In the previous instalment, we did exact calculations for a dummy line down a 650-foot entry straight, a 180-degree left-hander, and a 650-foot exit chute. Cornering radii vary from 150 feet to 200 feet, and the track is 100 feet wide all the way around. This dummy line carries constant speed around the entire left-hander. We did those calculations to provide reference times to compare against this month’s more sophisticated calculations, in which we unwind the steering wheel and accelerate at the same time. The baseline times for the dummy line over the whole course, as a function of cornering radius, are in the second-to-last column of the following table:

From this point on, we need only look at the last column. It’s after the apex and down the exit chute where we look for improvement; we actually drive the dummy line up to the apex. Many readers will be screaming that we could try to get on the gas before the apex for even more improvement. Others will be screaming “trail brake!,” that is, ease off the brakes at the same time as winding the steering wheel at turn in (thanks to reader Marc Sibilia for pointing this out to me). We leave those refinements to later articles.

The approach in this article is to find a line by building it up, step-by-step, honouring the traction circle and the sides of the track. This is one of the techniques we can use in computer simulations, so we get to kill two birds with one stone: previewing simulation and analysing a particular driving line. For convenience, we need a Cartesian coordinate system, that is, a square grid. Let’s turn the track around 180 degrees for this purpose, and put the centre of the coordinate system at the centre of the corner. Since the inside edge of the track and the outside edge of the track are concentric semicircles, there is only one identifiable centre of the corner.

We’ll work by measuring the position and heading of the centroid of the car with respect to this new coordinate system. We have a goal of arriving at the point x = 200, y = 650, measured in feet, in the least possible time, with a heading of as close to 90 degrees as we can get it, that is, heading straight down the track. We start at the apex, which measures from x = r₀ sin  , y = r₁ cos  . The following sketch illustrates:

I must note, at this point, if you haven’t already noticed, this instalment of The Physics of Racing is going to be more concentrated and intense than previous instalments. I’m just going to blurt out facts without the usual explanations and walkthroughs. The reasons are (1) that we have a lot to get through in a little space and (2) that we assume that if you’ve been following the series this far, you’ve got the fortitude to work through it. So, let’s get it on!

The initial heading is tangent to the inner edge of the track, that is, perpendicular to the line from the centre of the track’s corner to the apex. Therefore, it has the angle up from the horizontal x axis. We know the starting speed, v0, so we know its components in the x direction and in the y direction: v0_x =  v0 cos  , v0_y = v0 sin  .

We perform the entire manoeuvre whilst never exceeding the limits of the traction circle. We set those limits as 1g cornering and braking and 0.5g accelerating, with smooth transitions all way around, as in the following sketch (the horizontal cap shows a way of accounting for engine limitations with non-smooth transitions, which will allow us to accelerate harder with the wheel still turned but probably scare us in the seat. Also, we note that 0.5g is a plausible, if only approximate, number for acceleration. We leave it to the reader to show that 0.5g in the quarter mile results in a realistic 13-second elapsed time, if at an unrealistic speed of 150 mph):

In each step of the calculation, we keep track of the following information:

• the time, t
• the current position, x(t), y(t), which we check to make sure we’re still on the track (x < 200) and to see whether we’re done (y 650)
• the current velocity, v_x(t), v_y(t), which we use to update the current position: , and likewise for y
• the tangential and radial acceleration, a_t(t), a_r(t), that is, tangential and radial to the bit of racing line at each instant (the instantaneous line), which we check to make sure that we’re not cornering over the limit and that we’re not exceeding the capacity of the engine, i.e., that is inside the traction envelope
• the acceleration in the x and y directions, a_x(t), a_y(t), which we use to update the current velocity: , and likewise for v_y

We drive the whole simulation by feeding on the throttle linearly with time over a time span called k and by simultaneously increasing the instantaneous radius of the driving line over a potentially different time span called k_unwind. Feeding on the throttle allows us to increase the tangential acceleration, a_t at each time step, and unwinding allows us to decrease the radial acceleration, a_r so we can stay within the traction circle. Since we’ll still have centripetal traction available after the throttle is buried full on, we ought to be able to unwind more slowly, enabling us to stay on the track, but use it all up. In other words, we ought to look for solutions wherein k_unwind is larger than k, perhaps by twice.

Let’s look at the first few rows of this simulation in a spreadsheet and delve into the formulas more deeply:

[column 1]: increments by each row; we actually computed with = 0.05 sec and display here every fourth actual row; this is an independent column, meaning that it does not depend on data from any other column.

[column 2]: tangential acceleration,

,

accounting for squeezing on the throttle up to g / 2; depends only on column 1.

[column 3]: maximal radial acceleration,

,

accounting for the traction circle; more precisely, for the upper half of the circle treated as a flattened (oblate) ellipse with height g / 2; depends only on column 2.

[column 4]: radial

,

accounting for unwinding the steering wheel; in steps from the inner parentheses outwards: g(1 - t /  k_unwind) slowly decreases from g as time increases from 0, but, it is never allowed to exceed v² / r, by the min expression, as mandated by the traction circle, and then, never allowed to be negative, by the max expression, because we don’t want to start turning back toward the entry straight; depends on columns 1 and 3.

[column 5]:

;

just for amusement, it’s interesting to calculate the instantaneous radius of a circle we could be driving if we were not accelerating tangentially; depends on columns 4 and 12, but no other columns depend on this.

[column 6]:

,

this just selects out the x components of both the radial and tangential accelerations, but makes sure that we never turn the wheel so much that we start going to the left. Note that the radial acceleration always tries to pull the car to the left, hence the minus sign (centripetal: see part 4 of The Physics of Racing); depends on columns 2, 4, 10, 11, and 12.

[column 7]:

,

selecting the y components, this time always pointing down the track, the way we want to go; depends on columns 2, 4, 10, 11, and 12.

[column 8]:

,

just update the x coordinate by the velocity from the prior time step; depends on columns 8 (the prior row of itself) and 10.

[column 9]:

,

do likewise for the y coordinate; depends on columns 9 (prior row) and 11.

[column 10]:

,

for updating the x component of the velocity (but don’t let it go negative, checking yet again, and, yes, this is a hack); depends on columns 10 (prior row) and 6.

[column 11]:

,

likewise for the y coordinate of the velocity; depends on columns 11 and 7.

[column 12]: finally,

,

depends on columns 10 and 11.

I’ve packed all this in an Excel spreadsheet. The spreadsheet should be in the download package for readers who acquired this document electronically.

Enough talk! Let’s drive! Driving means playing with the values of r, k, and k_unwind, and possibly even , to find the lowest overall time at which columns 8 and 9 show 200 or less and 650 or more, respectively. In general, “playing with” should be a sophisticated process involving hill climbing, genetic search, simulated annealing, and other fancy strategies for finding the very best values. In a computer simulation, we’d do that. However, we can do a reasonable job, for the sake of demonstration, by just tweaking the numbers by hand in the spreadsheet.

I have to admit that as I did so, I got kinaesthetic feelings as if I where actually driving. When I ‘ran off the track,’ that is, picked numbers that gave me x > 200, I gritted my teeth and blushed. When I was still unwinding at the end, I got that panicky feeling of understeer, knowing that I wasn’t going to stay on after the end of the segment, and so on.

The best values I found by hand are shown in the following table at r = 167.5, k = 3.25, and k_unwind = 7.22. That means that we take 3.25 seconds to bury the gas and 7.22 seconds to unwind the wheel. There are solutions with lower segment times, but, since we’re still unwinding long after the segment is done, I reject these solutions as assuming too much about what’s going on after our segment is done. With more track to work with, however, we can find lots more time. In fact, it’s a slightly surprising fact that by taking 9 seconds to unwind at r = 167.5, k = 3.25, we lose hardly any time and stay 15 feet inside the outer edge. There is quite a bit of territory to investigate even in this simple model.

Since the best dummy time, with the widest possible circle, is 16.760, and the best time I found here was 16.466, the improvement by unwinding and accelerating simultaneously is 0.294 seconds. This is very significant. If the exit straight were longer, the improvement would be even more dramatic since it would continue to accumulate time down the straight.

Note that this does not involve changing the entry to the corner other than by slowing down! There is no trail braking or lifting-while-turning or other risk-taking going on at corner entry. There is a very important driving lesson, here: to go faster, it is not necessary to take risks on corner entry. It is, in fact, both safer and faster just to slow down on the entry. The improved exit will follow naturally from the combination of looking far ahead and of being smooth. And that’s not even fair!

There is no guarantee that this is the best possible improvement in the model. I found these numbers by ’seat-of-the-pants’ tweaking. A more systematic or algorithmic search would very likely find better ones. In other words, I was able to find almost three tenths by just driving a better line without trying very hard at all. There is another driving lesson, here: just driving a better line gives better times time without changing the driver’s margin for error, that is, without getting deeper into the g limits of the machine.

For the future, we can start taking more risks to get even more improvement. We can risk accelerating before the apex and we can risk deeper entry by trail braking, that is, easing off the brake and winding up the steering wheel at the same time. These manoeuvres do entail more driver risk since they are new opportunities for loss of car control.

Erratum: in part 17, I wrote “By driving a line just one foot larger than the minimum, one is able to apex more than fifteen degrees later!”. I should have written “…fifteen degrees earlier!” The point was that the tightest line does not apex until the geometric exit of the corner, and that’s way too late. The slip-of-the-pen occurred because one is so accustomed to talking about late apexing as preferable.

By Brian Beckman

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 90-degree right-hander 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-subject of part 7-to stay within the capabilities of the car. We also model a more complex segment than in part 5, including an all-important exit chute where we take advantage of improved corner-exit 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: “slow-in, fast-out.” 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 180-degree left-hander 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 right-hand 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 computation-the dummy line-that 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 right-hand 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 turn-in 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 circle-the driving line-apexes 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 / 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 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):

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:

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 post-apex 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.

By Brian Beckman

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