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Fourbar linkage kinematics


The equations modeled here are for a four-bar linkage with pinned revolute joints. This is a one-DOF system and is defined by an input angle θ2.  The necessary kinematics for the four-bar linkage can be found from the angular position, velocity and acceleration. A web applet of this model can be found here.

Kinematics are taken from the textbook Norton, R. "Design of Machinery" (2001) 2nd ed, McGraw-Hill.

Position

The angles can be found by solving a vector loop equation, which is written in terms of exponentials for simplicity:
a\mathbf{e}^{j\theta_2}+b\mathbf{e}^{j\theta_3}=d\mathbf{e}^{j\theta_1}+c\mathbf{e}^{j\theta_4}

Solving the vector loop equation in terms of θ2, where θ1=0, gives trigonometric relations for the remaining angles:
\theta_3 = 2 \arctan \left[ \frac{-E \pm \sqrt{E^2-4DF} }{2D} \right]

\theta_4 = 2 \arctan \left[ \frac{-B \pm \sqrt{B^2-4AC} }{2A} \right]

The expressions are simplified by collecting similar terms, which are given in the following relations:

k_1=\frac{d}{a}
k_2=\frac{d}{c}
k_3=\frac{a^2-b^2+c^2+d^2}{2ac}
k_4=\frac{d}{b}
k_5=\frac{c^2-a^2-b^2-d^2}{2ab}

A = \cos\theta_2 - k_1 - k_2\cos\theta_2 + k_3
D = \cos\theta_2 - k_1 + k_4\cos\theta_2 + k_5
B = -2\sin\theta_2
E = -2\sin\theta_2
C = k_1 - \left( k_2+1 \right)\cos\theta_2+k_3
F = k_1 + \left( k_4-1 \right)\cos\theta_2+k_5

It is important to note that the arctangent function has singular points and is often numerically implemented only over -pi/2 to pi/2. Adequate evaluation may require interjecting numerical noise to avoid the singularities if using previously implemented methods.

Velocity

In a similar fashion angular velocity for the driven angles can be found by solving the vector velocity loop equation:
ja\dot{\theta}_2\mathbf{e}^{j\theta_2}+jb\dot{\theta}_3\mathbf{e}^{j\theta_3}=jd\dot{\theta}_1\mathbf{e}^{j\theta_1}+jc\dot{\theta}_4\mathbf{e}^{j\theta_4}

The driven angular velocities are then given below in terms of the angles, which must be solved using the equations from above.
\dot{\theta}_3=\frac{a}{b}\dot{\theta}_2\frac{\sin\left(\theta_4-\theta_2\right)}{\sin\left(\theta_3-\theta_4\right)}

\dot{\theta}_4=\frac{a}{c}\dot{\theta}_2\frac{\sin\left(\theta_3-\theta_2\right)}{\sin\left(\theta_3-\theta_4\right)}

Acceleration

Finally, the angular acceleration of the driven angles can be solved for using the solutions to the angular position and velocity from the previous equations and the vector acceleration loop equation:

\left( j\ddot{\theta}_2-\dot{\theta}_2^2 \right) a\mathbf{e}^{j\theta_2}+\left( j\ddot{\theta}_3-\dot{\theta}_3^2 \right) b\mathbf{e}^{j\theta_3}=\left( j\ddot{\theta}_1-\dot{\theta}_1^2 \right) d\mathbf{e}^{j\theta_1}+\left( j\ddot{\theta}_4-\dot{\theta}_4^2 \right)c\mathbf{e}^{j\theta_4}

The angular accelerations are given below in terms of simplified variables:

\ddot{\theta}_3=\frac{WX-UZ}{UY-VX}

\ddot{\theta}_4=\frac{WY-VZ}{UY-VX}

The simplified terms used in the equations above are given below:

U=c\sin\theta_4

V=b\sin\theta_3

W=a\ddot{\theta}_2\sin\theta_2 + a\dot{\theta}_2^2\cos\theta_2 + b\dot{\theta}_3^2\cos\theta_3 - c\dot{\theta}_4^2\cos\theta_4

X=c\cos\theta_4

Y=b\cos\theta_3

Z=a\ddot{\theta}_2\cos\theta_2 - a\dot{\theta}_2^2\sin\theta_2 - b\dot{\theta}_3^2\sin\theta_3 + c\dot{\theta}_4^2\sin\theta_4
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Jeff Bingham,
Apr 8, 2013, 3:20 PM
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