Figure #3.1: Further crank design and hand calculations
After coming to a rough finalization of our crank arm design, basic calculations of the crank’s kinematic diagram showed that in order to open the bag, the force exerted on the bag by the L7L6 linkage needed to be twice of that required to pull the bag under normal conditions. In order to help increase the force of the L7L6 linkage, a rubber strip was applied to the inside of the wooden box’s hook mechanism in order to increase the friction between the hook and the bag, allowing for the crank to apply more force thanks to the associated friction helping pull one side of the bag in the opposite direction of the crank. After these hand calculations, we shifted our views to a virtual MATLAB simulation, where we performed the motion analysis for several significant parameters in the mechanism, mostly in regards to θ2, the angle of the motor gear. A complete document containing all of the MATLAB code can be found within the Appendix section of this project page.
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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, as seen in the annotated video below:
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Figure #3.4: Annotated animation of the full assembly as seen within SolidWorks (NOTE: Due to the large filesize of the video, the video may need to be downloaded to view.)
As seen in the video provided in Figure #3.4, the mechanism travels between four different parts over the course of one single cycle. Part A being the beginning of the counter-clockwise rotation with the motor pin starting at a 0° angle until the slider above the motor gear reaches its leftmost position in order to have the "claw" part of the arm clamp down on the bag. From here, Part B begins, consisting of the slider arm pulling backwards until the motor pin reaches an 180° angle. Part C and Part D are close to being inverses of Part A and Part B respectively, with Part C involving the motor gear slider moving back to its rightmost position in order to release the "claw" part of the arm from its clamping position and Part D moving the slider arm back forward. After Part D, Part A begins once again, repeating the entire process. These four parts are annotated within all eight graphs in Figure #3.3.
Analysis - Continued
The following details our motion analysis of the system.
The system's motion is divided up into 4 parts.
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. The mechanism acts as a slider crank. This pulls back the gripper until $d=s+l_{min}-a$ .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. The mechanism acts as a slider crank. This pushes the gripper forward until $b=l_{min}$ .
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 and can be analyzed using the slider crank motion equations.
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.
For 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:
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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 B
For Parts A and C, the vector loop analysis is the same, but the lengths will be different.
Position Analysis
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$$\vec{R_2}-\vec{R_3}-\vec{R_1}=0$$
$$ae^{j\theta_2}-be^{j\theta_3}-de^{j\theta_1}=0$$
$$a(\cos{\theta_2}-j\sin{\theta_2})-b(\cos{\theta_3}-j\sin{\theta_3})-d=0$$
$$\begin{bmatrix}
b\cos{\theta_3}=a\cos{\theta_2}-d \\
b\sin{\theta_3}=a\sin{\theta_2}
\end{bmatrix}$$
$$\begin{bmatrix}
\theta_3=\arctan(\frac{a\sin{\theta_2}}{a\cos{\theta_2}-d})\\
b=\sqrt{(a\cos{\theta_2}-d)^2+(a\sin{\theta_2})^2}
\end{bmatrix}$$ |
Velocity Analysis
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$$ae^{j\theta_2}-be^{j\theta_3}-de^{j\theta_1}=0$$
$$aj\omega_2e^{j\theta_2}-\dot{b}e^{j\theta_3}-b\omega_3je^{j\theta_3}-0=0$$
$$aj\omega_2(\cos{\theta_2}-j\sin{\theta_2})-\dot{b}(\cos{\theta_3}-j\sin{\theta_3})-b\omega_3j(\cos{\theta_3}-j\sin{\theta_3})=0$$
$$\begin{bmatrix}
a\omega_2\sin{\theta_2}-\dot{b}\cos{\theta_3}+b\omega_3\sin{\theta_3}=0\\
a\omega_2\cos{\theta_2}-\dot{b}\sin{\theta_3}-b\omega_3\cos{\theta_3}=0
\end{bmatrix}$$
$$a\omega_2\sin{\theta_2}\sin{\theta_3}-a\omega_2\cos{\theta_2}\cos{\theta_3}+b\omega_3\sin^2{\theta_3}+b\omega_3\cos^2{\theta_3}=0$$
$$-a\omega_2\cos(\theta_2-\theta_3)+b\omega_3=0$$
$$\begin{bmatrix}
\omega_3=\frac{a}{b}\omega_2\cos(\theta_2-\theta_3)\\
\dot{b}=\frac{a\omega_2\cos{\theta_2}-b\omega_3\cos{\theta_3}}{\sin{\theta_3}}
\end{bmatrix}$$ |
Result
For Part A:
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$$\theta_3=\arctan(\frac{a\sin{\theta_2}}{a\cos{\theta_2}-(a+l_{min}})$$
$$d=a+l_{min}$$
$$b=\sqrt{(a\cos{\theta_2}-(a+l_{min}))^2+(a\sin{\theta_2})^2}$$
$$\omega_3=\frac{a}{b}\omega_2\cos(\theta_2-\theta_3)$$
$$\dot{d}=0$$
$$\dot{b}=\frac{a\omega_2\cos{\theta_2}-b\omega_3\cos{\theta_3}}{\sin{\theta_3}}$$ |
For Part C:
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$$\theta_3=\arctan(\frac{a\sin{\theta_2}}{a\cos{\theta_2}-(s+l_{min}-a})$$ $$d=s+l_{min}-a$$ $$b=\sqrt{(a\cos{\theta_2}-(s+l_{min}-a))^2+(a\sin{\theta_2})^2}$$ $$\omega_3=\frac{a}{b}\omega_2\cos(\theta_2-\theta_3)$$ $$\dot{d}=0$$ $$\dot{b}=\frac{a\omega_2\cos{\theta_2}-b\omega_3\cos{\theta_3}}{\sin{\theta_3}}$$ |
Parts B and D
For Parts B and D, the equations for a simple slider crank can be used.
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$$\theta_3=\arcsin(\frac{a\sin{\theta_2}}{b})$$
$$d=a\cos{\theta_2}-b\cos{\theta_3}$$
$$\omega_3=\frac{a\cos{\theta_2}}{b\cos{\theta_3}}\omega_2$$
$$\dot{d}=-a\omega_2\sin{\theta_2}+b\omega_3\sin{\theta_3}$$ |
Result
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For Part B, $b=s+l_{min}$:
$$\theta_3=\arcsin(\frac{a\sin{\theta_2}}{s+l_{min}})$$
$$d=a\cos{\theta_2}-(s+l_{min})\cos{\theta_3}$$
$$b=s+l_{min}$$
$$\omega_3=\frac{a\cos{\theta_2}}{(s+l_{min})\cos{\theta_3}}\omega_2$$
$$\dot{d}=-a\omega_2\sin{\theta_2}+(s+l_{min})\omega_3\sin{\theta_3}$$
$$\dot{b}=0$$
For Part D, $b=l_{min}$:
$$\theta_3=\arcsin(\frac{a\sin{\theta_2}}{l_{min}})$$
$$d=a\cos{\theta_2}-l_{min}\cos{\theta_3}$$
$$b=l_{min}$$
$$\omega_3=\frac{a\cos{\theta_2}}{l_{min}\cos{\theta_3}}\omega_2$$
$$\dot{d}=-a\omega_2\sin{\theta_2}+l_{min}\omega_3\sin{\theta_3}$$
$$\dot{b}=0$$ |