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Researching this topic, I came across two methods that seem to contradict each other. Need to decide which to use.

I am a novice in this field so please be gentle. Also, I would have loved to use the fancy math signs in this post, but have no idea how that works. My apologies. As a filmmaker, I am trying to build a simple photometric calculator to calculate the output of different light sources under different circumstances. Now, I'm confronted with two different methods of calculating Lux (that give very different results) and I wonder which to choose.

  1. $Ev(lx) = Iv(cd) / (d(m))^2$ (see link): This one seems to be used in most of the Photometrics calculators I've seen.

However, Isn't this too simple? For one thing it doesn't take into account the apex angle of the light source. Many of the lights used in filmmaking have the ability to adjust the angle of the light coming out. And in my personal experience this has really changed the intensity of the light.

I've started to do some of my own calculations, before I found the formula mentioned above.

2a. First, I convert the Candela to lumen as follows:
2a. $cd \times sr$ | $\Phi v(lm) = Iv (cd) \times 2 \pi(1- cos(\thetaº/2))$

2b. Then I calculate the surface area of the base of the light beam.
2b. $r = distance \times \tan(\theta/2)$
2b. $\pi \times r^2$

2c. Then I simply divide the lumen by the surface area of the (base of the) beam.
2c. $\Phi v(lm)/m^2$

The 2 methods give very different results. Example Light A, has 9500 cd, a variable beam between 60º and 80º, and a distance of 10 meter.

Method 1 results in 95 Lux no matter the beam angle.

Method2 results in 76.36 Lux with a 60º angle and 63.13 Lux with an 80º angle.

Can anybody tell me what method is more accurate in my situation? All help is greatly appreciated!

Kind regards, Remco

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  • $\begingroup$ Get a copy of the RCA Electro-optics handbook. hardcopy at Amazon; soft copy by a simple google search. $\endgroup$ – Carl Witthoft Dec 4 '15 at 18:13
  • $\begingroup$ Note that this site has MathJax enabled, so you can replace your text equations to actual equations. $\endgroup$ – Kyle Kanos Dec 4 '15 at 18:46
  • $\begingroup$ Thank you both. I'll definitely look into the book. And as soon as I get the MathJax thing, I'll edit my post with the proper symbols. $\endgroup$ – Remco Hekker Dec 5 '15 at 10:35
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Both methods are doing more or less the same thing. But they both make different approximations, so the answers will be a little different. Both assume that the source is isotropic, that is, radiates light uniformly within the angle of interest. That is certainly not the case in practice. Furthermore:

Method 1 assumes that the size of the area you are illuminating is much smaller than the distance to the source ... roughly 10 times smaller (another approximation).

Method 2 doesn't make that assumption (although I'm not 100% certain about you equation 2a), but it does make the assumption that the luminance will be constant across the area to be lit up. When the distances between source and lit area is small, that's not correct: the parts of the area directly "under" (closest to) the source will be brighter than those at the edges.

It's not surprising at all that your two methods give roughly the same answer, but not exactly the same answer.

Neither method will brighten things up if you reduce the angle. Both of these methods simply ignore light that falls outside of the area of interest. The calculation does nothing to increase the actual light available ... that's determined by the lamps, mirrors, reflectors, etc. When you reduce the angle in the studio, you adjust mirrors and lenses to do so. That changes the intensity (candelas) of the source. No calculation can increase the actual amount of light available!!

To aproximately include the effect of decreasing the apex angle, you would have to increase the intensity in your calculations by the ratio of the solid angles before and after making the change. This will be yet another approximation because as I mentioned above, the intensity across the beam is not constant, and furthermore, the shape of the distribution of light will change as you change the settings on your source.

Sometimes luminaire (light fixture) manufacturers will have detailed information of intensity as a function of angle. You might check to see if such data are available for your sources. That would help. In any event, calculations of the type you are doing are usually done as a rough guide to lighting conditions. A designer would either do a full-on calculation using sophisticated and expensive software, or do the rough calculations as you have done, and then build in a margin or error by increasing the expected light needed by, say 30%. It's hard to judge which of your calculations will be more accurate without knowing more about the sources, but I doubt that your areas are very far away from the sources, so method 2 would seem to be the better choice. It is the more conservative choice.

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  • $\begingroup$ Thank you for such a complete answer. There is a lot in there to help me a long. I'll double check my candela to lumen calculations. Because now it seems strange that the lux amount changes so much when I adjust the angle. Anyway you've helped me a long greatly. Thank you! $\endgroup$ – Remco Hekker Dec 5 '15 at 10:34
  • $\begingroup$ For small angles, they should give the same answer. Try something like $\theta = 3^\circ$. That's not a practical value, of course, but it will serve as a check to see if there's some other mistake somewhere. I'm a little uncertain, because there are a lot of factors of 2 and $\pi$ floating around, and I did not check to see if they are all correct. $\endgroup$ – garyp Dec 5 '15 at 14:46

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