What is Tire Used

8/6/2019 What is Tire Used

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Whats the tire task?

The answer is easy, the tire has to perform two tasks:

- allowing the transfer ofthe drive force or brake forceon the ground;

- generating the necessary lateral forces to maintain the motorcycle equilibrium in the curve oralong a curvilinear path as for example those generated to avoid an obstacle, to carry out a linechange 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 bewhen 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. Highvertical load makes high lateral force. Understanding the dependence of lateral force upon thecamber angle and tire sideslip (tire side-slip expressed by side-slip angle which will be definedlater) is less intuitive.

The lateral force also depends on two more parameters which the bikers know very well: thetirepressure and thetire temperaturein the work condition.

First we considerthe camber angle effect.

In vertical position the tire imprint is elliptical and symmetric; the tire imprint showed in figure 1 ispainted with grey tones whose intensity is proportional to the pressure between the tire and theground.


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Figure 1. Vertical and tilted position tire imprints. The gray tone intensity is proportional to thepressure 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 doesnt follow the path that it would follow if there wasnt the contact between tire andground. 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 alateral force that increase when the camber angle increase.

Understanding better that phenomenon it is helpful to think about a shape-retaining tire, as if itwas metallic; in that case the contact patch becomes like a point, there is not carcassdeformation, hence the force due to the camber is null.

The camber force depends on shape and dimension of contact patch.

Contact patch depends on tires geometric characteristics (rolling radius and cross section radius)and carcass lateral/radial stiffness.


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Figure 2. Lateral and radial stiffness depend on pneumatic kind (front or rear), geometry, carcassconstructive 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 patchthe rubber particles tend to follow speed direction but, since the speed direction doesnt coincideto the wheel plane, the particles located inside contact patch are deformed with respect to tirecarcass. This is the contact patch part with adherence. When the deformation is al little more, theelastic 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 patchpressure give the lateral force due to the side-slip.


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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 thetire lateral slip by means of side-slip angle.

Usually the lateral force is plotted against side-slip angle for various roll angles as represented infigure 4 for a race front tire.


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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 thetires make the required lateral forces only by the lateral slip. Otherwise the motorcycle tires workfirst by the roll and second by lateral slip, at least for the production of the lateral force. Thatswhy for the motorcycle tires is better to represent the lateral force against roll angle for variousside-slip angles as showed in the figure 5.


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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 andside-slip angle, thinking the force as sum of two independent components: the roll componentand the side-slip component.

Force lateral = Kroll * + Kside-slip *

The K constants have a geometric meaning too; they represent the tangents of the curvesrespectively the normalized lateral force against roll angle with side-slip angle null and the lateralforce against side-slip angle with roll angle null.


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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 turningcondition with constant speed.

Neglecting the gyroscopic effects generated by the wheels during steady turning condition andthe fact that cross-section of tires is considerable (the tire cross diameter can be greater than 100mm for rear tires) the lateral force required to allow the equilibrium is obtained by multiplication ofthe vertical load by the tangent of the roll angle.


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Force lateral = load vertical * tang ()

Using his hypothesis is like to consider a virtual motorcycle with wheels having a very smallinertia moment (magnesium or carbon fiber wheels) with a very small cross section like the racebicycle tubular wheels.

For instance when the roll angle is equal to 45 the lateral force is equal exactly to the verticalload. The equilibrium condition is represented in figure 6 that shows that the centrifugal force isequilibrated exactly by the sum of the two lateral forces generated by the two tires.

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Figure 7. Motorcycle equilibrium in the curve. R indicate the curvature radius of the curve, mthe mass of the vehicle with driver and V the forward speed.


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Figure 8 shows lateral forces against roll angle for two tires, front and rear, related to therespective vertical loads. The figure shows also the trend of the following function:

tang ( )=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 theforce required in the equilibrium condition represented by the value of the tangent of the rollangle.

In the first case, with insufficient roll force, you need a lateral slip that is a positive side-slip anglethat 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 thanthe force in the equilibrium condition.


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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 isrunning 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 withside-slip angle null is about 0.73 (point A) while the required force is 0.85. The rear tire therefore


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will slip laterally out side and the side-slip angle will be such to increase the lateral force up to thevalue 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) isgreater then the force required for the equilibrium (point A).

Figure 9. Top view of a motorcycle in the curve.


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It would be possible to use a poor quality front tire and a good quality rear tire. In substance itwould be the case of large front tire side-slip and lightly negative or null rear tire side-slip. In thiscase the motorcycle behavior would be much under steering. The under steering motorcyclebehavior could be dangerous because, if it would will be necessary to force over on the bend, itwould 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 bikes handling and tires 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 generatedfor various roll angle, side-slip angle, temperature, pressure..

Maybe is better to talk with numbers soft rubber, hard, more grip..they only are voidwords.

What is Tire Used

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