I am working with a sensor that outputs data in forms of radial beams. In order for it to be useful for my application I have to compute a value of one side of the acute triangles that can be seen in the radial beams. Inside these triangles I can then check whether or not an obstacle is seen. Here's the diagram (sorry about the quality, I tried my best):

enter image description here

Figure 1.


  1. Assume α = β.
  2. Assume the segments and their angles aren't up to scale.
  3. Assume there are 16 radial segments (or arcs), first one perpendicular to the Y-axis, the last one is parallel with the X-axis, and that the angles subtended by the arcs are all equal.
  4. Assume the radii and the angles of the arcs are known.
  5. Assume TU is known.

The problem:

How would one calculate triangles' side created within the segments? For example, let's take triangles CAR and RAP (the names are accidental!), is it possible to calculate their sides, especially the opposites of the angles α and β, CR and RP?

My current idea is to use TU, form right angle triangles such as APJ (imagine if T was overlaying P) and use the angles known to calculate the remaining sides. The problem with that is it becomes increasingly complex for increasing number of segments.

Is there a better way of doing it, keeping in mind that it's a part of a obstacle detection algorithm?

EDIT 1: got rid of a redundant line, updated the diagram.

EDIT 2: extra question: imagine arc c being absent and the first arc is actually arc d (so no parallel segment with the Y-axis to begin with). How would you approach this problem in that case?


If $AO$ is perpendicular t0 $AU$: $$ CR=TU\tan\alpha,\quad PR=CP-CR=TU(\tan2\alpha-\tan\alpha),\quad \text{and so on.} $$


If line $AO$ is not perpendicular to $AU$, then we must have some other info, e.g. the angle between $AO$ (or $AB$) and $y$ axis.

  • $\begingroup$ Thank you for your reply. Could you elaborate in a slightly greater detail as to why you are doing it like this? $\endgroup$ – Shibalicious Feb 22 at 11:45
  • $\begingroup$ It's simple trigonometry, applied to right triangles $CRA$ and $CPA$. $\endgroup$ – Aretino Feb 22 at 11:47
  • $\begingroup$ Is that assuming that AO is parallel to the Y-axis and CRA and CPA are both right angle triangles? $\endgroup$ – Shibalicious Feb 22 at 11:51
  • $\begingroup$ Yes, exactly so. $\endgroup$ – Aretino Feb 22 at 11:52
  • $\begingroup$ I see, thank you! Could you see my second edit please? $\endgroup$ – Shibalicious Feb 22 at 11:53

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