Designing a Polynomial for an Asymmetrical Rise-Fall Single-Dwell Case.
Redefine the specification from Example 8-8 as:
rise-fall rise 1 in (25.4 mm) in 45° and fall 1 in (25.4 mm) in 135° over 180°
dwell at zero displacement for 180° (low dwell)
cam ω 15 rad/sec

Question

Accepted Answer

Figure 8-31 shows the minimum set of seven BCs for this problem that will give a sixth-degree polynomial. The dwell on either side of the combined rise-fall segment has zero values for S, V, A, and J. The fundamental law of cam design requires that we match these zero values, through the acceleration function, at each end of the rise-fall segment.

2 The endpoints account for six BCs; S = V = A = 0 at each end of the rise-fall segment.

3 We also must specify a value of displacement at the 1-in peak of the rise that occurs at θ = 45°. This is the seventh BC.

4 Simultaneous solution of this equation set gives a 3-4-5-6 polynomial whose equation is:

[latex]s=h\left[151.704\left(\frac{ heta}{\beta} ight)^{3}-455.111\left(\frac{ heta}{\beta} ight)^{4}+455.111\left(\frac{ heta}{\beta} ight)^{5}-151.704\left(\frac{ heta}{\beta} ight)^{6} ight][/latex] (a)

For generality we have substituted the variable h for the specified 1-in rise.

5 Figure 8-31 shows the S V A J diagrams for this solution with its maximum values noted. Observe that the derived sixth-degree polynomial has obeyed the stated boundary conditions and does in fact pass through a displacement of 1 unit at 45°. But note also that it overshoots that point and reaches a height of 2.37 units at its peak. The acceleration peak is also 2.37 times that of the symmetrical case of Example 8-8. Without any additional boundary conditions applied, the function seeks symmetry. Note that the zero velocity point is still at 90° when we would like it to be at 45°. We can try to force the velocity to zero with an additional boundary condition of V = 0 when θ = 45°.

Figure 8-32 shows the S V A J diagrams for a seventh-degree polynomial having 8 BCs, S = V = A = 0 at θ = 0°, S = V = A = 0 at θ =180°, S = 1, V = 0 at θ = 45°. Note that the resulting elsewhere. It now plunges to a negative displacement of –3.934, and the peak acceleration is much larger. This points out an inherent problem in polynomial functions, namely that their behavior between boundary conditions is not controllable and may create undesirable deviations in the follower motion. This problem is exacerbated as the degree of the function increases since it then has more roots and inflection points, thus allowing more oscillations between the boundary conditions.

7 Open the files Ex_08-09a and b in program Dynacam to see this example in greater detail.

Question 8.9: Designing a Polynomial for an Asymmetrical Rise-Fall Single-...

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