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After testing by opening bags by hand, we determined that the force exerted on the bag by the $L_3$ linkage needed to be twice of that required to pull the bag under normal conditions due to the clamping force required to prevent slip. In order to help increase the friction of the $L_3$ linkage (and thereby reduce the amount of clamping force required), a rubber strip was applied to the inside of the wooden box’s hook mechanism. Additionally, gearing for the input motor with a 1:7 ratio was added to ensure the motor would be able to apply enough torque for the given additional friction. After these informal tests, we shifted our views to a virtual MATLAB and SolidWorks simulation, where we performed the motion analysis for several significant parameters in the mechanism, mostly in regards to $\theta_2$, the angle of the crank gear. |
A complete document containing all of the MATLAB code and the SolidWorks assembly can be found within the Appendix section of this project page.
Figure #3.2: Diagrams showcasing a counter-clockwise rotation of the machine, demonstrating how the mechanism pulls the slider bar backwards, in addition to defining many of the significant parameters used during the MATLAB analysis
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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 using the Motion Analysis add-on, as seen in the annotated video below:
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Analysis - In Detail
The following section details our motion analysis of the system. The dimensions used in this analysis is shown below in Figure #3.5
Figure #3.5: Diagram showing the basic dimensions of the mechanism
The system's motion is divided up into 4 parts.
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For the boundary between Parts A and B:
Figure #3.6: Diagram showing the basic dimensions of the mechanism at the boundary between Parts A and B
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$$\theta_2=\arccos(\frac{a^2+(a+l_{min})^2-(s+l_{min})^2}{2a(a+l_{min})})$$ |
For the boundary between Parts C and D:
Figure #3.7: Diagram showing the basic dimensions of the mechanism at the boundary between Parts C and D
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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. |
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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.
Figure #3.8: Diagram showing the vector loop used for analyzing Parts A and C
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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