## 2 Materials and Methods
### Elastic Static Deformations and Errors of Robot Manipulators
\begin{array}{l}\Delta x_{\mathrm{M}}^{(0)}=\mathrm{e}_{11}\cdot\mathrm{F}_ {\mathrm{x}}+\mathrm{e}_{12}\cdot\mathrm{F}_{\mathrm{y}}+\mathrm{e}_{13}\cdot \mathrm{F}_{\mathrm{z}}+\mathrm{e}_{14}\cdot\mathrm{M}_{\mathrm{x}}+\mathrm{ e}_{15}\cdot\mathrm{M}_{\mathrm{y}}+\mathrm{e}_{16}\cdot\mathrm{M}_{z},\\ \Delta y_{\mathrm{M}}^{(0)}=\mathrm{e}_{21}\cdot\mathrm{F}_{\mathrm{x}}+ \mathrm{e}_{22}\cdot\mathrm{F}_{\mathrm{y}}+\mathrm{e}_{23}\cdot\mathrm{F}_{ \mathrm{z}}+\mathrm{e}_{24}\cdot\mathrm{M}_{\mathrm{x}}+\mathrm{e}_{25}\cdot \mathrm{M}_{\mathrm{y}}+\mathrm{e}_{26}\cdot\mathrm{M}_{z},\\ \theta_{\mathrm{y}}^{(0)}=\mathrm{e}_{51}\cdot\mathrm{F}_{\mathrm{x}}+\mathrm{ e}_{52}\cdot\mathrm{F}_{\mathrm{y}}+\mathrm{e}_{53}\cdot\mathrm{F}_{\mathrm{z}}+ \mathrm{e}_{54}\cdot\mathrm{M}_{\mathrm{x}}+\mathrm{e}_{55}\cdot\mathrm{M}_{ \mathrm{y}}+\mathrm{e}_{56}\cdot\mathrm{M}_{z},\\ \theta_{\mathrm{z}}^{(0)}=\mathrm{e}_{61}\cdot\mathrm{F}_{\mathrm{x}}+\mathrm{ e}_{62}\cdot\mathrm{F}_{\mathrm{y}}+\mathrm{e}_{63}\cdot\mathrm{F}_{\mathrm{z}}+ \mathrm{e}_{64}\cdot\mathrm{M}_{\mathrm{x}}+\mathrm{e}_{65}\cdot\mathrm{M}_{ \mathrm{y}}+\mathrm{e}_{66}\cdot\mathrm{M}_{z},\\ \end{array}  \tag{10}\]  
or in the matrix form of the record  
\[\delta_{\mathrm{M}}=\mathrm{E}\cdot\mathrm{Q}, \tag{11}\]  
where the \(\mathrm{E}\)--\(6\times 6\) matrix of elastic compliance of the gripper is:  
\[\mathrm{E}= \begin{bmatrix}\mathrm{e}_{11}&\mathrm{e}_{12}&\ldots&\mathrm{e}_{16 }\\ \mathrm{e}_{21}&\mathrm{e}_{22}&\ldots&\mathrm{e}_{26}\\ \ldots&\ldots&\ldots&\ldots\\ \mathrm{e}_{61}&\mathrm{e}_{63}&\ldots&\mathrm{e}_{66}\end{bmatrix} , \tag{12}\]  
which has the property of symmetry of (\(\mathrm{e}_{11}=\mathrm{e}_{12}=\ldots=\mathrm{e}_{65}=\mathrm{e}_{66}\)) and is the property of non-negativity of the elements \(\mathrm{e}_{11},\mathrm{e}_{12},\ldots,\mathrm{e}_{65},\mathrm{e}_{66}\) of the main diagonal. To determine the elements of the first column of matrix \(\mathbf{E}\), in accordance with (10), we assign \(\mathrm{F}_{\mathrm{x}}=1\) for zero values of all other loads. Then we have  
\[\Delta x_{\mathrm{M}}^{(0)}=\mathrm{e}_{11},\ \Delta y_{\mathrm{M}}^{(0)}= \mathrm{e}_{21},\ \ldots,\ \theta_{\mathrm{z}}^{(0)}=\mathrm{e}_{61}. \tag{13}\]  
Such a problem consists in determining the \(\delta_{\mathrm{M}}\) vector of errors under the action of a unit force \(\mathrm{F}_{\mathrm{x}}=1\). The problem is solved in two stages. The first stage. We determine the elastic forces of deformation \(\Delta q_{i}\) (\(i=1,2,3\)) in each kinematic pair under the action of \(\mathrm{F}_{\mathrm{x}}=1\) unit force. For a translational pair, the linear deformation will be  
\[\Delta q_{\mathrm{i}}=\left.\frac{\mathrm{F}_{\mathrm{x}}\cdot\mathrm{cos}\, a_{i}}{\mathrm{C}_{i}}\right|_{\mathrm{F}_{\mathrm{x}}}=1\quad, \tag{14}\]  
where \(\mathrm{C}_{5}\) is the stiffness of the drive of the \(i\)-th kinematic pair; \(a_{i}\) is the angle between the positive direction of \(q_{\mathrm{s}}\) translational movement and the direction of \(\mathrm{F}_{\mathrm{x}}\) force (note that \(\Delta q_{i}=0\) when \(a_{i}=90^{\circ}\)). For a rotational pair, the angular deformation will be  
\[\Delta q_{i}=\pm\left.\frac{\mathrm{M}_{\mathrm{x}}(\mathrm{F}_{\mathrm{x}})}{ \mathrm{C}_{i}}\right|_{\mathrm{F}_{\mathrm{x}}}=1\quad, \tag{15}\]  
where \(\mathrm{M}_{\mathrm{x}}(\mathrm{F}_{\mathrm{x}})\) is the moment of \(\mathrm{F}_{\mathrm{x}}\) force relative to the axis of rotation of the \(i\)-th kinematic pair. The \(\mathrm{M}_{\mathrm{x}}(\mathrm{F}_{\mathrm{x}})\) moment is considered positive if it acts counterclockwise, the moment is zero if the force and the axis of rotation are in the same plane.