Luminous Landscape Forum
Raw & Post Processing, Printing => Colour Management => Topic started by: JeremyLangford on September 15, 2009, 06:46:21 pm
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Is a 2d diagram of a color gamut such as this one.....
(http://www.colblindor.com/wp-content/images/CIE%201931%20color%20space.png)
basically an overhead view of an RGB cube with the white corner in the center?
(http://upload.wikimedia.org/wikipedia/commons/0/03/RGB_farbwuerfel.jpg)
(http://www.argyllcms.com/doc/cube.jpg)
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No, it is not. A 2D projection of a cube would not give you the smooth shape of your first diagram.
The term given to the object depicted in the first diagram is the spectral locus.
See http://en.wikipedia.org/wiki/CIE_1931_color_space (http://en.wikipedia.org/wiki/CIE_1931_color_space)
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No, it is not. A 2D projection of a cube would not give you the smooth shape of your first diagram.
The term given to the object depicted in the first diagram is the spectral locus.
See http://en.wikipedia.org/wiki/CIE_1931_color_space (http://en.wikipedia.org/wiki/CIE_1931_color_space)
Wikipedia also says this about Color Gamut Diagrams here (http://en.wikipedia.org/wiki/Gamut)
(http://upload.wikimedia.org/wikipedia/en/thumb/d/d3/CIExy1931_srgb_gamut.png/240px-CIExy1931_srgb_gamut.png)
"A typical CRT gamut:
The grayed-out horseshoe shape is the entire range of possible chromaticities. The colored triangle is the gamut available to a typical computer monitor; it does not cover the entire space. The corners of the triangle are the primary colors for this gamut; in the case of a CRT, they depend on the colors of the phosphors of the monitor. At each point, the brightest possible RGB color of that chromaticity is displayed, resulting in the bright Mach band stripes corresponding to the edges of the RGB color cube."
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Correct, the spectral locus is the shape and range of perceivable colors for an average viewer.
This is not the same thing as a RGB cube, which is a mathematical abstraction of an additive color space with three primaries.
As indicated above in the diagram you posted, if the primaries are visible (i.e., the vertices of the triangles lie within the horseshoe shape), then there will be physical colors that cannot be reproduced in that RGB space. This is clear, because no matter where you place the three vertices of the triangle within the horseshoe, there will be colors that lie outside the triangle.
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Jeremy,
If you want to see a visual description of how 2D color space relates to 3D color space, download ColorThink Pro. Even the free demo version allows you to choose between Lab space, Yxy, and Luv spaces. Then, clicking on the 2D graph or 3D graph buttons will automatically animate the change between the two. It's a great way to get your head around these concepts.
https://www.chromix.com/colorthink/download (https://www.chromix.com/colorthink/download)
(http://www.colorwiki.com/images/b/b6/2DLL.jpg)
(http://www.colorwiki.com/images/2/2a/3DLL.jpg)
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Both of these pictures seem to be showing the same thing: A 2d representation of a 3d RGB color space. The edges of the RGB cube even seem to show up in the color gamut diagram. Are they not doing the same thing: Mixing Red, Green, And blue together to show every possible color?
(http://www.sonycsl.co.jp/person/nielsen/visualcomputing/programs/colorcube-1.png)
(http://www.colblindor.com/wp-content/images/CIE%201931%20color%20space.png)
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.
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They are related, but definitely not the same. The spectral locus shows, approximately, the range of colors visible to humans. The cube shows the set of colors that can be generated by three primaries. These are not the same. Pick any set of real primaries. There are colors that are visible to us, that cannot be produced by those primaries, no matter how you add them up.
Furthermore, these shapes are quite different, which should be a straightforward giveaway. It's like comparing a square and a circle. They have different shapes, represent different things, and have different properties.
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These are not the same. Pick any set of real primaries. There are colors that are visible to us, that cannot be produced by those primaries, no matter how you add them up.
I didn't know that. Wouldn't that make the spectral locus pretty incomplete when viewed on an RGB monitor or printed with a printer that only uses CMYK?
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I didn't know that. Wouldn't that make the spectral locus pretty incomplete when viewed on an RGB monitor or printed with a printer that only uses CMYK?
That is correct. There is no display nor printer that can reproduce such saturated red, green and blue colors as depicted in the CIE (x,y)-chromaticity diagram (AKA spectral locus).
For the RGB cube and 2D x,y-chromaticity diagram, there is a way to transform the cube into spectral locus representation, but the opposite is not true. RGB values can be transformed into CIE-XYZ values using a linear transformation. XYZ values then can be further transformed into x,y coordinates which are the basis of the spectral locus diagram. However, along the way, the Y (luminance) component is lost and only 2D graph projected. Not knowing the luminance values, one cannot get back to XYZ, thus precluding calculation of RGB values. As Eric Chan pointed out, the RGB gamut (and cube) shown in your previous post (http://en.wikipedia.org/wiki/Gamut) would not match the x,y-diagram. However if you had a hypothetical RGB color space (gamut) that has primaries of the spectral locus, then there would be a mathematical translation going in direction RGB->XYZ>x,y. Prolem is that while RGB space is linear, x,y-chromaticity diagram is not. That means that same changes in the RGB cube would not correspond to perceived changes in the x,y diagram.
Some more details and pictures are at this webpage (http://www.marcelpatek.com/color.html). For transformation of sRGB space to XYZ (->y,y), see equation VII-inv.
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So the XYZ is a color space made in 1931 by CIE that encloses the visible gamut with three primaries that are actually outside the human spectral locus gamut, meaning that it will never be outdated and there will never be a color outside of it. And this makes it the standard reference space for viewing other color spaces.
Is that right?
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So the XYZ is a color space made in 1931 by CIE that encloses the visible gamut with three primaries that are actually outside the human spectral locus gamut, meaning that it will never be outdated and there will never be a color outside of it. And this makes it the standard reference space for viewing other color spaces.
Is that right?
The digital cinema initiative has incorporated that thinking into canonizing XYZ space. Otherwise, actually that "out of gamut" concept is more applicable when you are ready to display, and at that time you need to consider the gamut of colors that the display can accommodate. Any color, including out of gamut, and those outside of spectral locus may be expressed in any standard RGB model where the primaries are inside the spectral locus -- only that some of the values for the intensities of these primaries will be negative, and/or in normalized values greater than +1. And, when it comes to display, or for whatever reason, these negative or greater than +1 values may be dropped or mapped into the target gamut under consideration. Even in the XYZ space, since the primaries are physically non-realizable, when it comes to display then some sort of gamut conversion stage has to be employed to map the colors outside the gamut of the display device into colors that it can display, therefore some reduction happens even during XYZ->display gamut.
Think about it: the spectral locus is outside the CIE RGB primaries, i.e., the triangle formed by the location of these RGB primaries. So that means negative numbers for at least one of the CIE RGB intensities. But, CIE still measured those RGB numbers and then converted them to XYZ, since XYZ is always positive for the spectral locus. (CIE did that by adding some of the RGB primaries to the test color to make the overall system positive.)
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So the XYZ is a color space made in 1931 by CIE that encloses the visible gamut with three primaries that are actually outside the human spectral locus gamut, meaning that it will never be outdated and there will never be a color outside of it. And this makes it the standard reference space for viewing other color spaces.
Is that right?
I agree with your summary. XYZ system is quite nonintuitive, so other color systems were created based on it (Lab, xyY, RGB). Think of the XYZ color space as of computational rather than visual system. And yes, it is the underlying system of color science, digital photography, imaging ,..
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How is this
(http://stars.astro.illinois.edu/sow/emspectrum.jpg)
converted into this?
(http://upload.wikimedia.org/wikipedia/commons/thumb/f/f4/Spectrum_locus_12.png/720px-Spectrum_locus_12.png)
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How is this
(http://stars.astro.illinois.edu/sow/emspectrum.jpg)
converted into this?
(http://upload.wikimedia.org/wikipedia/commons/thumb/f/f4/Spectrum_locus_12.png/720px-Spectrum_locus_12.png)
The following tools offer even more transformations and mappings: Couleur.org (http://www.couleur.org/index.php?page=colorspace) (free download, install).
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How is this
converted into this?
Your second graph (the wiggly one) is actually the monochromatic stimuli of constant radiance for all wavelengths shown in the first graph. When this wiggly curve is projected on the plane x+y+z=1, you get that "standard" spectral locus diagram shown in the figure in the first message (post # 1) of this thread, that I copy below (the figure below is a further projection on the z-axis to show just x-y diagram, instead of an x-y-z diagram):
(http://www.colblindor.com/wp-content/images/CIE%201931%20color%20space.png)
I.e., from the origin draw a straight line connecting any point on the wiggly curve and extend it to the plane x+y+z=1. Where this line meets the plane draw a dot; do that for all points on the wiggly curve and you have now a bunch of dots on the plane. Connect the dots on the plane and you get your standard spectral locus diagram shown in post #1.
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I'm sorry but this stuff is really getting to be over my head.
Do you think someone here can give me an easy to understand, very basic idea on how these three diagrams are converted into each other?
This
(http://stars.astro.illinois.edu/sow/emspectrum.jpg)
Becomes This
(http://upload.wikimedia.org/wikipedia/commons/thumb/f/f4/Spectrum_locus_12.png/720px-Spectrum_locus_12.png)
Which Becomes This
(http://www.colblindor.com/wp-content/images/CIE%201931%20color%20space.png)
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Do you think someone here can give me an easy to understand, very basic idea on how these three diagrams are converted into each other?
Trying again. Pick any single color (monochromatic) from your first diagram and note its radiant power (used later). Then measure the amount of X, Y, and Z primaries, which are calibrated using a certain process, that you need to match that color, and this gives you 3 numbers. Plot the 3 numbers and you get a point on the wiggly curve in your second diagram. Now divide each of the X,Y,Z by their sum (X+Y+Z), and you get another set of 3 numbers. Plot them again and that gives you a point on spectral locus on the third diagram. Now go back to the first diagram and pick another color at the same radiant power as the first color picked and repeat the process. Keep on doing until you have done for all/enough colors in the first diagram. Now you have a whole bunch of points on the second and third diagrams and you just join them together to get the curves shown.
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Trying again. Pick any single color (monochromatic) from your first diagram and note its radiant power (used later). Then measure the amount of X, Y, and Z primaries, which are calibrated using a certain process, that you need to match that color, and this gives you 3 numbers. Plot the 3 numbers and you get a point on the wiggly curve in your second diagram. Now divide each of the X,Y,Z by their sum (X+Y+Z), and you get another set of 3 numbers. Plot them again and that gives you a point on spectral locus on the third diagram. Now go back to the first diagram and pick another color at the same radiant power as the first color picked and repeat the process. Keep on doing until you have done for all/enough colors in the first diagram. Now you have a whole bunch of points on the second and third diagrams and you just join them together to get the curves shown.
How do I find a color's radiant power?
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How do I find a color's radiant power?
Firstly, your second curve doesn't seem fully okay to me. I have a slight problem trying to reconcile it with one of the primaries. But that is fine, as the process I outlined will result in a wiggly nature of the curve and after that "beauty pass" of dividing by X+Y+Z your third diagram will emerge.
As far as measuring the radiant power you will need certain specialized hardware for that and even for setting up an experiment for matching the colors.
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I think I pretty much understand everything except for this:
"Now divide each of the X,Y,Z by their sum (X+Y+Z), and you get another set of 3 numbers. Plot them again and that gives you a point on spectral locus on the third diagram."
What exactly are the second set of 3 numbers? Do you ignore the Z number in order to plot this set into the 2d spectral locus?
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What exactly are the second set of 3 numbers? Do you ignore the Z number in order to plot this set into the 2d spectral locus?
Yep, sorry, you are right, to get the 3rd diagram you ignore the Z number and get the 2D spectral locus. If you don't want to ignore the Z number, then that is fine and you would get a 3D plot of spectral locus instead of 2D. The 3D spectral locus shape will be similar to 2D and won't look like the wiggly plot of the 2nd diagram.
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Trying again. Pick any single color (monochromatic) from your first diagram and note its radiant power (used later). Then measure the amount of X, Y, and Z primaries, which are calibrated using a certain process, that you need to match that color, and this gives you 3 numbers. Plot the 3 numbers and you get a point on the wiggly curve in your second diagram. Now divide each of the X,Y,Z by their sum (X+Y+Z), and you get another set of 3 numbers. Plot them again and that gives you a point on spectral locus on the third diagram. Now go back to the first diagram and pick another color at the same radiant power as the first color picked and repeat the process. Keep on doing until you have done for all/enough colors in the first diagram. Now you have a whole bunch of points on the second and third diagrams and you just join them together to get the curves shown.
Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?
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Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?
No, they are different, but people have tried to find a transformation relating the two responses to varying degree of success.
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No, they are different, but people have tried to find a transformation relating the two responses to varying degree of success.
Then what are the 3 primaries for the spectral locus? (Not the CIE XYZ space)
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Then what are the 3 primaries for the spectral locus? (Not the CIE XYZ space)
In theory any 3 primaries can be used as long as they are not in the same direction and all three don't lie in a plane. Recall, we want to define any color with 3 numbers, i.e. any color would be represented by a point in 3D, therefore we just need 3 vectors in a 3D space to represent any point. The primaries are these vectors. We don't want to point them in the same direction or lie in the same plane otherwise we would have effectively a 2D space (a plane), or a line (1D) embedded in a 3D space as we have suppressed one or two degrees of freedom, respectively. So if you want to use a different set of primaries it is just a reorientation of the original set of primaries, i.e., why we multiply primaries by a matrix to transform one set of primaries to another.
In practise it is advisable to have them spaced farther apart so that they cover as many colors that can be physically displayable (i.e., those with positive values). Primaries in the regions of Red, green and blue fulfill that criteria.
And, as I mentioned before, if we want to measure the values for the spectral locus, then for any set of "real" primaries (i.e., those we can display later) it would mean at least one primary has negative value. But that is fine, we are not going to display it at this stage, we are just measuring it. This is how CIE measured these numbers, including negative numbers, using RGB. But CIE did not want to work in negative numbers thinking that people would make mistakes with negative numbers. Therefore a transformation was found to convert RGB in such a way that resulted in all positive numbers, i.e., RGB->XYZ.
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Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?
I though that because of this picture.
(http://upload.wikimedia.org/wikipedia/commons/1/1e/Gamut_full.png)
But now this picture is just confusing me.
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I though that because of this picture.
(http://upload.wikimedia.org/wikipedia/commons/1/1e/Gamut_full.png)
But now this picture is just confusing me.
Please don't let the graph confuse you. I have to ascertain how close are LMS primaries to XYZ, however, typically, the shape of the spectral locus in that "beauty pass" plot (that horse-shoe like spectral locus) is going to look similar for various different primaries that are close by. It would seem to get stretched, elongated, etc., but overall similar shape.
BTW, which software did you use to generate that "wiggly" spectral locus shape?
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Please don't let the graph confuse you. I have to ascertain how close are LMS primaries to XYZ, however, typically, the shape of the spectral locus in that "beauty pass" plot (that horse-shoe like spectral locus) is going to look similar for various different primaries that are close by. It would seem to get stretched, elongated, etc., but overall similar shape.
BTW, which software did you use to generate that "wiggly" spectral locus shape?
I found it on wikipedia (http://en.wikipedia.org/wiki/Color_vision). I am still having trouble understanding how the "wiggly" 3d spectral locus shape is made. Heres another one.
(http://upload.wikimedia.org/wikipedia/commons/2/2b/Spectral-locus.png)
"Spectral locus in XYZ and xy (demonstration of chromaticity derivation)"
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Have a look at www.handprint.com (http://www.handprint.com), especially this section (http://www.handprint.com/HP/WCL/color1.html#trichrommix). The site goes into quite a lot of detail about chromaticity diagrams, etc.
Cliff
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Have a look at www.handprint.com (http://www.handprint.com), especially this section (http://www.handprint.com/HP/WCL/color1.html#trichrommix). The site goes into quite a lot of detail about chromaticity diagrams, etc.
Cliff
I actually found that site yesterday. It's great.
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Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?
Here is another diagram that seems to show that the 3d spectral locus dimensions correspond to the 3 human eye cones.
(http://graphics.stanford.edu/courses/cs248-03/color/color13a.gif)
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Nevermind, that was a dumb question.
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I'm confused about the green and purple lines in this spectrum locus diagram. I'm reading that the purple one is the V luminosity function which is L plus M and that the green line is the contrast between L and M which is L minus M. I don't really get what this means or what the lines are representing.
(http://handprint.com/HP/WCL/IMG/twopart.gif)
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I'm confused about the green and purple lines in this spectrum locus diagram. I'm reading that the purple one is the V luminosity function which is L plus M and that the green line is the contrast between L and M which is L minus M. I don't really get what this means or what the lines are representing.
(http://handprint.com/HP/WCL/IMG/twopart.gif)
Does the green line represent the hues that appear brightest to the human eye in photopic vision?
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Does the green line represent the hues that appear brightest to the human eye in photopic vision?
The colors that extend the farthest in the direction of the green line appear more luminous.
The blue line to 525nm is along the crease where the spectral locus seems to bend, marking a division of color perception. Quoting Handprint:
But if we set aside luminance perception defined by the L+M diagonal, then color perception is divided into two parts: • at wavelengths above 525 nm, changes in the relative excitation of the L and M cones define the color response; the S cones are silent.
• at wavelengths below 525 nm, the relative L,M excitations are approximately the same as they are at 525 nm (the dotted line and purple line are equivalent) so it is the relative excitation of the independent S cone that defines the color response.
Cliff
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The colors that extend the farthest in the direction of the green line appear more luminous.
The blue line to 525nm is along the crease where the spectral locus seems to bend, marking a division of color perception. Quoting Handprint:
Cliff
Thanks. I think I understand now.
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The colors that extend the farthest in the direction of the green line appear more luminous.
The blue line to 525nm is along the crease where the spectral locus seems to bend, marking a division of color perception. Quoting Handprint:
Cliff
Do the colors that extend the farthest in the direction of the green line appear more luminous than the white point?
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Do the colors that extend the farthest in the direction of the green line appear more luminous than the white point?
Yes, quoting further from Handprint:
A final observation is that the white point is not located on the luminosity function. This simply demonstrates that white is not the same as bright. The perception of white is a form of color sensation, whereas the perception of bright is a unique intensity sensation. The cone excitation space implies that a "bright" stimulus produces more than two times (http://www.handprint.com/HP/WCL/color2.html#grayness) the cone excitation of a "white" surface, and therefore visual "white" always has a lower luminosity than visual "bright" under the same viewing conditions.
I'm not sure, but I think this might be related to the Helmholtz-Kohlrausch effect, where in general, more chromatic colors will appear brighter than a white of the same luminance.
Cliff
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Yes, quoting further from Handprint:
I'm not sure, but I think this might be related to the Helmholtz-Kohlrausch effect, where in general, more chromatic colors will appear brighter than a white of the same luminance.
Cliff
Isn't brightness usually measured by the amount if white there mixed into the hue?
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Isn't brightness usually measured by the amount if white there mixed into the hue?
I think you're describing purity or chroma; the amount of white mixed with a spectral hue is a measure of purity.
Brightness is an absolute perception of a color's intensity.
Maybe a good book about color will help. "The Reproduction of Colour" by Hunt is an excellent resource, expensive but can probably be found in a local library.
Cliff
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Im confused because this quote from handprint:
A final observation is that the white point is not located on the luminosity function. This simply demonstrates that white is not the same as bright. The perception of white is a form of color sensation, whereas the perception of bright is a unique intensity sensation. The cone excitation space implies that a "bright" stimulus produces more than two times the cone excitation of a "white" surface, and therefore visual "white" always has a lower luminosity than visual "bright" under the same viewing conditions.
doesn't seem to match up with this HSV color model. In the HSV model, the value or brightness goes up when the color starts containing more white and less black. So I've always thought of color as a variety of hues that have a brightness determined by the shade of gray from black to white present in the color, and a saturation level determined by the amount of that shade of gray compared to the amount of the actual hue present in the color. Is this view wrong?
(http://upload.wikimedia.org/wikipedia/commons/e/e0/HSV_cylinder.png)
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Im confused because this quote from handprint:
doesn't seem to match up with this HSV color model. In the HSV model, the value or brightness goes up when the color starts containing more white and less black. So I've always thought of color as a variety of hues that have a brightness determined by the shade of gray from black to white present in the color, and a saturation level determined by the amount of that shade of gray compared to the amount of the actual hue present in the color. Is this view wrong?
You're mixing pieces of models from computer graphics with models of color perception.
The HSV model is a device-dependent RGB model from computer graphics that has little to do with perception-based color models like CIELAB, CIELUV, or color appearance models like CIECAM02. It's not clear to me what the "V" or value in HSV is supposed to correspond to: lightness, brightness, Munsell Value, or (probably) something else.
CIELAB in cylindrical coordinates, which is Lightness, Chroma, and hue, is similar to your HSV model. Note that the white-gray-black axis in CIELAB is Lightness, not Brightness. There is a difference between Lightness and Brightness, (and saturation and Chroma) and you should be careful to differentiate the two while delving into Handprint.
Cliff
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Why are there extratraspectral hues mixed by Red and Violet between 620 nm and 445 nm?
(http://www.handprint.com/HP/WCL/IMG/twopart.gif)
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Why are there extratraspectral hues mixed by Red and Violet between 620 nm and 445 nm?
(http://www.handprint.com/HP/WCL/IMG/twopart.gif)
Those are the purple hues that we can see, but don't exist anywhere as a monochromatic color on the spectrum/spectral locus. Those hues can only be produced by mixing red and violet.
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Those are the purple hues that we can see, but don't exist anywhere as a monochromatic color on the spectrum/spectral locus. Those hues can only be produced by mixing red and violet.
So is the reason they're different than other mixed colors because they make up an edge that is past monochromatic colors in the cone excitation space? Or is this completely wrong?
I thought I was beginning to understand the spectral locus as the monochromatic wavelengths curving around in a 3d space but the extraspectral colors are confusing me because they look like they're outside the spectral locus.
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So is the reason they're different than other mixed colors because they make up an edge that is past monochromatic colors in the cone excitation space? Or is this completely wrong?
I thought I was beginning to understand the spectral locus as the monochromatic wavelengths curving around in a 3d space but the extraspectral colors are confusing me because they look like they're outside the spectral locus.
Pretty much. Take magenta as an example of an extraspectral color. There is no single pure wavelength of light that looks magenta - there is no magenta on the spectral locus. You can only get that color by mixing at least two different wavelengths from opposite ends of the spectrum, such as red and blue. The extraspectral colors are different because you can't match them with a single pure spectral color (or pure spectral color plus an amount of white light), like you can with the "non-extraspectral" ones.
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Pretty much. Take magenta as an example of an extraspectral color. There is no single pure wavelength of light that looks magenta - there is no magenta on the spectral locus. You can only get that color by mixing at least two different wavelengths from opposite ends of the spectrum, such as red and blue. The extraspectral colors are different because you can't match them with a single pure spectral color (or pure spectral color plus an amount of white light), like you can with the "non-extraspectral" ones.
So all the other colors that are not monochromatic are mixtures of monochramitic colors and grayscale colors, except for the extraspectral colors?
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So all the other colors that are not monochromatic are mixtures of monochramitic colors and grayscale colors, except for the extraspectral colors?
Yes, that is the concept of "Dominant Wavelength" (http://en.wikipedia.org/wiki/Dominant_wavelength) .
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So when you take extraspectral colors into consideration, this doesn't show every color we can see right?
(http://stars.astro.illinois.edu/sow/emspectrum.jpg)
But this does?
(http://www.colblindor.com/wp-content/images/CIE%201931%20color%20space.png)
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So when you take extraspectral colors into consideration, this doesn't show every color we can see right?
(http://stars.astro.illinois.edu/sow/emspectrum.jpg)
But this does?
(http://www.colblindor.com/wp-content/images/CIE%201931%20color%20space.png)
The first one - the spectrum - shows the visible frequencies of light and their approximate hues at each frequency. It doesn't show the hues like magenta that you can only get by mixing.
The second one - the spectrum locus - shows not only a similar spectrum along the curved edge, but also all the less saturated colors and mixtures that can be made. The straight line between 380 and 700 is called the purple line, along which you can see magenta and the other extraspectral hues.
The spectrum locus doesn't show "every color we can see" because it leaves out the lightness dimension. Actually, I think the spectrum locus doesn't define a gamut of what can be seen, but rather what can be "physically realized." In fact, there are colors outside the spectrum locus that could be seen if there were a way to stimulate only one or two of the LMS cone types without stimulating the others. It's possible to do it by using adaptation - stare at a bright magenta for a while to suppress the L and S cones, then look at a spectral green and it should look "super-saturated," or more saturated than a pure spectral green.
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The first one - the spectrum - shows the visible frequencies of light and their approximate hues at each frequency. It doesn't show the hues like magenta that you can only get by mixing.
The second one - the spectrum locus - shows not only a similar spectrum along the curved edge, but also all the less saturated colors and mixtures that can be made. The straight line between 380 and 700 is called the purple line, along which you can see magenta and the other extraspectral hues.
The spectrum locus doesn't show "every color we can see" because it leaves out the lightness dimension. Actually, I think the spectrum locus doesn't define a gamut of what can be seen, but rather what can be "physically realized." In fact, there are colors outside the spectrum locus that could be seen if there were a way to stimulate only one or two of the LMS cone types without stimulating the others. It's possible to do it by using adaptation - stare at a bright magenta for a while to suppress the L and S cones, then look at a spectral green and it should look "super-saturated," or more saturated than a pure spectral green.
Is Magenta kind of like white, in that white has no actual wavelength but is seen as a mixture of all wavelengths?
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Is Magenta kind of like white, in that white has no actual wavelength but is seen as a mixture of all wavelengths?
Yes, you can say that you need at least two wavelengths to make either magenta or white. You can make white from two complimentary dominant wavelengths, you don't need all wavelengths to make white. This is shown in the following diagram from page 179 of Wyszecki & Stiles:
[attachment=18022:chromaticitydiag.png]
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Does a glass prism or a rainbow show the extraspectral hues?
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Does a glass prism or a rainbow show the extraspectral hues?
No, prisms and rainbows both display spectrums (http://en.wikipedia.org/wiki/Spectrum), purple and magenta (http://en.wikipedia.org/wiki/Purple) are non-spectral (http://en.wikipedia.org/wiki/Magenta) = extra-spectral hues.
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This might be a really stupid question, but, can you take the spectrum locus and connect the wavelengths to make a cage around every single color possibility like this?
(http://img694.imageshack.us/img694/3971/3dsl.jpg)
If (0,0,0) was black and all the colors made a white point somewhere, then wouldn't this be a good representation of all colors that can be seen by humans in a way that looks like a normal 3d color space like this?
(http://www.cambridgeincolour.com/tutorials/graphics/tut_colormanagement_aRGB1.jpg)
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This might be a really stupid question, but, can you take the spectrum locus and connect the wavelengths to make a cage around every single color possibility like this?
...
If (0,0,0) was black and all the colors made a white point somewhere, then wouldn't this be a good representation of all colors that can be seen by humans in a way that looks like a normal 3d color space like this?
That should work.
The "cone excitation space" you are showing is not as intuitive as other color spaces, because it is far from perceptually uniform, and isn't shaped like "normal" color spaces.
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As Im still reading through handprint, I'm to the part where the "cone excitation space" is converted in the "chromaticity plane" by taking away luminance or dividing each of the three cone types by the sum of all three. I am trying to understand exactly what is happening here but I'm really confused. Are we turning a 3d graph into a 2d graph? How does this diagram show brightness?
(http://www.handprint.com/HP/WCL/IMG/twopart.gif)
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As Im still reading through handprint, I'm to the part where the "cone excitation space" is converted in the "chromaticity plane" by taking away luminance or dividing each of the three cone types by the sum of all three. I am trying to understand exactly what is happening here but I'm really confused. Are we turning a 3d graph into a 2d graph? How does this diagram show brightness?
Chromaticity shows the proportion that each component (L,M,S or X,Y,Z etc., depending on color space) contributes to the whole stimulus. You can think of it like a percentage.
Here's an example using XYZ. Let's say you're looking at someone's face, and the X,Y,Z of the skin in a certain light is 182, 38, 28. The total is 182+38+28 = 248. The chromaticity of the skin would be:
x = 182/248 = 0.7339 (or 73.4%)
y = 38/248 = 0.1532 (or 15.3%)
z = 28/248 = 0.1129 (or 11.3%)
Then a cloud passes across the sun and the light is cut in half. The X,Y,Z are also cut in half, to 91, 19, 14 respectively. The total now is 124, and the chromaticity is:
x = 91/124 = 0.7339
y = 19/124 = 0.1532
z = 14/124 = 0.1129
So the chromaticity doesn't change. It is independent of exposure. Chromaticity represents that part of the color that doesn't change with lightness/brightness: the combination of hue and saturation.
If you plot chromaticities x,y,z in the 3-dimensional X,Y,Z space, they all lie on a triangular plane. The standard CIE chromaticity diagram only plots the two dimensions x and y, as z is implied (since x + y + z = 1). Brightness or lightness is not shown on the chromaticity diagram.
I will add that, unlike a lot of the concepts in this thread, chromaticity is actually pretty relevant for photographers. When you change exposure in the camera, chromaticity does not change in the raw file. But when you adjust exposure in your image editor, chromaticity is not necessarily preserved. Some editing tools change the hue or saturation while you change lightness. Luminosity Blend Mode doesn't preserve chromaticity. Some color spaces don't preserve chromaticy, either, such as in CIELAB when editing the lightness channel. When chromaticity isn't preserved, you get those often unwanted changes in saturation and hue that make images look less realistic.
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Let me back up a little bit. I understand that this diagram represents the hues that our brain interprets from single wavelengths of light.
(http://stars.astro.illinois.edu/sow/emspectrum.jpg)
I also understand that Magenta is a mix of both ends of the spectrum. Where do black and white come into play with these other colors? Is white simply a mixture of complementary hues and/or all hues? Is white considered an extra-spectral color like Magenta or is it in a completely separate category? Is black simply what we see in space when light is absent and is white just the color of light that is sent to us from the sun?
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So the sun sends us white light which is a mixture of all the monochromatic colors that our eyes are sensitive to and when we mix both sides of the monochromatic spectrum, we get the extra spectral purples. We can then mix all of the monochromatic colors/extra spectral purples with with white to create less saturated colors. And Black is simply what we see when there is no light present at all. Does this explain how every single color is made? What about mixing the monochromatic colors/extra spectral purples with different grays or even black?
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If this question doesn't make sense then could you tell me how and what is wrong with it?
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Isn't brightness usually measured by the amount if white there mixed into the hue?
You have to be careful about what sort of experiment each diagram is describing. If you take a certain amount of optical power (watts)and put it all into the wavelength of maximum luminance (with a laser operating at that wavelength), that will be a bright yellow green and will be brighter than a white color with the same amount of power (watts). Therefore, white does not correspond to the maximum luminance function. Another way to see this is that white must be composed of at least two different wavelengths of light, and therefore at least part of the optical power in white has less luminance than the maximum possible, because it is not that particular wavelength that has maximum lumens per watt.
When you look at the gamut of colors for a display, however, you see that the maximum luminance that the display can produce is for white, because it is the case where all three primary colors are turned on to the maximum available. The display cannot take all that optical power and put it into the single wavelength with the maximum luminance, it can only produce varying amounts of its three primary colors.
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So the sun sends us white light which is a mixture of all the monochromatic colors that our eyes are sensitive to and when we mix both sides of the monochromatic spectrum, we get the extra spectral purples.
Yes
We can then mix all of the monochromatic colors/extra spectral purples with with white to create less saturated colors.
True, but you can also create a less saturated color by mixing a spectral color with its complementary color. And if you mix the right amounts of a spectral color and its complement, you will make gray or white.
And Black is simply what we see when there is no light present at all.
Yes.
Does this explain how every single color is made?
Sort of, but you have to remember that any non-spectral color can be created by many different spectra, provided that they all give the same stimulation to the three types of cone cells in the eye. This is the fundamental basis of color "reproduction" (meaning, color matching) by use of three primary colors. The wonderful thing is that the spectrum of the reproduction does not have to match the original spectrum wavelength by wavelength; only the net stimulation of each of the three types of cones has to match.
What about mixing the monochromatic colors/extra spectral purples with different grays or even black?
Remember we're talking additive color here (mixing light), so mixing with gray is just mixing with less white, and black is the special case of mixing with nothing, getting the starting color as a result.
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If you take a certain amount of optical power (watts)and put it all into the wavelength of maximum luminance (with a laser operating at that wavelength), that will be a bright yellow green and will be brighter than a white color with the same amount of power (watts). Therefore, white does not correspond to the maximum luminance function.
That would bring out an interesting point. The radiant power amounts required for matching equi-energy white using a usual set of CIE primaries (R=700, G=546.1, B=435.8) would be in the ratio R:G:B = 72.1:1.4:1, which in photometric units of luminance would be in the ratio of R:G:B = 1:4.6:0.06. Hence, to match 74.5 watts of equi-energy white one would require 72.1 watts of red, 1.4 watts of green, and 1 watt of blue. Most of the contribution in wattage is coming from Red, i.e., 72.1 watts. Therefore, perhaps, one can make a statement that at least in this case the wattage of equi-energy white is close to a certain red as opposed to some green.
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That would bring out an interesting point. The radiant power amounts required for matching equi-energy white using a usual set of CIE primaries (R=700, G=546.1, B=435.8) would be in the ratio R:G:B = 72.1:1.4:1, which in photometric units of luminance would be in the ratio of R:G:B = 1:4.6:0.06. Hence, to match 74.5 watts of equi-energy white one would require 72.1 watts of red, 1.4 watts of green, and 1 watt of blue. Most of the contribution in wattage is coming from Red, i.e., 72.1 watts. Therefore, perhaps, one can make a statement that at least in this case the wattage of equi-energy white is close to a certain red as opposed to some green.
Hmmm - could you rephrase that last statement? I'm not following what it means, "the wattage is close to a certain red." Red is a sensation, so I don't follow what you mean by watts being close to a certain red.
This particular choice of red primary is far down the sensitivity curve of even the long wavelength cones, and therefore lots of watts are needed to get a long-wavelength cone stimulus equal to that of the equi-energy white source. Using lots of power for this deep red primary does not imply that the color being reproduced is subjectively "close" to that primary. Another color that is "closer," for example a shade of pink, will require even more watts of this red primary to match it. In other words, with this choice of primaries, the 72.1 is just a scaling factor required because you are so far down the slope of the cone response. In the published chromaticity diagrams, these power factors of 72.1, 1.4, and 1 are divided out so that the amounts of the primaries can be stated as R=1, G=1, and B=1 for the match to equi-energy white. If the red wavelength is moved to something shorter and more reasonably up on the long cone sensitivity curve, these ratios are different and much less extreme.
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Hmmm - could you rephrase that last statement? I'm not following what it means, "the wattage is close to a certain red." Red is a sensation, so I don't follow what you mean by watts being close to a certain red.
Just saying that for that particular choice of primaries (which is not a random set and one of those actually used in colorimetry publications) most of the actual physical power in matching an equi-energy white is coming from the red color. Of course one can derive the equivalent photometirc luminance units and they are found to be in ratio of R:G:B = 1:4.6:0.06. Now one can see the green having a sensation of higher luminance. Further more, one can do a rescaling of the size of units for red, green, and blue so that equal-energy white is matched by unit quantities of R,G and B, and derive tristimulus values.
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I learned a lot from this thread, the main thing being how much I don't know. I only started this thread because I was planning on buying a new monitor to use with my new film scanner and because I'm such a perfectionist, I tried to learn everything I could about color before letting my self decide on a monitor. I now see how ridiculously complicated this stuff really is. I am really curious about why and how some of you guys know as much as you do about color? Do most people here have careers that require it?
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Just saying that for that particular choice of primaries (which is not a random set and one of those actually used in colorimetry publications) most of the actual physical power in matching an equi-energy white is coming from the red color.
Oh, OK, totally agree. Just didn't understand your original phrasing.
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I learned a lot from this thread, the main thing being how much I don't know. I only started this thread because I was planning on buying a new monitor to use with my new film scanner and because I'm such a perfectionist, I tried to learn everything I could about color before letting my self decide on a monitor. I now see how ridiculously complicated this stuff really is. I am really curious about why and how some of you guys know as much as you do about color? Do most people here have careers that require it?
I usually don't like to tout myself, but "how are you qualified" is a fair question. My curriculum vitae is here:
http://www.bretl.com/curricvit.htm (http://www.bretl.com/curricvit.htm)
Some details that aren't in the CV: I learned the basics as a necessary adjunct of doing TV set chroma IC design. I became the go-to guy in Zenith's advanced development lab for colorimetry issues, in conjunction with their phosphor and color instrumentation specialists. Later I participated in the SMPTE monitor colorimetry work and also in the camera standardization for the system comparison shoots used by the FCC Advisory Committee on Advanced Television Systems (ACATS).
I am still learning a lot about the "non-ideal" (or at least, not simple) aspects of color processing that are necessary to make a camera's results really look good.
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I am really curious about why and how some of you guys know as much as you do about color? Do most people here have careers that require it?
Hi Jeremy, my interest in color science developed when I was a student in computer vision and pattern recognition while working on a certain problem in high-dimensional multimedia databases requiring color-based query mechanism. Later I held jobs in HD video compression and communications, and currently in making advanced digital cameras for low-noise scientific imaging.
I have an interest in color science as derived from traditional colorimetry, color TV, analog and digital communications, and also along the lines of connections with digital signal processing and pattern recognition.
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My curriculum vitae is here:
http://www.bretl.com/curricvit.htm (http://www.bretl.com/curricvit.htm)
Hi, welcome here. You have great qualifications and experience. Did you by chance ever meet Donald Fink?
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That would bring out an interesting point. The radiant power amounts required for matching equi-energy white using a usual set of CIE primaries (R=700, G=546.1, B=435.8) would be in the ratio R:G:B = 72.1:1.4:1, which in photometric units of luminance would be in the ratio of R:G:B = 1:4.6:0.06. Hence, to match 74.5 watts of equi-energy white one would require 72.1 watts of red, 1.4 watts of green, and 1 watt of blue. Most of the contribution in wattage is coming from Red, i.e., 72.1 watts. Therefore, perhaps, one can make a statement that at least in this case the wattage of equi-energy white is close to a certain red as opposed to some green.
Where did you get these figures 72.1:1.4:1 ? That seems out of line with the usual plots of the CIE standard observer spectral response functions; unless their normalizations are not as usually plotted.
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So the sun sends us white light which is a mixture of all the monochromatic colors that our eyes are sensitive to and when we mix both sides of the monochromatic spectrum, we get the extra spectral purples. We can then mix all of the monochromatic colors/extra spectral purples with with white to create less saturated colors. And Black is simply what we see when there is no light present at all. Does this explain how every single color is made? What about mixing the monochromatic colors/extra spectral purples with different grays or even black?
The eye (barring issues like color blindness) has three separate types of photoreceptors. Each has a distinct response to the visible spectrum:
http://upload.wikimedia.org/wikipedia/comm...nctions.svg.png (http://upload.wikimedia.org/wikipedia/commons/thumb/8/8f/CIE_1931_XYZ_Color_Matching_Functions.svg/446px-CIE_1931_XYZ_Color_Matching_Functions.svg.png)
If one takes the intensity of a light source as a function of frequency, multiplies it separately by each response function and adds up the result over frequency, one derives three numbers usually called X,Y, and Z. Then given one more piece of information -- the X0,Y0,Z0 values of white -- there is then a well-defined map between (X/X0,Y/Y0,Z/Z0) and the RGB values in your favorite color space, say sRGB. For details, see
http://www.brucelindbloom.com/index.html?W...gSpaceInfo.html (http://www.brucelindbloom.com/index.html?WorkingSpaceInfo.html)
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Where did you get these figures 72.1:1.4:1 ? That seems out of line with the usual plots of the CIE standard observer spectral response functions; unless their normalizations are not as usually plotted.
I think you are talking about photometric response curves, in which case I said the Luminances for that particular set of R, G, and B are in ratio of R:G:B = 1:4.6:0.06. The corresponding radiometric response is based upon R:G:B=72.1:1.4:1. You can find these numbers (72.1:1.4:1) in the book, "Color Science: Concepts and Methods, Quantitative Data and Formulae, 2nd ed," by Gunter Wyszecki and Walter Stanley Stiles, among several other sources.
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I think you are talking about photometric response curves, in which case I said the Luminances for that particular set of R, G, and B are in ratio of R:G:B = 1:4.6:0.06. The corresponding radiometric response is based upon R:G:B=72.1:1.4:1. You can find these numbers (72.1:1.4:1) in the book, "Color Science: Concepts and Methods, Quantitative Data and Formulae, 2nd ed," by Gunter Wyszecki and Walter Stanley Stiles, among several other sources.
A simple explanation would help more than sending me off to the library.
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A simple explanation would help more than sending me off to the library.
Can't believe you said that Emil. A physics professor at a major university wants to avoid going to a library I've right now over a hundred technical books checked out from the library .
Here is the deal: In the color matching experiments the actual physical amounts of R=700 nm, G=546.1 nm, and B=435.8 nm primaries used to match an equi-energy white are in the ratio of R:G:B = 72.1:1.4:1. In such experiments you have 3 different radiometric response curves for matching spectral colors. Now you can transform this set of curves using the Luminosity curve (http://en.wikipedia.org/wiki/Luminosity_function) to the photometric set of numbers, which will give you another 3 curves. You can further renormalize each curve so that equal amount of primaries are needed for matching the white point -- and you have your tristimulus curves, which were perhaps what you had in mind when you mentioned the CIE standard observer. In the photometric set of numbers the luminances are in the ratio of R:G:B=1:4.6:0.06 when matching the equi-energy white.
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I will be starting college next semester at the University of Tennessee. I plan to major in Journalism and Electronic Media and either minor or double-major in Studio Art. I love film photography more than anything but I guess I hope that I can get a good job in some type of video.
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Hi, welcome here. You have great qualifications and experience. Did you by chance ever meet Donald Fink?
Nope, never met him. It might have been possible, but never happened. I was born in 1944, so he was the preceding generation, really. He was writing his first TV engineering books before I had ever seen television. His last notable contribution to TV was a paper arguing that there was no practical way to get to analog high definition in a reasonable bandwidth, which spurred others to develop our digital HDTV systems.
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Nope, never met him. It might have been possible, but never happened. I was born in 1944, so he was the preceding generation, really. He was writing his first TV engineering books before I had ever seen television. His last notable contribution to TV was a paper arguing that there was no practical way to get to analog high definition in a reasonable bandwidth, which spurred others to develop our digital HDTV systems.
What about the NHK analog HDTV in Japan? Though, I understand that it was discontinued.
At my previous job with HD video communications, during development phase of our HDMI signals we more more successful in displaying them on LG HDTVs compared to more expensive Sharps.
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What about the NHK analog HDTV in Japan? Though, I understand that it was discontinued.
"MUSE" worked reasonably well in satellite bandwidths, but the extreme subsampling required to fit it into 6 MHz for terrestrial use made the picture quality inferior. It had only one motion vector for the whole image, which helped for camera panning, but resulted in inferior rendition of individually moving objects. When the digital proposals were introduced, MUSE was effectively eliminated from the U.S. competition.
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Nope, never met him. It might have been possible, but never happened. I was born in 1944, so he was the preceding generation, really. He was writing his first TV engineering books before I had ever seen television. His last notable contribution to TV was a paper arguing that there was no practical way to get to analog high definition in a reasonable bandwidth, which spurred others to develop our digital HDTV systems.
While still off-topic (sorry), I found in Fink's bio here:
http://www.ieeeghn.org/wiki/index.php/Donald_Fink (http://www.ieeeghn.org/wiki/index.php/Donald_Fink)
that he was instrumental in setting up LORAN, the nav system that was just recently shut down in favor of GPS.
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