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)
** This vector does not represent the coordinates of point D, but rather the vector from point C to point D
Position Plot
I input the following values for the parameters:
Parameter | Value [in] |
---|---|
a | 0.9375 - j0.3125 |
b | 2.5 |
cy | 0.25 |
d | 2.1875 |
f | 2.1875 |
g | 3.75 |
h | 3.75 |
k | 0.625 |
Table 1, parameter values
Figure 5, all position responses solved for.
Here we can note that the claws intersect when θ2 is 205.7 degrees.