# Let us assume that the tire forces can be modeled

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Let us assume that the tire forces can be modeled through the following linear approximations: F_x,i(t) = C_x*s_i(t) F_y,i(t) = C_y*alpha_i(t) for i = {FL, FR, RL, RR} where C_x and C_y are the longitudinal and lateral tire stiffness, respectively. Here we have assumed that these stiffness parameters are the same for all 4 tires. s_i(t) is the so-called (longitudinal) slip of tire i and alpha_i(t) a tire slip angle. For a front-wheel driven vehicle (as considered here), the slips s_FL(t) and s_FR(t) are derived from the individual wheel speeds (measured) by assuming that the rear wheels do not show any slip (i.e., s_RL(t) = s_RR(t) = 0). Hence the slips are inputs to our model structure. For the front wheels, the tire slip angles alpha_Fj(t) can be approximated by (when v_x(t) > 0) alpha_Fj(t) = delta(t) - arctan((v_y(t) + a*r(t))/v_x(t)) ~ delta(t) - (v_y(t) + a*r(t))/v_x(t) for j = {L, R} For the rear wheels, the tire slip angles alpha_Rj(t) are similarly derived and computed as alpha_Rj(t) = - arctan((v_y(t) - b*r(t))/v_x(t)) ~ - (v_y(t) - b*r(t))/v_x(t) for j = {L, R} With J = 1/((0.5*(a+b))^2*m) we can next set up a state-space structure describing the vehicle dynamics. Introduce the states: x1(t) = v_x(t) Longitudinal velocity [m/s]. x2(t) = v_y(t) Lateral velocity [m/s]. x3(t) = r(t) Yaw rate [rad/s]. the five measured or derived input signals u1(t) = s_FL(t) Slip of Front Lef tire [ratio]. u2(t) = s_FR(t) Slip of Front Right tire [ratio]. u3(t) = s_RL(t) Slip of Rear Lef tire [ratio]. u4(t) = s_RR(t) Slip of Rear Right tire [ratio]. u5(t) = delta(t) Steering angle [rad]. and the model parameters: m Mass of the vehicle [kg]. a Distance from front axle to COG [m]. b Distance from rear axle to COG [m]. Cx Longitudinal tire stiffness [N].

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Cy Lateral tire stiffness [N/rad]. CA Air resistance coefficient [1/m]. The outputs of the system are the longitudinal vehicle velocity y1(t) = x1(t), the lateral vehicle acceleration (measured by an accelerometer): y2(t) = a_y(t) = 1/m*( (F_x,FL(t) + F_x,FR(t))*sin(delta(t)) + (F_y,FL(t) + F_y,FR(t))*cos(delta(t)) + F_y,RL(t) + F_y,RR(t)) and the yaw rate y3(t) = r(t) (measured by a gyro). Put together, we arrive at the following state-space model structure: d -- x1(t) = x2(t)*x3(t) + 1/m*( Cx*(u1(t)+u2(t))*cos(u5(t)) dt - 2*Cy*(u5(t)-(x2(t)+a*x3(t))/x1(t))*sin(u5(t)) + Cx*(u3(t)+u4(t)) - CA*x1(t)^2) d -- x2(t) = -x1(t)*x3(t) + 1/m*( Cx*(u1(t)+u2(t))*sin(u5(t)) dt + 2*Cy*(u5(t)-(x2(t)+a*x3(t))/x1(t))*cos(u5(t)) + 2*Cy*(b*x3(t)-x2(t))/x1(t)) d -- x3(t) = 1/((0.5*(a+b))^2)*m)*( a*( Cx*(u1(t)+u2(t)*sin(u5(t)) dt + 2*Cy*(u5(t) - (x2(t)+a*x3(t))/x1(t))*cos(u5(t))) - 2*b*Cy*(b*x3(t)-x2(t))/x1(t)) y1(t) = x1(t) y2(t) = 1/m*( Cx*(u1(t)+u2(t))*sin(u5(t)) + 2*Cy*(u5(t)-(x2(t)+a*x3(t))/x1(t))*cos(u5(t)) + 2*Cy*(b*x3(t)-x2(t))/x1(t)) y3(t) = x3(t) Gear Ratio calculation: Choose mean value for initial design, ω5 = 85 rev/min. e = ω5 /ω2 = 85 /1750 = 1/ 20.59 For a compound reverted geartrain, e = 1 /20.59 = N2 N3/ N4 N5
For smallest package size, let both stages be the same reduction. Also, by making the two stages identical, the in-line condition on the input and output shaf will automatically be satisfied N2 /N3 = N4 /N5 = 1/4.54 For this ratio, the minimum number of teeth N2 = N4 = 16 teeth N3 = 4.54(N2) = 72.64 Try rounding down and check if ω5 is within limit ω5 =(16/72)^2 * 1750 = 86.42 rev/min Proceed with N2 = N4 = 16 teeth N3 = N5 = 72 teeth e = 1/20.25 ω5 = 86.42 rev/min ω3 = ω4 = (16/72)* 1750 = 388.9 rev/ min For Torque, T2 = 60.0 lbf · f T3 = 270 lbf · f T5 = 1215 lbf · f

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