We have a normalized orthogonal co-ordinate frame travelling through the curve as in figure 1 below, from one end to other. Let us call starting end as A and ending end as B. What we know is initial frame $\psi(0)$,length of the curve L,a parameter $s \in[0,L]$ for length($s=0$ for A and $s = L$ for B,mentioned in the second picture)
Given Data in the question
Each co_ordinate axis at each $s$ is represented by $d_i(s)=\left(\begin{array}{c}d_{ix}(s)\\ d_{iy}(s)\\ d_{iz}(s)\end{array}\right)$ and all together in frame represented by $d(s)=(d_3(s),d_2(s),d_1(s))$
We have a starting Co-ordinate frame say $\psi(0)$. It is a $3\times 3$ matrix and each entries are defines by $\psi(0) = \begin{bmatrix} d_{3x}(0)& d_{2x}( 0)&d_{1x}( 0)\\d_{3y}( 0)& d_{2y}(0)&d_{1y}(0)\\d_{3z}(0)& d_{2z}(0)&d_{1z}(0)\end{bmatrix} \tag 2 $ where $d_3$ is the tangent
- $\psi(s) = \begin{bmatrix} d_{3x}(s)& d_{2x}( s)&d_{1x}( s)\\d_{3y}( s)& d_{2y}(s)&d_{1y}(s)\\d_{3z}(s)& d_{2z}(s)&d_{1z}(s)\end{bmatrix} \tag 3$
- $\chi_1 = \left(\begin{array}{c}\alpha_0\\\beta_0 \\ \gamma_0\end{array}\right)\tag 4$$\chi_2 = \left(\begin{array}{c}\alpha_1\\\beta_1 \\ \gamma_1\end{array}\right)\tag 5$ $\chi(s)=(1-\frac{s}{L})\chi_1+\frac{s}{L}\chi_2 \tag 6$, where $\alpha,\beta,\gamma$ are twist , bend coefficients(2nos) respectively, in other words curvature vectors. It is constant for A and B, but in between these two points we interpolate it as in above eqaution (6)
- $\psi^{'}(s)=[\chi(s)]_{\times}\psi(s)\tag 7$
- $[\hspace{.2cm} ]_{\times}$ is defined as follows if $Y= \left(\begin{array}{c}a_0\\ b_0\\ c_0\end{array}\right)$ then $[Y ]_{\times}=\left( \begin{array}{ccc} 0 & -c_0 & b_0 \\ c_0 & 0 & -a_0 \\ -b_0 & a_0 & 0 \\ \end{array} \right)$
Question
** Can we write $\psi^{'}(s)=[\chi(s)]_{\times}\psi(s) $ in terms of Quaternion calculus ? Means an equation in rotations only? Or else as second thought,can we find a rotation matrix $R(s,\chi1,\chi2)$ such that $\psi(0)R(s,\chi1,\chi2)=\psi(s)\tag 8$