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.):

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 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.