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/s2
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/s2 is the acceleration of Earth's gravity. The force of the spring
acts upward on the mass with a magnitude k(y0 - y),
where k is the spring constant and (y - y0)
is the height of the spring above its resting height y0 (the
force term is positive--that is, upward--when y - y0
is negative--that is, when the mass has compressed the spring and the spring
pushes back upwards). We can avoid schlepping y0 around our
math by simply defining our coordinate system so that y0 = 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
(dvy / dt) = m
(d2y / dt2),
where vy is the up-and-down velocity of the mass and
dvy / 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,
vy = (dy / dt).
So, the acceleration is the second rate of change of y, and we
write it as d2y / dt2
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/sec2,
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 B1 and B2 are determined by two
initial conditions, that is, the value of y0 = A
+ B1 + B2
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.
, 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/s2
and that a pound weight is (1/32) slug. So, we're looking at
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 B1 and B2 are determined by two
initial conditions, that is, the value of y0 = A
+ B1 + B2
and 
.
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:
and 
.
If, as before,
for the
solution of a quadratic equation. This gives us the answer for C:
, 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.