In the last two
articles, we plunged right into some relatively complex
issues, namely weight transfer and tire adhesion. This month,
we regroup and review some of the basic units and dimensions
needed to do dynamical calculations. Eventually, we can work
up to equations sufficient for a fullblown computer
simulation of car dynamics. The equations can then be
`doctored' so that the computer simulation will run fast
enough to be the core of an autocross computer game.
Eventually, we might direct this series of articles to show
how to build such a game in a typical microcomputer
programming language such as C or BASIC, or perhaps even my
personal favorite, LISP. All of this is in keeping with the
spirit of the series, the Physics of Racing, because so much
of physics today involves computing. Software design and
programming are essential skills of the modern physicist, so
much so that many of us become involved in computing full
time.
Physics is the science of
measurement. Perhaps you have heard of highly abstract
branches of physics such as quantum mechanics and relativity,
in which exotic mathematics is in the forefront. But when
theories are taken to the laboratory (or the race course) for
testing, all the mathematics must boil down to quantities that
can be measured. In racing, the fundamental quantities are
distance, time, and mass. This month, we will review basic
equations that will enable you to do quick calculations in
your head while cooling off between runs. It is very valuable
to develop a skill for estimating quantities quickly, and I
will show you how.
Equations that don't involve
mass are called kinematic. The first kinematic
equation relates speed, time, and distance. If a car is moving
at a constant speed or velocity, ,
then the distance
it travels in time
is
or velocity times time. This equation really expresses nothing
more than the definition of velocity.
If we are to do mental
calculations, the first hurdle we must jump comes from the
fact that we usually measure speed in miles per hour (mph),
but distance in feet and time in seconds. So, we must modify
our equation with a conversion factor, like this
If you ``cancel out'' the
units parts of this equation, you will see that you get feet
on both the left and right hand sides, as is appropriate,
since equality is required of any equation. The conversion
factor is 5280/3600, which happens to equal 22/15. Let's do a
few quick examples. How far does a car go in one second
(remember, say, ``oneonethousand, twoonethousand,'' etc.
to yourself to count off seconds)? At fifteen mph, we can see
that we go
or about 1 and a half car lengths for a 14 and 2/3 foot car
like a latemodel Corvette. So, at 30 mph, a second is three
car lengths and at 60 mph it is six. If you lose an autocross
by 1 second (and you'll be pretty good if you can do that with
all the good drivers in our region), you're losing by
somewhere between 3 and 6 car lengths! This is because the
average speed in an autocross is between 30 and 60 mph.
Everytime you plow a little
or get a little sideways, just visualize your competition
overtaking you by a car length or so. One of the reasons
autocross is such a difficult sport, but also such a pure
sport, from the driver's standpoint, is that you can't make up
this time. If you blow a corner in a road race, you may have a
few laps in which to make it up. But to win an autocross
against good competition, you must drive nearly perfectly. The
driver who makes the fewest mistakes usually wins!
The next kinematic equation
involves acceleration. It so happens that the distance covered
by a car at constant acceleration from a standing start is
given by
or 1/2 times the acceleration times the time, squared. What
conversions will help us do mental calculations with this
equation? Usually, we like to measure acceleration in s.
One
happens to be 32.1 feet per second squared. Fortunately, we
don't have to deal with miles and hours here, so our equation
becomes,
roughly. So, a car accelerating from a standing start at ,
which is a typical number for a good, stock sports car, will
go 8 feet in 1 second. Not very far! However, this picks up
rapidly. In two seconds, the car will go 32 feet, or over two
car lengths.
Just to prove to you that
this isn't crazy, let's answer the question ``How long will it
take a car accelerating at
to do the quarter mile?'' We invert the equation above (recall
your high school algebra), to get
and we plug in the numbers: the quarter mile equals 1320 feet,
,
and we get
which is about 13 seconds. Not too unreasonable! A real car
will not be able to keep up full
acceleration for a quarter mile due to air resistance and
reduced torque in the higher gears. This explains why real
(stock) sports cars do the quarter mile in 14 or 15 seconds.
The more interesting result
is the fact that it takes a full second to go the first 8
feet. So, we can see that the launch is critical in an
autocross. With excessive wheelspin, which robs you of
acceleration, you can lose a whole second right at the start.
Just visualize your competition pulling 8 feet ahead
instantly, and that margin grows because they are `hooked up'
better.
For doing these mental
calculations, it is helpful to memorize a few squares. 8
squared is 64, 10 squared is 100, 11 squared is 121, 12
squared is 144, 13 squared is 169, and so on. You can then
estimate square roots in your head with acceptable precision.
Finally, let's examine how
engine torque becomes force at the drive wheels and finally
acceleration. For this examination, we will need to know the
mass of the car. Any equation in physics that involves mass is
called dynamic, as opposed to kinematic. Let's say we
have a Corvette that weighs 3200 pounds and produces 330
footpounds of torque at the crankshaft. The Corvette's
automatic transmission has a first gear ratio of 3.06 (the
auto is the trick set up for vettesjust ask Roger Johnson or
Mark Thornton). A transmission is nothing but a set of
circular, rotating levers, and the gear ratio is the leverage,
multiplying the torque of the engine. So, at the output of the
transmission, we have
of torque. The differential is a further levermultiplier, in
the case of the Corvette by a factor of 3.07, yielding 3100
foot pounds at the center of the rear wheels (this is a lot of
torque!). The distance from the center of the wheel to the
ground is about 13 inches, or 1.08 feet, so the maximum force
that the engine can put to the ground in a rearward direction
(causing the ground to push back forwardremember part 1 of
this series!) in first gear is
Now, at rest, the car has about 50/50 weight distribution, so
there is about 1600 pounds of load on the rear tires. You will
remember from last month's article on tire adhesion that the
tires cannot respond with a forward force much greater than
the weight that is on them, so they simply will spin if you
stomp on the throttle, asking them to give you 2870 pounds of
force.
We can now see why it is
important to squeeeeeeeze the throttle gently when launching.
In the very first instant of a launch, your goal as a driver
is to get the engine up to where it is pushing on the tire
contact patch at about 1600 pounds. The tires will squeal or
hiss just a little when you get this right. Not so
coincidentally, this will give you a forward force of about
1600 pounds, for an
(part 1) acceleration of about ,
or half the weight of the car. The main reason a car will
accelerate with only
to start with is that half of the weight is on the front
wheels and is unavailable to increase the stiction of the
rear, driving tires. Immediately, however, there will be some
weight transfer to the rear. Remembering part 1 of this series
again, you can estimate that about 320 pounds will be
transferred to the rear immediately. You can now ask the tires
to give you a little more, and you can gently push on the
throttle. Within a second or so, you can be at full throttle,
putting all that torque to work for a beautiful hole shot!
In a rear drive car, weight
transfer acts to make the driving wheels capable of
withstanding greater forward loads. In a front drive car,
weight transfer works against acceleration, so you have to be
even more gentle on the throttle if you have a lot of power.
An allwheel drive car puts all the wheels to work delivering
force to the ground and is theoretically the best.
Technical people call this
style of calculating ``back of the envelope,'' which is a
somewhat picturesque reference to the habit we have of writing
equations and numbers on any piece of paper that happens to be
handy. You do it without calculators or slide rules or
abacuses. You do it in the garage or the pits. It is not
exactly precise, but gives you a rough idea, say within 10 or
20 percent, of the forces and accelerations at work. And now
you know how to do backoftheenvelope calculations, too.
