Is there a formula that relates the two angles in a crank rocker?

Question

I'm working on a mechanism that requires a crank rocker - is there a formula that relates the angle (to some fixed reference; horizontal, vertical, whatever) of the driving "crank" to the angle of the driven "rocker"? I've tried to work the function out geometrically, but I'm not able to. All the other equations that I can find seem to be based on finding the limits of the system, and not any one point in time.

To illustrate; I'd like to find the relation between the angle of "s" and that of "p", relative to i.e. the horizon - something that I can, say, plot on a graph of theta(s) to theta(p). Obviously, it would be dependent on variables for the bar lengths.

Thanks.

Answer 1

I started writing this, as a solution, but I didn't end up on a neat solution (although I am certain it exists). In any case you can find a solution with numerical methods at end. If I find the neater way I will update.

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Assuming that :

• the end of bar s is point S
• the end of bar p is point P
• the length of bar p,q,s are respectively $L_p, L_q, L_s$
• the angles of bars p,s are respectively $\theta_p, \theta_s$
• the beginning of the coordinate system is at the bottom left end of bar s.

Then the coordinates for :

• point S are:

$$\vec S = \begin{bmatrix}L_s \cos \theta_s \\L_s \sin \theta_s \\0\end{bmatrix}$$

• point P are:

$$\vec P = \begin{bmatrix}L_q + L_p \cos \theta_p \\L_s \sin \theta_p \\0\end{bmatrix}$$

Then the distance between S and P as a vector is $$ \vec P -\vec S = \begin{bmatrix}L_q + L_p \cos \theta_p \\L_s \sin \theta_p \\0\end{bmatrix} - \begin{bmatrix}L_s \cos \theta_s \\L_s \sin \theta_s \\0\end{bmatrix} = \begin{bmatrix}L_q + L_p \cos \theta_p - L_s \cos \theta_s \\L_s \sin \theta_p -L_s \sin \theta_s \\0\end{bmatrix}$$

Therefore the distance between points P and S should be equal to the length of rod l:

$$(L_q + L_p \cos \theta_p - L_s \cos \theta_s)^2 + (L_s \sin \theta_p -L_s \sin \theta_s)^2 = L_l^2$$

if we expand and we collect all $L_p$ and $L_s$:

$$L_p^2 \left(\sin ^2(\theta_p)+\cos ^2(\theta_p)\right)+L_p (2 L_q \cos (\theta_p)+L_s (-2 \sin (\theta_p) \sin (\theta_s)-2 \cos (\theta_p) \cos (\theta_s)))+L_q^2-2 L_q L_s \cos (\theta_s)+L_s^2 \left(\sin ^2(\theta_s)+\cos ^2(\theta_s)\right) = L_l^2$$

Substituting $\left(\sin ^2(\theta_p)+\cos ^2(\theta_p)\right) = 1 =\left(\sin ^2(\theta_s)+\cos ^2(\theta_s)\right) $, simplifies the above to

$$L_p^2 +L_p (2 L_q \cos (\theta_p)+L_s (-2 \sin (\theta_p) \sin (\theta_s)-2 \cos (\theta_p) \cos (\theta_s)))+L_q^2-2 L_q L_s \cos (\theta_s)+L_s^2 = L_l^2$$

$$L_p (2 L_q \cos (\theta_p)-2 L_s ( \sin (\theta_p) \sin (\theta_s)+ \cos (\theta_p) \cos (\theta_s)))= L_l^2 - (L_p^2 - L_q^2-2 L_q L_s \cos (\theta_s)+L_s^2) $$

Also because $ \sin (\theta_p) \sin (\theta_s)+ \cos (\theta_p) \cos (\theta_s)) = \cos(\theta_p+\theta_s)$

$$L_p (2 L_q \cos (\theta_p)-2 L_s ( \cos(\theta_p+\theta_s)))= L_l^2 - (L_p^2 - L_q^2-2 L_q L_s \cos (\theta_s)+L_s^2) $$

$$L_q \cos (\theta_p)-L_s ( \cos(\theta_p+\theta_s))=\frac{1}{2 L_p }\left( L_l^2 - (L_p^2 - L_q^2-2 L_q L_s \cos (\theta_s)+L_s^2) \right) $$

At this point you can probably find a clevel trigonometrical way to solve this (I am not that good), but in this form it would be easy enough to find a numerical solution to the equation by assuming a value for $\theta_s$.