Writing Constraint Equations for Two-Point and Three-Point Mounts

We describe how to obtain the constraint equations for a two point pivot and three point pivot.  Designing a mechanism which can obtain a desired set of constraints is often an important step in kinematic or exact constraint machine design.

We begin with the simple lever mechanism shown in the figure below constraining the motion of two points A and C using the pivot at O.

two point constraint lever

We treat the bar as rigid and are only interested in small amplitude rotation of the pivot about O.   For a rotation of angle \theta about point O, the displacement at point C is

    \[ u_C=L_A \sin \theta \approx L_C \theta \]

and the displacement at point A is

    \[ u_A=-L_A \sin \theta \approx -L_A \theta \]

where the approximation is due to the assumption of small angles.  Eliminating \theta from both equations, we obtain a single constraint equation relating u_A and u_C

    \[ \frac{u_A}{L_A}+\frac{u_C}{L_C}=0 \]

Another method of deriving this set of constraints is to fit a line to the points O and C and then derive the constraint equation A must satisfy to also fall on this line.  Let the (x,y) coordinates of X be (0,0) and of C be (L_C,u_C).  The line connecting the two is given by

    \[ y=\frac{u_C}{L_C} x \]

Then, substitute the coordinates of point A (x=-L_A,y=u_A) to obtain

    \[ \frac{u_A}{L_A}=-\frac{u_C}{L_C} \]

Now consider a constraint of three points A, B, and C at the vertices of a rigid, triangular plate with a pivot O in the middle of the plate shown in the figure below which allows displacement of A, B, and  C in the z direction but not in the x and y (there is no \theta_z motion of the plate).

three point constraint

Again, there are multiple ways to obtain the constraint equations.  One could define angles of rotation \theta_x and \theta_y about to derive the constraints, but here we focus on using the equation of a plane to derive the constraints similar to the method above of fitting a line to two points.

Let the (x,y,z) coordinates of the pivot O be (0,0,0) and the coordinates of A (x_A,y_A,u_A), B (x_B,y_B,u_B), and C (x_C,y_C,u_C).  The equation of the plane can be written setting the determinant equal to zero (note: there are a number of equivalent matrix determinants which can be used to define the plane)

    \[ \left| \begin{matrix} x & y & z & 1  \\ x_A & y_A & u_A  & 1\\ x_B & y_B & u_B & 1 \\ x_C & y_C & u_C & 1 \\ \end{matrix} \right| =0 \]

Because x=0, y=0, z=0, this simplifies greatly to 

    \[ \left| \begin{matrix} x_A & y_A & u_A \\ x_B & y_B & u_B \\ x_C & y_C & u_C \\ \end{matrix} \right| =0 \]

which becomes

    \[ \alpha u_A + \beta u_B + \gamma u_C = 0 \]

where

    \[ \alpha=x_B y_C - x_C y_B \]

    \[ \beta= x_C y_A - x_A y_C \]

    \[ \gamma=x_A y_B - x_B y_A \]