...
Figure 1: Motion of the Main Mechanism.
We can consider the mechanism to be divided into two parts:
...
Both links can be modeled as a slotted link with the physical slots being grounded pins on which the links rotate and slide.
First Mechanism
...
Figure 2: First Part of Main Mechanism.
By solving the loop equation for the first part of the mechanism we get 2 equations to define the rotation θ3 and length of b1.
...
Figures 3 & 4: Length b1 (cm) and Angle θ3(degrees).
Second Mechanism
...
The second part of the mechanism depends on the angle and length of link b. The top part of link b extends into the body of the bear and drives the motion of link d. We define the angle of the top part of link b as θ32 and the length that extends into the body as b2.
Figure 5: Second Part of Main Mechanism.
In general the angle of b2 is the same as b1, the Matlab code, however, outputs the angle as a negative, since it is calculating the vector going down. We can simply use the absolute value of the angle to get θ32. The length b2 is then defined as the difference between length b and length b1. The length changes depending on the angle of the crank.
| Latex formatting |
|---|
\begin{equation}
b_2 = b - b_1\\
\end{equation}
\begin{equation}
\theta_{32} = |\theta_3|
\end{equation} |
The height, f, and offset, g, of the top slot remains remain constant so we can find the length of d1 with simple trigonometry.
| Latex formatting |
|---|
\begin{equation}
d_{1x} = g - b_2\cos\theta_{32}
\end{equation}
\begin{equation}
d_{1y} = f - b_2\sin\theta_{32}
\end{equation}
\begin{equation}
d_1 = \sqrt{d_{x}^2 + d_{y}^2}
\end{equation} |
Then we can find theta4 to finish the math for the main mechanism.
| Latex formatting |
|---|
\begin{equation} \theta_4 = \arctan({b_2\sin\theta_{32}-f}{b_2\cos\theta_{32}-g}) \end{equation} |
Matlab outputs the following plots for d1 and θ4:
Figures 6 & 7: Length d1 (cm) and Angle θ4(degrees).