Driving a car is a classic problem in control. Here, we
mean control in the technical sense of control theory, an established branch
of engineering science (once again, I find http://www.britannica.com/
to have a very nice, brushup article on that term). In a moreorless continuous fashion,
the driver compares desired direction, speed, and acceleration with actual direction,
speed, and acceleration. The driver uses visual input to sense actual direction and speed;
and uses visceral, inertial feedbackthe butt sensorfor actual acceleration. When the
actual differs too much from the desired, the driver applies throttle, brake, steering,
and gear selection to change the actual. These inputs cause the tyres to react with the
ground, which pushes back against the tyres, and through the suspension, pushes the body
of the car and driver. Drivers in highspeed circumstances can also generate desired
aerodynamic forces, as in slipstreaming, in the "slingshot pass," and in the
Earnhardt TIP manoeuvre, where the driver "takes the air off" the spoiler of the
car in front of him. Tyres generate forces by sticking and sliding and everything in
between. They transmit these forces to the wheels by elastic deformation. The elastic
deformation is extremely complex and theoretical computation requires numerical solution
of finiteelement equations. However, despite fierce trade secrecy, industry and academia
have reached apparent consensus in recent years on a formula that summarizes experimental
and theoretical data. This socalled magic formula is not a solution to
equations of motiona solution in such a form is not feasible. It's just a convenient
fitting of commonplace mathematical functions to data. It allows one to compute forces at
a higher precision than something like RARS (see parts 16 and 19 of the Physics of Racing
[PhOR]), but without integrating equations. Therefore, forces can be computed
within a reasonable time, say in a realtime simulation program.
To understand the magic formula, we need first to define its inputs, which include slip.
Slip is an indirect measure of the fraction of the contact patch that is sticking. It is
frequently asserted in the literature that a tyre with no slip at all cannot create
forces. It has taken me a very long time to accept this assertion. Why can I steer a
tintoy car with metal tyres on a hard surface like Formica? If there is any slip in such
tyres, it is microscopic, yet there are sufficient forces to brake and steer, even if just
a little. I finally caved in when I realized that the forces are minute, also. If there is
any friction between the tyre and the surface, there MUST be slip, as it is defined below.
Though to a very small degree, the Formica and the tiny contact patches of the tin tyres
actually twist and stretch each other. The only way to eliminate slip completely is to
eliminate GRIP completely. Any grip, and you will have slip.
There are two, slightly different flavours of the magic formula. The longitudinal
one is the subject of this entire instalment of PhOR, and we cover the lateral
one in the next instalment. Longitudinal slip is along the mean plane of the wheel and
might also be called circumferential or tangential. It creates braking and
accelerating forces. Lateral slip is our old friend grip angle [PhOR10], and it generates
cornering forces.
We write longitudinal slip as . It's defined by the actual angular velocity, , of a wheel plus tyre
under braking or acceleration, compared to the corresponding angular velocity of the same
wheel plus tyre when rolling freely. We write the freerolling angular velocity as , where
V is the current, instantaneous velocity of the hub centreline of the wheel with
respect to the ground, and R_{e} is the effective radius, a constant
defined below. Since the dimensions of V are length/time, and the dimensions of any
radius are length, the ratio, , has dimensions of inverse time. In fact, it should be viewed
as measuring radians per unit time, radians being the natural, dimensionless measure of
angular rotation. There are radians in one rotation or one circumference of a circle, just as
the length of the circumference is times the radius.
Let's begin the discussion of longitudinal slip with a question. Consider a wheeltyre
combination with 13inch radius or 26inch diameter, say a 25550/16 tyre on a 16inch
rim. The "50" in the tyre specification is the ratio of the sidewall HEIGHT to
the tread WIDTH, which is also written into the specification as 255, millimetres
understood. We get a sidewall HEIGHT of 50 percent of 255 mm, which is 5.02 inch.
Therefore, the total, unloaded radius, half of the treadtotread diameter,
is about 5 + 16/2 = 13 Inch. Now consider a rigid tyre of the same radius, made,
say, of steel or of wood with an iron tread like old Western wagon wheels. The question is
whether, given a certain constant hub velocity, pneumatic tyres spin faster than, slower
than, or at the same speed as equivalent rigid tyres?
At first glance, one might say, "Well, faster, obviously. Since the pneumatic tyre
compresses radially under the weight of the car, its radius is actually smaller than the
unloaded radius at the point of contact, where it sticks and acquires linear velocity
equal in magnitude and opposite in direction to the hub velocity. Since smaller wheels
spin faster than larger ones at the same speed, the pneumatic tyre spins faster than the
equivalent rigid tyre of the same unloaded radius. Let the unloaded, natural radius of the
pneumatic tyre be R, also the radius of the equivalent solid tyre. If the hub has
velocity V, the solid tyre spins with angular velocity . Since the loaded
radius, of the pneumatic tyre, R_{l}, is smaller than R, V/R_{l},
the angular velocity of the loaded pneumatic tyre, must be larger than V/R."
This is partly correct. The pneumatic tyrewheel combination does spin faster
than a rigid wheel of the same unloaded radius, but it does not spin as fast as a
rigid wheel of the same loaded radius, which is the HEIGHT of the hub centre off
the ground under load. The reason is that the tyre also compresses circumferentially
or tangentially, setting up complex longitudinal twisting in the sidewall. The
tangential speed of a particle of tread varies as the particle goes around the
circumference of the tyre.
Let's mentally follow a piece of tread around as the wheel, not necessarily the
tyre, turns at a constant radial velocity, . Imagine a plug
of yellow rubber embedded in the tread, so that you could visually track it or photograph
it with a movie camera or strobe system as it moves around the circumference. The rubber
of the tread does not travel at constant speed, even though the wheel supporting the tyre
does. At the top of the tyre, the radius is almost exactly R, the unloaded radius,
so the tread moves with tangential velocity . As the yellow plug
rolls around and approaches the contact patch from the front, it slows down in the bunched
up area at the leading edge of the contact patchjust forward of it. There is
a bunchedup area, because the tyre is made up of elastic material that gets squeezed and
stretched out of the contact patch and piles up ahead of the contact patch as it rolls
into it from the direction of the leading edge. Eventually, the plug enters the patch, in
the centre of which it must move at speed relative to the hub
centre, that is, backwards at a speed dictated by the loaded radius and the wheel
velocity. We've assumed that the plug is not slipping on the ground at the point where it
has speed with respect to the hub. This means that it has speed zero with respect to the
ground at that point.
The average of the tangential velocities around the wheel defines the effective radius,
R_{e}, as follows. Let measure the angular position, from 0 to , around the wheel.
Suppose we knew the tangential velocity with respect to the hub centre, V(), at every . We could easily measure this with
our strobe light and cameras. V() gives us the radius at every angular position via the equation V()/ = R(), where is the constant
angular velocity of the wheel. The average would be computed by the following integral:
Let's run some numbers. 10 mph is 14 feet/second or 176 inches/second. With an unloaded
circumference of inch/revolution, we get 176/= 2.154 revs per
second, or 129 RPM for each 10 MPH. Under ordinary circumstances, the effective radius
will be no more than a few percent less than then the unloaded radius, and the RPMs should
be, then, a few percent more than 129 RPM per 10 MPH. At 100 MPH, the tyre is under
considerable stress and spins at something over 1,300 RPM.
Now we're in a position to define longitudinal slip, written . We want a quantity
that vanishes when the wheel rolls at constant speed, increases when the wheel accelerates
the car by pulling the contact patch backwards, and decreases below zero when then wheel
brakes the car by pushing the contact patch forward. Under acceleration, the wheel and
tyre combination will tend to spin a little faster than it would do while free rolling. We
already know that, for a given V, the freerolling angular velocity is , by
definition. The actual angular velocity, , then, is higher under acceleration. So, if we know V,
, and
the constant R_{e}, then we can define the longitudinal slip as the ratio,
minus 1, so that it's zero under freerolling conditions:
Just looking at this formula, a freerolling wheel has , = 0 a
lockedup wheel under braking has = 0, = 1 and an accelerating wheel has a positive of
any value.
The magic formula yields the longitudinal force, in Newtons, given some constants and
dynamic inputs. The formula takes eleven empirical numbers that characterize a particular
tyre {b_{0}, b_{1}...b_{10}}. The dynamic
parameters are F_{z}, or weight, in KiloNewtons on the tyre,
and the instantaneous slip, . The eleven numbers are measured for each tyre. We borrow an
example from Motor Vehicle Dynamics by Giancarlo Genta. On page 528, he offers the
following numbers for a car that appears to be a Ferrari 308 or 328, to which I have added
dimensions:
b_{0} 
1.65 
dimensionless 
b_{1} 
0 
1/MegaNewton 
b_{2} 
1688 
1/Kilo 
b_{3} 
0 
1/MegaNewton 
b_{4} 
229 
1/Kilo 
b_{5} 
0 
1/KiloNewton 

b_{6} 
0 
1/(KiloNewton)^{2} 
b_{7} 
0 
1/KiloNewton 
b_{8} 
10 
dimensionless 
b_{9} 
0 
1/KiloNewton 
b_{10} 
0 
dimensionless 

Though the majority of these values are zero for the tyres on this car,
it is by no means always the case. In fact, the 'largesaloon' example just before the
(alleged) Ferrari in Genta's book has no zeros.
We build up the magic formula in stages. The first helper quantity is µ_{p } = b_{1}F_{z} + b_{2}.
This is an estimate of the peak, longitudinal coefficient of friction,
fitted as a linear function of weight (see Part 7 of PhORs). From this definition, we
begin to see what's going with the dimensions. A typical, streetable sports car might
weigh in at 3,000 lbs, which is about
3,000 / 2.2 = 1,500 * 0.9 = 1,350 kg, which is
about 1,350 * 9.8 = 13,200 Newtons, or 13.2 KiloNewtons (look,
ma, no calculator!). Let's assume each tyre gets a quarter of that to start off with, or
3.3 KN. b_{1} multiplies that number to give us something with dimensions
of KiloNewton/MegaNewton, which we write simply as 1/Kilo (inventing units onthefly, one
Mega = 1 Kilo squared). b_{2} has the same dimensions, so it's kosher to
add it in, yielding µ_{p} = 1688 / Kilo in this
case. The next step is the helper D = µ_{p}F_{z}, which
will be in Newtons. We now see the reason for the 1/Kilo unit. In our case, we get about D = (1700  12) * 3.3 = 5610  40 = 5570
N. The important point is that D is linear in F_{z}, so µ_{p}
acts, mathematically, like a coefficient of friction, as promised. b_{2} is
a pretty direct measurement of stickiness, times 1,000 for convenience. This model tyre
has a coefficient of friction of almost 1.7! Not my data, man.
The next step is to compute the product of a new helper, B, times b_{0}
and the aforecomputed D. The magicians who created the formula tell us that Bb_{0}D = (b_{3}F_{z}^{2} + b_{4}F_{z}) exp(b_{5}F_{z}).
This slurps up a few more of the magical eleven empirical numbers, and a pattern emerges.
These b_{i} numbers serve as coefficients in polynomial expressions over F_{z}.
So, b_{5}F_{z} is dimensionless, as must be the argument of
the exponential function. b_{3}F_{z}^{2} + b_{4}F_{z}
has dimensions of Newtons, as does the entire product. Therefore, B must be
dimensionless. We need B in the next step, so let's solve for it now:
,
Where we've been able symbolically to divide out one factor of F_{z},
convenient especially for numerical computation, where overflow is an everpresent hazard.
Continuing with our numerical sample, b_{3}F_{z} + b_{4} = 229 / Kilo,
the exponential is unity, and the numerator is
yielding B = 229 / 2786 = 0.0822. Most
importantly, B depends only weakly on F_{z}. In the sample
case, not at all, because b_{1} = b_{3} = b_{5} = 0,
but there are lots of other ways to characterize the algebraic dependence of B on F_{z}.
The next step is to account for the longitudinal slip with another helper, S = (100+ b_{9}F_{z} + b_{10});
in our sample case, this reduces to just S = 100.
Only one more helper is needed, and that's E = (b_{6}F_{z}^{2} + b_{7}F_{z} + b_{8}),
very straightforward. The final formula is
Once again, don't try to find any physics in here: it's just a convenient formula that
fits the data reasonably well. Plugging in numbers for = 0, because that's
an easy sanity check to do in our heads, we see immediately the result is zero. Let's try S = 10,
ten percent slip. SB = 0.822, tan^{1}(0.822) = 0.688, E = 10,
so the argument of the outer arctangent is SB  10 * (0.266) = SB + 2.66 = 3.48,
tan^{1}(3.48) = 1.29, 1.29 b_{0} = 2.13,
sin(2.13) = 0.848, and, finally, D * 0.848 = 4720 Newtons.
Lots of longitudinal force for a 3,300 N vertical load!
Let's plot the whole formula:
The horizontal axis measures S = 100, which is really just
slip in percent. The deep axis, going into the page, measures F_{z} from 5
KN, nearest us, to zero in the back. The vertical axis measures the result of applying the
formula to our model tyre, so it's longitudinal forceforce of launching or braking.
Notice that for a load of 5 KN, the model tyre can generate almost 8 KN of force. Very
sticky tyre, as we've already noticed! Also notice that the generated force peaks at
around = 0.08,
or 8 percent. The peak would be something one could definitely feel in the driver's seat.
Overcooking the throttle or brakes would produce a palpable reduction in gforces as the
tyres start letting go. Worse than that, increasing braking or throttle beyond the peak
leads to reduced grip. This is an instability area, where increasing slip leads to
decreasing grip.
Finally, note that the function behaves roughly linearly with F_{z},
showing that it acts like a Newtonian coefficient of friction, albeit a different one for
each value of slip. 