You may remember way back in part 5 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
circlesubject of part 7to stay within the capabilities of the car. We also model a more
complex segment than in part 5, 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. 