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 happening.
However, in these articles I'll try and explain the basics with the absolute minimum of mathematics,
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
opposite."
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 reaction.
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
characteristics.
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
carcass.
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
forces,
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.
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, whilst
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 allowed
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 approximately
1000 G. (the peak value would be higher than this). This means than if the wheel and brake
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 and
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 BMW Fairing Kits S1000RR 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 previous.
· 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,