Suppose we have a 3D vector that can be rotated around two axis (a, b in fig.). So let's going to say we just made an small rotation to left in the axis $a$, the vector will be pointing to lower $z$-coordinate. And what I want to know is, if is there a way to calculate the increment in $b$ that I should do depending the amount of rotation in $a$ and the initial orientation of the vector, for drawing a imaginary straight line (keeping $z$-coordinate constant as shown in fig. with green line).
What I have found is that given the initial orientation and the increment that will be made on $a$, it's going to be the increment I need to add to $b$. But this little increment I found was a guessed value (not properly calculated).
A fast example would be:
Initial orientation of "b": 59.21º
Increments on "a": 0.01º
Estimated guessed compensation in $b$ in order to keep $z$-coordinate constant: $\tan(0.0036º)$
I have an scenario where I'm simulating this case, and hence tested the above example. Thanks in advance guys.
We can assume that the parametric equation of the line to be traced is given by $p(t) = p_0 + t u \hspace{24pt} (1)$
where $u$ is the direction vector of the line. If the tracing vector has its origin at point $P$, then vector pointing from $P$ to $p(t)$ is given by
$v(t) = p(t) - P = p_0 - P + t u \hspace{24pt} (2) $
Let's normalize this vector as $w(t) = v(t) / | v(t) | \hspace{24pt} (3) $
Before $t = 0$, this vector is pointing in the direction given by the unit vector $w_1 = \dfrac{p_0 - P}{| p_0 - P| }$
We want to find $\phi_1$ and $\phi_2$ (functions of $t$ ) such that
$w(t) = R_b(\phi_2) R_a(\phi_1) w_1 \hspace{24pt} (4)$
where the axes $a$ and $b$ are fixed vectors in space. Note that $w(t)$ and $w_1$ are both unit vectors.
From equation $(4)$, it follows that
$ q = R_b^T (\phi_2) w(t) = R_a (\phi_1) w_1 \hspace{24pt} (5) $
Rotation preserves the angle between the rotated vector and the axis of rotation.
Therefore, we want the angle between $R_a (\phi_1) w_1$ and axis $b$ to be the same as the angle between $w(t)$ and $b$
Using the dot product this can be expressed as
$ b^T R_a(\phi_1) w_1 = b^T w(t) \hspace{36pt} (6) $
Recall that the rotation matrix about an axis $a$ is given by
$R_a (\phi_1) = a a^T + (I - a a^T) \cos \phi_1 + S_a \sin \phi_1 \hspace{36pt} (7)$
where $S_a y = a \times y $ for any vector $y$, i.e.
$S_a = \begin{bmatrix} 0 && - a_z && a_y \\ a_z && 0 && -a_x \\ -a_y && a_x && 0 \end{bmatrix} $
Premultiplying $(7)$ by $b^T$ and post-multiplying by $w_1$, and plugging that in $(6)$ generates a scalar trigonometric equation,
$ c_1 \cos \phi_1 + c_2 \sin \phi_1 = c_3 \hspace{36pt} (8) $
where
$c_1 = b^T (I - {a a}^T) w_1 ,\hspace{6pt} c_2 = b^T S_a w_1 , \hspace{6pt} c_3 = b^T w(0) - b^T {a a}^T w_1 $
Equation $(8)$ can be solved quite easily (by combining the $\cos$ and $\sin$ terms ) and will generally have two solutions.
Choose one of the solutions $\phi_1$ and compute $R_a(\phi_1)$, and plug it in equation $(5)$ to compute the rotated vector $q$.
We still have to find $\phi_2$. This can be done in the same way as we found $\phi_1$, namely, from (5), we want the angle that $R_b^T (\phi_2) w(t) $ with vector $a$ , to be the same as the angle that vector $w_1$ makes with vector $a$. Thus
$ a^T R_b^T (\phi_2) w(t) = a^T w_1 \hspace{36pt} (9) $
And the same procedure as before is used to find the two values of $\phi_2$. Now that we have two values of $\phi_1$ and $\phi_2$ we have to mix and match the values to obtain only two possible solutions (not four), using equation $(5)$.
Now, we'll differentiate the above obtained equations with respect to the parameter $t$.
So we have, from equations (1) and (3),
$w(t) = v(t) / | v(t) | = v(t) / \sqrt(v^T v)$
Differentiate this
$\begin{equation} \begin{split} w'(t) &= v'(t) / |v(t) | + v(t) (-1/(v^T v) )(1/(2 \sqrt{v^T v} ) (2 v(t)^T v'(t) ) \\ &= v'(t) / |v(t) | - v(t) v(t)^T v'(t) / (v^T v)^(3/2)\\ &= 1/|v(t)| ( I - v v^T / v^T v ) v'(t) \\ &= 1/| v(t) | ( I - v v^T/ v^T v ) u \hspace{36pt} (10) \end{split} \end{equation}$
From equation (6),
$ b^T R_a(\phi_1) w_1 = b^T w(t) \hspace{36pt} (6) $
Differentiating,
$b^T ( \dfrac{dR_a}{d\phi_1} \dfrac{ d\phi_1}{dt} ) w_1 = b^T w'(t) \hspace{36pt} (11) $
from which we can find $\dfrac{d \phi_1}{dt} $
similarly
$ a^T R_b^T (\phi_2) w(t) = a^T w_1 \hspace{36pt} (9)$
Differentiating,
$ a^T ( \dfrac{dR_b^T}{d\phi_2} \dfrac{d\phi_2}{dt} w(t) + R_b^T w'(t) ) = 0 \hspace{36pt} (12) $
Note that $R_b^T(\phi_2) = R_b(- \phi_2 ) $
So, in summary, to find the incremental $d \phi_1 $ and $d \phi_2 $ :
Find $w(t)$, $\phi_1$ and $\phi_2$ as outlined above.
Find the derivative $w'(t)$, as given by equation $(10) $
Find $\dfrac{ d\phi_1}{dt} $ from equation $(11)$
Find $\dfrac{ d\phi_2}{dt} $ from equation $(12) $
Use $ d\phi_1 = \dfrac{ d\phi_1}{dt} dt $ and $ d \phi_2 = \dfrac{ d\phi_2}{dt}dt $