Talk:White point

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Re Merge - yes, good idea PAR 14:36, 9 December 2005 (UTC)

Smart idea indeed. Genji

Exactly. I agree. Jwgu

[edit] No!

A white point and a standard illuminant are two completely different things and should not be confused!

A white point is exactly that: a point. It is a set of (usually three) "tristimulus coordinates" (eg: CIE XYZ).

If the tristimulus coordinates are normalised (ie Y=100 in the CIE system) then the point can be expressed in one less "chromacity coordinates". (eg (x,y), (a*,b*) ).

But a white point is still just that, a single point expressing very little information. As well as that it means nothing outside its coordinate system (eg CIE 1931).

It is possible to alter the white point of a colour by performing a colour correction transform, but this is only a very rough estimate of how an object of that colour might appear under a different illuminant. Multispectral imaging is required to reasonably estimate the appearance of an object under a differernt illuminant.

So leave it alone. Certainly these articles need to be explained better, but I haven't got time to do it now. If anyone wants to do it well, I suggest they have a read of:

"Billmeyer and Saltzman's Principles of Color Tecnology" by Roy S. Berns

TomViza 16:18, 17 March 2006 (UTC)

I've removed the merge tag. It looked ugly and no one said anything for it. TomViza 09:47, 4 April 2006 (UTC)

[edit] White?

The displayed colours seem a bit "off" to me. I did not check all of them, but if x=y=1/3, then R=G=B=1, so the value used to display the colour seems wrong to me. Odedee 22:54, 30 August 2006 (UTC)

To understand the answer in full, I suggest you read this Color FAQ. Maybe this article should include that as a reference, as the underlying concepts are not simple.

But basically, the issue is defining what R, G and B are. If you tell a monitor to set its red, green and blue guns to equal intensity, the "white" or "grey" you will get is determined by the precise colour of the 3 components, and their relative intensities. Various standards have defined this for TVs; and the sRGB standard basically adopts similar values for computer colour specifications.

This page is written on the assumption that your web browser and system follows sRGB (which it probably will, or close enough), which will mean that a notional white/grey specified with equal R,G,B will produce illuminant D65, or close. Any other illuminant will appear "off-white", but that's purely a perception relative to the other stuff on your screen.

What you need to realise is that the "white" produced by TVs and monitors is really rather blue-ish. This can be seen quite clearly if you pass a house at night with a darkened room illuminated only by a TV. The light is rather a vivid blue, compared to the warmer light produced by light bulbs. Most other standard illuminants are yellower than D65, and thus on a computer monitor will appear to be absolutely yellow, or brown at the low intensity used in the table.

(The low intensity is a necessary result of trying to represent multiple colours on a D65 device; if the D65 illuminant was brighter, you wouldn't be able to output enough red to represent A at the same brightness. A computer monitor can produce D65 brighter than any other colour in its repertoire, by setting its RGB to maximum. Any other colour can only produced at lower levels.)

And the equal-energy illuminant E, does not represent any conventional light source seen in real-life. You wouldn't regard an illuminant E source as "white" if you saw it in an artificially-illuminated or sunlit environment. --KJBracey 15:22, 31 August 2006 (UTC)

You misunderstood me entirely, but it is probably my fault. What I meant is that the RGB values used for displaying the colour in the table seem dubious to me. Unless I misunderstand something, then for x=y=z=1/3 R, G and B should equal 1. However the color currently defined in the table for this entry is not #FFFFFF but #d3beba. Does that make sense? Odedee 16:04, 31 August 2006 (UTC)

Now I read the article again, more closely, and noticed that you used Y=0.54 which explains the coordinates. But why not use Y=1? We're looking for white, not gray. Odedee 16:20, 31 August 2006 (UTC)

I don't think I misunderstood you. I've answered both those questions above. I think you may need to read that FAQ to fully understand the answers though.

But addressing them briefly again, the white on a monitor is illuminant D65, not illuminant E, so it's D65 that has equal RGB values, not E. And then, the maximum luminance we can use is limited by the illuminant furthest from D65, in this case A. You cannot display illuminant A brighter than Y=0.54 on a monitor because you can't get the R value any higher. So all the entries have that Y value to aid comparison. --KJBracey 18:15, 31 August 2006 (UTC)

Thanks, indeed it seems I need to do some reading. I'll do that in a few days, as I am leaving to Burning Man shortly. You may want to consider including your explanation in the article - I believe it will BE helpful to readers. Odedee 19:07, 31 August 2006 (UTC)