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I would like to convert a rectangular beam shape given in Horizontal and Vertical beam angle, into solid angle representing the surface area in steradians of projected light. For example a light projection that has a 120° H beam by 5° Vertical beam pattern.

It would be helpful if this could be represented in a geometric form that can be used in an Excel Spreadsheet that allows one to input a horizontal beam angle and a vertical beam angle and then output the area in steradians. Given that a sphere has $4\pi$ sr and a hemisphere is $2\pi$ sr I was approaching this for example above that 120°=3.1414 sr or 1/3 of a sphere, and 5°= 0.00597 sr, I thought if there are 120 sections of 5° vertical sections this becomes a total area of 0.7164 sr. Then second guess I thought that there are only 24 vertical sections that are 5 degrees wide in the total horizontal area of 120° area. That yields only 0.1435 sr which seems unbelievably small. I came to the conclusion I need help and found this forum trying to find the solution.

I reviewed many questions that are similar. This is my first post, and I'd appreciate any help and guidance trying to figure where I have gone wrong in the application. Thank you in advance

Widawensen
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luminary
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1 Answers1

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Consider a beam with a rectangular profile, having horizontal & vertical beam angles $\theta_{H}$ & $\theta_{V}$, emitted by a uniform point source S (as shown in the figure below). Then consider an imaginary rectangular plane with center O, length $l$ & width $b$ at a normal distance $h$ from the point-source S, then the solid angle ($\omega$) by the rectangular plane at the point-source S is given by $$\omega=4\sin^{-1}\left(\frac{lb}{\sqrt{(l^2+4h^2)(b^2+4h^2)}}\right)$$ (Note: For derivation of the above formula, kindly go through HCR's Theory of Polygon)

Using geometry in right $\Delta SOP$ from the figure, we get $$l=2h\tan\frac{\theta_{H}}{2}$$ Similarly, in right $\Delta SOM$, we get $$b=2h\tan\frac{\theta_{V}}{2}$$

Rectangular shape beam

Now, substituting these values in the formula, we get $$\omega=4\sin^{-1}\left(\frac{2h\tan\frac{\theta_{H}}{2}2h\tan\frac{\theta_{V}}{2}}{\sqrt{\left(\left(2h\tan\frac{\theta_{H}}{2}\right)^2+4h^2\right)\left(\left(2h\tan\frac{\theta_{V}}{2}\right)^2+4h^2\right)}}\right)$$ $$=4\sin^{-1}\left(\frac{\tan\frac{\theta_{H}}{2}\tan\frac{\theta_{V}}{2}}{\sec\frac{\theta_{H}}{2}\sec\frac{\theta_{V}}{2}}\right)=4\sin^{-1}\left(\sin\frac{\theta_{H}}{2}\sin\frac{\theta_{V}}{2}\right)$$ Where, $0\leq\theta_{H}, \theta_{V}\leq\pi$

If $\theta_{H}, \theta_{V}>\pi$ apply the following formula to calculate solid angle $$\omega=4\pi-4\sin^{-1}\left(\sin\frac{\theta_{H}}{2}\sin\frac{\theta_{V}}{2}\right)$$

(Note: For more detailed explanation, kindly go through Radiometry & Photometry by H.C. Rajpoot)

As per numerical values provided in your question, the solid angle $\omega$ (i.e. area covered\intercepted by rectangular beam shape on a unit spherical surface), subtended by the rectangular beam shape having horizontal & vertical beam angles $\theta_{H}=120^o=\frac{2\pi}{3}$ & $\theta_{V}=5^o=\frac{\pi}{36}$ at the point-source S, is given as follows $$\omega=4\sin^{-1}\left(\sin\frac{\frac{2\pi}{3}}{2}\sin\frac{\frac{\pi}{36}}{2}\right)=4\sin^{-1}\left(\sin\frac{\pi}{3}\sin\frac{\pi}{72}\right)\approx 0.15113795 \space sr$$

Harish Chandra Rajpoot
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  • Thank you for your answer, I appreciate your response and now will try to work through it. I also appreciate the reference links. I can appreciate the rectangular reference you made, but was unsure how to translate that to a spherical system. I will respond again, but for now I wanted to say thank you for your effort to help me through this. – luminary May 19 '15 at 00:49
  • Again Thank you, I read most parts of your paper HCR's Theory of Polygon and was fascinated by it. I worked through your solution and have some additional questions: – luminary May 19 '15 at 04:44
  • @ Harish, What are the beam limits that can be inputted? For example using your solution 120° x 120° yielded 3.000 sr but as I increased above 180° the value of sr began to get smaller. I have not been able to figure where I made a mistake. I'd appreciate additional thoughts. I'm very impressed though by your work and wish you success at the University. – luminary May 19 '15 at 05:18
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    @ Luminary, In case, $\theta_{H}, \theta_{V}>\pi$ you can subtract the smaller value obtained by formula from $4\pi$. You can also see the correction for limits I put in the answer. – Harish Chandra Rajpoot May 19 '15 at 13:15
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    @ Luminary, As you have mentioned that the profile of beam is rectangular. Thus no beam with rectangular profile can subtend a solid angle greater than $2\pi\space sr$ at the point-source. The reason is that for angles greater that $180^o$, the rectangular profile (beam front similar to a wave front) traverses on other side of the point-source sweeping an imaginary half sphere that you are consider as the rear part of beam which is actually not associated with rectangular beam profile. So if you still want to apply the formula for angles more than $180^o$ then it can be modified. – Harish Chandra Rajpoot May 19 '15 at 14:00
  • @ Harish, Thank you again for further explanations. Most of my applications are either very narrow or to the extreme Omni-directional. Typically I would not need to go below the horizon but have seen some applications for now I am happy to have the 180° limit. Under the scenario that it is not a point-source, say multiple sources or a linear source, would any modification be needed? Thank you again this has been a good education for me and I appreciate your input. – luminary May 19 '15 at 14:25
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    @ Luminary, For your kind attention, the solid angle is subtended by an object (2-D or 3-D) only at a point in the space. How to define the solid angle subtended by a beam at multiple sources or a linear source as you mentioned under the scenario? – Harish Chandra Rajpoot May 19 '15 at 14:34
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    In actual practice the point-sources are not possible but the extended sources for longer distances & the smaller sources for usual distances may be assumed to be the point-sources for analysis purpose & practical applications. – Harish Chandra Rajpoot May 19 '15 at 14:46
  • I was instructed to limit chat, but I can't contribute or vote because not enough reputation has been accumulated. I could continue with some variations, but will post them as separate questions tomorrow after doing more research. Thank you for your help so far. – luminary May 20 '15 at 02:35
  • The formula for the sine of the solid angle shown in the answer above is for example published by A. Khadjavi in J. Opt. Soc. Am. 58 (1968) 1417 https://doi.org/10.1364/JOSA.58.001417 . – R. J. Mathar Feb 03 '22 at 12:14