Ball joint connecting link

The aim of this post deals with ball joint connecting link. The following model is considered:

Drawing

Kinematic

This part explains how to calculate \Theta_2=f(\Theta_1).

DrawingKinematic

The angle \Theta_2 will be calculated with the following relation:

 \Theta_2=\pi- \widehat{P_2MP_1} -\widehat{P_1MO}

First, let’s calculate the coordinates of P_1:

 P_1=\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} l_1.cos(\Theta_1) \\ l_1.sin(\Theta_1) \end{pmatrix}

Based on the previous result, let’s calculate the distance between P_1 and M:

 \| P_1M \| = \sqrt{ (x_1-X)^2 + y_1^2 }

In the triangle HP_1M, rectangle in H, it becomes easy to calculate the angle \widehat{P_1MO}:

 \widehat{P_1MO}=atan2(\|HP_1\|,\|MH\|)=atan2(y_1,X-x_1)

Now we can calculate the angle \widehat{P_2MP_1} in the triangle P_1MP_2 thanks to the Al Kashi’s theorem:

 \|P_1P_2\| = \|MP_1\|^2 + \|MP_2\|^2 -2.\|MP_1\|.\|MP_2\|.cos(\widehat{P_2MP_1})

 L = \|MP_1\|^2 + {l_2}^2 -2.\|MP_1\|.l_2.cos(\widehat{P_2MP_1})

and finally:

 cos(\widehat{P_2MP_1}) = \frac{\|MP_1\|^2 + {l_2}^2 - L} {2.\|MP_1\|.l_2}

In conclusion:

 \Theta_2=\pi \pm acos\left(\frac{ (x_1-X)^2 + y_1^2  + {l_2}^2 - L} {2.l_2.\sqrt{ (x_1-X)^2 + y_1^2 }} \right) - atan2(y_1,X-x_1)

where:

 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} l_1.cos(\Theta_1) \\ l_1.sin(\Theta_1) \end{pmatrix}

Torque

This second part explains how to estimate the torque transmitted. Let’s assume that the actuated rod is the left one (centered on O). The torque provided on this rod is T_1. The aim is to calculate the torque transmitted (T_2) on the second rod (centered on M). The purpose of this second part is to express T_2=g(T_1).

DrawingTorque

First, let’s calculate the coordinates of P_2:

 P_2=\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} X+l_2.cos(\Theta_1) \\ l_2.sin(\Theta_1) \end{pmatrix}

Let’s calculate the amplitude of the force \vec{F_1}:

 \| \vec{F_1}\| = \frac{T_1}{l_1}

Now, calculate the angle \alpha between \vec{F_1} and \vec{F_b}:

 \alpha = ( \widehat{ \vec{F_1} ; \vec{F_b} } )= ( \widehat{ \vec{OM} ; \vec{F_1} } ) - ( \widehat{ \vec{OM} ; \vec{F_2} } ) = ( \widehat{ \vec{OM} ; \vec{F_1} } ) -( \widehat{ \vec{OM} ; \vec{P_1P_2} } )

Let’s calulate the angles:

 ( \widehat{ \vec{OM} ; \vec{F_1} } ) = \Theta_1 - \frac{\pi}{2}

And :

 ( \widehat{ \vec{OM} ; \vec{P_1P_2} } ) = atan2(y_2-y_1,x_2-x_1)

Now \alpha can be calculated:

 \alpha = \Theta_1 - \frac{\pi}{2} - atan2(y_2-y_1,x_2-x_1)  = \Theta_1 - atan2(x_2-x_1,y_1-y_2)

and F_b can also now be calculated:

 \| \vec{F_b} \| = \| \vec{F_1} \|.cos(\alpha)

As it was done for \alpha, let’s calculate the angle \beta between \vec{F_b} and \vec{F_2}:

 \beta = ( \widehat{ \vec{F_2} ; \vec{F_b} } )= \Theta_2 - \frac{\pi}{2} - atan2(y_2-y_1,x_2-x_1)  =  \Theta_2 - atan2(x_2-x_1,y_1-y_2)

The force F_2 can now also be calculated:

 \| \vec{F_2} \| = \| \vec{F_b} \|.cos(\beta)

The outpout torque is :

 T_2=\| \vec{F_2} \|.l_2

In conclusion:

 T_2= \frac{l_2}{l_1}. T_1.cos(\beta).cos(\alpha)

where:

 \alpha = \Theta_1 - atan2(x_2-x_1,y_1-y_2)

and

 \beta =  \Theta_2 - atan2(x_2-x_1,y_1-y_2)

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