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I have a line $L$ in the plane expressed as the points in $L = \{(x,y) \in {\mathbb{R}}^2 : x \cos \theta + y \sin \theta = r \; \wedge \; 0 > \theta > \pi/2 \}$ (note that the line cannot be fully horizontal or fully vertical). This line can possibly intersect the $y$-axis in $w-m \leq y \leq w+m$ for a fixed frame width $w > 0$ and a margin $0 \leq m << w$ (the line is generally "stuck" to a certain distance from the origin).

Let's call the point of intersection $Q$. I need to rotate line L in either direction around Q with angle $\phi$ (generally quite a small rotation; $\phi < \lvert\pi/20\rvert$). After the rotation I need to translate the line by a vector $\mathbf{t}$ perpendicular to the now rotated line. Again the distance $\lVert \mathbf{t} \rVert$ is generally small but its direction is always in the direction of the previous rotation.

Question: Was is the relationship between the original line $L$'s parameters $\theta$ and $r$ and the new rotated and translated line's parameters ${\theta}_\text{new}$ and $r_\text{new}$?

EDIT - Feb 8th 2012: Major changes. Original posing of the question was entirely wrong. The geometrical situation is now quite different and not quite as trivial as hinted at below in the comments.

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If you realize that $\theta$ is the direction of the normal vector of the line, and $r$ is the distance from the origin to the line, I think you should find the answer quite obvious. It's a matter of simple addition. –  Harald Hanche-Olsen Feb 7 '12 at 21:10
@HaraldHanche-Olsen - Thank you for the comment. Yes, I am aware of the geometrical interpretation. And I have a piece of paper filled with triangles and lines :) My current guess must be ${\theta}_D = \theta - \phi$ and $r_D = r_R - \lVert \mathbf{t} \rVert$, where $r_R$ is the distance to the rotated line (before translation). –  Ole Thomsen Buus Feb 8 '12 at 8:35
May I suggest posting your solution as an actual answer instead of as an edit to your question? –  J. M. Feb 8 '12 at 13:13
@J.M. : Oh yes. I will do that. –  Ole Thomsen Buus Feb 8 '12 at 15:49
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1 Answer 1

up vote 0 down vote accepted

After some fidling around with triangles and trigonometry I found the solution myself:

  • ${\theta}_\text{new} = \theta \pm \phi$,
  • $r_\text{new} = r \cos \phi [1 \pm (\tan \phi / \tan \theta)] \pm \lVert \mathbf{t} \rVert$.

Not trivial but also not very dificult. Seems to be working beautifully in Mathematica.

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