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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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  • $\begingroup$ 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. $\endgroup$ Feb 7, 2012 at 21:10
  • $\begingroup$ @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). $\endgroup$ Feb 8, 2012 at 8:35
  • $\begingroup$ May I suggest posting your solution as an actual answer instead of as an edit to your question? $\endgroup$ Feb 8, 2012 at 13:13
  • $\begingroup$ @J.M. : Oh yes. I will do that. $\endgroup$ Feb 8, 2012 at 15:49

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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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lets' go the following graph of a line: General equation of a line is: Y= mx +b where m= the slope of the line and -b = the x intercept of the line. How how ever your question is not complete, in the sense that, the axes of rotation of the line is not mention. if it is rotated around it self it will remain the same line withe same slope and x intercept. But if it is rotated around the x axis it will generate two cones whose vertexes are joined at the same point. and so on. But if it is translated to another point in the plane for example (n,p) as its new origin then coordinate of all points with respect to the original axes are: new x= x-n and new y=y-p

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  • $\begingroup$ Thanks - but this was already solved two years ago :-) $\endgroup$ Feb 10, 2015 at 18:33

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