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
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.
Matlab Equations for Position Analysis, where L5a and L5b are two halves of linkage 5
I squared the first two equations to obtain equations with only one unknown for theta3 and theta4
These two equations returned a very long logarithmic expression for both theta3 and theta4
Therefore, the following plots are provided.
Kinematic Plots of Slider
Kinematic Plots of Linkage 5
Kinematic Plots of Linkage 4
Kinematic Plots of Linkage 3
The following animation is provided to show simulated movement
