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These are various useful notes from chapters of the book.
3.5 Tyre Rolling Resistance
Rolling Friction Coefficient (Page 119)
ur = u0 + u1*v^2
where u0 = 0.015 & u1 = 7*10-6 s^2/m^2
Fourth degree equation
Fr = C0 + C1*v + C2*v^4 (Page 119)
where C0 = 9.91*10-3, C1 = 1.95*10-5, C2 = 1.76*10-9
Power consumption can be estimated as approximately 2.4 kW for a vehicle moving at 100 km/h with each tyre at 2.2 Bar and 220 kg mass on each tyre (880 kg sprung mass).
Motorcycle tyres have a relatively higher friction coefficient, and to draw a comparison, 4 of these would absorb 5.7 kW.
Table 3.1 - The value of u0 (friction) on different pavements
Notes: "Tarmac" is a process invented by Edgar Purnell Hooley for a Tar Macadam road construction
3.6 Tyre Longitudinal Force
Some empirical models with slip (s) as a variable:
1) Pacejka (1991): Fx = c1*sin(c2*tan^-1(c3*s-c4*(c3*s-tan^-1(c3*s))))
2) Burckhardt (1987): Fx = c1*(1-e^-c2s)-c3*s
3) Burckhardt velocity dependent (1997): Fx = (c1*(1-e^-c2s)-c3*s)*e^-c4v
4) Kiencke & Daviss: Fx = ks * s / (1+c1*s+c2*s^2) where ks is slope of Fx at zero slip
5) De-Wit: Fx = c1*s^0.5-c2*s
3.6 Tyre Lateral Force
A third degree function for high sideslips:
It is useful to note that Radial tyres generate lower camber force - about half - of crossply tyres, due to their higher flexibility. Any camber control vehicle can potentially achieve an additional 1 kN of lateral force with approximately 12 degrees of camber angle.
7.1 Kinematic Steering
Ackerman (kinematic) condition: cot do - cot di = w/l
do = steer angle of outer wheel
di = steer angle of inner wheel
w = track
l = wheelbase
12.0 Applied Vibrations (page 730)
Most of the optimisation methods for vehicle suspensions are based on frequency responses.
There are four types of applied excitations: harmonic, periodic, transient, random.
Example of three springs in series: k1x1=k2x2=k3x3
Because the following relationship holds: fs/keq=fs/k1+fs/k2+fs/k3
The three springs can be substituted by a single one of stiffness:
1/keq=1/k1+1/k2+1/k3
If the springs are in parallel then keq=k1+k2+k3
A one degree of freedom model typically has the equation:
m*a = -c*v -k*x + f(x,v,t)
The number of masses times the degrees of freedom for each mass makes the total degrees of freedom of the vibrating system.
So for example a 1/4 car model with wheel mass is a 2-DOF.
A 3-DOF system would include the driver mass and seat springs
For a 2-DOF base-excited system we can write the following notes alongside the free body diagram:
ks*(xs - xu)
cs*(xdot_s - xdot_u)
ks*(xs-xu)
cs*(xdot_s - xdot_u)
ku*(xu - y)
cu*(xdot_u - y_dot)
If we take the upwards direction as positive sign, the equations of motions are:
ms*as = -ks*(xs - xu) -cs*(vs - vu)
mu*au = ks*(xs - xu) +cs*(vs - vu) -ku*(xu - y) -cu*(vu - vy)
If we write it in matrix form:
[M]*a + [c]*v + [k]*x = F
12.3 Frequency Response (page 744)
In frequency response analysis we are looking for the steady state amplitude of oscillation as a function of the excitation frequency.
There are 4 types of excitation:
1) base e.
2) eccentric e. (model of engine on mounts)
3) eccentric base e.
4) forced e. (no practical application)
We use the Base Excitation because it is the most useful for suspension systems.
The excitation is y = Y*sin(omega*t)
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