To perform our kinematic analysis, we simply set up a vector loop equation and solved for the unknowns using position analysis, velocity analysis and finally acceleration analysis.
We made sure to solve for all of our variables symbolically at first so that we could easily input them into MATLAB for graphing and further analysis. The hand derivation of the variables of interest are shown below:
Here, we set up the vector loop equation, solved for degrees of freedom and derived the position analysis to get results for velocity. The next few steps are essentially just deriving and reorganizing terms to find the values of interest. The acceleration analysis is as shown below:
After solving for these values symbolically, we can input them into MATLAB, changing the crank's angle value to get all of the necessary graphs and kinematic information. The code is as shown below:
clear all;
theta1 = atand(25/125); % degrees
AB = 65.91; % mm
BC = 205.234; % mm
AD = sqrt(125^2 + 25^2); % mm
w2 = 10.5; % rad/s
for i = 1:361
theta2(i) = i;
theta3(i) = asind((AB*sind(theta2(i))-AD*sind(theta1))/BC);
DC(i) = AB*cosd(theta2(i)) + BC*cosd(theta3(i)) - AD*cosd(theta1);
w3(i) = (AB*w2*cosd(theta2(i))/(BC*cosd(theta3(i))));
DCdot(i) = -BC*w3(i)*sind(theta3(i)) - AB*w2*sind(theta2(i));
a3(i) = (-AB*(w2^2)* sind(theta3(i)) + BC*(w3(i)^2)*sind(theta3(i)))/(BC*cosd(theta3(i)));
DCdotdot(i) = -AB*(w2^2)*cosd(theta2(i)) - BC*a3(i)*sind(theta3(i)) - BC*(w3(i)^2)*cosd(theta3(i));
end
The code above allowed us to plot the graphs below, changing the variable on the y-axis to a different variable of interest for each case. The plots are as shown below:
The main graphs of interest are the angular position of link 3, angular velocity of link 3 and angular acceleration of link 3. These graphs confirmed the motion we needed for our sauté machine given that the crank moves at a constant and smooth rate. From the acceleration and velocity graphs, we can see that link 3 undergoes a quick change in direction, giving the mechanism the "flick" required.