This is the first in a series of articles on the basics of motorcycle tires and various basic dynamic characteristics of motorcycle behavior. In general, this is a very difficult question.
and needs a good level of mathematics and physics to properly understand what is happening.
However, in these articles I will try to explain the basics with the absolute minimum of mathematics,
but where it is inevitable, I do not go beyond simple trigonometry. For those who are unhappy
with any math at all, don't worry, just skip these parts, and the rest will still be useful.
I will try to illustrate the mechanisms with a lot of sketches and graphs.
It's incredible that only two small rubber contact spots can support our cars and
manages to deliver large amounts of power to the road, while at the same time supporting turns
strength, at least as much as the weight of the bike and rider. Thus, tires provide the likelihood that one
the most important influence on the general characteristics of the treatment, therefore it seems appropriate to study their
performance over other various aspects of chassis design.
When Newton first outlined his theories of mechanics to the world, he undoubtedly meant
in addition to the interaction of motorcycle tires with the road surface. Still his
assumptions are equally valid for this situation. In particular, his third law states: “For every power
is an equal and opposite force to withstand it. "" Or otherwise: "Action and reaction are equal and
opposite. "
Assigning it to a tire action means that when a tire presses the road, the road approaches
back equally hard on the tire. This is equally good, regardless of whether we are looking at
supporting the weight of the bike or resisting turns, braking or driving loads.
What is this particular Newton's law does not apply to, is there any source force, and
whether it really matters for many analysis purposes. However, as a guide to understanding
Some physical systems are often helpful in mentally separating action from reaction.
The forces that occur between the ground and the tires determine that our behavior
but they are so often taken for granted. tires really carry so many different
tasks and their apparent simplicity hide the degree of engineering complexity that goes into them
design and workmanship. Originally, pneumatic tires were installed to improve comfort and reduce the load on
wheels. Even with modern suspension systems, there are still tires that provide the first line.
protection to absorb road shocks.
In order to explore the carcass design, the tread compound and tread design are very detailed beyond
this book. Rather, here we are dealing with some basic principles and their influence on the processing
characteristics.
Weight support
The most obvious function of the tire is to support the weight of the car, whether vertical or
leaving in the corner. However, the actual mechanism by which air and tire pressure passes
wheel load on the road is often misunderstood. Consider rice. 1, this sketch is a slice
through the bottom of the rim and the tire thickness of the unit with inflation pressure P. Left hand
side shows that the wheel is unloaded, and the right side shows that it supports weight F. When
the loaded tire is compressed vertically and the width increases, as shown, perhaps surprisingly,
The internal air pressure does not change significantly with the load, the internal volume varies slightly.
In the widest section (X1) of the unloaded tire, the inner half of the width W1, and therefore the force normal to
This section due to internal pressure is simply 2.P.W1. This force acts up to the wheel.
rims, but since the pressure and tire widths are evenly distributed around the circumference, the total
The effect is completely balanced. This force must also resist the equal voltage (T) in the bus.
carcass.
The loaded tire has half the width W2 in the widest section (X2), and therefore the normal force is 2.P.W2.
Therefore, the additional effort on this section when loading is 2.P. (W2 - W1), but since the tire is only
extended over a small portion of the lower part of the circle, this force supports the load F.
The above describes how increasing tire pressure and increasing tire width create strengths to counteract
vertical loading of the wheel, but does not fully explain the details of the mechanism by which these
Forces are transferred to the rim. Bead with installed tire is a noise-resistant fit over the bead seat
wheel rim that compresses this area, built-in side wall tension component
due to the pressure of inflation it reduces this contraction somewhat. This component appears as F1 on
unloaded half F1 = T.cos (U1). A larger angle U2 side wall when loading means
that the linear component of the tension decreases, thereby also restoring the rim portion for the tire
ball compression. This occurs only in the lower part of the tire circumference, where the extension
going on. Thus, an increase in compressive force on the lower rim acts upwards, this
supports the weight of the bike. Nett power is the difference between unloaded and loaded into a line.
strength
F = T. (cos (U1) -cos (U2))
The left side shows half inflated, but
unloaded tire, tension is created in the frame (T)
internal pressure. Right compressed and
The enlarged shape of the loaded tire is shown.
Suspension action
When performing this function, the pneumatic tire is the first object that senses any road shocks and therefore acts
as the most important element in the car's suspension system. To the extent
inconvenient, it would be perfectly possible to ride a bike on the roads within a reasonable time without
another form of absorption. In fact, the rear suspension was not at all common until the 1940s or 50s.
Whereas, despite the complexity of the traditional suspension system, it would be perfectly
it is impractical to use wheels without pneumatic tires or some other form of tire that allows
consider deflection from beats. Loads loaded into wheels without such tires will be intense when
all but slow speeds and continuous wheel failure will be the norm.
A few numbers illustrate what I mean: - Suppose a bike with a normal-sized front wheel hits 25
mm, sharp relief relief at a speed of 190 km / h. It is not a big hit.
In the absence of a tire, the wheel was then subjected to average vertical acceleration of approximately
1000 G. (peak value will be higher than this). This means that if the wheel and brake
the assembly had a weight of 25 kg. then the average point load on the rim will be equal to 245 kN. Egypt about 25
tons. What wheel could stand it? If the wheel was loaded with a normal tire, then it would have
soil level, spring speed, sharp edge, approx. 17-35 N / mm. Then the maximum force
transmitted to the wheel by 25 mm. the step will be about 425-875 N. that is, less than four thousandth
the previous figure, and this load will be more evenly distributed around the rim. No bus
shock loads transferred back to the sprung part of the bike would be much higher. Vertical wheel
the speed will be much greater, and therefore the damping forces of the shock wave, which depend on the wheel
speed, it would be generous. These high forces will be transmitted directly by bike and rider.
The following five diagrams show some results of computer simulations of accelerations and
movements on a typical motorcycle and illustrate the importance of a tire for comfort and road
holding The bike travels at a speed of 100 km / h. And the front wheel reaches 0.025 meters by 0.1
seconds Note that the time scales vary from schedule to schedule.
Three cases are considered:
· With typical vertical tire stiffness and typical suspension and suspension damping.
· With identical properties of a tire, but with a spring of a suspension spring of 100 X, than at previous.
· With a tire stiffness of 100 X higher and with a normal suspension.
Thus, we mainly consider a typical case, another case when there is practically no suspension of the suspension and
in the latter case with the correct hard tire. Structural load, comfort and traffic will all be negative
without initial depreciation of the tire. Please note that the above charts do not all refer to the same time scale,
This is just to better illustrate the relevant points.
It shows the vertical displacement of the front wheel. There is a slight difference between the maximum
displacement for two bodies with a normal tire; for a small pitch, the front tire absorbs most of the impact. But,
in the case of a very stiff tire, the wheel movement increases about 10 times. Obviously the tire
leaves the ground in this case, and the loss of departure can be seen after 0.5 seconds.
These curves show the vertical movement C of G of the bicycle and rider. As shown in fig. 1, it is clear that the hard tire
causes much higher movements on the bike, to an apparent deterioration in comfort.
By showing the different accelerations transmitted to the bike and rider, these curves show the vertical
accelerations on C G. Both tougher tires and tougher suspension cases show similar values around 5 or 6 times.
normal case, but the shape of the two curves is completely different. With stiff suspension little
damping, and we can see that it takes several cycles to calm down. The second blow is about 0.155 seconds - this is when
the rear wheel hits this step; this rear wheel signal is not shown on the other graphs for clarity.
Vertical acceleration of the front wheel for two bodies with a normal tire. The early part is similar for the two cases
The suspension has a slight effect here, it is a tire deflection, which is most important for this step height. As in FIG. 5
the lack of damping suspension allows the tire to bounce for several cycles before collapsing.
Since these curves have wheel acceleration, the values of the normal case are overloaded with hard
tire cover, with a peak value of about 600 G, compared with about 80 G normal. Note again the effects of the landing.
rebounds in 0.5 seconds. This high acceleration will lead to a very high structural load.
Because the tire is so good at removing most of the road shocks, right on the site of the application, it is possible that
it would be advisable to consider developing it in order to absorb even more and eliminate the need for other
suspension. Unfortunately, we encountered other problems. We all saw a big construction
cars bounce along the road on their own balloons, sometimes it gets so strong that
the wheels actually leave the ground. The pneumatic tire acts like an air spring, and the rubber acts like
damper when it bends, but when the tire is made larger, the spring effect suppresses
damping, and then we get uncontrollable bouncing. Thus, there are practical limits on the number
depreciation that can be built into the tire for any particular application.
Influence of tire pressure
Obviously, the spring characteristics mentioned above are highly dependent on tire inflation.
but there are other effects. Materials and frame design and properties
and the tread pattern of the outer rubber layer affects both the spring properties and the
ground contact area (contact patch). Insufficient inflation allows the tire
imply non-optimal cross-sectional shapes, additionally the inflation pressure extracts the effect
over side tire flexibility and this property is extremely important for a motorcycle
stability. Manufacturers and # 39; recommendations should always be followed.
The effect of tire pressure on the vertical stiffness of an inflated tire under load
flat surface. These curves are taken from actual measured data. Note that the spring speed is close to
linear across the entire load range and varies from 14 kgf / mm. with a pressure of 1.9 bar to 19 kgf / mm. at
2.9 bar. The effective spring speed when the tire is loaded on a sharp edge, such as a brick, is
considering this below, and is more non-linear due to the changing shape of the contact area as
the tire is wrapped around the object.
This spring speed acts in series with the suspension springs and is an important part of the overall
suspension. An interesting property of rubber is that when compressed and released
usually does not return exactly to its original position, this is known as hysteresis. This effect is shown.
only for 1.9 bar. case, the curve drawn during the loading phase is not observed during
unloading stage. The area between these two curves represents the loss of energy that leads to
tire and also acts as a damping suspension. In this particular case, the energy lost
in one cycle of loading and unloading, approximately 10% of the total energy stored in
compressed tire, and is a significant parameter controlling tire rebound.
The vertical stiffness of a standard road tire is relatively flat with different inflation pressures. This data is taken from
Avon Azaro Sport II 170/60 ZR17. Up arrows estimate tire compression, and the second with
the down arrow (shown only at 1.9 bar for clarity) shows the behavior of the tire when the load is released.
the shaded area between the two lines is the loss of energy, called hysteresis. It acts as a source of suspension.
damping and also heats the tire. (From data provided by Avon tires).
The lateral stiffness of the same tire shown in fig. 9. The vertical load was constant at 355 kgf. and the wheel was
vertical. As expected, the tire is somewhat tougher with a higher inflation pressure, but loses grip or saturation on
lower lateral load 460 kgf. compared to 490 kgs. with lower pressure. (From data provided by Avon tires).
Contact area
The tire must fully support the bike through a small area of rubber in contact with
and, thus, “patch contact area = vertical force — average contact contact surface pressure”. This
applied in all conditions.
However, the surface pressure of the contact patch is NOT the same as the inflation pressure, as well as
sometimes claimed. They are related, but there are at least four factors that change relationships.
The stiffness of the frame, the shape of the frame, the depth and softness of the surface rubber and the conformity of the road surface. If
we have a very high rigidity of the frame, then the pressure of inflation will decrease.
Let's look at this in a bit more detail and see why:
If the tire was made in the same way as the inner tube, that is, from fairly thin rubber and with a slight stiffness without
inflated, then the internal air pressure will be the only means to support the weight of the bike. In that
if the pressure of the contact layer is equal to the pressure of the internal air pressure. For air
pressure 2 bar and vertical load 1.0 kN. Then the contact area will be 5003 square meters. Mm If we
air pressure has now increased to say that the 3 bar area will drop to 3335 sq. m. mm
Now imagine that we are replacing a hard steel tubular wrap for our rim and tire, the contact area
with the earth will be quite small. If we now inflate a hoop with some air pressure, it does not take
lots of imagination to see that, unlike the inner tube, this internal pressure will have a negligible effect
on the outer surface of the contact. Obviously, the tire is not entirely similar to the steel hoop, nor to the inner tube,
but it shows that the rigidity of the framework can reduce the contact surface area, calculated purely
only from inflationary pressure.
I have done 2 sets of tests. For the first, I kept the tire inflation pressure constant at 2.4 bar and changed the tire
load between 178 and 1210 N. (taking into account the weight of glass and wooden beams). Secondly, I
maintain a constant load of 1210 N. and try to change the inflation pressure from 2.4 to 1 bar.
Even with a generous account of experimental error, the effects are clear. Graphs show that
the results turned out to be quite acceptable for a smooth line, there was not a large scatter.
Point (1) on the curve with a constant inflation pressure shows how the actual pressure of the contact formation
lower (slightly more than half) than inflation pressure, or in other words, the contact area is larger. it
because of the compliance of the rubber surface, therefore it is more important at low vertical loads, whereas
the rigidity of the frame has become more important as the load has risen, as shown in (3) - (6), where
the actual contact pressure is higher than the air pressure, i.e. the reduced contact area.
Measurement setup. At the end of the beam, different weights were set, which also loaded the tire through
thick plate of glass. Луч выполнен с возможностью нагрузки на шину с утечкой 4: 1. Итак, 25
кгс. вес будет загружать шину 100 кгс. Проследив за стеклом площадь контакта
был определен.
Верхний график показывает измеренное давление контактного пласта при различных нагрузках на колесо для постоянного давления
2,4 бар. Нижние кривые показывают контактное давление при различных давлениях инфляции при фиксированной нагрузке 1210 Н.
номера в точках данных соответствуют трассировке области контакта в предыдущем эскизе. Простая линия на каждом участке
показывает, что давление контактного пласта равно инфляционному давлению.
Жесткость каркаса помогает поддерживать машину, поскольку давление воздуха
уменьшено, давление контактного пласта значительно выше, чем давление надувания. It looks
хотя две линии будут пересекаться при давлении воздуха около 3,5 бар. (хотя это не было проверено
измерение), в этот момент наибольшее значение будет иметь поверхностное сжатие резины.
Это соответствует аналогии стального обруча выше.
Мы можем легко увидеть два отдельных эффекта соответствия поверхности и жесткости каркаса и
относительная важность этих вариантов с нагрузкой и / или инфляционным давлением.
Эти тесты проводились только с одной конкретной шиной, другие типы будут показывать разные детали, но
общие эффекты должны следовать аналогичной схеме.
Площадь под углом
Отражает ли угол касание зоны контакта с шинами?
Предположим, что горизонтальная поверхность и боковое ускорение 1G. В этих условиях велосипед / всадник
CoG будет находиться на линии под углом 45 ° к горизонтали и проходит через контактный патч. Там будет
которая действует вдоль этой линии через контактный патч, в 1,4 раза превышающий поддерживаемый вес.
Эта сила является результатом поддерживаемого веса и силы поворота, которые имеют одинаковые
в этом примере 45 ° мяса. Сила, нормальная к поверхности, просто из-за
поддерживаемый вес и не зависит от силы поворота. Сила поворота реагирует на
горизонтальная сила трения, создаваемая поверхностью шины / дороги, и эта сила трения «разрешена»
в силу нормальной силы.
Поэтому в первом приближении угловая сила не будет влиять на зону контакта шины, и на самом деле это
случай может быть приближен, если мы просто рассматриваем внутреннюю трубку без реальной шины.
Однако на самом деле боковая сила вызовет некоторое дополнительное искажение шины на
дорожного покрытия / шины и в зависимости от характеристик шины, упомянутых выше, площадь контакта может
хорошо изменение.
Другим аспектом этого является, конечно, профиль поперечного сечения шины. Старые треугольные гонки Dunlop
шина, например, была спроектирована так, чтобы прикладывать больше резины на дороге, когда она опиралась, так что даже без шины
искажение области контактных патчей увеличилось, просто в силу угла наклона.
Рэй Тейлор
http://www.CarsNet.com/motorcycle
