Before moving to the more complex combined slip case, pure side slip will be ana- lyzed first. In this case the longitudinal slip𝜅 will be zero and same applies to the longitudinal bristle deflection 𝜀x(x). Using (34) the lateral bristle deflection then
becomes
𝜀y(x) = (a−x) tan(𝛼). (43) Considering the limit case, when the side slip angle𝛼is very small, the assumption can be made that all bristle elements in the contact region are in the adhesion state.
To calculate the lateral forceFythe lateral force per unit of lengthqy(x)has to be integrated over the contact length
Fy=
a
−a∫
qy(x)dx=kb
a
−a∫
𝜀y(x)dx=kbtan(𝛼)
a
−a∫
(a−x)dx (44)
Since𝛼is small,tan(𝛼) ≈𝛼, and solving the integral gives
Fy= 2kba2𝛼=CF𝛼𝛼. (45) Note that we have obtained an expression for the cornering stiffnessCF𝛼 of the tire, being dependent on the bristle stiffnesskband contact length. To calculate the self- aligning momentMzthe moment armxhas to be taken into account and the expres- sion becomes
Mz=
a
−a∫
qy(x)xdx=kb
a
−a∫
𝜀y(x)xdx=kbtan(𝛼)
a
−a∫
(a−x)xdx. (46)
Solving the integral gives
Mz= − 2
3kba3𝛼= −CM𝛼𝛼. (47) The expression for the pneumatic trailtpbecomes
tp= CM𝛼 CF𝛼 = 1
3a. (48)
So far it has been assumed that the side slip angle𝛼is infinitely small. To handle larger values of the side slip angle both sticking and slipping of bristles in the contact region has to be considered. A generic expression for the transition pointxt, (41), has already been derived in the previous section. For pure side slip it reduces to
xt= 4kba3
3𝜇Fz|tan(𝛼)|−a. (49) Sliding of all bristle elements will start to occur whenxt=a. The corresponding side slip angle𝛼slidingis given by
Tire Characteristics and Modeling 73
|tan(𝛼sliding)|= 3𝜇Fz
2kba2. (50)
To determine the lateral forceFyand self-aligning momentMz, similar to (44) and (46), the lateral force per unit of lengthqy(x)has to be integrated over the contact length. However part of the bristles in the contact region are in the adhesion state betweenx=xtandx=a, and the other bristles are in the sliding state betweenx=
−aandx=xt. For both regions different expressions forqy(x)are applicable. For the sliding part we haveqy(x) =𝜇qz(x), for the adhesion partqy(x) =kb𝜀y(x). The integral to calculateFythus becomes
Fy= 3𝜇Fz 4a
xt
−a∫ (
1 − (x
a )2)
dx+kbtan(𝛼)
a
∫
xt
(a−x)dx. (51)
For the self-aligning momentMzthe expression is very similar, but the moment arm xneeds to be taken into account,
Mz= 3𝜇Fz 4a
xt
−a∫ (
1 − (x
a )2)
xdx+kbtan(𝛼)
a
∫
xt
(a−x)xdx. (52)
Note that when|𝛼|≥𝛼slidingthatxt=aand second part of the integrals (51) and (52) disappears. For this condition the following expressions are obtained for the lateral forceFyand self-aligning momentMz:
Fy=𝜇Fz, (53)
and
Mz= 0 (54)
respectively. When|𝛼|≤𝛼sliding the integrals (51) and (52) have to be solved, and expression (49) is used to determinext. To simplify the expressions, the parameter 𝜃is introduced, it is defined as
𝜃= 2kba2
3𝜇Fz = 1
|tan(𝛼sliding)|. (55)
The theoretical lateral slipsyfollows from (36), given that𝜅= 0
sy= tan(𝛼). (56)
By solving the integrals the expressions forFyandMzbecome
tan(α) Fy
0
μFz
−μFz
tan(α) Mz
0 25627aμFz
−25627aμFz
tan(α) tp
0 13a
A B C D
V Fy
tp α
A
V Fy
α
B max. bristle deflection
V Fy
α
C α=αsliding
V Fy
α D
α > αsliding
Fig. 20 The brush tire model results for pure side slip
Fy= 3𝜇Fz𝜃sy (
1 −|𝜃sy|+ 1 3(𝜃sy)2
)
(57)
Mz= −𝜇Fza𝜃sy (
1 − 3|𝜃sy|+ 3(𝜃sy)2−|𝜃sy|3 )
. (58)
Using these results, the following expression can be obtained for the pneumatic trail tp,
tp= a 3
(1 − 3|𝜃sy|+ 3(𝜃sy)2−|𝜃sy|3 1 −|𝜃sy|+13(𝜃sy)2
)
. (59)
A graphical representation of the lateral forceFy, self-aligning momentMzand pneumatic trailtp is shown in Fig.20. As the side slip angle𝛼increases the bristle deflection in the contact region increases, and thus the lateral forceFyincreases, up to the level where all bristles have reached the maximum deflection. The self-aligning moment Mz is a result of the non-symmetric bristle deflection in the leading and trailing parts of the contact zone. For small side slip angles an increase of the side slip angle will result in a larger bristle deflection and increase of the self-aligning moment. For large values of side slip the asymmetry in the bristle deflection in the leading and trailing part of the contact zone is reduced and vanishes completely when all bristles are sliding. The pneumatic trailtpcorresponds to the centroid of the bristle deflection. It decreases monotonically with increasing side slip angle, until it becomes zero when all bristles in the contact zone are sliding.
Tire Characteristics and Modeling 75 The discussion of the pure longitudinal slip case is very similar to the pure lateral slip case. In this case𝛼is equal to zero thus𝜀y(x)will be zero too, the expression for 𝜀x(x)is given by (33). The expression for the transition pointxtbecomes
xt= 4kba3 3𝜇Fz||
|| 𝜅 1 +𝜅||
||−a= 2𝜃a||
|| 𝜅 1 +𝜅||
||−a. (60)
Again full sliding of all bristle elements in the contact zone will occur whenxt=a, thus the following condition has to be met
|||| 𝜅sliding
1 +𝜅sliding
||||= 1
𝜃. (61)
Two solutions exist for𝜅sliding, for positive𝜅 𝜅sliding = 1
𝜃− 1 (62)
and for negative𝜅
𝜅sliding = − 1
𝜃+ 1. (63)
Note that the longitudinal slip 𝜅 where full sliding occurs is different for driving (𝜅 >0) and braking (𝜅 <0). The expressions for the bristle deflections𝜀y(x)(34) and𝜀x(x)(33) are very similar, the only difference being thatsyis replaced bysx. This also implies that the functionFx(sx)will be identical toFy(sy). Thus the equation for the lateral forceFy(57) can also be used to calculate the longitudinal forceFxwhen replacingsybysx. The theoretical slipsxwill be equal to the longitudinal slip𝜅for small values of slip. It allows to determine the longitudinal slip stiffnessCF𝜅, which equals
CF𝜅= 2kba2. (64) Note that the expression for the longitudinal slip stiffness CF𝜅 is identical to the cornering stiffnessCF𝛼 (45).