$q = \begin{pmatrix} x \\ y \\ \theta \\ \varphi_1 \\ \varphi_2 \\ \varphi_3 \end{pmatrix}$ $\textcolor{red}{\dot x \cdot \sin(\varphi) - \dot y \cdot \cos(\varphi) = 0}$ BPP
$\begin{cases} x_2 = x+r\cdot \cos(\theta) + l \cdot \cos(\theta + \varphi_2)\\
y_2 =y+r\cdot \sin(\theta) + l \cdot \sin(\theta + \varphi_2) \end{cases}$
$\begin{cases} \dot x_2 = \dot x-r\cdot \sin(\theta) \cdot \dot \theta - l \cdot \sin(\theta + \varphi_2) \cdot (\dot \theta + \dot \varphi_2)\\
\dot y_2 = \dot y+r\cdot \cos(\theta) \cdot \dot \theta + l \cdot \cos(\theta + \varphi_2) \cdot(\dot\theta + \dot\varphi_2) \end{cases}$
$\dot x_2 \cdot \sin(\theta + \varphi_2) - \dot y_2 \cdot \cos(\theta + \varphi_2) = 0$
$\left(
\dot x-r\cdot \sin(\theta) \cdot \dot \theta - l \cdot \sin(\theta + \varphi_2) \cdot (\dot \theta + \dot \varphi_2)
\right) \cdot \sin(\theta + \varphi_2) - \left(
\dot y+r\cdot \cos(\theta) \cdot \dot \theta + l \cdot \cos(\theta + \varphi_2) \cdot(\dot\theta + \dot\varphi_2)
\right) \cdot \cos(\theta + \varphi_2) = 0$
$\textcolor{red}{\sin(A)\sin(B)+\cos(A)\cos(B) = \cos(A-B) = \cos(B-A)}$
\dot y_2 \cdot \sin(\theta + \varphi_2) -r\cdot \dot \theta \cdot \left[ \sin(\theta) \cdot \sin(\theta + \varphi_2)+ \cos(\theta) \cdot \cos(\theta + \varphi_2) \right] -\left( l(\dot \theta + \dot \varphi_2) \right) \left( \cos^2(\theta+\varphi_2)+ \sin^2(\theta+\varphi_2) \right) = 0$
\dot \theta \cdot r \cdot \cos(\varphi_2)
\dot \theta \left( r \cdot \cos(\varphi_2) -l \right)
$\begin{bmatrix} \sin(\theta+\varphi_2) & -\cos(\theta + \varphi_2) &
$\begin{bmatrix} \sin(\alpha_1+\varphi_2+\theta) & -\cos(\alpha_1+\varphi_2+\theta) & -l- r \cdot \cos(\varphi_2)& 0 & -l & 0 \end{bmatrix} \cdot \dot q = 0$
Na podstawie drugiego ogona domyślamy się jak wyglądają pozostałe dwa.
$\begin{bmatrix} \sin(\alpha_1+\varphi_1+\theta) & -\cos(\alpha_1+\varphi_1+\theta) & -l- r \cdot \cos(\varphi_1)& -l & 0 & 0 \\
\sin(\alpha_2+\varphi_2+\theta) & -\cos(\alpha_2+\varphi_2+\theta) & -l- r \cdot \cos(\varphi_2)& 0 & -l & 0 \\
\sin(\alpha_3+\varphi_3+\theta) & -\cos(\alpha_3+\varphi_3+\theta) & -l- r \cdot \cos(\varphi_3)& 0 & 0 & -l \end{bmatrix} \cdot \dot q = 0$
$\alpha_1 = 0,\ \ \ \ l = r = 1$
$\begin{bmatrix} \sin(\alpha_1+\varphi_1+\theta) & -\cos(\alpha_1+\varphi_1+\theta) & -1-\cos(\varphi_1)& -1 & 0 & 0 \\
\sin(0+\varphi_2+\theta) & -\cos(0+\varphi_2+\theta) & -1-\cos(\varphi_2)& 0 & -1 & 0 \\
\sin(\alpha_3+\varphi_3+\theta) & -\cos(\alpha_3+\varphi_3+\theta) & -1-\cos(\varphi_3)& 0 & 0 & -1 \end{bmatrix} \cdot \dot q = 0$
$n=6,\ \ \ l=3, \ \ \ m=n-l=3$ → więc szukamy 3 wektorów $g_i$ . $A\cdot g_i = 0$ $i \in \{1, 2, 3\}$
$\begin{bmatrix} \sin(\alpha_1+\varphi_1+\theta) & -\cos(\alpha_1+\varphi_1+\theta) & -1-\cos(\varphi_1)& -1 & 0 & 0 \\
\sin(\varphi_2+\theta) & -\cos(\varphi_2+\theta) & -1-\cos(\varphi_2)& 0 & -1 & 0 \\
\sin(\alpha_3+\varphi_3+\theta) & -\cos(\alpha_3+\varphi_3+\theta) & -1-\cos(\varphi_3)& 0 & 0 & -1 \end{bmatrix} \cdot \begin{pmatrix} g_{11}\\g_{12}\\ g_{13}\\g_{14}\\ g_{15}\\g_{16} \end{pmatrix} = 0$
$\begin{cases}
\sin(\alpha_1+\varphi_1+\theta) \cdot g_{11} - \cos(\alpha_1+\varphi_1+\theta) \cdot g_{12} - \left( 1+\cos(\varphi_1)\right) \cdot g_{13} - g_{14} = 0 \\
\sin(\varphi_2+\theta) \cdot g_{11} - \cos(\varphi_2+\theta) \cdot g_{12} - \left( 1+\cos(\varphi_2)\right) \cdot g_{13} - g_{15} = 0 \\
\sin(\alpha_3+\varphi_3+\theta) \cdot g_{11} - \cos(\alpha_3+\varphi_3+\theta) \cdot g_{12} - \left( 1+\cos(\varphi_3)\right) \cdot g_{13} - g_{16} = 0
\end{cases}$
$\begin{cases}
g_{14} = \sin(\alpha_1+\varphi_1+\theta) \cdot g_{11} - \cos(\alpha_1+\varphi_1+\theta) \cdot g_{12} - \left( 1+\cos(\varphi_1)\right) \cdot g_{13} \\
g_{15} = \sin(\varphi_2+\theta) \cdot g_{11} - \cos(\varphi_2+\theta) \cdot g_{12} - \left( 1+\cos(\varphi_2)\right) \cdot g_{13} \\
g_{16} = \sin(\alpha_3+\varphi_3+\theta) \cdot g_{11} - \cos(\alpha_3+\varphi_3+\theta) \cdot g_{12} - \left( 1+\cos(\varphi_3)\right) \cdot g_{13}
\end{cases}$