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.Rather, we get the lowest overall time by cornering moreslowly so we can get back on the gas earlier.It's always tempting to corner a littlefaster, 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 linewe will drive up to the apex, but just a reference line after the apex.In the nextinstalment, we improve on the dummy line by accelerating and unwinding, predictingthe 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 toa 180-degree left-hander with outer radius 200 feet and inner radius 100 feet, connectedto an exit chute of 650 feet.In the following sketch, we show the segment twice withdifferent lines.The line on the left contains the widest possible inscribed corneringradius, and therefore the greatest possible cornering speed.The sketch on the rightshows the line with the lowest overall time.Although its cornering speed is slower thanin the line on the left, it includes a lengthy acceleration and unwinding phase on exitthat more than makes up for it.74 Line with Line withFastest Cornering Speed Lowest Overall TimeNote that both lines begin on the extreme right-hand side of the entry straight.Such willbe a feature of every corner we analyse.Lines that begin elsewhere across the entrystraight may be valid in scenarios like passing.However, we focus here on lines that aremore obvious candidates for lowest times.Also, throughout, we ignore the width of thecar, working with the 'bicycle line'.If we were including the width, w, of the car, wewould get the same final results on a track with outer radius of 200 + w / 2 feet andinner radius of 100 - w / 2 feet.First, we compute exact times where we can on the course: the entry straight, thebraking zone, and the corner up to the apex.To have a concrete baseline forcomparison, we also do a 'suboptimal' exit computation-the dummy line-that includescompleting the corner without unwinding and then running down the exit chute deadstraight somewhere in the middle of the track.In the next instalment of The Physics ofRacing, we compare the dummy line to the more sophisticated exit that includessimultaneously accelerating and unwinding to use up the entire width of the track in theexit chute.Let us enter the segment in the right-hand chute at 100 mph = 146.667 fps (feet persecond).We want the total times for a number of different cornering radii between twoextremes.The largest extreme is a radius of 200 feet, which is the same as the radius ofthe outer margin of the track.It should be obvious that it is not possible to drive a circlewith a radius greater than 200 feet and still stay on the track.This extreme is depicted inthe following sketch:75 We take the opportunity, here, to define a number of parameters that will servethroughout.First, let us call the radius of the outer edge of the track r1; this is obviously200 feet, but, by giving it a symbolic name, we retain the option of changing its numericvalue 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.Inthe extreme case of the widest possible line, r is the same as r1, namely, 200 feet.In theother extreme case, that of the tightest inscribed circle, r is 150 feet, as shown in thefollowing 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:76 h indicates the point where we must be done with braking.More precisely, h is thedistance of the turn-in point below the geometric start of the corner.Its value, byinspection, is (r - r0) cos.is the angle past the geometric top where the inscribedcircle-the driving line-apexes the inner edge of the track.We see two values for thehorizontal distance between the centre of the inscribed circle and the centre of the inneredge, 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.):InscribedCornerRadius (ft) (deg) h (ft)150 90.00 0.00151 73.90 14.14152 67.38 20.00153 62.47 24.49154 58.41 28.28155 54.90 31.62160 41.81 44.72165 32.58 54.77170 25.38 63.25175 19.47 70.71180 14.48 77.46185 10.16 83.67190 6.38 89.4477 195 3.02 94.87200 0.00 100.00There are a couple of interesting things to notice about these numbers.First, they matchup 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 drivingline.By driving a line just one foot larger than the minimum, one is able to apex morethan fifteen degrees later!With these data, we're now equipped to compute all the times up to the apex andbeyond.First, let's compute the speed in the corner by assuming that our car can cornerat 1g = 32.1 ft / s2 = v2 / 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 theconversions explicitly, but just remind ourselves once and for all that there are 22 feetper second for every 15 miles per hour.Now that we have the maximum cornering speed, we can compute how much brakingdistance we need to get down to that speed from 100 mph.Let's assume that our car canbrake at 1g also.We know that braking causes us to lose a little velocity for each littleincrement of time.Precisely, dv / dt = g.However, we need to understand how thevelocity changes with distance, not with time.Recall that dx / dt = v, dt = dx / v, so we getdx = vdv / g.Those who remember differential and integral calculus will immediately seethat is the required formula for braking distance.In any event, thebraking distance goes as the square of the speed, that is, like the kinetic energy, andthat's intuitive.However, there's a factor of two in the numerator that's easy to miss (theorigin 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, togive 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:
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