Cam-Driven Fourbar Slider With Motion Functions Applied to the Input Link Given: A cam-driven fourbar crank-slider with the geometry shown in Figure 16-3 is driven by a cam with a constant velocity motion program similar to that developed in Example 8-12 and Figure 8-42. The slider is the end

Question

Cam-Driven Fourbar Slider With Motion Functions Applied to the Input Link
Given: A cam-driven fourbar crank-slider with the geometry shown in Figure 16-3 is driven by a cam with a constant velocity motion program similar to that developed in Example 8-12 and Figure 8-42. The slider is the end effector and must chase a constant velocity conveyor for at least 6 in at 10 in/sec with minimal velocity error and return to the start position to repeat the cycle.* The motion functions are applied to the cam-follower arm. The linkage was designed in a CAD system, which determined the geometry of Figure 16-3 and that 30° of crank rotation gives 6.435 in of constant velocity motion at 1 Hz.
Compare the displacement and velocity functions applied to the cam follower (link 2) and the resulting functions at the end effector of the linkage (link 4).

Accepted Answer

1 The equations for the position of the crank-slider linkage were developed in Section 4.6, its velocity in Section 6.7, and acceleration in Section 7.3 and are repeated here for your convenience.

[latex]\begin{gathered} heta_{3_{1}}=\arcsin \left(\frac{a \sin heta_{2}-c}{b} ight) (4.16a) \d=a \cos heta_{2}-b \cos heta_{3} (4.16b)\\omega_{3}=\frac{a}{b} \frac{\cos heta_{2}}{\cos heta_{3}} \omega_{2} (6.22a)\\dot{d}=-a \omega_{2} \sin heta_{2}+b \omega_{3} \sin heta_{3} (6.22b)\\alpha_{3}=\frac{a \alpha_{2} \cos heta_{2}-a \omega_{2}^{2} \sin heta_{2}+b \omega_{3}^{2} \sin heta_{3}}{b \cos heta_{3}} (7.16d)\\ddot{d}=-a \alpha_{2} \sin heta_{2}-a \omega_{2}^{2} \cos heta_{2}+b \alpha_{3} \sin heta_{3}+b \omega_{3}^{2} \cos heta_{3} (7.16e)\end{gathered}[/latex]

The terminology used in the equations is shown in Figure 16-3.

2 The cam program has two segments: 220° of constant velocity at 49.066 05°/s for 30° of follower arm (crank) motion, and a fifth-degree polynomial return motion over 140°. Total angular displacement of the crank is 33.2° in segment 2.

3 The boundary conditions for the polynomial in segment 1 are shown in Table 16-3.
The boundary conditions for the polynomial in segment 2 are shown in Table 16-4.
The normalized displacement equations for the cam functions as derived with program Dynacam with x = θ/β are as follows:†

Segment 1:

[latex]\begin{aligned}&S=30.00 x \operatorname{deg} \&V=\frac{\omega}{\beta}(30)=\frac{2 \pi}{3.8397}(30)=49.066 deg / sec \&A=\frac{\omega^{2}}{\beta^{2}}(0) deg / sec ^{2}\end{aligned}[/latex] (16.1)

Segment 2:

[latex]\begin{aligned}&S=-294.546 x^{5}+736.364 x^{4}-490.909 x^{3}+19.091 x+30.00 deg \&V=\frac{\omega}{\beta}\left(-1472.73 x^{4}+2495.46 x^{3}-1472.73 x^{2}+19.091 ight) deg / sec \&A=\frac{\omega^{2}}{\beta^{2}}\left(-5890.91 x^{3}+7486.38 x^{2}-2945.46 x ight) deg / sec ^{2}\end{aligned}[/latex] (16.2)

4 These S, V, A equations are the driving functions for the input crank of the linkage where w = 2p rad/sec for a 1-Hz cycle, and b is the angular duration of the motion in radians—3.8397 rad for segment 1 and 2.4435 rad for segment 2.

5 The equations for the kinematic behavior of the linkage from step 1 must be solved at a set of discrete points (e.g., every degree of the cam cycle) to determine the behavior of the slider d. Equation 16.1 applies to the first 220° (constant velocity) and equation 16.2 to the last 140°.

6 This is obviously a job for a computer and one was used to solve these equations. A good approach is to use an equation solver such as Matlab, Mathcad, or TKSolver. Plots of the polynomial motions from program Dynacam are shown in Figure 16-4.

7 Figure 16-5a shows the displacement of the follower arm and the slider superposed and normalized to percent for comparison. The nonlinear behavior of the linkage distorts the entire function. Figure 16-5b shows the normalized velocities of follower arm and slider superposed. Note that the follower arm has constant angular velocity during the first segment but the slider does not have constant linear velocity.

8 Figure 16-6 shows the percent error between the crank arm and slider in both displacement and velocity. The differences in this example are not extreme, mainly because the crank only rotates through 30° during the constant velocity motion. The larger the angle of crank rotation, the larger the distortion in the output functions.

9 If the magnitude of the deviation from constant velocity is small enough to still allow proper function, then this solution will be adequate as is. If the error is not tolerable, then the solution is to apply the desired motion function to the slider and calculate the linkage’s deviation in reverse to redefine the cam function to correct it. The next example will do this.

TABLE 16-3
Boundary
Conditions

Example 16-1, Segment 1

When θ = 0:
s = 0°,
v = 49.066 05 °/sec

TABLE 16-4
Boundary
Conditions

Example 16-1, Segment 2

When θ = 0:
s = 30.000 05°,
v = 49.066 05 °/sec,
a = 0
When θ = 140°
s = 0°,
v = 49.066 05 °/sec
a = 0

Question 16.1: Cam-Driven Fourbar Slider With Motion Functions Applied to t...

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Cam-Driven Fourbar Slider With Motion Functions Applied to the Output Link Given: A cam-driven fourbar crank-slider with the geometry shown in Figure 16-3 is driven by a cam with a constant velocity motion program similar to that developed in Example 8-12 and Figure 8-42. The slider is the end ...

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