BLUEBOX SOLUTIONS FOR SESSION 2.
6 - TYRES
sx
1 Show that x =
1 + sx
2 What is the maximum achievable Fx predicted by the brush model?
Cx sx
From the equation Fx = C x x = , as s x → , Fx → Cx . Clearly the maximum force is
1 + sx
achieved at maximum slip.
3Fz
3 Show that is proportional to the vertical tire load according to the formula =
4a 3
4 Modify the script brush1.m to allow different values for static and sliding. Check results
for the above case.
mu is set the same as mu_static, except for in the equations for the sliding region when
mu_sliding is used – obviously!! (see brush2.m also, which has additional commands for
setting changing the contact patch length depending on vertical load)
suggestion: run brush 1 with and without the sliding friction term to see the effect
5 For the data given in the first plot, evaluate cex and Cx
[Ans. 1.05 10 7 , 3.36 10 4 ]
Note – according to the above this implies a very high limiting force of around 34kN –
equivalent to roughly twice the weight of a medium-sized passenger car, just from one tire!
(of course, this initial version of the brush model does not include any friction limit)
6 Check directly by integration that in the pure sliding case, Fx = Fz
(three methods are given! the second is probably the most appropriate)
A more direct method is to integrate under the condition that all elements are sliding:
Finally, since each tread element is sliding, the integration can be done completely
painlessly:-
7 Why is the longitudinal stiffness constant? What effect is being ignored?
The longitudinal stiffness comes from the elastic properties of the tyre, and is not related to
friction limits etc. So there is nothing in the model that will change when vertical load is
altered. In reality the extra load will change the length of the contact patch, so there are more
brush elements in contact with the road and the longitudinal stiffness will increase.
8 Sketch the above function for the case E = 0 , B = 1 , C = 1 , D = 1 . Add two extra curves:
(i) B = 1.1 (ii) C = 1.1 .
9 Use Matlab to investigate the influence of the four parameters on the shape of the graph
y ( x ) . Start with B = 1, C = 1.5, D = 1, E = 0 and vary one parameter at a time.
Use the Matlab file posted. B and D are just scale factors, C and E are shape factors.
Generally the conditions E 0 , 1 C 2 are sufficient for the magic formula to have one
peak as required for modeling lateral or longitudinal forces.
dFy
10 Check that in the case s x = 0 : Fx = 0 , Fy = Fp , = C
max
d =0
11 Briefly check the above equations are represented in the Matlab function tm1.m
open tm1.m and highlight the equations in the code.
It’s always a good idea to have the theory/equations written out separate from a Matlab function,
unless the Matlab function is very simple; in that case the comments in the code are sufficient to
explain what it’s doing.
12 Check and run tireplot1.m to recreate the above plots
run the script!!
The anisotropic case is not required reading but the bluebox solutions are included for
completeness.
13 Plot this function
Write a little matlab script or use a plotting calculator etc. E.g. for eta_0 =0.5
0.8
0.6
0.4
0.2
0
0 2 4 6 8 10 12
sample script:
e0=0.5;
x1=0:0.01:2*pi;
x2=x1+2*pi;
y1=(1+e0-(1-e0)*cos(x1/2))/2
y2=1*ones(size(x2))
x=[x1,x2];y=[y1,y2];
plot(x,y)
axis([0 4*pi 0 1.2])
14 Briefly check the above equations are represented in the Matlab function tm2.m
see comment for 11
15 Check and run tireplot2.m to create plots equivalent to those of the isotropic tire model.
see comment for 12
