Motorcycle Tire Basics by aihaozhe2


									This is the first in a series of articles exploring motorcycle tire basics and various
basic dynamic characteristics of the handling behavior of motorcycles. Overall this is
a very complex subject
and needs a good level of mathematics and physics to properly understand what's
However, in these articles I'll try and explain the basics with the absolute minimum of
but where this is unavoidable I'll not go beyond simple trigonometry. For those that
are unhappy
with any mathematics at all, don't worry, just skip those parts and the rest should still
prove useful.
I'll try and illustrate the mechanics with many sketches and graphs.

It seems incredible that just two small contact patches of rubber, can support our
machines and
manage to deliver large amounts of power to the road, whilst at the same time
supporting cornering
forces at least as much as the weight of the bike and rider. As such the tires exert
perhaps the single
most important influence over general handling characteristics, so it seems
appropriate to study their
characteristics before the other various aspects of chassis design.
When Newton first expounded to the world his theories of mechanics, no doubt he
had on his mind,
things other than the interaction of motorcycle tires with the road surface.
Never-the-less his
suppositions are equally valid for this situation. In particular his third law states, "For
every force there
is an equal and opposite force to resist it." or to put it another way "Action and
reaction are equal and

Relating this to tire action, means that when the tire is pushing on the road then the
road is pushing
back equally hard on the tire. This applies equally well regardless of whether we are
looking at
supporting the weight of the bike or resisting cornering, braking or driving loads.
What this particular law of Newton does not concern itself with, is which force is the
originating one nor
indeed does it matter for many purposes of analysis. However, as a guide to the
understanding of
some physical systems it is often useful to mentally separate the action from the
The forces that occur between the ground and the tires determine so much the
behaviour of our
machines, but they are so often taken for granted. tires really perform such a multitude
of different
tasks and their apparent simplicity hides the degree of engineering sophistication that
goes into their
design and fabrication. Initially pneumatic tires were fitted to improve comfort and
reduce loads on
the wheels. Even with modern suspension systems it is still the tires that provide the
first line of
defence for absorbing road shocks.
To explore carcass construction, tread compound and tread pattern in great detail is
beyond the scope
of this book. Rather we are concerned here with some basic principles and their
effects on handling

Weight Support

The most obvious function of the tire is to support the weight of the machine, whether
upright or
leaning over in a corner. However, the actual mechanism by which the air pressure
and tire passes
the wheel load to the road is often misunderstood. Consider fig. 1, this sketch
represents a slice
through the bottom of a rim and tire of unit thickness with an inflation pressure of P.
The left hand
side shows the wheel unloaded and the right hand side shows it supporting the weight
F. When
loaded the tire is compressed vertically and the width increases as shown, perhaps
surprisingly the
internal air pressure does not change significantly with load, the internal volume is
little changed.
At the widest section (X1) of the unloaded tire the internal half width is W1, and so
the force normal to
this section due to the internal pressure is simply 2.P.W1 . This force acts upwards
towards the wheel
rim, but as the pressure and tire width are evenly distributed around the circumference
the overall
effect is completely balanced. This force also has to be resisted by an equal tension (T)
in the tire

The loaded tire has a half width of W2 at it's widest section (X2) and so the normal
force is 2.P.W2 .
Therefore, the extra force over this section, when loaded, is 2.P.(W2 - W1) but as the
tire is only
widened over a small portion of the bottom part of the circumference, this force
supports the load F.
The above describes how the inflation pressure and tire width increase produce forces
to oppose the
vertical wheel loading, but does not completely explain the detail of the mechanism
by which these
forces are transferred to the rim. The bead of a fitted tire is an interference fit over the
bead seat of
the wheel rim, which puts this area into compression, the in-line component of the
side-wall tension
due to the inflation pressure reduces this compression somewhat. This component is
shown as F1 on
the unloaded half of F1 = T.cos(U1). The greater angle U2 of the side-wall when
loaded means
that the in-line component of the tension is reduced, thereby also restoring some of
the rim to tire
bead compression. This only happens in the lower part of the tire circumference,
where the widening
takes place. So there is a nett increase in the compressive force on the lower rim
acting upward, this
supports the bike weight. The nett force is the difference between the unloaded and
loaded in-line

F = T.(cos( U1) -cos(U2))

The left hand side shows half of an inflated but
unloaded tire, a tension (T) is created in the carcass by
the internal pressure. To the right, the compressed and
widened shape of the loaded tire is shown.

The in-line components (F1 & F2) of the side-wall
tension are reduced by factors equal to the cosines of
the angles of the side-wall. This reduction is greater
with the loaded tire resulting in a greater compressive
force on the lower part of the rim.
This is the principle but not the only mechanism which passes force from the wheel to
the ground, the
above ignores the effects of the flexure stiffness of the carcass itself, in addition to
supporting the
tension forces as outlined, the side-walls also have some bending resistance which can
resist small
wheel loads without any internal air pressure.

Suspension Action

In performing this function the pneumatic tire is the first object that feels any road
shocks and so acts
as the most important element in the machine's suspension system. To the extent that,
uncomfortable, it would be quite feasible to ride a bike around the roads, at reasonable
speeds with no
other form of bump absorption. In fact rear suspension was not at all common until
the 1940s or 50s.
Whereas, regardless of the sophistication of the conventional suspension system, it
would be quite
impractical to use wheels without pneumatic tires, or some other form of tire that
considerable bump deflection. The loads fed into the wheels without such tires would
be enormous at
all but slow speeds, and continual wheel failure would be the norm.
A few figures will illustrate what I mean:--Assume that a bike, with a normal size
front wheel, hits a 25
mm, sharp edged bump at 190 km/h. This not a large bump.
With no tire the wheel would then be subject to an average vertical acceleration of
1000 G. (the peak value would be higher than this). This means than if the wheel and
assembly had a mass of 25 kg. then the average point load on the rim would be 245
kN. or about 25
tons. What wheel could stand that? If the wheel was shod with a normal tire, then this
would have at
ground level, a spring rate, to a sharp edge, of approx. 17-35 N/mm. The maximum
force then
transmitted to the wheel for a 25 mm. step would be about 425-875 N. i.e. less than
four thousandths
of the previous figure, and this load would be more evenly spread around the rim.
Without the tire the
shock loads passed back to the sprung part of the bike would be much higher too. The
vertical wheel
velocity would be very much greater, and so the bump damping forces, which depend
on wheel
velocity, would be tremendous. These high forces would be transmitted directly back
to bike and rider.
The following five charts show some results of a computer simulation of accelerations
displacements on a typical road motorcycle, and illustrate the tire's significance to
comfort and road
holding. The bike is traveling at 100 km/h. and the front wheel hits a 0.025 metre high
step at 0.1
seconds. Note that the time scales vary from graph to graph.
Three cases are considered:

· With typical vertical tire stiffness and typical suspension springing and damping.
· With identical tire properties but with a suspension spring rate of 100 X that of the
· With tire stiffness 100 X the above and with normal suspension springing.

So basically we are considering a typical case, another case with almost no suspension
springing and
the final case is with a virtually rigid tire. Structural loading, comfort and roadholding
would all be adversely
affected without the initial cushioning of the tire. Note that the above charts are not all
to the same time scale,
this is simply to better illustrate the appropriate points.

This shows the vertical displacement of the front wheel. There is little difference
between the maximum
displacements for the two cases with a normal tire, for a small step the front tire
absorbs most of the shock. However,
in the case of a very stiff tire, the wheel movement is increased by a factor of about 10
times. It is obvious that the tire
leaves the ground in this case and the landing bounces can be seen after 0.5 seconds.

These curves show the vertical movement of the C of G of the bike and rider. As in
Fig 1 it is clear that the stiff tire
causes much higher bike movements, to the obvious detriment of comfort.

Demonstrating the different accelerations transmitted to the bike and rider, these
curves show the vertical
accelerations at the C of G. Both of the stiffer tire or stiffer suspension cases show
similar values of about 5 or 6 times
that of the normal case, but the shape of the two curves is quite different. With the
stiff suspension there is little
damping and we can see that it takes a few cycles to settle down. The second bump at
around 0.155 seconds is when the
rear wheel hits the step, this rear wheel response is not shown on the other graphs for

Front wheel vertical acceleration for the two cases with a normal tire. The early part is
similar for the two cases,
the suspension has little effect here, it is tire deflection that is the most important for
this height of step. As in Fig 5 the
lack of suspension damping allows the tire to bounce for a few cycles before settling

As in these curves are of the wheel acceleration, the values of the normal case are
overwhelmed by the stiff
tire case, with a peak value of close to 600 G compared with nearly 80 G normally.
Again note the effects of the landing
bounces after 0.5 seconds. This high acceleration would cause very high structural

As the tire is so good at removing most of the road shocks, right at the point of
application, perhaps it
would be worth while to consider designing it to absorb even more and eliminate the
need for other
suspension. Unfortunately we would run into other problems. We have all seen large
machinery bouncing down the road on their balloon tires, sometimes this gets so
violent that the
wheels actually leave the ground. A pneumatic tire acts just like an air spring, and the
rubber acts as
a damper when it flexes, but when the tire is made bigger the springing effect
overwhelms the
damping and we then get the uncontrolled bouncing. So there are practical restraints
to the amount of
cushioning that can be built into a tire for any given application.

Effects of Tire Pressure

Obviously, the springing characteristics mentioned above are largely affected by the
tire inflation
pressure, but there are other influences also. Carcass material and construction and the
and tread pattern of the outer layer of rubber all have an effect on both the springing
properties and
the area in contact with the ground (contact patch). Under and over inflation both
allow the tire to
assume non-optimum cross-sectional shapes, additionally the inflation pressure exerts
an influence
over the lateral flexibility of a tire and this is a property of the utmost importance to
stability. Manufacturers' recommendations should always be adhered to.
The influence of tire pressure on the vertical stiffness of an inflated tire, when loaded
a flat surface. These curves are from actual measured data. Note that the spring rate is
close to
linear over the full range of loading and varies from 14 kgf/mm. at 1.9 bar pressure to
19 kgf/mm. at
2.9 bar. The effective spring rate when the tire is loaded against a sharp edge, such as
a brick, is
considerably lower than this, and is more non-linear due to the changing shape of the
contact area as
the tire "wraps" around the object.

This spring rate acts in series with the suspension springs and is an important part of
the overall
suspension system. An interesting property of rubber is that when compressed and
released it
doesn't usually return exactly to it's original position, this is known as hysteresis. This
effect is shown
only for the 1.9 bar. case, the curve drawn during the loading phase is not followed
during the
unloading phase. The area between these two curves represents a loss of energy which
results in
tire heating and also acts as a form of suspension damping. In this particular case the
energy lost
over one loading and unloading cycle is approximately 10% of the total stored energy
in the
compressed tire, and is a significant parameter controlling tire bounce.

Vertical stiffness of a standard road tire against a flat surface at different inflation
pressures. This data is from an
Avon Azaro Sport II 170/60 ZR17. The upward arrows indicate the compression of
the tire and the 2nd line with the
downward arrow (shown only at 1.9 bar for clarity) shows the behaviour of the tire
when the load is released. The
shaded area between the two lines represents a loss of energy called hysteresis. This
acts as a source of suspension
damping and also heats the tire. (From data supplied by Avon tires.)

The lateral stiffness of the same tire measured at two different pressures. In both cases
the tire was loaded vertically with it's maximum rated capacity of 355 kgf. The lateral
spring rate is
less than half that of the vertical rate at 7.7 and 7.3 kgf/mm. at 2.9 and 2.5 bar
respectively. It is
interesting to note that at the higher pressure the tire saturates or loses adhesion at the
lower figure
of 460 kgf. compared to 490 kgf. at the lower pressure. Saturation is indicated when
the curve more
or less becomes horizontal, this is when the tire cannot support an more lateral force
and it displaces
or slides sideways, with an approximately constant force. The contact patch area and
produced at the lower air pressure has allowed more static grip. However, these tests
are done with
the artificial case of an upright and non-rotating wheel and hence it would be risky to
extrapolate this
grip characteristic to a moving machine. Although not shown, the lateral deformation
would also be
subject to some hysteresis and this damping and the lateral flexibility exert an
important influence over
the weave stability.

Lateral stiffness of the same tire shown in fig. 9. The vertical load was constant at 355
kgf. and the wheel was
kept vertical. As expected the tire is somewhat stiffer with the higher inflation
pressure but loses grip or saturates at the
lower lateral load of 460 kgf. compared to 490 kgf. at the lower pressure. (From data
supplied by Avon tires.)

Contact Area

The tire must ultimately give it's support to the bike through a small area of rubber in
contact with the
ground, and so "contact patch area = vertical force ÷ average contact patch surface
pressure". This
applies under ALL conditions.

The contact patch surface pressure is NOT however, the same as the inflation pressure,
as is
sometimes claimed. They are related but there are at least four factors which modify
the relationship.
Carcass stiffness, carcass shape, surface rubber depth and softness, and road surface
compliance. If
we have an extremely high carcass stiffness then inflation pressure will have a
reduced influence.
Let's look at this in a little more detail and see why:

If a tire was made just like an inner tube, that is from quite thin rubber and with little
stiffness unless
inflated, then the internal air pressure would be the only means to support the bike's
weight. In this
case the contact patch pressure would be equal to that of the internal air pressure. For
an air
pressure of 2 bar and a vertical load of 1.0 kN. Then the contact area would be 5003 If we
now increased the air pressure to say 3 bar the area would fall to 3335

Let's now imagine that we substitute a rigid steel tubular hoop for our rim and tire, the
area in contact
with the ground will be quite small. If we now inflate the hoop with some air pressure,
it doesn't take
much imagination to see that, unlike the inner tube, this internal pressure will have a
negligible effect
on the external area of contact. Obviously, a tire is not exactly like the steel hoop, nor
the inner tube,
but this does show that the carcass rigidity can reduce the contact surface area as
calculated purely
from inflation pressure alone.

On the other hand, let's now imagine that we cover the previous steel hoop with a
layer of thick soft
rubber. Now, the actual contact area will be considerably increased and the average
contact patch
pressure will be reduced. Substitute this mental picture back to a real tire and we see
that the tread
layer of rubber will give us a greater contact area and lower contact pressure than that
of the inflation
pressure alone. It is this compliance of the surface rubber that gives us more contact
with the road
when we increase tire width and diameter, but this must be balanced against the
opposing effects of
the carcass stiffness. Radial and bias or cross ply tires exhibit quite different
characteristics in this

The properties of the road surface are also important, a soft surface, mud and sand for
example, will
obviously give support over a wider area of the tire and so reduce the contact pressure.
On a hot day
with softened tarmac, even a normal road will deflect significantly enough to affect
the contact patch.
To get a feeling for the degree of departure of the contact patch pressure from the
inflation pressure,
consider a completely flat tire, in this case the rubber area will probably be no more
than 3 or 4 times,
at most, the area when inflated correctly. Based on the notion that rubber pressure =
pressure, we would expect the rubber area to be much higher, infinite in fact. Another
extreme case
to consider: Imagine a knobbly tire with very few knobs such that only one knob
supports the bike. In
this example the rubber pressure is simply the weight divided by the area of the one
knob, this is
regardless of the inflation pressure. These are extreme examples of course but should
demonstrate the lie in the proposition that rubber contact pressure = inflation pressure.
The following
describes some simple measurements that I made to check out the actual relationship
between load,
inflation pressure and contact area.

I did 2 sets of tests. For the first I kept the tire inflation pressure constant at 2.4 bar
and varied the tire
load between 178 and 1210 N. (allowing for the weight of the glass and wooden
beams). Secondly, I
keep a constant load of 1210 N. and tried varying the inflation pressure between 2.4 to
1 bar.
Even with a generous allowance for experimental error the effects are clear. The
graphs show that
the results appeared to fit reasonably well to a smooth line, there wasn't much scatter.

Point (1) on the curve with constant inflation pressure, shows how the actual contact
patch pressure is
lower (just over half) than the inflation pressure, or in other words the contact area is
greater. This is
due to the rubber surface compliance, thus this is more important at low vertical loads,
carcass stiffness became more important as the load rose as shown by points (3) to (6)
where the
actual contact pressure is higher than the air pressure, i.e. reduced area of contact.

Measurement setup. Various weights were placed on the end of a beam, which also
loaded the tire via a
thick plate of glass. The beam was arranged to apply the load to the tire with a 4:1
leverage. So a 25
kgf. weight would load the tire with 100 kgf. By tracing over the glass the contact
was determined.

Tracings of tire footprint for different loads and pressures. The numbers relate to the
data points below.

The top plot shows the measured contact patch pressure at various wheel loads for a
constant inflation pressure
of 2.4 bar. The lower curves show the contact pressure at various inflation pressures
for a fixed load of 1210 N. The
numbers at the data points correspond with the contact area tracings in the previous
sketch. The plain line on each plot
shows the case of the contact patch pressure being equal to the inflation pressure.

The carcass stiffness helps to support the machine as the air pressure is
reduced, the contact patch pressure being considerably higher than the inflation
pressure. It looks as
though the two lines will cross at an air pressure of about 3.5 bar. (although this was
not tested by
measurement), at which point the surface rubber compression will assume the greatest
This is as per the steel hoop analogy above.

We can easily see the two separate effects of surface compliance and carcass stiffness
and how the
relative importance of these varies with load and/or inflation pressure.

These tests were only done with one particular tire, other types will show different
detail results but
the overall effects should follow a similar pattern.

Area Under Cornering

Does cornering affect tire contact area?
Let's assume a horizontal surface and lateral acceleration of 1G. Under these
conditions the bike/rider
CoG will be on a line at 45° to the horizontal and passing through the contact patch.
There will a
resultant force acting along this line through the contact patch of 1.4 times the
supported weight.

This force is the resultant of the supported weight and the cornering force, which have
the same
magnitude, in this example of a 45° lean. The force normal to the surface is simply
that due to the
supported weight and does NOT vary with cornering force. The cornering force is
reacted by the
horizontal frictional force generated by the tire/road surface and this frictional force is
"allowed" by
virtue of the normal force.

Therefore, to a first approximation cornering force will NOT affect the tire contact
area, and in fact this
case could be approximated to, if we were just considering the inner tube without a
real world tire.
However in reality, the lateral force will cause some additional tire distortion to take
place at the
road/tire interface and depending on the tire characteristics, mentioned above, the
contact area may
well change.

Another aspect to this is of course the tire cross-sectional profile. The old Dunlop
triangular racing
tire, for example, was designed to put more rubber on the road when leant over, so
even without tire
distortion the contact patch area increased, simply by virtue of the lean angle.

Next month we'll look at other aspects of tires, such as friction, grip, drifting, under-
and over-steer,
and tire construction and materials.

by Tony Foale


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