Lumens and actual power of luminous flux

[sophiecentaur]

I'm involved with a query about Transmittance in this other thread and I need a way to convert lumens to actual power. there are many pages relating lumens out of a light bulb to the electrical power input but all I can find about the lumen is self-referential and uses terms like 'perceived' light. It's not trivial if you want to include a particular illuminant because the spectrum and the spectral response of the eye. Does anyone have a link that unravels the whole business?
Cheers

 

[Rap]

Hi sophiecentaur - Thank you so much for posting part of my problem here. I think I can unravel some of it. If you have a beam of light with radiant intensity (eg, power per unit area) J[w] where w is wavelength, then the luminous intensity is J[w]V[w] where V[w] is the luminous efficiency function, which has a peak value of unity around 550 nm and drops to near zero around 400 and 700 nm. I'm just trying to figure out what the "luminous transmittance of a filter as a function of wavelength" means.

 

[Andy Resnick]

I'm involved with a query about Transmittance in this other thread and I need a way to convert lumens to actual power. there are many pages relating lumens out of a light bulb to the electrical power input but all I can find about the lumen is self-referential and uses terms like 'perceived' light. It's not trivial if you want to include a particular illuminant because the spectrum and the spectral response of the eye. Does anyone have a link that unravels the whole business?
Cheers

That other thread looks like a mess.... I can try an untangle this specific issue, tho.

First- there is no simple way to interconvert lumens and Watts because while Watts is a radiometric quantity (independent of human vision), lumens is a photometric quantity (depends very much on the eye spectral response).

So, to convert watts to lumens you need to take into account both the human eye spectral response, call it V(λ), and the spectral power output of the source, call it Φ(λ). A functional form of V(λ) is given here:

https://laser.physics.sunysb.edu/_carolyn/report/index.html

And note there are two distinct vision curves for day vision and night vision.

To convert Watts to lumens, simply compute the integral

Lumens = ∫Φ(λ) V(λ) dλ

Reversing the computation (lumens to Watts) can't easily be done unless you have information about the spectral properties of the source. It's possible for multiple sources, each with different Φ(λ), to provide the same amount of luminous power because of the factor V(λ).

Does that help?

 

[sophiecentaur]

That other thread looks like a mess..

Lol!

First- there is no simple way to interconvert lumens and Watts because while Watts is a radiometric quantity (independent of (individual?) human vision), lumens is a photometric quantity (depends very much on the eye spectral response).

Thanks for that!!! It's what I suspected and it implies everything is relative and varies from case to case. The lesson seems also to be that Power (W) doesn't come into it. Something for the OP to take on board.
Reading around in the links and elsewhere, it get the feeling that pretty much all the colourimetry used in imaging technology (TVs, cameras, printing ) hangs on statistics with wide margins. But it's the best that we can do and we just have to deal with the fact that three people watching the same TV at the same time are watching entirely different pictures.

 

[Gleb1964]

To convert Watts to lumens, simply compute the integral

Lumens = ∫Φ(λ) V(λ) dλ

Reversing the computation (lumens to Watts) can't easily be done unless you have information about the spectral properties of the source.

Exactly,
if you know or can assume the spectral distribution of the source and definite spectral range, you can calculate converging coefficient as (∫Φ(λ) V(λ) dλ)/(∫Φ(λ) dλ) within defined spectral range, that would be [Lumens/W]. That coefficient can be used for conversion back and forward.

 

[sophiecentaur]

Exactly,
if you know or can assume the spectral distribution of the source and definite spectral range, you can calculate converging coefficient as (∫Φ(λ) V(λ) dλ)/(∫Φ(λ) dλ) within defined spectral range, that would be [Lumens/W]. That coefficient can be used for conversion back and forward.

This makes me think of the dB scale for acoustics. That all hinges on the 'threshold of hearing' which is, of course, different for all of us. So we relate it all to a Standard Ear. But we don't seem to go as far as having a Standard Eye; the power and the perception are a bit more 'honestly' adrift from one another.

 

[Rap]

Thank you everybody for your input, but I still am a bit puzzled by the original problem. I'm trying to find out the meaning of "luminous transmittance" at a particular wavelength when characterizing an optical filter. Here's the quote:

 

[sophiecentaur]

I'm trying to find out the meaning of "luminous transmittance"

I thought we'sorted this out. It's just the percentage of light that gets through at a particular wavelength.
All the rest of that paragraph refers to a particular situation when they use it. For a single peaked filter they seem to reference the throughput at any wavelength to the peak level. For two peaked filters (complementary range of colours like purple) there is a problem because there are two maxima. What is written about that is not clear but you may see what they mean by looking at some actual results of some purple filters. It strikes me that they quote Illuminant C because the components of two-peaked filters will be more affected by the peak wavelength of the illuminant as they lie on the upper and lower slopes of the C spectrum.

If it's not relevant to your particular work then I'd recommend not getting too involved; the basic definition is very straightforward. If you are using more than one filter then you need to use one maximum as a reference.

 

[Rap]

Forgive me, but it's not sorted out in my mind. Can we define "radiant" to mean energy, or power, or power/cm^2/steradian/hz, i.e. NOT involving the the luminous efficiency function for the human eye V(λ). Can we define "spectral" to mean "as a function of wavelenth". So your definition of transmittance as power out/power in at a particular wavelength is the spectral radiant transmittance. It does not depend on the particular illuminant being used to make the measurement and it has nothing to do with V(λ).

Looking at the numbers in the article, they do not seem to reference any peak. 100% is never found in the list of "luminous transmittances".

When I see "luminous", I expect something involving V(λ). So when I see "luminous transmittance as a function of wavelength" and "using Illuminant C", then either:

1) "luminous transmittance" involves V(λ), in which case your statement that it is spectral radiant transmittance is wrong.
2) "luminous transmittance" does not involve V(λ), in which case the statement in the article is misleading and/or wrong and/or nonsense.
3) The article is correct, and I am baffled and confused.

I just need help figuring out which case is true. And thanks for any insight you can give!

 

[sophiecentaur]

The term “mean energy” (Joules) cannot apply when Power Flow or amount of light is involved. Afaics it’s lumens per wavelength interval that applies here for spectral density.
The spectrum of the illuminant is used if you can’t assume a flat spectrum. I can see no reason for any different treatment of two peaked filters if everything is normalised.
If you need more opinions about that quoted snippet then you should post more or all of it.

 

[sophiecentaur]

100% is never found in the list of "luminous transmittances".

Wouldn't 100% represent no filter in place or a filter with zero loss at its maximum? All filter gels have some loss - they are not multicoated as with high quality lenses.

1) "luminous transmittance" involves V(λ),

Transmittance is nothing to do with the source or detector. It's the fraction of light that 'gets through'

2) "luminous transmittance" does not involve V(λ), in which case the statement in the article is misleading and/or wrong and/or nonsense.

I have not seen the original article so I couldn't say but it has confused you.

 

[Rap]

Hi - thanks sophiecentaur -

"Wouldn't 100% represent no filter in place or a filter with zero loss at its maximum? All filter gels have some loss"
Yes, but the suggestion was made that the transmittance was somehow referenced to a maximum, which would mean every filter would have one or more wavelengths registering 100% (of maximum)

"Transmittance is nothing to do with the source or detector. It's the fraction of light that 'gets through'".
Yes, Jout(λ)/Jin(λ) - but then why call it "luminous transmittance"? "luminous" implies that V(λ) is involved, like "luminous intensity" or "total luminous transmittance" which is the integral of Jout(λ) V(λ)/Jin(λ) over all λ.

I have attached an image of the statement in the article that is bothering me. Please let me know if you cannot view it.

 

[sophiecentaur]

I have attached an image of the statement in the article that is bothering me. Please let me know if you cannot view it.

The only way for you to resolve this is to look elsewhere for the definitions of quantitles. You have quoted from just the one source and only one paragraph of that, even. Do they discuss or justify what they have written here? To my mind, it's like specifying the attenuation of a linear RF circuit in terms of the RF source signal spectrum, which makes no sense at all.

If the text you are using doesn't agree with mainstream work (which it appears not to and it's pretty ancient) then perhaps you should question it; things may have moved on. You have briefly described your project (in the other thread) but is your quoted source the only way in? The facilities in the 30's didn't include TV type phosphor displays; using filtered tungstan light as a source is maybe not the best approach in 2024.

You seem to be implying that all the work that led to the CIE chromaticity chart hasn't been re-run and checked but that seems unlkely to be the case, bearing in mind the huge commercial interests in improving image processing and TV displays. Is there no other way to access what you want to know?

 

[Rap]

The image of the text was copied from the CRC Handbook of Chemistry and Physics, 90th Edition (2009-2010).

What I want to do is to take MacAdam's raw data from 1942 and convert this into the LMS color space, rather than the XYZ color space as MacAdam did. The LMS color space is based on the latest measured response functions of each cone type in the human eye, and is the basis for any theoretical investigation of color discrimination.

I want to do this without all the approximations MacAdam had to make due to the comparatively limited state of computational resources available in 1942, and without the few calculation errors he made. To do this, I need to know the *spectral radiant transmission* of the Kodak Wratten filters that he used. Kodak published this data every few years starting around 1920, with apparently increasing accuracy, clearly stating that it was spectral radiant transmission. They published much less frequently with the advent of digital photography, culminating in the 2009 article above by Allie Peed, an employee of Kodak, but which confused me. I expect it is the most accurate description.

Thanks for your attention to this!

Attached is a larger image of the first page of the 2009 article by Peed:

 

[sophiecentaur]

To do this, I need to know the *spectral radiant transmission* of the Kodak Wratten filters that he used.

To be certain what it's all about, you would need details of the experimental setup.

I could suggest that they didn't measure input and output levels at each wavelength but relied on measuring the peak (or something) of the illuminant C spectrum and a 'corrected' value for the filter input for each measured output wavelength. That would give a more accurate measurement for the Wratten Filter characteristics relative to a flat input spectrum.

 

[Rap]

Hmm - That might be the case. In other words, knowing the SPD of Illuminant C to within an constant, they would only have to measure it at one wavelength to determine that contant, and then could scan the output from the filter to get transmittance.

MacAdam's 1942 paper gives loads of detail on the experimental setup, but states that the Wratten transmittances were measured for each filter, not trusting the published data for any individual filter. He published his measured transmittance data for maybe 15 filters as examples, but there are about 120 filters that he used, so I have to rely on published data to fill in the blanks.

I think I am going to have to compare Peed's data to earlier measurements which I know are intensity out/intensity in, and see if they differ greatly, in which case I will suspect Peed's data.

 

[sophiecentaur]

Hmm - That might be the case. In other words, knowing the SPD of Illuminant C to within an constant, they would only have to measure it at one wavelength to determine that contant, and then could scan the output from the filter to get transmittance.

It makes sense. Things were just not that convenient in the thirties. I look forward to reading that all that stuff has been done with a set of bright, calibraed monochromatic sources.
You could just compare the luminosities needed for an equal visual response from two (or more) sources (like the magnitude comparisons of stars) and let the computer take care of all the number crunching.