Linkages and Mechanisms

Grashof's law

In a planar four-bar linkage, if the sum of the shortest and longest links is less than or equal to the sum of the other two links, then the shortest link can make a full rotation relative to the other links. If the sum is greater, no link can rotate fully (all inversions are double-rockers).

$$s + l \le p + q$$

where $s$ = shortest, $l$ = longest, $p, q$ = the other two links.

Check

$s = 30$ mm (AB), $l = 80$ mm (CD), $p = 60$ mm, $q = 70$ mm.

$$s + l = 30 + 80 = 110\ \text{mm} \quad < \quad p + q = 60 + 70 = 130\ \text{mm}$$

The linkage satisfies Grashof's law, so it is a Grashof chain and the shortest link AB can rotate fully.

Inversions

• Shortest link fixed: both links adjacent to it rotate completely.
• Shortest link as the frame's neighbour: the shortest link is the crank and the link opposite the frame is the rocker.
• Shortest link as coupler (opposite to frame): both side links only oscillate, the coupler makes full turns.

Answer: $s + l = 110$ mm $< p + q = 130$ mm, so Grashof; double-crank (AB fixed), crank-rocker (BC or DA fixed), double-rocker (CD fixed).

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 $\mu = 71.03^\circ$.

Loop-closure equation

Represent the links as vectors $\vec r_1$ (frame), $\vec r_2$ (crank), $\vec r_3$ (coupler), $\vec r_4$ (follower) with angles $\theta_1 = 0, \theta_2, \theta_3, \theta_4$ measured from the frame. The closed loop gives

$$\vec r_2 + \vec r_3 - \vec r_4 - \vec r_1 = 0$$

Splitting into components:

$$\begin{aligned} r_2\cos\theta_2 + r_3\cos\theta_3 - r_4\cos\theta_4 - r_1 &= 0 \\ r_2\sin\theta_2 + r_3\sin\theta_3 - r_4\sin\theta_4 &= 0 \end{aligned}$$

These are two equations in the two unknowns $\theta_3$, $\theta_4$ for a given input $\theta_2$.

Freudenstein's equation (closed form)

Eliminating $\theta_3$ by squaring and adding gives

$$K_1\cos\theta_2 - K_2\cos\theta_4 + K_3 = \cos(\theta_2 - \theta_4)$$

with $K_1 = r_1/r_4$, $K_2 = r_1/r_2$, $K_3 = \dfrac{r_1^2 + r_2^2 + r_4^2 - r_3^2}{2r_2r_4}$.

Using the half-angle substitution $t = \tan(\theta_4/2)$ gives a quadratic $A t^2 + Bt + C = 0$, whose two roots are the open and crossed configurations. Then $\theta_3$ follows from the component equations.

Newton-Raphson iteration

Write the two component equations as $f_1(\theta_3,\theta_4) = 0$ and $f_2(\theta_3,\theta_4) = 0$.

1. Guess $\theta_3^{(0)}$, $\theta_4^{(0)}$ (from a sketch or the previous crank position).
2. Form the Jacobian

$$J = \begin{bmatrix} -r_3\sin\theta_3 & r_4\sin\theta_4 \\ r_3\cos\theta_3 & -r_4\cos\theta_4 \end{bmatrix}$$

3. Solve $J\,\Delta = -\begin{bmatrix} f_1 \\ f_2 \end{bmatrix}$ for the corrections $\Delta = [\Delta\theta_3,\ \Delta\theta_4]^T$.
4. Update $\theta_3 \leftarrow \theta_3 + \Delta\theta_3$, $\theta_4 \leftarrow \theta_4 + \Delta\theta_4$.
5. Repeat until $|f_1|$, $|f_2|$ are below a tolerance (about $10^{-6}$); it usually converges in 3 to 5 steps.

Comparison

The iterative method is preferred in computer programs because the previous solution gives an excellent starting guess for the next crank angle, and the same Jacobian is reused for velocity analysis.