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For this section, the input gear reduction ratio will be ignored so the input is assumed to be the 'L'  link 2 position.

Figure 3, 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 4, 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 5, 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:

ParameterValue [in]
a0.9375 - j0.3125
b2.5
cy0.25
d

2.1875

f2.1875
g3.75
h3.75
k0.625
Table 1, parameter values


Figure 6,  all position responses solved for.

Here we can note that the claws intersect when θ2 is 205.7 degrees.



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