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

Question

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 effector and must chase a constant velocity conveyor for at least 6 inches at 10 in/sec with minimal velocity error and then return to the start position to repeat the cycle. The motion functions are applied to the slider and converted to modified functions to be applied to the cam-follower arm with a modified contour cam running at 1 Hz.
Compare the displacement, velocity, and acceleration functions between the cam follower (link 2) and the end effector of the linkage (link 4) and compare the cam profile for this solution to that of Example 16-1.

Accepted Answer

1 The linkage geometry is the same as shown in Figure 16-3.

2 The equations that define the motion of the slider-crank when driven at the slider rather than at the crank take on a different form than those shown in Example 16-1. The vector loop equation for both cases is the same, so equations 4.14a, b, c, and 4.15a, b apply to both cases. The derivation starts with the above noted equations for either mode of operation. The simultaneous solution of equations 14.15a, b for the crank-driven mode is done by simple substitution as shown in equations 4-16a, b. However, solving for the slider-driven mode with slider position d as the input is more complex because both variables now sought, [latex] heta_{2} ext { and } heta_{3} [/latex], are contained in transcendental expressions in both equations 4.15a and 4.14b. This requires the derivation done in equations 4.18 to 4.21 to find [latex] heta_{2} ext { and } heta_{3} [/latex] as functions of slider position d.

[latex] R _{2}- R _{3}- R _{4}- R _{1}=0 [/latex] (4.14a)

[latex] a e^{j heta_{2}}-b e^{j heta_{3}}-c e^{j heta_{4}}-d e^{j heta_{1}}=0 [/latex] (4.14b)

[latex] a \cos heta_{2}-b \cos heta_{3}-c \cos heta_{4}-d=0 [/latex] (4.15a)

[latex] heta_{3_{1}}=\arcsin \left(\frac{a \sin heta_{2}-c}{b} ight) [/latex] (4.16a)

[latex] d=a \cos heta_{2}-b \cos heta_{3} [/latex] (4.16b)

[latex]\begin{array}{r}a \cos heta_{2}-b \cos heta_{3}-d=0 \a \sin heta_{2}-b \sin heta_{3}-c=0\end{array}[/latex] (4.18)

[latex] a^{2}-b^{2}+c^{2}+d^{2}-2 a c \sin heta_{2}-2 a d \cos heta_{2}=0 [/latex] (4.19)

[latex] K_{1}+K_{2} \sin heta_{2}+K_{3} \cos heta_{2}=0 [/latex] (4.20)

[latex] heta_{2_{1,2}}=2 \arctan \left(\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A} ight) [/latex] (4.21)

3 The derivation for crank velocity is done in equations 6.24 and for crank acceleration in equations 7.17. The results of these derivations for [latex] heta_{2}, \omega_{2}, ext { and } \alpha_{2} [/latex], as a function of slider position d, velocity [latex] \dot{d}, ext { acceleration } \ddot{d} [/latex] , and linkage geometry a, b, c (shown in Figure 16-3) are used in the following steps.

[latex] \omega_{3}=\frac{a \omega_{2} \cos heta_{2}}{b \cos heta_{3}} [/latex] (6.24a)

[latex] \omega_{2}=\frac{\dot{d} \cos heta_{3}}{a\left(\cos heta_{2} \sin heta_{3}-\sin heta_{2} \cos heta_{3} ight)} [/latex] (6.24b)

[latex] \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}} [/latex] (7.17a)

[latex] \alpha_{2}=\frac{a \omega_{2}^{2}\left(\cos heta_{2} \cos heta_{3}+\sin heta_{2} \sin heta_{3} ight)-b \omega_{3}^{2}+\ddot{d} \cos heta_{3}}{a\left(\cos heta_{2} \sin heta_{3}-\sin heta_{2} \cos heta_{3} ight)} [/latex] (7.17b)

[latex]\begin{aligned}K_{1} &=a^{2}-b^{2}+c^{2}+d^{2} \K_{2} &=-2 a c \K_{3} &=-2 a d \A &=K_{1}-K_{3} \B &=2 K_{2} \C &=K_{1}+K_{3}\end{aligned}[/latex]

[latex] heta_{2_{1,2}}=2 \arctan \left(\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A} ight) [/latex] (16.3)

4 The angular velocities [latex] \omega_{2} ext { and } \omega_{3} [/latex] as a function of slider velocity and link geometry are:

[latex] \begin{aligned}&\omega_{2}=\frac{\dot{d} \cos heta_{3}}{a\left(\cos heta_{2} \sin heta_{3}-\sin heta_{2} \cos heta_{3} ight)} \&\omega_{3}=\frac{a \omega_{2} \cos heta_{2}}{b \cos heta_{3}}\end{aligned}[/latex] (16.4)

5 The angular accelerations [latex] \alpha_{2} ext { and } \alpha_{3} [/latex] as a function of slider acceleration and link geometry are:

[latex] \begin{aligned}&\alpha_{2}=\frac{b \omega_{3}^{2}-a \omega_{2}^{2} \cos \left( heta_{2}- heta_{3} ight)-\ddot{d} \cos heta_{3}}{a \sin \left( heta_{2}- heta_{3} ight)} \&\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}}\end{aligned}[/latex] (16.5)

6 The cam program has two segments: 220° of constant velocity at 10in/s over 30° of follower arm (crank) motion, and a fifth-degree polynomial return motion over 140°. Total displacement is 6.767 in.

7 The boundary conditions for the polynomial in segment 1 are shown in Table 16-5. The boundary conditions for the polynomial in segment 2 are shown in Table 16-6. The s, v, a functions as derived with program Dynacam are:

Segment 1:

[latex]\begin{aligned}S &=6.114 x ext { in } \V &=\frac{\omega}{\beta}(6.114)=\frac{2 \pi}{3.8397}(6.114)=10 in / sec \A &=0\end{aligned}[/latex] (16.6)

Segment 2:

[latex]\begin{aligned}&S=-60.029 x^{5}+150.073 x^{4}-100.049 x^{3}+3.891 x+6.114 ext { in } \&V=\frac{\omega}{\beta}\left(-300.145 x^{4}+600.292 x^{3}-300.147 x^{2}+3.891 ight) ext { in/sec } \&A=\frac{\omega^{2}}{\beta^{2}}\left(-1200.58 x^{3}+1800.88 x^{2}-600.294 x ight) ext { in/sec }^{2}\end{aligned}[/latex] (16.7)

8 Equation 16.6 applies to the first 220° and equation 16.7 to the last 140°. The normalized variable x = θ/β runs from 0 to 1. The S, V, A diagrams for these polynomial motions as applied to the slider are shown in Figure 16-7. Note that the displacement is now in length units rather than degrees as was the case in the previous example (Figure 16-4).

9 The equations for the kinematic behavior of the linkage from equations 16.3 through 16.5 must be solved at a set of discrete points (e.g., every degree of the cam cycle) using the functions of equations 16.7 and 16.8 as input to the slider to determine the behavior of the crank arm. The crank arm displacement is then used to calculate the required cam profile.

10 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. The problem can be solved numerically using discrete representations of the motion functions calculated at every degree or fraction thereof.

11 Figure 16-8a shows the displacement of the slider and the crank arm superposed and normalized to percent for comparison. The nonlinear behavior of the linkage distorts the entire function. Figure 16-8b shows the normalized velocities of follower arm and slider superposed.

Note that the slider has constant linear velocity during the first segment but the crank arm does not have constant angular velocity.

12 Figure 16-9 shows the percent difference between the crank arm and slider in both displacement and velocity. The distorted crank arm functions serve to correct the error introduced by linkage geometry and give the designed motions at the slider.

13 Figure 16-10 shows the difference in radius to the roller center for the cam of Example 16-1 and that of Example 16-2. The first has the desired functions applied to its follower. The second has the functions applied to the slider with the cam profile recalculated to remove the linkage geometry error. The largest radial difference between the cams is nearly 0.2 in.

TABLE 16-5
Boundary
Conditions

Example 16-2, Segment 1

When θ = 0:
S = 0,
V = 6.114 in/sec

TABLE 16-6
Boundary
Conditions

Example 16-2, Segment 2

When θ = 0:
s = 6.114 210 in,
v = 10 in/sec,
a = 0
When θ = 140°
s = 0,
v = 10 in/sec,
a = 0

Question 16.2: Cam-Driven Fourbar Slider With Motion Functions Applied to t...

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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 ...

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