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

## Kinematic

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

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 $T_2=g(T_1)$.

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

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