficiently long that it will take two instalments of this series. In this, the first instalment, we do exact
calculations on a dummy line, which is the actual line we will drive up to the apex, but just a reference line after the apex. In
the next instalment, we improve on the dummy line by accelerating and unwinding, predicting the times for a line we would actually
drive, but entailing some small inexactitude.
Let's first describe the track segment. Imagine an entry straight of 650 feet, connected to a 180degree lefthander with outer
radius 200 feet and inner radius 100 feet, connected to an exit chute of 650 feet. In the following sketch, we show the segment
twice with different lines. The line on the left contains the widest possible inscribed cornering radius, and therefore the greatest
possible cornering speed. The sketch on the right shows the line with the lowest overall time. Although its cornering speed is slower
than in the line on the left, it includes a lengthy acceleration and unwinding phase on exit that more than makes up for it.
Line with
Fastest Cornering Speed Line with
Lowest Overall Time
Note that both lines begin on the extreme righthand side of the entry straight. Such will be a feature of every corner we analyse.
Lines that begin elsewhere across the entry straight may be valid in scenarios like passing. However, we focus here on lines that are
more obvious candidates for lowest times. Also, throughout, we ignore the width of the car, working with the 'bicycle line'. If we
were including the width, w, of the car, we would get the same final results on a track with outer radius of 200 + w / 2 feet and
inner radius of 100  w / 2 feet.
First, we compute exact times where we can on the course: the entry straight, the braking zone, and the corner up to the apex. To
have a concrete baseline for comparison, we also do a 'suboptimal' exit computationthe dummy linethat includes completing the
corner without unwinding and then running down the exit chute dead straight somewhere in the middle of the track. In the next
instalment of The Physics of Racing, we compare the dummy line to the more sophisticated exit that includes simultaneously
accelerating and unwinding to use up the entire width of the track in the exit chute.
Let us enter the segment in the righthand chute at 100 mph = 146.667 fps (feet per second). We want the total times for a number of
different cornering radii between two extremes. The largest extreme is a radius of 200 feet, which is the same as the radius of the
outer margin of the track. It should be obvious that it is not possible to drive a circle with a radius greater than 200 feet and
still stay on the track. This extreme is depicted in the following sketch:
We take the opportunity, here, to define a number of parameters that will serve throughout. First, let us call the radius of the
outer edge of the track r1; this is obviously 200 feet, but, by giving it a symbolic name, we retain the option of changing its
numeric value some other time. Likewise, let's call the radius of the inner circle r0, now 100 feet. Let's use the symbol r to
denote the radius of the inscribed circle we intend to drive. In the extreme case of the widest possible line, r is the same as r1,
namely, 200 feet. In the other extreme case, that of the tightest inscribed circle, r is 150 feet, as shown in the following sketch:
We're now ready to discuss the two remaining parameters you may have noticed: h and (Greek letter alpha). Consider the following
figure illustrating the general case:
h indicates the point where we must be done with braking. More precisely, h is the distance of the turnin point below the geometric
start of the corner. Its value, by inspection, is (r  r0) cos
.
is the angle past the geometric top where the inscribed circlethe
driving lineapexes the inner edge of the track. We see two values for the horizontal distance between the centre of the inscribed
circle and the centre of the inner edge, and those values are (r  r0) sin
and
r1  r. Their equality allows us to solve for
:
The following table shows numeric values of h and
for a number of inscribed radii
(Note that if we varied r0 and r1 we would have a much larger 'book' of values to show. For now, we'll just vary r.):
Inscribed
Corner
Radius (ft) 
(deg) 
h (ft) 
150 
90.00 
0.00 
151 
73.90 
14.14 
152 
67.38 
20.00 
153 
62.47 
24.49 
154 
58.41 
28.28 
155 
54.90 
31.62 
160 
41.81 
44.72 
165 
32.58 
54.77 
170 
25.38 
63.25 
175 
19.47 
70.71 
180 
14.48 
77.46 
185 
10.16 
83.67 
190 
6.38 
89.44 
195 
3.02 
94.87 
200 
0.00 
100.00 

There are a couple of interesting things to notice about these numbers.
First, they match up with the visually obvious values of
h = 0,
= 90
and
h = 100, = 0 when
r = 150,
r = 200 respectively. This is a good check that we haven't
made a mistake. Secondly,
changes
very rapidly with corner radius, and this fact has
major ramifications on
driving line.
By driving a line just one foot larger than the minimum, one
is able to apex more than fifteen degrees later!
With these data, we're now equipped to compute all the times up to the apex
and beyond. First, let's compute the speed in the corner by assuming that our
car can corner at 1g = 32.1 ft / s^{2}
= v^{2} / r,
giving us .
We express all speeds in miles per hour, but other lengths in
feet. We won't take the time and space to write out all the conversions
explicitly, but just remind ourselves once and for all that there are 22 feet
per second for every 15 miles per hour.
Now that we have the maximum cornering speed, we can compute how much braking
distance we need to get down to that speed from 100 mph. Let's assume that our
car can brake at 1g also. We know that braking causes us to lose a little
velocity for each little increment of time. Precisely, dv / dt = g.
However, we need to understand how the velocity changes with distance, not with
time. Recall that dx / dt = v,
dt = dx / v, so we get dx = vdv / g.
Those who remember differential and integral calculus will immediately see that
is
the required formula for braking distance. In any event, the braking distance
goes as the square of the speed, that is, like the kinetic energy, and that's
intuitive. However, there's a factor of two in the numerator that's easy to miss
(the origin of this factor is in the calculus, where we compute limit
expressions like ).
We next subtract the braking distance from the entry straight, and also
subtract h, to give us the distance in which we can go at 100 mph, top
speed, before the braking zone.
Now, we need the time spent braking, and that's easy:
. All the
other times are easy to compute, so here are the times for a variety of
cornering lines up to the apices (or apexes for those who aren't Latin
majors):
Inscribed Corner Radius (ft) 
Cornering speed @1g in mph 
Braking Distance (ft) @1g from 100 mph 
Straight Distance (ft) prior to braking 
Time (sec) in straight @100 mph prior to braking 
Time (sec) in braking zone 
Time (sec) in corner prior to apex 
Total time (sec) up to the apex 
150 
47.24 
261.11 
388.89 
2.652 
2.418 
6.802 
11.872 
152 
47.55 
260.11 
369.89 
2.522 
2.404 
5.987 
10.912 
154 
47.86 
259.11 
362.60 
2.472 
2.390 
5.682 
10.544 
155 
48.02 
258.61 
359.77 
2.453 
2.382 
5.566 
10.401 
160 
48.79 
256.11 
349.17 
2.381 
2.347 
5.144 
9.872 
170 
50.29 
251.11 
335.64 
2.288 
2.278 
4.641 
9.208 
180 
51.75 
246.11 
326.43 
2.226 
2.212 
4.325 
8.762 
190 
53.16 
241.11 
319.45 
2.178 
2.147 
4.099 
8.424 
200 
54.55 
236.11 
313.89 
2.140 
2.083 
3.927 
8.150 
At first glance, it appears that the widest line is a huge winner, but
we must realize that these times include only driving up to the apex, and that
is far earlier on the widest line, where = 0. Suppose we continued driving all
the way around the corner at constant speed and then accelerated out the exit
chute at 0.5g? This is the dummy line. We won't really drive this line
after the apex, but discuss it nonetheless to provide a reference time. It's
very easy to compute and provides a foundational intuition for the more advanced
exit computation to follow in the next instalment:
Inscribed Corner Radius
(ft) 
Total time (sec) up to
the apex 
Time (sec) in corner
after apex 
Time for entrance and
complete corner 
Exit speed from chute
(mph) @ g/2 accel 
Time in exit chute (sec) 
Combined segment time 
Combined postapex time
and exitchute time 
150 
11.872 
0.000 
11.872 
109.091 
5.670 
17.541 
5.670 
152 
10.912 
0.860 
11.773 
107.857 
5.528 
17.301 
6.388 
154 
10.544 
1.209 
11.754 
107.422 
5.460 
17.213 
6.669 
155 
10.401 
1.348 
11.750 
107.260 
5.430 
17.180 
6.779 
160 
9.872 
1.881 
11.753 
106.697 
5.308 
17.061 
7.189 
170 
9.208 
2.600 
11.808 
106.101 
5.116 
16.924 
7.716 
180 
8.762 
3.126 
11.888 
105.806 
4.955 
16.844 
8.082 
190 
8.424 
3.556 
11.980 
105.666 
4.813 
16.792 
8.369 
200 
8.150 
3.927 
12.077 
105.627 
4.682 
16.760 
8.609 
So, we see that, driving the dummy line, the widest line yields the slowest
time from the entrance up through the complete semicircle, but the quickest overall
time when the exit chute is included. The widest line has lower (better) times
than the tightest line in
 the entry straight by about half a second, because h is large and
the entry straight is shorter for wider circles
 in the braking zone by about three tenths because the cornering speed is
higher and less braking is needed
 and in the exit chute by almost a second, again because is h large
and the exit chute is thereby shorter
The widest line has higher (worse) times by about a second in the circle
itself because the wider circle is also longer. When the balances are all added
up, the widest line is about eight tenths quicker than the tightest line, but
it's all because of the effects of the corner on the straights before and
after.
Recall once again that the dummy line is not a line we would actually drive
after the apex. But, with that as a framework, we're in position to introduce
the next improvement. Everything we do from here on improves just the postapex
portion of the corner and the exit chute. We will actually drive the dummy line
up to the apex. So, from this point on, we need only look at the last column in
the table above, where we are shocked to see that there are almost three
seconds' spread from the slowest to the quickest way out. A good deal of this
ought to be available for improvement by accelerating and unwinding.