[Math] Angle between two line segments

geometrytrigonometry

Here's a question? I have two line segments which have an angle between them. For the line segments, I know coordinates of their points. I need to be able to rotate one segment for one angle (for example, $\alpha$ needs to be $60^\circ$) – it doesn't matter in which direction is angle calculated, as long as it stays persistent. Because that I only know point of line segment, idea is to rotate other point around point of intersection for desired angle, and to get coordinates for that point after rotation. But I can not come to solution which works for every angle and where rotation direction (clockwise or counter-clockwise) is persistent. $(0, 0)$ is in top left corner, as shown on picture. You don't have to bother with theory (if you don't want), I need to apply this as algorithm. Thanks
enter image description here

Best Answer

The usual rotation matrix approach works. To rotate by an angle $\theta$, you have $x'=x \cos \theta - y \sin \theta, y'=y \cos \theta + x \sin \theta$ If you apply it to all the points, the angles between lines will stay the same.

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[Math] Calculate Angle between Two Intersecting Line Segments

What you are looking for is an angle $\theta$ such that \begin{align} x &= r\cos\theta \\ y &= r\sin\theta \end{align} where $r = \sqrt{x^2+y^2}$ is the length of segment $B.$

If you are doing this on a computer, you may be in luck, because many programming environments offer a function (usually named $\mathrm{atan2}$) which is defined so that $$ \mathrm{atan2}(y,x) $$ gives the exact angle you want, in radians, except that the result of $\mathrm{atan2}$ is always in the range $-\pi < \mathrm{atan2} \leq \pi,$ whereas you want $0 \leq \theta < 2\pi$ (in radians). So if $\mathrm{atan2} < 0$ you will want to add $2\pi.$ Multiply by $180/\pi$ to get the answer in degrees.

The function $\tan^{-1}$ can only give an angle $\theta$ such that $-\frac\pi2 < \theta < \frac\pi2.$ So this will be the correct angle only if $x > 0.$ But if $x < 0$ you can use the fact that $$ \tan\frac yx = \tan\frac{-y}{-x} $$ and the fact that $(-x,-y)$ is what you get when you rotate $(x,y)$ by $180$ degrees in either direction around $(0,0).$ In short, if $x<0$ then you need to add $180$ degrees to your result; if $x > 0$ but $y < 0$ you will need to add $360$ degrees to compensate for the fact that $\tan^{-1}\frac yx < 0$; and if $x=0$ you need to look at the sign of $y$ to decide whether the angle is $90$ degrees or $270$ degrees.

Given n line segments and a circular sector, find the leftmost and the rightmost segments

Note: Here we want a line segment $(s_k, e_k)$ to be oriented such that the point $c$ is to the right of the colinear line through $s_k$ to $e_k$.

(BTW, "colinear line" is redundant.) This is a key property. As long as you choose $s_k$ and $e_k$ this way (so $(s_{ky}-c_y)(e_{kx}-c_x) \ge (s_{kx}-c_x)(e_{ky}-c_y)$), the problem has a simple solution. If travelling from $s_k$ to $e_k$ has the circle center on your right, then you are travelling clockwise, and the angle should be decreasing. That is, the argument for $e_k$ should always be less than or equal to the argument for $s_k$.

If this isn't true, there are two ways we can correct it: either add $2\pi$ to the argument of $s_k$ or subtract $2\pi$ from the argument of $e_k$. Which technique makes sense depends on whether we are looking for the rightmost or leftmost segment. And just to be clear, I assume that these mean:

"Rightmost segment" is, among all segments with at least one end in the sector, the one with the most clockwise argument of all ends (include one outside the sector, provided the other end is in).
"Leftmost segment" is, among all segments with at least one end in the sector, the one with the most counter-clockwise argument of all ends (include one outside the sector, provided the other end is in).

Note that if the sector happens to open downward (instead of upward as in your examples), the "Rightmost segment" will be on the left and the "Leftmost segment" will be on the right (this is why "clockwise" and "counter-clockwise" are the better terminology for rotational phenomena).

Initialize variables

MinIndex = -1
MaxIndex = -1
MinAngle = 13   # > 4 * pi
MaxAngle = -7   # < -2 * pi

Pass through your segments, skipping over any segment without at least one endpoint in the sector. For the rest,

If $\arg s_k \ge \arg e_k$, then
- if $\arg s_k >\text{ MaxAngle}$, then $\text{MaxAngle } = \arg s_k, \text{ MaxIndex } = k$.
- if $\arg e_k <\text{ MinAngle}$, then $\text{MinAngle } = \arg e_k, \text{ MinIndex } = k$.
If $\arg s_k < \arg e_k$, then
- if $\arg s_k + 2\pi >\text{ MaxAngle}$, then $\text{MaxAngle } = \arg s_k + 2\pi, \text{ MaxIndex } = k$.
- if $\arg e_k - 2\pi <\text{ MinAngle}$, then $\text{MinAngle } = \arg e_k - 2\pi, \text{ MinIndex } = k$.

When you are finished, MinIndex will point to the Leftmost segment and MaxIndex will point to the Rightmost.

Best Answer

Related Solutions

[Math] Calculate Angle between Two Intersecting Line Segments

Given n line segments and a circular sector, find the leftmost and the rightmost segments

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