Question 16.4: Improvement to the Design of Example 16-3 Given: The design ...
Improvement to the Design of Example 16-3
Given: The design of Example 16-3 proved to be dynamically poor. In that example the required position and velocity constraints were applied to the slider motion and the resulting function was transformed by equations relating slider motion to crank motion. The constraints require that the slider move through a 2-in stroke ending at TDC with crank angle \theta_{2}=0 . The slider needs to be brought to zero velocity for an instant at 1.588-in stroke then advanced into the material. It is then pulled back to dwell at 1.400 in before returning to zero. Rather than apply those constraints directly to the slider motion as was done in Example 16-3, we will instead transform the constraints from linear parameters to angular parameters first and then apply those angular parameters to the motion of the servo-driven crank arm. Machine speed is 120 cycles per min or 2 Hz.
Find: The motion function needed to program a servomotor to drive a crank-slider linkage that provides the required linkage output motion parameters as defined for specific positions of the slider.
Assume: The position and velocity accuracy required at intermediate points in the stroke only requires that those specific parameters be transformed to corresponding parameters at the servo axis. The shape of the functions between those points is not critical.
1 The functions chosen to provide the motions will be similar to those defined in step 1 of Example 16-3 but will instead be applied directly to the crank arm motion.
2 The start and end angles \theta_{\text {start }}=71.564^{\circ}, \theta_{\text {end }}=0^{\circ} required to achieve the 2-in stroke were calculated in Step 2 of Example 16-3 and will be the same here.
3 The crank angles corresponding to the intermediate positions can be calculated by the same method used for the start angle in step 2 of Example 16-3. There must be a point d_{1} having instantaneous zero velocity when the slider has moved through 1.588 in of its 2-in stroke. The start position of the slider as referenced to the crank pivot was calculated in step 2 of Example 16-3 to be 7.8 in. The point d_{1} will be at 7.8 + 1.588 = 9.388 in. The corresponding crank angle for that position is found from equations 16.4 to be 29.721°.
\begin{aligned}\omega_{2} &=\frac{\dot{d} \cos \theta_{3}}{a\left(\cos \theta_{2} \sin \theta_{3}-\sin \theta_{2} \cos \theta_{3}\right)} \\\omega_{3} &=\frac{a \omega_{2} \cos \theta_{2}}{b \cos \theta_{3}}\end{aligned} (16.4)
4 The dwell at a slider stroke of 1.4 in is 7.8 + 1.4 = 9.6 in from the crank pivot. Equations 16.4 show this slider position to correspond to a crank angle of 36.10°.
5 The boundary conditions for crank motion then become:
Start at crank angle of 71.564°.
Segment 1: B-spline over 60° with the boundary conditions of Table 16-8.*
Segment 2: Dwell at zero crank degrees for 20°.
Segment 3: Cycloidal rise to 36.19° over 20°.
Segment 4: Dwell at 36.19° for 20°.
Segment 5: Cycloidal rise to 71.564° over 60°.
Segment 6: Dwell at 71.564° for 180°.
6 The resulting function superposed on the solution for Example 16-3 is shown in Figure 16-15. Note the smoother contour approaching 60 and leaving 80 timing degrees. The slope discontinuity is gone but the critical points are still being maintained. This is a dynamically superior design to that of Example 16-3.
| TABLE 16-8 Boundary Conditions |
| Example 16-4, Segment 1 |
| When θ = 0: s = 71.546°, v = 0, a = 0 When θ = 35°: s = 29.721°, v = 0 When θ = 60°: s = 0°, v = 0, a = 0 |
