Position analysis of a four-bar mechanism

Method

Take A as origin and AD along the $x$-axis, so D = (300, 0). The diagonal BD is used to split the quadrilateral into triangles ABD and BCD and solved with the cosine rule (the loop-closure solution).

Given: $a = AB = 100$, $b = BC = 250$, $c = CD = 200$, $d = AD = 300$ mm, $\theta_2 = 60^\circ$.

Step 1: Triangle ABD

$$B = (a\cos\theta_2,\ a\sin\theta_2) = (50.00,\ 86.60)\ \text{mm}$$ $$BD^2 = a^2 + d^2 - 2ad\cos\theta_2 = 100^2 + 300^2 - 2(100)(300)\cos 60^\circ = 70000$$ $$BD = 264.58\ \text{mm}$$ $$\sin\angle ADB = \frac{a\sin\theta_2}{BD} = \frac{86.60}{264.58} \Rightarrow \angle ADB = 19.11^\circ$$

Step 2: Triangle BCD

$$\cos\angle BDC = \frac{c^2 + BD^2 - b^2}{2c\,BD} = \frac{200^2 + 70000 - 250^2}{2(200)(264.58)} \Rightarrow \angle BDC = 63.33^\circ$$ $$\cos\angle CBD = \frac{b^2 + BD^2 - c^2}{2b\,BD} \Rightarrow \angle CBD = 45.63^\circ$$

Step 3: Angles of the links

Since A and C lie on opposite sides of BD:

$$\angle ADC = \angle ADB + \angle BDC = 19.11 + 63.33 = 82.44^\circ$$

The rocker CD makes this angle with DA, so measured from the $+x$ axis:

$$\theta_4 = 180^\circ - 82.44^\circ = 97.56^\circ$$

Coordinates of C: $C = (300 + 200\cos\theta_4,\ 200\sin\theta_4) = (273.68,\ 198.26)$ mm.

$$\theta_3 = \tan^{-1}\frac{C_y - B_y}{C_x - B_x} = \tan^{-1}\frac{198.26 - 86.60}{273.68 - 50.00} = 26.53^\circ$$

(Check: $100\cos 60^\circ + 250\cos\theta_3 = 300 + 200\cos\theta_4$ is satisfied, and the vertical components also balance.)

Step 4: Transmission angle

The transmission angle $\mu$ is the angle between coupler BC and rocker CD:

$$\cos\mu = \frac{b^2 + c^2 - a^2 - d^2 + 2ad\cos\theta_2}{2bc} \Rightarrow \mu = 71.03^\circ$$

This is greater than $40^\circ$, so force transmission is good.

Answer: coupler angle $\theta_3 = 26.53^\circ$, rocker angle $\theta_4 = 97.56^\circ$ (both from AD), transmission angle $\mu = 71.03^\circ$.