I'm back after a hiatus of nine years. Time does fly, doesn't it? For those
counting articles, the last one published was part 12.

After such a long time away, it might be worthwhile to repeat the motivation and goals
of this "Physics of Racing" series. I am a physicist (the "PhD" after
my name is from my Union card). I'm also an active participant in motorsports. It would be
almost impossible for me not to use my professional training to analyse my hobby. So, I've
been thinking for some time about the physics of racing cars.

Part of the fun for me is to do *totally original* analyses. As such, they won't
have the specifics of a hardcore engineering analysis. You can look that up in books by
Fred Puhn, William Milliken, and Carrol Smith, amongst many others. I want to find the
bare-bones physics behind the engineering--at the risk of bypassing some detail. In sum, I
analyse things completely from scratch because:

- I want the depth of understanding that can only come only from figuring things out from
first principles,
- "peeking at the answer" from someone else's work would spoil the fun for me,
- I hope to give a somewhat fresh outlook on things.

In 1990, one of my fellow autocrossers asked me to write a monthly column for the SCCA
CalClub newsletter. After receiving lots of encouragement, I released the columns to the
Internet via Team Dot Net. Back then, the Internet was really small, so I was just sharing
the articles in a convenient way with other autocrossers. Since then, the Internet got big
and my articles have acquired a life of their own. I have received thousands of
happy-customer emails from all over the world, plus a few hate mails (mostly about article
#4, in case you're wondering).

So, here we go again. This month, I'd like to understand, from first principles, why
it's so important to be smooth on the controls of a racing car. To me, "smooth"
means avoiding jerkiness when applying *or releasing* the brakes, the gas, or
steering. Most of the time, you want to roll on and off the gas, squeeze on and off the
brakes, slither in and out of steering. It's just as important to avoid jerkiness at the
end of a manoeuvre as at the beginning. For example, when steering, not only should you
start turning the steering wheel with a gradual, smooth push, but you want to complete the
wind-up with a gradual, smooth slowing of the push. Likewise, when you unwind the wheel,
you want to start and stop the unwinding smoothly. Thus, a complete steering manoeuvre
consists of *four* gradual, slithery start-and-stop mini-manoeuvres. A complete
braking event has four little mini-slithers: one each for the start and stop of the
application and the releasing of the pedal. Same for the throttle.

Ok, great, but why? At first blush, it seems one would be able to get back on the gas *more
quickly* by snapping the throttle on or get into a corner more quickly by whipping the
wheel. Furthermore, supposing we can justify smoothness, are there exceptions to the rule?
Are there times when it is best to snap, whip, or jerk? And exactly how smooth should one
be? Smoothing means slowing the control inputs down, in a particular way, so it's
obviously possible to be too smooth, as in not quick enough, as in not getting as much out
of the car as it's capable of delivering.

Let's tackle "why", first. As usual, physics has technical meanings for
everyday words. One of the "physically correct" meanings of "smooth"
is *sinusoidal*. A sinusoid is a curve that looks like this:

If we think of, say, steering-wheel winding angle as proportional to the vertical axis
and time in seconds along the horizontal axis, then this picture describes a really smooth
windup taking one second followed immediately by a really smooth unwinding taking another
second. In fact, you can easily see the four mini-slithers discussed above as the
head-and-tail-sections of the bumps and valleys of the curves. So, the question
"why", in technical terms, amounts to asking why such a curve represents better
steering input than a curve like the following, "upside-down-hat" curve:

Now, here's the reason: sinusoidal inputs are better because they match the natural
response of the car! The suspension and tyres perform, approximately, as *damped
harmonic oscillators*, or DHOs. A DHO can be in one of three conditions: *underdamped*,
*critically damped*, or *overdamped*. In the underdamped condition, a DHO
doesn't have a strong *damper*, which is another term for shock absorber. An
underdamped DHO responds sinusoidally. We've all seen cars with broken shocks bouncing up
and down on the springs. In the critically damped and overdamped conditions, the car
bounces just once, because the damper provides some friction to quiet down continued
bouncing. However, even in these conditions, the one bounce has an approximate sinusoidal
shape.

The most important parameter of any DHO is its *frequency*. In the underdamped
condition, the frequency corresponds to the number of bounces per second the DHO performs.
In the critically damped and overdamped conditions--as well as in the underdamped
condition, the frequency corresponds to the *resonance* frequency or natural
frequency of the system! In other words, if you provide so many inputs per second, back
and forth, as in a slalom, at the resonance frequency, the car will have maximal response.
If the inputs are faster, they will be too fast for the DHO to catch up and rebound before
you've reversed the inputs. If the inputs are slower, the DHO will have caught up and
started either to bounce the other way or to settle, depending on condition, when the
reverse input comes in.

So here's the bottom line: to maximize the response of a car, you want to provide
steering, braking, and throttle inputs with sinusoidal shapes at the resonance frequency
of the DHOs that constitute the suspension and tyre systems. Inputs that are more jerky
just dump high-frequency energy into the system that it must dissipate at lower
frequencies. In other words, jerky inputs *upset *the car, which what drivers say all
the time. By matching the shape and frequency of your control inputs to the car's natural
response curve, you're telling the car to do something it can actually do. By giving the
car an "instruction" like the upside-down hat, you're telling it to do something
it can't physically do, so it responds by flopping and bouncing around some approximation
of your input. Flopping and bouncing means not getting optimum traction; means wasting
energy in suspension oscillation; means going slower. Now, there is an exception: if the
front tyres are *already* sliding, a driver may benefit from quickly steering them
into line, hoping to "catch" the car. Likewise, a jerky blip on the throttle
with the clutch engaged to bring up the revs to match the gears on a downshift is usually
the right thing to do. But, when the car is hooked up, getting the most out of the car
means *simulating* the response of the various DHOs in the system with steering,
braking, and throttle inputs.

Now we know the physics behind it. Let's do some math!

The frequency turns out to be , as we show below. *k* is the *spring constant*,
typically measured in pounds per inch, and *m* is the mass of the sprung weight,
typically measured in pound-masses. Suppose our springs were 1,000 lb/in, supporting about
800 lb of weight on one corner of the car. First, we note that a pound *force* is
roughly (1/32) *slug* - ft/s^{2} and that a pound *weight*
is (1/32) *slug*. So, we're looking at

Notice that we've used the back-of-the-envelope style of computation discussed in part
3 of this series. We've found that the resonance frequency of one corner of a car is about
4 bounces per second! This matches our intuitions and experiences: if one pushes down on
the corner of a car with broken shocks, it will bounce up and down a few times a second,
not very quickly, not very slowly. We can also see that the frequency varies as the square
root of the spring constant. That means that to double the frequency, say, to 8 bounces
per second, we must quadruple the spring strength to 4,000 lb/in or quarter the sprung
weight to 200 lb. [Note added in proof: My friend, Brad Haase, has pointed out that 4 Hz,
while in the "ballpark", is much too fast for a real car. Now, this series of
articles is only about fundamental theory and ballpark estimates. Nonetheless, he wrote
convincingly "can you imagine a 4-Hz slalom?" I have to admit that 4 Hz seemed
too fast to me when I first wrote this article, but I was unable to account for the
discrepancy. Brad pointed out that the suspension linkages supply leverage that reduces
the effective spring rate and cited the topic "installation ratio" in Milliken's
book *Race Car Vehicle Dynamics*. Since I have not peeked at that book, on purpose,
as stated in the opening of this entire series and reiterated in this article, I can only
confidently refer you there. Nonetheless, intuition says that 1 Hz is more like it, which
would argue for an effective spring rate of
1000 / 16 = 62 lb/in.]

How do we derive the frequency formula? Let's work up a sequence of approximations in
stages. By improving the approximations gradually, we can check the more advanced
approximations for mistakes: they shouldn't be too far off the simple approximations. In
the first approximation, ignore the damper, giving us a mass block of sprung weight
resting on a spring. This model should act like a corner of a car with a broken shock.

Let the mass of the block be *m*. The force of gravitation acts downwards on the
block with a magnitude *mg*, where *g *= 32.1 ft/s^{2} is the
acceleration of Earth's gravity. The force of the spring acts upward on the mass with a
magnitude *k*( *y*_{0} - *y*), where *k* is the
spring constant and (*y* - *y*_{0}) is the HEIGHT of the
spring above its resting HEIGHT *y*_{0} (the force term is positive--that is,
upward--when *y* - *y*_{0} is negative--that is, when the mass
has compressed the spring and the spring pushes back upwards). We can avoid schlepping *y*_{0}
around our math by simply defining our coordinate system so that *y*_{0} = 0.
This sort of trick is very useful in all kinds of physics, even the most advanced.

It's worth noting that the model so far ignores not only the damper, but the weight of
the wheel and tyre and the spring itself. The weight of the wheel and tyre is called the *unsprung
weight*. The weight of the spring itself is partially sprung. We don't add these
effects in the current article. Today, we stop with just adding the damper back in, below.

Newton's first law guides us from this point on. The total force on the mass is *-ky - mg*.
The mass times the acceleration is *m*(*dv*_{y }/ *dt*)* = m*(*d*^{2}y / *dt*^{2}),
where *v*_{y} is the up-and-down velocity of the mass and *dv*_{y} / *dt*
is the rate of change of that velocity. That velocity is, in turn, the rate of change of
the *y* coordinate of the mass block, that is, *v*_{y} = (*dy* / *dt*).
So, the acceleration is the *second* rate of change of *y*, and we write it as *d*^{2}*y* / *dt*^{2}
because that's the way Newton and Leibniz first wrote it 350 years ago. We have the
following *dynamic equation* for the motion of our mass block.

Let's divide the entire equation by *m* and rearrange it so all the terms are on
the left:

If we're careful about units, in particular about *slugs* and lbs (see article 1),
then we can note that *k/m* has the dimensions of 1/sec^{2}, which is a
frequency squared. Let's define

yielding

We need to solve this equation for *y*(*t*) as a function of time *t*.
To follow the rest of this, you'll need to know a little freshman calculus. Take, as *ansatz*,

then

and

therefore

So, we see there are two solutions, and . In fact, the
time-dependent parts of these solutions can operate simultaneously, so we *must*
write in all generality. The values of the two unknowns *B*_{1} and *B*_{2}
are determined by two initial conditions, that is, the value of *y*_{0 }= *A* + *B*_{1} + *B*_{2}
and .

Let's get out of the complex domain by writing

This definition makes our initial conditions simpler, too:

It's easy, now, to add the damper. Damping forces are proportional to the velocity;
that is, there is no damping force when things aren't moving. Each corner approximately
obeys the equation

where is the damper response in lb - force / (ft / s).
The three rightmost terms represent forces, and they are all negative when *y* and *dy */ *dt*
are positive. That is, if you pull the sprung weight up, the spring tends to pull it down.
Likewise, if the sprung weight is moving up, the damper tends to pull it down. The force
of gravitation always pulls the weight down. Let's rewrite, as before:

where and . If, as before,

then

and

therefore

You may remember the little high-school formula for the solution
of a quadratic equation. This gives us the answer for *C*:

and I'll leave the simple arithmetic for *A* and the initial conditions to the
reader. The critically damped condition obtains when , overdamped when , and
underdamped when . In the underdamped condition, *C* has an imaginary
component and the exponentials oscillate. Otherwise, they just take one bounce and then
settle down.

It will be fun and easy for anyone who has followed along this far to plot out some
curves and check out my math. If you find a mistake, please do let me know (I just wrote
this off the top of my head, as I always do with these articles).

We could improve the approximation by writing down the coupled equations, that is,
treating all four corners of the car together, but that would just be a lot more math
without changing the basic physics that the car responds more predictably to smooth inputs
and less predictably to jerky inputs. Another improvement would be to add in the effect of
the unsprung and partially sprung weight.