Most autocrossers and race drivers learn early in their careers the
importance of balancing a car. Learning to do it consistently and automatically
is one essential part of becoming a truly good driver. While the skills for
balancing a car are commonly taught in drivers' schools, the rationale behind
them is not usually adequately explained. That rationale comes from simple
physics. Understanding the physics of driving not only helps one be a better
driver, but increases one's enjoyment of driving as well. If you know the deep
reasons why you ought to do certain things you will remember the things better
and move faster toward complete internalisation of the skills.
Balancing a car is controlling weight transfer using throttle, brakes, and
steering. This article explains the physics of weight transfer. You will often
hear instructors and drivers say that applying the brakes shifts weight to the
front of a car and can induce oversteer. Likewise, accelerating shifts weight to
the rear, inducing understeer, and cornering shifts weight to the opposite side,
unloading the inside tyres. But why does weight shift during these manoeuvres?
How can weight shift when everything is in the car bolted in and strapped down?
Briefly, the reason is that inertia acts through the centre of gravity (CG) of
the car, which is above the ground, but adhesive forces act at ground level
through the tyre contact patches. The effects of weight transfer are
proportional to the height of the CG off the ground. A flatter car, one with a
lower CG, handles better and quicker because weight transfer is not so drastic
as it is in a high car.
The rest of this article explains how inertia and adhesive forces give rise
to weight transfer through Newton's laws. The article begins with the elements
and works up to some simple equations that you can use to calculate weight
transfer in any car knowing only the wheelbase, the height of the CG, the static
weight distribution, and the track, or distance between the tyres across the
car. These numbers are reported in shop manuals and most journalistic reviews of
cars.
Most people remember Newton's laws from school physics. These are fundamental
laws that apply to all large things in the universe, such as cars. In the
context of our racing application, they are:
The first law: a car in straight-line motion at a constant speed will keep
such motion until acted on by an external force. The only reason a car in
neutral will not coast forever is that friction, an external force, gradually
slows the car down. Friction comes from the tyres on the ground and the air
flowing over the car. The tendency of a car to keep moving the way it is moving
is the inertia of the car, and this tendency is concentrated at the CG point.
The second law: When a force is applied to a car, the change in motion is
proportional to the force divided by the mass of the car. This law is
expressed by the famous equation F = ma, where F is
a force, m is the mass of the car, and a is the
acceleration, or change in motion, of the car. A larger force causes quicker
changes in motion, and a heavier car reacts more slowly to forces. Newton's
second law explains why quick cars are powerful and lightweight. The more
F and the less m you have, the more a you
can get.
The third law: Every force on a car by another object, such as the ground,
is matched by an equal and opposite force on the object by the car. When you
apply the brakes, you cause the tyres to push forward against the ground, and
the ground pushes back. As long as the tyres stay on the car, the ground pushing
on them slows the car down.
Let us continue analysing braking. Weight transfer during accelerating and
cornering are mere variations on the theme. We won't consider subtleties such as
suspension and tyre deflection yet. These effects are very important, but
secondary. The figure shows a car and the forces on it during a "one g" braking
manoeuvre. One g means that the total braking force equals the weight of the
car, say, in pounds.
In this figure, the black and white "pie plate" in the centre is the CG.
G is the force of gravity that pulls the car toward the centre of the
Earth. This is the weight of the car; weight is just another word for the force
of gravity. It is a fact of Nature, only fully explained by Albert Einstein,
that gravitational forces act through the CG of an object, just like inertia.
This fact can be explained at deeper levels, but such an explanation would take
us too far off the subject of weight transfer.
Lf is the lift force exerted by the ground on the
front tyre, and Lr is the lift force on the rear
tyre. These lift forces are as real as the ones that keep an airplane in the
air, and they keep the car from falling through the ground to the centre of the
Earth.
We don't often notice the forces that the ground exerts on objects because
they are so ordinary, but they are at the essence of car dynamics. The reason is
that the magnitude of these forces determine the ability of a tyre to stick, and
imbalances between the front and rear lift forces account for understeer and
oversteer. The figure only shows forces on the car, not forces on the ground and
the CG of the Earth. Newton's third law requires that these equal and opposite
forces exist, but we are only concerned about how the ground and the Earth's
gravity affect the car.
If the car were standing still or coasting, and its weight distribution were
50-50, then Lf would be the same as Lr.
It is always the case that Lf plus Lr
equals G, the weight of the car. Why? Because of Newton's first
law. The car is not changing its motion in the vertical direction, at least as
long as it doesn't get airborne, so the total sum of all forces in the vertical
direction must be zero. G points down and counteracts the sum of
Lf and Lr, which point
up.
Braking causes Lf to be greater than Lr.
Literally, the "rear end gets light," as one often hears racers say. Consider
the front and rear braking forces, Bf and Br,
in the diagram. They push backwards on the tyres, which push on the wheels,
which push on the suspension parts, which push on the rest of the car, slowing
it down. But these forces are acting at ground level, not at the level of the
CG. The braking forces are indirectly slowing down the car by pushing at ground
level, while the inertia of the car is 'trying' to keep it moving forward as a
unit at the CG level.
The braking forces create a rotating tendency, or torque, about the CG.
Imagine pulling a table cloth out from under some glasses and candelabra. These
objects would have a tendency to tip or rotate over, and the tendency is greater
for taller objects and is greater the harder you pull on the cloth. The
rotational tendency of a car under braking is due to identical physics.
The braking torque acts in such a way as to put the car up on its nose. Since
the car does not actually go up on its nose (we hope), some other forces must be
counteracting that tendency, by Newton's first law. G cannot be
doing it since it passes right through the centre of gravity. The only forces
that can counteract that tendency are the lift forces, and the only way they can
do so is for Lf to become greater than Lr.
Literally, the ground pushes up harder on the front tyres during braking to try
to keep the car from tipping forward.
By how much does Lf exceed Lr?
The braking torque is proportional to the sum of the braking forces and to the
height of the CG. Let's say that height is 20 inches. The counterbalancing
torque resisting the braking torque is proportional to Lf
and half the wheelbase (in a car with 50-50 weight distribution), minus Lr
times half the wheelbase since Lr is helping the
braking forces upend the car. Lf has a lot of work to
do: it must resist the torques of both the braking forces and the lift on the
rear tyres. Let's say the wheelbase is 100 inches. Since we are braking at one
g, the braking forces equal G, say, 3200 pounds. All this is
summarized in the following equations:
3200 lbs times 20 inches = Lf times 50 inches - Lr times 50 inches
Lf + Lr = 3200 lbs (this is always true)
With the help of a little algebra, we can find out that
Thus, by braking at one g in our example car, we add 640 pounds of load to
the front tyres and take 640 pounds off the rears! This is very pronounced
weight transfer.
By doing a similar analysis for a more general car with CG height of h,
wheelbase w, weight G, static weight distribution
d expressed as a fraction of weight in the front, and braking with
force B, we can show that
Lf = dG + Bh / w, Lr = (1 - d)G - Bh / w
These equations can be used to calculate weight transfer during acceleration
by treating acceleration force as negative braking force. If you have
acceleration figures in gees, say from a G-analyst or other device,
just multiply them by the weight of the car to get acceleration forces (Newton's
second law!). Weight transfer during cornering can be analysed in a similar way,
where the track of the car replaces the wheelbase and d is always
50% (unless you account for the weight of the driver). Those of you with science
or engineering backgrounds may enjoy deriving these equations for yourselves.
The equations for a car doing a combination of braking and cornering, as in a
trail braking manoeuvre, are much more complicated and require some mathematical
tricks to derive.
Now you know why weight transfer happens. The next topic that comes to mind
is the physics of tyre adhesion, which explains how weight transfer can lead to
understeer and oversteer conditions.
