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We can carry out the velocity analysis on the mechanism by simply taking the derivative of its positional constitutive equations. As such we have two sets of vector loop equations for each component part of the mechanism.

Part 1: $d_1j\omega_3e^{j\theta_3}-L_2j\omega_2e^{j\theta_2}-\dot{h}j=0$

Part 2: $d_4j\omega_3e^{j\theta_3}+L_6j\omega_6e^{j\theta_6}+(h_2j\omega_6+\dot{h_2})e^{j(\theta_6-\frac{\pi}{2})}+L_5j\omega_5e^{j\theta_5}=0$

Thanks to the linearity of these equations in their respective unknown variables, we can solve explicitly for the solution which makes to code for velocity analysis much more efficient. Solving each of these equations using MATLAB symbolic toolbox we get the following results.

For the first part:

$\omega_2 = \frac{\dot{h_1}sin(\theta_3)}{L_2sin(\theta_2-\theta_3)}$

$\omega_3 = \frac{\dot{h_1}sin(\theta_2)}{d_1sin(\theta_2-\theta_3)}$

And for the second part:

$\omega_5 = -\frac{h_2\dot{h_2}-L_6d_4sin(\theta_3-\theta_6)-d_4h_2cos(\theta_3-\theta_6)}{L_5(L_6sin(\theta_5-\theta_6)+h_2cos(\theta_5-\theta_6))}$

$\omega_6 = \frac{\dot{h_2}sin(\theta_5-\theta_6)+d_4\omega_3sin(\theta_3-\theta_5)}{L_6sin(\theta_5-\theta_6)+h_2cos(\theta_5-\theta_6)}$

Keeping h₂ constant and applying an input sliding motion of -0.3 m/s, we get the following velocity profiles. As expected we can see that links 3, 5, and 6 all move as a unit since they are all part of the folded umbrella and as such should have about the same angular velocities.

For the velocity analysis, we have an additional layer of velocity calculation since for the real mechanism the inputs are not the velocities of h₁ and h₂, but instead the velocity of h₁ and the angular rate of the hand crank ω_c. Therefore we must relate the value θ_c to h₂. Since they are connected by a tight string connected to a spool inside the crank mechanism, the relationship between the two is

\begin{equation}
\dot{h_2} = r_2\omega_c
\end{equation}
where $r_2$ is the radius of the spool inside of the mechanism. 

Keeping h₁ constant at 0.8m and turning the crank clockwise at 2π rad/s, we get the following velocity profiles

We can see that ω₂ and ω₃ are, as expected, unaffected by the unfolding of the umbrella. Both ω₅ and ω₆ start with low velocities and then accelerate to rapidly expand the umbrella at the end of the motion.

Now the only step is to calculate the linear velocities of each point on the mechanism. This is trivial since we already have all of the vector equations for the positions of each of the points. By simply taking the derivatives of these vectors we can solve for the velocities, which can then be plotted. This is well exemplified simply by looking at the code used for calculation of velocities; the process is recursive and starts at point A and radiates out.

%% Solve for Point Velocities
self.Vel.A = 0;
self.Vel.B = self.Vel.A - 1j*h_dot(1);
self.Vel.C = self.Vel.A + 1j*self.Omega(2)*self.L(2)*exp(1j*self.T(2));
self.Vel.G = self.Vel.B + 1j*self.Omega(3)*self.L(3)*exp(1j*self.T(3));
self.Vel.F = self.Vel.G + 1j*self.Omega(6)*self.L(6)*exp(1j*self.T(6));
self.Vel.E = self.Vel.F + (1j*self.Omega(6)*self.h(2) + self.h_dot(2))*exp(1j*(self.T(6) - pi/2));
self.Vel.D = self.Vel.E + 1j*self.Omega(5)*self.L(5)*exp(1j*self.T(5));

Combining all of this information, the following animation shows the umbrella being opened while the slider is simultaneously moved upwards, both at a constant rate. The arrows show the direction and relative magnitude of the velocities of key points on the mechanism, and the subplots show the linear and angular velocities directly on their own scales. view-fileThe schematic of the mechanism is shown here to reference link and angle labels.