Brakes Calculations

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Vehicle Dynamics & Control - 17 Acceleration and braking performance
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%MatLab Code
clear all;
clc;
% GIVENS
a = 4.91;             % braking deceleration (abs val, m/s^2)
m_r = 20;                   % mass of rear wheel (kg)
m_c = 120;                  % mass of chassis (kg)
m_f = 20;                   % mass of front wheel (kg)
m_tot = m_r + m_c + m_f;
r_w = 0.20;                 % radius of wheel (m)
I_r = 0.5*m_r*(r_w^2);      % moment of inertia (rear wheel)
I_f = 0.5*m_r*(r_w^2);      % moment of inertia (front wheel)
mu = 0.0055;
d_rear = 1;                      % horizontal distance from CoG to rear wheel
d_front = 1;                      % horizontal distance from CoG to front wheel
h = 0.5;                    % height from chassis to CoG
thetaDeg = 0;              % incline (degrees)
theta = thetaDeg*pi/180;    % incline (radians)
alpha = a/r_w;
g = 9.81;
% SOLUTION 1: SOLVING JUST WEIGHT TRANSFER, NORMAL FORCES, AND TORQUE
% (STEP-BY-STEP SOLUTION)
A = [
    1, 1;
    -d_rear, d_front;
    ];
B = [
    m_tot*g*cos(theta);
    m_tot*a*h;
    ];
sol = linsolve(A,B);
F_N_r = sol(1);
F_N_f = sol(2);
F_rc_y = F_N_r - m_r*g*cos(theta);
F_fc_y = m_c*g*cos(theta) - F_rc_y;
F_r_r = F_N_r * mu;
F_r_f = F_N_f * mu;
F_rc_x = I_r*a/(r_w^2)+m_r*a;
F_fc_x = m_c*a+ F_rc_x;
T = I_f*a/r_w + m_f*a*r_w + F_fc_x*r_w;
fprintf("SOLUTION 1\n");
fprintf("F_N_r: %f\n", F_N_r);
fprintf("F_N_F: %f\n", F_N_f);
fprintf("F_rc_y: %f\n", F_rc_y);
fprintf("F_fc_y: %f\n", F_fc_y);
fprintf("F_rc_x: %f\n", F_rc_x);
fprintf("F_fc_x: %f\n", F_fc_x);
fprintf("T: %f\n\n", T);
% SOLUTION 2: SOLVING ENTIRE SYSTEM AT ONCE (9 EQNS, 9 UNKNOWNS)
% F_rc_x, F_rc_y, F_N_r, F_f_r, F_fc_x, F_fc_y, F_N_f, F_f_f, T
eqns = [
        1,      0,      mu,    1,     0,      0,      0,     0,     0;
        0,      -1,     1,     0,     0,      0,      0,     0,     0;
        1,      0,      0,     0,     0,      0,      0,     0,     0;
        1,      0,      0,     0,     -1,     0,      0,     0,     0;
        0,      1,      0,     0,     0,      1,      0,     0,     0;
        0,      -d_rear,0,     0,     0,      d_front,0,     0,     0;
        0,      0,      0,     0,     -1,     0,      mu,    1,     0;
        0,      0,      0,     0,     0,      -1,     1,     0,     0;
        0,      0,      0,     0,     -r_w,   0,      0,     0,     1;
       ];
constants = [
            m_r*a + m_r*g*sin(theta);                   % eqn 1 RHS
            m_r*g*cos(theta);                           % eqn 2 RHS
            I_r*a/(r_w^2) + m_r*a + m_r*g*sin(theta);   % eqn 3 RHS
            -m_c*a - m_c*g*sin(theta);                  % eqn 4 RHS
            m_c*g*cos(theta);                           % eqn 5 RHS
            m_c*a*h;                                    % eqn 6 RHS
            m_f*a + m_f*g*sin(theta);                   % eqn 7 RHS
            m_f*g*cos(theta);                           % eqn 8 RHS
            I_f*a/r_w + m_f*a*r_w + m_f*g*sin(theta);   % eqn 9 RHS
            ];
solution = linsolve(eqns, constants);
fprintf("SOLUTION 2\n");
fprintf("F_rc_x: %f\n", solution(1));
fprintf("F_rc_y: %f\n", solution(2));
fprintf("F_N_r: %f\n", solution(3));
fprintf("F_f_r: %f\n", solution(4));
fprintf("F_fc_x: %f\n", solution(5));
fprintf("F_fc_y: %f\n", solution(6));
fprintf("F_N_f: %f\n", solution(7));
fprintf("F_f_f: %f\n", solution(8));
fprintf("T: %f\n", solution(9));
7:14
Sorry for the delay, that's the matlab code. Solution 1 is the step-by-step one, where you just need 3 eqns to get the torque, and solution 2 is the 9-eqns at the same time solution.
7:14
Here's the output:
7:14
SOLUTION 1
F_N_r: 588.400000
F_N_F: 981.200000
F_rc_y: 392.200000
F_fc_y: 785.000000
F_rc_x: 147.300000
F_fc_x: 736.500000
T: 176.760000
SOLUTION 2
F_rc_x: 147.300000
F_rc_y: 441.300000
F_N_r: 637.500000
F_f_r: -52.606250
F_fc_x: 736.500000
F_fc_y: 735.900000
F_N_f: 932.100000
F_f_f: 829.573450
T: 176.760000
Google Sheet Link: https://docs.google.com/spreadsheets/d/1emCQpfTXff3dAr8s4W90xkS2jfdpFnwxZciVqpC97YI/edit?usp=sharing