Motorcycle Tire Basics

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Motorcycle Tire Basics

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

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 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

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 clarity.

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 down.

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 loading.

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 construction

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 properties

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 motorcycle

stability. Manufacturers’ recommendations should always be adhered to.

The influence of tire pressure on the vertical stiffness of an inflated tire, when loaded on

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

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.)

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 sq.mm. If we

now increased the air pressure to say 3 bar the area would fall to 3335 sq.mm.

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.

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, whereas

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 area

was determined.

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 importance.

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.

by Ray Taylor

http://www.CarsNet.com/motorcycle

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