Kinematics

This linkage system is unique in that it is a six-bar mechanism that must be divided and analyzed as a five bar then a four bar mechanism. Since we did not cover how to analyze such a five bar mechanism (from L2, L3, L5/2, L4, to f), I attempted to obtain equations for the unknowns in this system (theta3, theta4, theta5, and R6) in a similar way to how equations for a four bar linkage system are found, by manipulating the position vector loop equations and replacing the variable of interest with its 1/2 angle tangent equivalent. The following graphs show the position, velocity, and acceleration for the linkages in this system

Diagram showing vector loops

The position vector loop equations are as follows

eqn1 = L2*cos(theta2) - L5/2*cos(theta5) - (O2O4)*cos(thetaf) - L3*cos(theta3) -L4*cos(theta4) == 0;
eqn2 = L2*sin(theta2)- L5/2*sin(theta5) - (O2O4)*(sin(thetaO2O4)) + L3*sin(theta3) -L4 * sin(theta4)== 0;
eqn3 = -L5/2*sin(theta5) + R4*sin(theta4) == 0;
eqn4 = -R5*cos(theta5) + R6 + f1 - R4 * cos(theta4) == 0;

I was unable to solve for the linkage variables by hand since these vector loop equations resulted in a polynomial to the fourth degree. My hand notes can be found in the appendix. I utilized Matlab's solve() function to attempt to solve for these unknowns.