For this section, the input gear reduction ratio will be ignored so the input is assumed to be the 'L' link 2 position.
Figure 2, animation of the mechanism in motion.
Derivation
The system as mentioned earlier consists of a driving crank slider that is coupled with a driven clank slider that manipulates the rest of the structure.
Figure 3, Initial linkages of the clank sliders.
In order to find the angles of links 4 and 5, I need to find the position of point C, the sliding linkage.
Driver Vector Loop
.................................................(Eq. 1)
...........................(Eq. 2)
Where Cy is a constant value and Cx is a function of θ2. This loop was put into a math solver to find Cx .
Driven Vector Loop
.................................................(Eq. 3)
...........................(Eq. 4)
Which can again be put into a solver to find the proximal angle, θ4, and the distal angle, θ5
These angles are important because the rest of the mechanism is parallel to links 4 and 5. This means no other vector loops are necessary.
End Effector Position
Figure 4, all linkages of the mechanism
Solving for the end effector is done by using links 4 and 5 as building block vectors, defining point A as origin, and defining point C as a function of θ2.
....................................................................................(Eq. 5)
...................................................................................(Eq. 6)
.............................(Eq. 7)
...........................................(Eq. 8)
Position Plot
Figure 6, all position responses solved for.
Here we can note that the claws intersect when θ2 is 205.7 degrees.