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Overview...

I'm a novice in the very basics of light physics and calculation between various units. I'm sure this is a very simple problem to those who know the process. And, before we continue, this is not a homework assignment. I am a computer programmer working on a current research project that involves the visual effects industry and post process rendering. To make sure that our processes are as physically accurate as possible, we are researching physical units of light intensity and how best to replicate that in a visual effects renderer. I have found lux to lumens calculators but am quite sure that I'm missing some very basic information on the geometry of light propogation and the best units to measure this.

What I have Tried...

Yes... I have found online lux to lumens calculators. Why do I not trust them? Because they don't discuss the parameters of our test nor do they explain why their results are achieved. Thirdly, they do not define reprogrammability for someone who has no experience with steradians, foot-candles, candelas, etc. So while the formulas may be present, a clear, graphical explanation, that involves the exact data I've presented, is not present.

The Problem...

We have measured a variety of lights (LEDs) at a given distance from each light (8"). The light intensity we measured was, on average, 120 LUX (lumens per meter*meter... as far as I can tell). The angle of the light cone from the LED is given at 180 degrees, and the radius of the LED is .002 meters. The measuring unit was a hemisphere who's surface area (2 * pi * r * r) was 0.00893 sq feet (where the radius of said hemisphere is 0.0115 meters, or... 0.452 inches). With all this information, a few questions arise:

  1. Given this information, what is the known amount of lumens projected from the light source?
  2. Do I have all the information needed to successfully yield an answer to the first question?
  3. Is there a gap, hole, missing or incomplete element in my process and data?

A Final Request...

Would someone be able to explain not only the answer, but the steps to create the answer? The answer isn't important if I can't recreate it.

Please feel free to ask for more details; I hope I can provide them.

Thank you!

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    $\begingroup$ Just go here: rapidtables.com/calc/light/lux-to-lumen-calculator.htm It's a lux to lumens calculator, found by a google search of your question title. $\endgroup$
    – Dave Coffman
    Commented May 12, 2015 at 0:02
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    $\begingroup$ @DaveCoffman, Perhaps you missed the part of my post where I mentioned the following: I have found lux to lumens calculators but am quite sure that I'm missing some very basic information on the geometry of light propogation and the best units to measure this. I'll edit my question to reflect the requirements of my question and the operable guidelines I've already been through $\endgroup$
    – Andrew
    Commented May 12, 2015 at 1:04
  • $\begingroup$ @DaveCoffman I am also still seeking answers to questions 2 & 3. Can you assist with those? $\endgroup$
    – Andrew
    Commented May 12, 2015 at 1:06
  • $\begingroup$ You measured 120 lux at what distance? 8" or 0.00893 sq feet? 0.002m or 0.0115m? I'm confused. $\endgroup$
    – LDC3
    Commented May 12, 2015 at 1:59
  • $\begingroup$ @LDC3, Taken from the initial question: we have measured a variety of lights (LEDs) at a given distance from each light (8"). The other numbers you reference are as follows: 0.002 meters = radius of the LED :: 0.00893 sq ft = surface area of the hemispherical measuring unit :: 0.0115 meters = radius of the measuring hemisphere. $\endgroup$
    – Andrew
    Commented May 12, 2015 at 16:30

1 Answer 1

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First of all I think a suggestion is mandatory: please don't mix units of measure. You are using lux and lumens, so it's better if you stick to SI units. So distances are measured in meters, areas in square meters and angles in radians. This is meant to help you, not to annoy. ;)

Lux measure how much light (lumens) hits a square meter of surface. So if you know the flux $\Phi = 120$ lux, the distance from the source $r \simeq 0.20$ m and the solid angle $\Omega = 2 \pi$, you can calculate the total amount of light emitted as: $$\Phi\, \Omega\, r^2$$ Inserting the numbers into the formula gives 31.13 lumen.

The problem is not really a physics problem, the solution lies in the realm of geometry. You just need to know what each piece of data represents (wikipedia is a good source in this case) and combine everything in order to obtain the desired result. No need to be an expert of luminosity.

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  • $\begingroup$ Thank you. I'll look into this today. I immediately want to upvote your response because it was thorough and informative. I wish I could upvote your response but as I'm new to this forum I can't even seem to do that, despite being the asker. The mixing of units was a result of the measuring tools available on our production stage. It is recognized as less than ideal and your feedback is understood and welcome. $\endgroup$
    – Andrew
    Commented May 12, 2015 at 16:23
  • $\begingroup$ understood on how to mark an answer correct. I will certainly do this after I've investigated the numerical calculations into the final result. I just need to verify that I'm not missing anything else and that it all works correctly. Again, thank you for reading the entire question and answering thoroughly. It is refreshing. I will be back to confirm everything later today. $\endgroup$
    – Andrew
    Commented May 12, 2015 at 16:36
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    $\begingroup$ Right now, there is one aspect of your answer that I'm confused on. When I run the math through with units we end up with $31.13 lumen*sr^2$. However, given that sr are unitless, do these just disappear? $\endgroup$
    – Andrew
    Commented May 12, 2015 at 17:52
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    $\begingroup$ Oops... That should have read: $31.13 candela*sr^2$ or $31.13 lumen*sr$. We get this because lux = $lm/(m^2)$ which leads to: $(lm * sr * m^2)/(m^2)$. The two $m^2$ cancel out and leave: $lm*sr$. $\endgroup$
    – Andrew
    Commented May 12, 2015 at 18:03
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    $\begingroup$ As you said radians and steradians are unitless, so you don't really need to use them in calculation. It is only useful because of its geometrical meaning. $\endgroup$
    – GRB
    Commented May 12, 2015 at 18:10

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