1. Position Analysis

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

$\vec{R}_2 + \vec{R}_3 - \vec{R}_1 = \vec{0}$ .................................................(Eq. 1)

$a e^{j\theta_2} + b e^{j\theta_3} - c_x - c_y = 0$ ...........................(Eq. 2)

Where C_yis a constant value and C_x is a function of θ₂. This loop was put into a math solver to find C_x .

Driven Vector Loop

$\vec{R}_4 - \vec{R}_5 - \vec{R}_1 = \vec{0}$ .................................................(Eq. 3)

$de^{j\theta_4} - fe^{j\theta_5} - c_x - c_y = 0$ ...........................(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.

test

test2

$\vec{E} = ge^{j\theta_4}$ ...................................................................(Eq. 5)

$\vec{D} = he^{j\theta_5}$ ..............................(Eq. 6)

$\vec{M} = \vec{C} + \vec{E} + 2\vec{D} + ke^{j(\theta_5 - 90)}$ ..............................(Eq. 5)

$\vec{N} = 2\vec{E} + \vec{D} + ke^{j(\theta_4 + 90)}$ ...................................................................(Eq. 5)