RGB - (wavelength, intensity)

Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity
?

Second question: how about if I put a IR-pass visible-block filter
such as the Hoya R72 on my digital camera? Can I compute
infrared wavelength and intensity from remaining RGB data?

Thanks!

On 25 Nov 2006 18:05:05 -0800, "Yeppers" wrote:

Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity



No. Too simplistic a model. What's the wavelenght of
white? Or gray, or black?

Not all visible colors (or RGB combinations) correspond
to a specific wavelength. In the general case, the
perception of color depends on the a spectrum of
illumination (eg., skylight) multiplied by a spectrum of
absorption (eg. the pigments of a leaf) selectively
absorbing portions of the illuminant.

To get really technical, there's no "pure" RGB color
space -- you'd have to specify which one.


rafe b
www.terrapinphoto.com

Yeppers wrote:
Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity

Yes. You want to convert from the RGB to the HLV color space. That
will isolate the color from the luminance and the saturation. Mapping
the color to a wavelength is a straightforward process.

This will tell you more than you want to know.
http://en.wikipedia.org/wiki/HSV_color_space

--
Ray Fischer

Yeppers wrote:
Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity
?

Second question: how about if I put a IR-pass visible-block filter
such as the Hoya R72 on my digital camera? Can I compute
infrared wavelength and intensity from remaining RGB data?

Thanks!

The answer is probably yes and no, though no in reality. Let me explain.
The RGB filters used in a digital camera have pretty broad spectral
coverage (see the Kodak graph in one of my IR articles:
http://www.dimagemaker.com/article.php?articleID=466). What this means
is that the RGB values represent the light intensity over a pretty broad
part of the spectrum. What this means is that, in reality, you cannot
convert this into exact wavelength and intensity values.

However, you could produce a 'simulated' wavelength value by using some
maths but it would not, IMHO, be meaningful in any real way.

Same goes for the R72. See my above mentioned article for how IR
sensitivity works, but the same issue applies. The filter are fairly
broad in their coverage, so you can't pin something down to a particular
wavelength.

Cheers,

Wayne

--
Wayne J. Cosshall
Publisher, The Digital ImageMaker, http://www.dimagemaker.com/
Blog http://www.digitalimagemakerworld.com/

In article , Ray Fischer
writes
Yeppers wrote:
Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity

Yes. You want to convert from the RGB to the HLV color space. That
will isolate the color from the luminance and the saturation. Mapping
the color to a wavelength is a straightforward process.

However, that is merely a conversion of the output data and bears almost
no relationship at all to the wavelengths and intensity of the source:
the original scene spectral content.

Say, for example, the green pixels in your camera respond from 500nm to
600nm. Wherever the source spectrum lies in that range you still get
the same output. An almost pure green light at 550nm, or a light at
525nm or even a broad spectrum between 475 and 525 would give the same
RGB output from the camera.

The camera convolves the source spectral content with its own filter
bandwidths before the signal is sampled by the sensor. Consequently
there is no way to get back from the spectral content (wavelength &
intensity for each pixel) with any more accuracy than the filter
bandwidths themselves.

The error in the original question is the assumption that the original
light can be uniquely defined by a set of red, green and blue values, it
can't. Even worse, those red green and blue values will only produce a
certain range of wavelengths and intensity if they are input to a
display with a particular spectral output of each of the red green and
blue channels. In most cases the rgb output is merely an approximation
achieved by calibration, but doesn't need to be any better because
another spectral convolution occurs in your eye.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's ****ed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)

In article ,
"Wayne J. Cosshall" wrote:

The RGB filters used in a digital camera have pretty broad spectral
coverage (see the Kodak graph in one of my IR articles:
http://www.dimagemaker.com/article.php?articleID=466). What this

And this leads to an accidental experimental confirmation of your
statement that I once encountered.

The graph on your page shows that the green and blue filters actually
"turn on" again (and the red filter stays on, and the underlying sensor
retains some sensitivity) as you move beyond the visible, well out into
the near IR.

As a result, sending 1.06 micron YAG laser radiation into a digital
camera produces a bright *purple* image, at least in the digital camera
I was involved with -- and it's not due to any harmonic generation.

Kennedy McEwen wrote:
Ray Fischer
Yeppers wrote:
Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity

Yes. You want to convert from the RGB to the HLV color space. That
will isolate the color from the luminance and the saturation. Mapping
the color to a wavelength is a straightforward process.

However, that is merely a conversion of the output data and bears almost
no relationship at all to the wavelengths and intensity of the source:
the original scene spectral content.

True enough. Finding the original scene's spectral content is not
possible without a spectrograph. You need more than just three
different color sensors.

[...]
The error in the original question is the assumption that the original
light can be uniquely defined by a set of red, green and blue values, it
can't. Even worse, those red green and blue values will only produce a
certain range of wavelengths and intensity if they are input to a
display with a particular spectral output of each of the red green and
blue channels. In most cases the rgb output is merely an approximation
achieved by calibration, but doesn't need to be any better because
another spectral convolution occurs in your eye.

What the RGB-HLV conversion will do is give you an approximation of a
single wavelength that will approximate the perception of the color of
the original scene.

--
Ray Fischer

AES wrote:
In article ,
"Wayne J. Cosshall" wrote:

The RGB filters used in a digital camera have pretty broad spectral
coverage (see the Kodak graph in one of my IR articles:
http://www.dimagemaker.com/article.php?articleID=466). What this

And this leads to an accidental experimental confirmation of your
statement that I once encountered.

The graph on your page shows that the green and blue filters actually
"turn on" again (and the red filter stays on, and the underlying sensor
retains some sensitivity) as you move beyond the visible, well out into
the near IR.

As a result, sending 1.06 micron YAG laser radiation into a digital
camera produces a bright *purple* image, at least in the digital camera
I was involved with -- and it's not due to any harmonic generation.

Yup, and it is that rise in transmission that allows IR photography with
digital cameras, providing the separate IR blocking filter is not too
strong.

Cheers,

Wayne

--
Wayne J. Cosshall
Publisher, The Digital ImageMaker, http://www.dimagemaker.com/
Blog http://www.digitalimagemakerworld.com/

In article , Ray Fischer
writes

What the RGB-HLV conversion will do is give you an approximation of a
single wavelength that will approximate the perception of the color of
the original scene.

Whether or not that colour in the original scene was produced by a
single wavelength or not.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's ****ed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)

"Yeppers" wrote in news:1164506705.856175.226690
@l12g2000cwl.googlegroups.com:

Hi all,

Is there a simple straightforward set of equations
for converting normal visible light
* red
* green
* blue
values from my digital camear valuesto
* wavelength
* intensity
?

Second question: how about if I put a IR-pass visible-block filter
such as the Hoya R72 on my digital camera? Can I compute
infrared wavelength and intensity from remaining RGB data?

Thanks!

In addition to the answer posted by others, I'd like to direct
you to the following page,

http://www.efg2.com/Lab/Graphics/Col...romaticity.htm

You should read up on concepts such as "color space" and "color
gamut".

Many years ago I spent considerable time on this problem. I
eventually emperically derived an approximation curve by
comparing the spectrum of the sun off a grating to views on
the monitor. The result was used for a chart of laser lines.

You can see it here,

http://www.skywise711.com/lasers/reference.html

Look for the "Visible Laser Spectrum Chart".

Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Seismic FAQ: http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
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