Part 14: Why Smoothness?

**Brian Beckman PhD©Copyright April 2000**

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; there is no Part 13.

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 }*/

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.