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Figure #3.3: Complete compilation of all MATLAB motion analysis graphs, featuring θ3, θp, d, b, ω3, ddot, and bdot as a function of θ2, in addition to the position of point P (the end of the claw) in relation to O2
Due to the nature of our machine having two degrees of freedom rather than simply just having one, the mechanism's full range linkage operation could not be rendered in a tool such as Planar Mechanism Kinematic Simulator. Instead, the linkage operation was animated within SolidWorks, as seen in the annotated video below:
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Latex formatting Part A: The internal slider is the only linearly moving component and effectively lengthens the connecting rod. The slider is stationary. This opens the gripper until $b=s+l_{min}$ .Latex formatting Part B: The internal slider bottoms out at its maximum, and the connecting rod’s length becomes constant at $b=s+l_{min}$. The mechanism acts as a slider crank. This pulls back the gripper until $d=s+l_{min}-a$ when the crank is at $180\degree$.
Latex formatting Part C: The internal slider is the only linearly moving component and effectively shortens the connecting rod. The slider is stationary. This closes the gripper until $b=l_{min}$ .Latex formatting Part D: The internal slider bottoms out at its minimum, and the connecting rod’s length becomes constant at $b=l_{min}$. The mechanism acts as a slider crank. This pushes the gripper forward until $b$d=a+l_{min}$ when the crank is at $0\degree$.
Thus, for Parts A and C, length b changes, and length d is constant. For Parts B and D, the mechanism is a simple slider crank with length b constant and length d changing. Thus, Parts B and D can be analyzed using the slider crank motion equations while Parts A and C can be solved using vector loop analysis.
Boundary Conditions
The boundary conditions governing the transitions between Parts A and B and between Parts C and D are based on the geometry of the mechanism when the internal slider bottoms out. For this, the Law of Cosines can be used.
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$$\theta_2=360\degree-\arccos(\frac{a^2+(s+l_{min}-a)^2-(l_{min})^2}{2a(s+l_{min}-a)})$$
Note that the boundaries between Parts D and A and between Parts B and C are 0 degrees and 180 degrees respectively since the joints $O_2$ and $O_4$ are collinear with the global X axis. |
Vector Loop Analysis
Parts A and
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C
For Parts A and C, the vector loop analyses are the same, but the lengths are different. So, to more easily solve them, a generic system can be solved once and the dimensions for Parts A and C applied to it.
Position Analysis
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$$\vec{R_2}-\vec{R_3}-\vec{R_1}=0$$
$$ae^{j\theta_2}-be^{j\theta_3}-de^{j\theta_1}=0$$
$$a(\cos{\theta_2}-j\sin{\theta_2})-b(\cos{\theta_3}-j\sin{\theta_3})-d=0$$
$$\begin{bmatrix}
b\cos{\theta_3}=a\cos{\theta_2}-d \\
b\sin{\theta_3}=a\sin{\theta_2}
\end{bmatrix}$$
$$\begin{bmatrix}
\theta_3=\arctan(\frac{a\sin{\theta_2}}{a\cos{\theta_2}-d})\\
b=\sqrt{(a\cos{\theta_2}-d)^2+(a\sin{\theta_2})^2}
\end{bmatrix}$$ |
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