It has more grip...it has less grip
on the wet road…they are soft tires, medium, are not they hard tires?.......these are terms frequently used to define the
behavior and the quality of the tires.
But what do they mean? What does it mean to ask how much
grip the tire has? How much is it soft or hard?
What does it mean to define
soft a tire?.....they are terms better used to speak about the beef!
It would be the same if you comparing different
engines say: it is more powerful or it is less powerful, it has more torque or
less torque, without to know the torque and power concepts and without to
associate the numerical values expressed in kW and Nm to those contents.
We clear up it giving the answers to some easy and
intuitive questions.
What’s
the tire task?
The answer is easy, the tire has to perform two tasks:
- allowing the transfer of the
drive force or brake force on the ground;
- generating the necessary lateral forces to maintain the
motorcycle equilibrium in the curve or along a curvilinear path as for example
those generated to avoid an obstacle, to carry out a
line change or to run on a S shape curve.
Right now we try to understand how the
lateral force is generated and how much it has to be when the motorcycle is
running on steady turning condition with constant speed (stable condition) and
which parameter the force depends from.
It is easy to understand that lateral force depends on vertical load
applied on the wheel. High vertical load makes high lateral force.
Understanding the dependence of lateral force upon the camber angle and tire sideslip
(tire side-slip expressed by side-slip angle which will be defined later) is
less intuitive.
The lateral force also depends on two more parameters
which the bikers know very well: the tire pressure and the tire temperature in the work condition.
First we consider the
camber angle effect.
In vertical position the tire imprint is elliptical
and symmetric; the tire imprint showed in figure 1 is painted with grey tones
whose intensity is proportional to the pressure between the tire and the
ground.
Figure 1. Vertical and tilted position tire imprints. The gray tone
intensity is proportional to the pressure between the tire and the ground.
As you can
observe in figure 1, when the wheel is tilted, the rubber particle that through
the imprint doesn’t follow the path that it would follow if there
wasn’t the contact between tire and ground. Because of there is contact
with the ground the particle has to follow a different path, therefore the
ground contact cause a deformation of tire carcass;
this deformation generates a lateral force that increase when the camber angle
increase.
Understanding better that phenomenon it is helpful to
think about a shape-retaining tire, as if it was metallic; in that case the
contact patch becomes like a point, there is not carcass deformation, hence the
force due to the camber is null.
The camber force depends on shape and dimension of
contact patch.
Contact patch depends on tire’s geometric
characteristics (rolling radius and cross section
radius) and carcass’ lateral/radial stiffness.
Figure 2. Lateral and radial stiffness depend
on pneumatic kind (front or rear), geometry, carcass constructive
characteristics (radial, cross), pressure and normal load.
Right now we consider the lateral slip effect that is
the side-slip
angle.
The side-slip angle is the angle between the forward
direction and the central plane of the wheel.
The contact patch is asymmetrical when the there is
lateral slip. In the first part of contact patch the rubber particles tend to
follow speed direction but, since the speed direction doesn’t coincide to
the wheel plane, the particles located inside contact patch are deformed with
respect to tire carcass. This is the contact patch part with adherence. When
the deformation is al little more, the elastic recall forces due to the
deformation of the rubber are greater then the adherence force so the particles
start to slider. This is the contact patch part with slide. The integral of
contact patch pressure give the lateral force due to the side-slip.
Figure 3. Side-slip force
origin. The lateral pressure resultant is the lateral force due to the
side-slip angle.
We saw the lateral force depends both roll angle and lateral slide of the tire; we
expressed the tire lateral slip by means of side-slip angle.
Usually the lateral force is plotted against side-slip
angle for various roll angles as represented in figure 4 for a race front tire.
Figure 4. Lateral force against side-slip
angle for various roll angles.
This kind of representation is used in the car field
because the roll angle is very small and the tires make the required lateral
forces only by the lateral slip. Otherwise the motorcycle tires work first by
the roll and second by lateral slip, at least for the production of the lateral
force. That’s why for the motorcycle tires is
better to represent the lateral force against roll angle for various side-slip
angles as showed in the figure 5.
Figure 5. Lateral force against roll angle
for various side-slip angles.
The lateral force can be expressed analytically by
means of a linear function of roll angle and side-slip angle, thinking the
force as sum of two independent components: the roll component and the
side-slip component.
Force lateral = Kroll * j + K
side-slip * l
The K
constants have a geometric meaning too; they represent the tangents of the
curves respectively the normalized lateral force against roll angle with
side-slip angle null and the lateral force against side-slip angle with roll
angle null.
Figure 6. Roll and
side-slip stiffness geometric meaning. Stiffness’ are expressed in
1/rad.
To understand how much is the
lateral force we consider a motorcycle running on
steady turning condition with constant speed.
Neglecting the gyroscopic effects generated by the
wheels during steady turning condition and the
fact that cross-section of tires is considerable (the tire cross diameter can
be greater than 100 mm for rear tires) the lateral force required to allow the
equilibrium is obtained by multiplication of the vertical load by the tangent
of the roll angle.
Force lateral = load vertical * tang (j)
Using his hypothesis is like to consider a virtual
motorcycle with wheels having a very small inertia moment (magnesium or carbon
fiber wheels) with a very small cross section like the race bicycle tubular
wheels.
For instance when the roll angle is equal to 45° the
lateral force is equal exactly to the vertical load. The equilibrium condition
is represented in figure 6 that shows that the centrifugal force is equilibrated
exactly by the sum of the two lateral forces generated by the two tires.
.
Figure 7. Motorcycle
equilibrium in the curve. “R” indicate
the curvature radius of the curve, “m” the mass of the vehicle with
driver and “V” the forward speed.
Figure 8 shows lateral forces against roll angle for
two tires, front and rear, related to the respective vertical loads. The figure
shows also the trend of the following function:
tang (j )=Force
lateral / Load vertical = V2/ g*R
that represents
the equilibrium condition in the curve.
The lateral force generated by roll, related to the
vertical load, can be greater or smaller than the force required in the
equilibrium condition represented by the value of the tangent of the roll
angle.
In the first case, with insufficient roll force, you
need a lateral slip that is a positive side-slip angle that makes the part of
the force missing to the equilibrium; in the second case, instead, the
side-slip angle has to be negative to decrease the lateral force due to the
only roll that is greater than the force in the equilibrium condition.
Figure 8. Example of front
and rear lateral forces versus roll angle. The dotted lines represented
the lateral force required for the equilibrium in the curve.
We consider the tire whose characteristics are
represented in figure 7 and hypothesize the bike is running in the curve with
roll angle 40°.
For the rear tire the roll force is insufficient to
make the equilibrium. In fact the lateral force with side-slip angle null is
about 0.73 (point A) while the required force is 0.85. The rear tire therefore
will slip laterally out side and the side-slip angle will be such to increase the
lateral force up to the value for the equilibrium (point B).
In the equilibrium condition for the front tire case
you will have instead a negative side-slip angle because, for roll angle equal
to the 40° degrees, the lateral force only due to the roll (point B) is greater
then the force required for the equilibrium (point A).
Figure 9. Top view of a
motorcycle in the curve.
It would be possible to use a poor quality front tire
and a good quality rear tire. In substance it would be the case of large front
tire side-slip and lightly negative or null rear tire side-slip. In this case
the motorcycle behavior would be much under steering. The under steering
motorcycle behavior could be dangerous because, if it would will be necessary
to force over on the bend, it would be required to the front tire a larger
lateral force that is a larger side-slip angle.
What would happen if the tire would not able to
generate this larger force?
The front tire would slide and the fall would be sure.
The tire problem is not all here.
For instance, how do temperature and pressure
influence roll and side-slip force?
The forces depend on vertical load in proportional or
not proportional mode?
How much is the lateral force when it is required a
braking or driving longitudinal force?
Why changing the tires you feel a different driving
even if you feel a grip feeling?
What is the correlation between bike’s handling
and tire’s characteristics?
The questions are very many; we will give scientific
answers on the next articles.
In the meanwhile we will start to
asked to the tire companies the values of the forces generated for various roll
angle, side-slip angle, temperature, pressure…..
Maybe is better to talk with numbers……
soft rubber, hard, more grip…..they only are void words.