Slip Angle/Slip Ratio relation with Lateral/Longitudinal Force by using MATLAB Scripting
AIM:
Write an Octave script to show:
• How different slip angles (0 deg, 2 deg, 4 deg and 6 deg) influence the force (Fx and Fy) vs slip ratio relationship (sweep from 0 to 60%).
• How different slip ratios (0%, 3%, 6% and 9%) influence the force (Fx and Fy) vs slip angle relationship (sweep from 0 to 20 deg).
Use the modified slip approach. Develop your script based on the algorithm we looked at in class. These are your deliverable's:
1. Plot (use sub-plots) showing the effect of different slip ratios on lateral force vs slip angle relationship (5 points)
2. Plot showing the effect of different slip ratios on longitudinal force vs slip angle relationship (5 points)
3. Plot showing the effect of different slip angles on longitudinal force vs slip ratio relationship (10 points)
4. Plot showing the effect of different slip angles on lateral force vs slip ratio relationship (10 points)
5. Error free Octave code (5 points) 6. Explanation of the concept using the generated plots (5 points)
THRORY & EXPLANATION:
The Longitudinal Force is in two forms, one is in Traction Force during acceleration and other is Braking force during deaccelerating condition.
These forces are generated when there is a slip propagation in the form of Slip ratio and Slip Angle.
Slip Ratio defined as the basic difference in wheel angular velocity and wheel forward velocity. When there is an addition or reduction in wheel angular velocity which creates a deviation between angular velocity and forward velocity resulting in generation of slip, this percentage of deviation is known as slip ratio.
Slip Angle is the angle between a wheel's heading direction and the wheel travel direction. This slip angle is generated due to the moment caused by Lateral force and Pneumatic trail which multiplies together. Generally slip angle is caused during turning, as as we steer the wheel the friction force holds the tyre tread generating lateral force and deflecting the tread in the direction of turn.
The Fx is the Longitudinal Force related with Slip Ratio. Plot shows how much force you get from your tyres is mainly dependent upon slip ratio means that the more deviation in velocities the more force produced in longitudinal direction. But it is not purely linear dependent as the maximum force you will get is upto a point after which the force saturates at approx constant values. The point where the max force is generated is called peak slip ratio.
The Fx also defers with slip angle during combined slip conditions where Fy(Lateral force) also generated at the time of accelerating/braking maneuver. In this condition what happens is that sideway slip is also being generated alongside longitudinal slip which reduces the amount of Traction/braking force at the same slip ratio. During this event, lateral force starts to dominate the longitudinal force as it starts to deviate the contact patch sideways reducing the adhesion region longitudinally.
1. How different slip angles (0 deg, 2 deg, 4 deg and 6 deg) influence the force (Fx and Fy) vs slip ratio relationship (sweep from 0 to 60%).
Input Data for both Observations:
#Pac model for Fx and Fy vs Slipratio w.r.t different slipangles
clear all;
#-----------------------------------------------------------------------------#
sr_peak=0.09; #As given in data i.e. Peak Slip Ratio = 9%
alpha_peak=4; #As given in data i.e. Peak Slip Angle = 4 deg
Cx=1.15; #As given in data i.e. Cx = 1.15
Dx=3200; #As given in data i.e. Dx = 3200
Bx=25; #As given in data i.e. Bx = 25
Ex=-0.4; #As given in data i.e. Ex = -0.4
Cy=1.2;
Dy=2900;
By=0.27;
Ey=-1.6;
Svx=0; #As given in data i.e. Svx=0
Shx=0; #As given in data i.e. Shx=0
Svy=0; #As given in data i.e. Svy=0
Shy=0; #As given in data i.e. Shy=0
for j=1:4
alpha = [0 2 4 6];
alpha_n=(alpha(j))/alpha_peak;
for i=1:61
sr(i)=(i-1)*0.01; #slip ratio
sr_n=sr(i)/sr_peak;
slip_rs=sqrt((sr_n.^2)+(alpha_n.^2));
slip_mod=slip_rs*sr_peak;
SR_mod=slip_mod+Shx;
alpha_mod=slip_rs*alpha_peak;
SA_mod=alpha_mod+Shy;
fx(i,j)= Dx*sin(Cx*atan(Bx*SR_mod-Ex*(Bx*SR_mod-atan(Bx*SR_mod))));
Fx(i,j)=((fx(i,j))*(sr_n/slip_rs))+Svx;
fy(i,j)= Dy*sin(Cy*atan(By*SA_mod-Ey*(By*SA_mod-atan(By*SA_mod))));
Fy(i,j)=((fy(i,j))*(alpha_n/slip_rs))+Svy;
end
end
subplot(1,2,1);
plot(sr,Fx,'linewidth',1);
set(gca,'fontsize',13);
grid on;
xlabel('Slip Ratio');
ylabel('Longitudinal Force');
grid on;
subplot(1,2,2);
plot(sr,Fy,'linewidth',1);
set(gca,'fontsize',13);
grid on;
legend('0 deg Slip Angle','2 deg Slip Angle','4 deg Slip Angle','6 deg Slip Angle');
xlabel('Slip Ratio');
ylabel('Lateral Force');
grid on;This is also shown through the graphs plotted btw Fx and Fy v/s slip ratio with respect to different slip angles.
DELIVERABLES:
- The Lateral Force also known as cornering force is the force generated due to the shift of your contact patch laterally.
- When the tyre is steered it causes a twist of carcass and tread which results in a force laterally in the direction of the steering.
- Lateral Force increases with Slip angle and peak_slip_angle is the value at which you will get maximum lateral force or cornering force a tyre can offer.
- But, it doesn't mean that at zero slip angle you will not get cornering force, if you are driving on a banked road profile then the tyre will generate some lateral force due to the wheel camber/camber thrust.
- But in this plot you can't observe the camber effect because the PAC Model which is behind this plot is more generic or basic formulation of magic formula.
- As already discussed above about slip angle and what it is, here we will talk about the relation of slip ratio and its dependency over lateral force.
- In the case of combined slip scenario, Lateral Force decreases with increase of slip ratio as it increases the Fx value. So, It's just a trade off between lateral and longitudinal force during the combined slip approach. One has to select the range of slip angle and slip ratio during modelling ABS and TC systems.
2. How different slip ratios (0%, 3%, 6% and 9%) influence the force (Fx and Fy) vs slip angle relationship (sweep from 0 to 20 deg).
#Pac model for Fx and Fy vs Slipangle w.r.t different slipratios
clear all;
#-----------------------------------------------------------------------------#
sr_peak=0.09; #As given in data i.e. Peak Slip Ratio = 9%
alpha_peak=4; #As given in data i.e. Peak Slip Angle = 4 deg
Cx=1.15; #As given in data i.e. Cx = 1.15
Dx=3200; #As given in data i.e. Dx = 3200
Bx=25; #As given in data i.e. Bx = 25
Ex=-0.4; #As given in data i.e. Ex = -0.4
Cy=1.2;
Dy=2900;
By=0.27;
Ey=-1.6;
Svx=0; #As given in data i.e. Svx=0
Shx=0; #As given in data i.e. Shx=0
Svy=0; #As given in data i.e. Svy=0
Shy=0; #As given in data i.e. Shy=0
for j=1:4
sr = [0 0.03 0.06 0.09];
sr_n=(sr(j))/sr_peak;
for i=1:201
alpha(i)=(i-1)*0.1; #slip ratio
alpha_n=alpha(i)/alpha_peak;
slip_rs=sqrt((sr_n.^2)+(alpha_n.^2));
slip_mod=slip_rs*sr_peak;
SR_mod=slip_mod+Shx;
alpha_mod=slip_rs*alpha_peak;
SA_mod=alpha_mod+Shy;
fx(i,j)= Dx*sin(Cx*atan(Bx*SR_mod-Ex*(Bx*SR_mod-atan(Bx*SR_mod))));
Fx(i,j)=((fx(i,j))*(sr_n/slip_rs))+Svx;
fy(i,j)= Dy*sin(Cy*atan(By*SA_mod-Ey*(By*SA_mod-atan(By*SA_mod))));
Fy(i,j)=((fy(i,j))*(alpha_n/slip_rs))+Svy;
end
end
subplot(1,2,1);
plot(alpha,Fx,'linewidth',1);
set(gca,'fontsize',13);
legend('0 Slip Ratio','0.03 Slip Ratio','0.06 Slip Ratio','0.09 Slip Ratio');
grid on;
xlabel('Slip Angle');
ylabel('Longitudinal Force');
grid on;
subplot(1,2,2);
plot(alpha,Fy,'linewidth',1);
set(gca,'fontsize',13);
grid on;
xlabel('Slip Angle');
ylabel('Lateral Force');
grid on;
"Here, only four to five variables are taken into consideration which are quite less to mirror the behaviour of tyres completely as it requires over 100 variables which goes into play to generate more realistic empirical curves. But, behind all this the theory remains the same and logic remains solid"