1. **Problem Statement:** We need to find the values of $x$, $y$, $\theta$, and $\Phi$ at three precision points for the function $$y = x^{1.6}$$ over the range $1 \leq x \leq 4$. The input link moves from $30^\circ$ to $120^\circ$, and the output link moves from $90^\circ$ to $180^\circ$. 2. **Identify the precision points:** Since we have three precision points, we select three $x$ values evenly spaced in the range $[1,4]$: $$x_1 = 1, \quad x_2 = 2.5, \quad x_3 = 4$$ 3. **Calculate corresponding $y$ values:** Using the function $y = x^{1.6}$: $$y_1 = 1^{1.6} = 1$$ $$y_2 = 2.5^{1.6} = e^{1.6 \ln(2.5)}$$ Calculate $\ln(2.5) \approx 0.9163$: $$y_2 = e^{1.6 \times 0.9163} = e^{1.466} \approx 4.33$$ $$y_3 = 4^{1.6} = e^{1.6 \ln(4)}$$ Calculate $\ln(4) \approx 1.3863$: $$y_3 = e^{1.6 \times 1.3863} = e^{2.218} \approx 9.19$$ 4. **Calculate $\theta$ values for input link:** Input link moves from $30^\circ$ to $120^\circ$ over $x$ from 1 to 4. Assuming linear relation: $$\theta = 30^\circ + \frac{x - 1}{4 - 1} (120^\circ - 30^\circ) = 30 + \frac{x - 1}{3} \times 90$$ Calculate for each $x$: $$\theta_1 = 30 + \frac{1 - 1}{3} \times 90 = 30^\circ$$ $$\theta_2 = 30 + \frac{2.5 - 1}{3} \times 90 = 30 + \frac{1.5}{3} \times 90 = 30 + 45 = 75^\circ$$ $$\theta_3 = 30 + \frac{4 - 1}{3} \times 90 = 30 + 90 = 120^\circ$$ 5. **Calculate $\Phi$ values for output link:** Output link moves from $90^\circ$ to $180^\circ$ over $y$ from $y_1$ to $y_3$. Assuming linear relation: $$\Phi = 90^\circ + \frac{y - y_1}{y_3 - y_1} (180^\circ - 90^\circ) = 90 + \frac{y - 1}{9.19 - 1} \times 90$$ Calculate for each $y$: $$\Phi_1 = 90 + \frac{1 - 1}{8.19} \times 90 = 90^\circ$$ $$\Phi_2 = 90 + \frac{4.33 - 1}{8.19} \times 90 = 90 + \frac{3.33}{8.19} \times 90 \approx 90 + 36.6 = 126.6^\circ$$ $$\Phi_3 = 90 + \frac{9.19 - 1}{8.19} \times 90 = 90 + 90 = 180^\circ$$ 6. **Summary of precision points:** | Point | $x$ | $y = x^{1.6}$ | $\theta$ (deg) | $\Phi$ (deg) | |-------|-----|---------------|----------------|--------------| | 1 | 1 | 1 | 30 | 90 | | 2 | 2.5 | 4.33 | 75 | 126.6 | | 3 | 4 | 9.19 | 120 | 180 | This completes the calculation of the three precision points for the four-bar mechanism design.