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>> Also it would be surprising to me if the
>> human visual range came out exactly as a cylinder (after a brief glance
>> of the maths involved), are you sure about this?
>>
>
> The human visual range has a very irregular shape.
>
> https://en.wikipedia.org/wiki/File:Cie_Chart_with_sRGB_gamut_by_spigget.png
>
> There's a lot of blank empty space around it in every color space. (The
> one in the picture is called CIExyY I think.
>
> I would like to learn how to plot this irregular shape in the near future.
You just need the XYZ colour-matching functions in terms of wavelength.
Look here (just click the first "Submit" to get a basic table):
http://cvrl.ioo.ucl.ac.uk/cmfs.htm
This table then gives you the exact XYZ values for each pure wavelength.
XYZ is a linear representation of absolute colour, so you can do a lot
of math with them (eg adding, averaging, mixing etc).
To get the graph you linked to, you just need to calculate x and y for
each of those wavelengths.
x = X/(X+Y+Z)
y = Y/(X+Y+Z)
You'll then find that the x,y pairs give the outline of the graph you
linked to, which are the pure wavelengths. The internal area is formed
by mixing pure wavelengths, so roughly speaking the further you are away
from the boundary the more "wideband" the light is.
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On 11/28/2016 3:53 AM, scott wrote:
>>> Also it would be surprising to me if the
>>> human visual range came out exactly as a cylinder (after a brief glance
>>> of the maths involved), are you sure about this?
>>>
>>
>> The human visual range has a very irregular shape.
>>
>> https://en.wikipedia.org/wiki/File:Cie_Chart_with_sRGB_gamut_by_spigget.png
>>
>>
>> There's a lot of blank empty space around it in every color space. (The
>> one in the picture is called CIExyY I think.
>>
>> I would like to learn how to plot this irregular shape in the near
>> future.
>
> You just need the XYZ colour-matching functions in terms of wavelength.
> Look here (just click the first "Submit" to get a basic table):
>
> http://cvrl.ioo.ucl.ac.uk/cmfs.htm
>
> This table then gives you the exact XYZ values for each pure wavelength.
> XYZ is a linear representation of absolute colour, so you can do a lot
> of math with them (eg adding, averaging, mixing etc).
>
> To get the graph you linked to, you just need to calculate x and y for
> each of those wavelengths.
>
> x = X/(X+Y+Z)
> y = Y/(X+Y+Z)
>
> You'll then find that the x,y pairs give the outline of the graph you
> linked to, which are the pure wavelengths. The internal area is formed
> by mixing pure wavelengths, so roughly speaking the further you are away
> from the boundary the more "wideband" the light is.
>
Well, I don't understand that. Clipka did most of the work for me last
time...
Maybe you could take a look at the code on the LCH image's page on
Wikimedia and make some suggestions.
Mike
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On 11/28/2016 3:53 AM, scott wrote:
> You just need the XYZ colour-matching functions in terms of wavelength.
> Look here (just click the first "Submit" to get a basic table):
>
> http://cvrl.ioo.ucl.ac.uk/cmfs.htm
>
> This table then gives you the exact XYZ values for each pure wavelength.
> XYZ is a linear representation of absolute colour, so you can do a lot
> of math with them (eg adding, averaging, mixing etc).
>
> To get the graph you linked to, you just need to calculate x and y for
> each of those wavelengths.
>
> x = X/(X+Y+Z)
> y = Y/(X+Y+Z)
>
> You'll then find that the x,y pairs give the outline of the graph you
> linked to, which are the pure wavelengths. The internal area is formed
> by mixing pure wavelengths, so roughly speaking the further you are away
> from the boundary the more "wideband" the light is.
>
Also, I'm more interested in the 3-dimensional solid than the 2D
chromaticity diagram.
Mike
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> Also, I'm more interested in the 3-dimensional solid than the 2D
> chromaticity diagram.
The 3D solid of the Yxy shape is just the 2D shape you linked to
extruded upwards as a prism. That's why you don't normally see Yxy or
Yuv shown in 3D - it doesn't give any benefit over the 2D shape.
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On 28/11/2016 09:11, Mike Horvath wrote:
> On 11/28/2016 3:53 AM, scott wrote:
>>>> Also it would be surprising to me if the
>>>> human visual range came out exactly as a cylinder (after a brief glance
>>>> of the maths involved), are you sure about this?
>>>>
>>>
>>> The human visual range has a very irregular shape.
>>>
>>> https://en.wikipedia.org/wiki/File:Cie_Chart_with_sRGB_gamut_by_spigget.png
>>>
>>>
>>>
>>> There's a lot of blank empty space around it in every color space. (The
>>> one in the picture is called CIExyY I think.
>>>
>>> I would like to learn how to plot this irregular shape in the near
>>> future.
>>
>> You just need the XYZ colour-matching functions in terms of wavelength.
>> Look here (just click the first "Submit" to get a basic table):
>>
>> http://cvrl.ioo.ucl.ac.uk/cmfs.htm
>>
>> This table then gives you the exact XYZ values for each pure wavelength.
>> XYZ is a linear representation of absolute colour, so you can do a lot
>> of math with them (eg adding, averaging, mixing etc).
>>
>> To get the graph you linked to, you just need to calculate x and y for
>> each of those wavelengths.
>>
>> x = X/(X+Y+Z)
>> y = Y/(X+Y+Z)
>>
>> You'll then find that the x,y pairs give the outline of the graph you
>> linked to, which are the pure wavelengths. The internal area is formed
>> by mixing pure wavelengths, so roughly speaking the further you are away
>> from the boundary the more "wideband" the light is.
>>
>
> Well, I don't understand that.
Which bit?
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On 11/28/2016 4:24 AM, scott wrote:
>> Also, I'm more interested in the 3-dimensional solid than the 2D
>> chromaticity diagram.
>
> The 3D solid of the Yxy shape is just the 2D shape you linked to
> extruded upwards as a prism. That's why you don't normally see Yxy or
> Yuv shown in 3D - it doesn't give any benefit over the 2D shape.
>
>
I thought there was some tapering as the Y increases? This image
suggests that is the case.
http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/bajwa/images/cie_3d.gif
Mike
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Am 28.11.2016 um 23:21 schrieb Mike Horvath:
> On 11/28/2016 4:24 AM, scott wrote:
>>> Also, I'm more interested in the 3-dimensional solid than the 2D
>>> chromaticity diagram.
>>
>> The 3D solid of the Yxy shape is just the 2D shape you linked to
>> extruded upwards as a prism. That's why you don't normally see Yxy or
>> Yuv shown in 3D - it doesn't give any benefit over the 2D shape.
>>
>>
>
> I thought there was some tapering as the Y increases? This image
> suggests that is the case.
>
> http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/bajwa/images/cie_3d.gif
That's not the limit of the xyY space.
That's the theoretical limit of _reflective_ colours illuminated by
standard illuminant C, plotted in xyY space.
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On 11/28/2016 6:05 PM, clipka wrote:
>> I thought there was some tapering as the Y increases? This image
>> suggests that is the case.
>>
>> http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/bajwa/images/cie_3d.gif
>
> That's not the limit of the xyY space.
>
> That's the theoretical limit of _reflective_ colours illuminated by
> standard illuminant C, plotted in xyY space.
>
Yes, that is fine. Not sure whether to use C or D65, though.
https://en.wikipedia.org/wiki/Illuminant_D65
These images on Wikipedia use D65, so I may just stick with that.
https://en.wikipedia.org/wiki/File:Cie_Chart_with_sRGB_gamut_by_spigget.png
https://en.wikipedia.org/wiki/File:CIE1931xy_gamut_comparison.svg
https://en.wikipedia.org/wiki/File:Lab_color_space.png (I think)
Mike
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Am 29.11.2016 um 00:33 schrieb Mike Horvath:
> On 11/28/2016 6:05 PM, clipka wrote:
>>> I thought there was some tapering as the Y increases? This image
>>> suggests that is the case.
>>>
>>>
http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/bajwa/images/cie_3d.gif
>>>
>>
>> That's not the limit of the xyY space.
>>
>> That's the theoretical limit of _reflective_ colours illuminated by
>> standard illuminant C, plotted in xyY space.
>>
>
> Yes, that is fine. Not sure whether to use C or D65, though.
>
> https://en.wikipedia.org/wiki/Illuminant_D65
>
> These images on Wikipedia use D65, so I may just stick with that.
>
> https://en.wikipedia.org/wiki/File:Cie_Chart_with_sRGB_gamut_by_spigget.png
> https://en.wikipedia.org/wiki/File:CIE1931xy_gamut_comparison.svg
> https://en.wikipedia.org/wiki/File:Lab_color_space.png (I think)
The first two /show/ the D65 whitepoint (or, more precisely, its xy
coordinates), but other than that they're entirely independent of any
whitepoint. The first image even shows various other whitepoints.
As for the third image, I can't see where you got the idea that it has
anything to do with any whitepoint (except maybe for the fact that the
sRGB colour space, of which a few slices are shown, is defined such that
its "upper right" corner coincides with D65).
(Of course the image files themselves may also happen to be using a
colour encoding scheme that uses D65 as its nominal whitepoint, but I
don't think that's relevant in this context.)
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On 11/28/2016 6:59 PM, clipka wrote:
> Am 29.11.2016 um 00:33 schrieb Mike Horvath:
>> On 11/28/2016 6:05 PM, clipka wrote:
>>>> I thought there was some tapering as the Y increases? This image
>>>> suggests that is the case.
>>>>
>>>>
http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/bajwa/images/cie_3d.gif
>>>>
>>>
>>> That's not the limit of the xyY space.
>>>
>>> That's the theoretical limit of _reflective_ colours illuminated by
>>> standard illuminant C, plotted in xyY space.
>>>
>>
>> Yes, that is fine. Not sure whether to use C or D65, though.
>>
>> https://en.wikipedia.org/wiki/Illuminant_D65
>>
>> These images on Wikipedia use D65, so I may just stick with that.
>>
>> https://en.wikipedia.org/wiki/File:Cie_Chart_with_sRGB_gamut_by_spigget.png
>> https://en.wikipedia.org/wiki/File:CIE1931xy_gamut_comparison.svg
>> https://en.wikipedia.org/wiki/File:Lab_color_space.png (I think)
>
> The first two /show/ the D65 whitepoint (or, more precisely, its xy
> coordinates), but other than that they're entirely independent of any
> whitepoint. The first image even shows various other whitepoints.
>
> As for the third image, I can't see where you got the idea that it has
> anything to do with any whitepoint (except maybe for the fact that the
> sRGB colour space, of which a few slices are shown, is defined such that
> its "upper right" corner coincides with D65).
>
> (Of course the image files themselves may also happen to be using a
> colour encoding scheme that uses D65 as its nominal whitepoint, but I
> don't think that's relevant in this context.)
>
Okay, sorry, I didn't think it was possible to compare different color
spaces like sRGB or Adobe RGB without selecting a white point.
Still, I would like to plot the horseshoe in 3D, anyway, using a
specific white point. (Or maybe multiple images, each with a different
white point.)
I don't think I'll figure out the math though.
Mike
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