Sky Blue

[Gernot Hoffmann]

In touristic books and on postcards the sky is often
too blue and too dark, IMHO.

1. Which color (Lab) has a nice blue sky at noon ?

2. Which color (Lab) should this sky have in an image ?
My suggestion: 65 / -5 / -40 or so.
This is a kind of cyanish-blue.

Best regards --Gernot Hoffmann

[Mike Russell]

Here are some Lab numbers collected by Jim Hamilton. These values are
provided to curvemeister customers as a reference for a blue sky.

The values are from photographs, taken at different times of year, in
Scotland.

The values of the first one are extremely close to your suggested value.
Cyan-blue is generally acceptable, whereas Magenta-blue looks unnatural.
The sky should also decrease in saturation toward the horizon.

Glencoe Sky Lab(68,-2,-45) pin hue and saturation ;by Jim Hamilton
Inverness Sky Lab(81,-11,-29) pin hue and saturation ;by Jim Hamilton
Brodie Sky Lab(75,-7,-38) pin hue and saturation ;by Jim Hamilton

--
Mike Russell - http://www.curvemeister.com

[Gerhard Fuernkranz]

Mike Russell wrote:
> On Wed, 2 Jul 2008 11:33:49 -0700 (PDT), Gernot Hoffmann wrote:
>
>
>> In touristic books and on postcards the sky is often
>> too blue and too dark, IMHO.
>>

Gernot, well, whether the sky is reproduced darker or brighter on a
photo is IMO a matter of the selected exposure. I have also seen photos,
where the opposite was the case, i.e. where the sky looked too bright
and washed out, in order that other parts of the scene don't become too
dark. Unless one captures the full dynamic range of a scene as HDR
image, the exposure is always a trade-off which must be chosen by the
photographer according to artistic aspects. And for touristic books or
postcards I guess that photographers often use tricks like polarization
filters in order to make the sky "bluer" and darker on the photo than it
appears originally, since people like this :-)

Also keep in mind, that prints are usually not a colorimetric
reproduction, but they are the result of a perceptual color
transformation, which also includes gamut mapping. If for instance blue
tones are turning purple on the reproduction, then a potential reason is
gamut mapping being carried out in CIELAB space, which has a bad hue
constancy in some regions (including the blue one).

I'm also not sure, whether CIELAB is good metric for specifying the sky
color, since CIELAB numbers are always relative to a reference white
point. But which chromaticity should be chosen as reference white? And
which luminance should be chosen as reference white luminance? (which
brings us back to the issue above - what's the "right" exposure?)
Wouldn't it be better, to specify just the [x,y] chromaticity?

Regards,
Gerhard

[Gernot Hoffmann]

Gerhard Fuernkranz schrieb:

Thanks, Mike and Gerhard.
I'm cooperating with a professional landscape photographer.
His daylight skies are are often too dark and have too little
cyan, IMHO. Obviously his publisher doesn't worry about.
I've never seen such a dark sky, not in Africa, not in South-
East-Asia, neither in Germany nor in Mallorca.
Therefore I would like to give some guidelines how to improve
the images. The skies (whatever exposure) are manipulated
in Lab without selection until the result is pleasing.
Nice methods in Dan Margulis' book about Photoshop Lab !

Best regards --Gernot Hoffmann

[Roger Breton]

Gerard,

Please correct me if I am wrong but, [x,y] chromaticity will not give us a
more illuminant independent measurement. Suppose I point my Minolta CS-100
to the sky and take a reading (I'll report those numbers shortly I hope --
it is raining here today). I think you can argue that, since this type of
instrument is calibrated using an illiminant A type of source, the problem
remains, the chromaticity is not truly independent of some illuminant. So
maybe using CIELAB is not so bad after all?

Gernot, sorry I don't have those numbers for you yet :(

What do you think Gerhard?

Roger

[Dieter Michel]

Hi Roger,

> I think you can argue that, since this type of instrument
> is calibrated using an illiminant A type of source, the
> problem remains, the chromaticity is not truly independent
> of some illuminant.

as I unterstand it, the XYZ (and derived Yxy) chromaticities
shoud be independent of of any reference illuminant.

The problem with tristimulus meters like the CS-100, however,
is that the internal filters mimic the color matching function
only to a certain degree. (In principle, it's possible to manufacture
these to the desired precision by splitting up the spectral ranges,
but that would render the instrument much more expensive.)

So there will be deviations in some spectral regions and
calibration with a broadband source (such as Illuminant A)
will avarage out these local deviations. Measuring a light
source with significant peaks (such as CCFLs in an LCD-Monitor)
may result in noteworthy deviations in case you are unlucky
enough to have deviations in the filter response in spectral
regions where your illuminant has significant peaks.

In case of the CS-100, x1(lambda) (the short-wavelength
portion of X(lambda) has no filter of it's own, but is
produced by scaling z(lambda). With broadband sources,
this will probably not be a problem, but with narrowband
sources, you may see a difference in cases.

Best regards

Dieter Michel

[Alessandro]

Dieter Michel wrote:
> Hi Roger,
>
>> I think you can argue that, since this type of instrument is
>> calibrated using an illiminant A type of source, the problem
>> remains, the chromaticity is not truly independent of some
>> illuminant.
>
> as I unterstand it, the XYZ (and derived Yxy) chromaticities shoud be
> independent of of any reference illuminant.

In which sense are you saying that they are independent of any reference
illuminant? As far as I know the X coordinate is computed as

X = K integral S(lambda)R(lambda) xbar(lambda) d(lambda)

with S(lambda)spectrum pf the source, R(lambda) reflectance of the
sample and xbar(lambda) being the tristimuls function for the X channel.
Y and Z are computend in the same way. K is a normalisation factor
computed as

K = integral S(lambda)ybar(lambda) d(lambda)

The illuminant enters heavily in the results via the integration
operation. Going from one illuminant to another the values of XYZ are
changing quite a lot.

Regards,
Alessandro

> The problem with tristimulus meters like the CS-100, however, is that
> the internal filters mimic the color matching function only to a
> certain degree. (In principle, it's possible to manufacture these to
> the desired precision by splitting up the spectral ranges, but that
> would render the instrument much more expensive.) So there will be
> deviations in some spectral regions and calibration with a broadband
> source (such as Illuminant A) will avarage out these local
> deviations. Measuring a light source with significant peaks (such as
> CCFLs in an LCD-Monitor) may result in noteworthy deviations in case
> you are unlucky enough to have deviations in the filter response in
> spectral regions where your illuminant has significant peaks.
>
> In case of the CS-100, x1(lambda) (the short-wavelength portion of
> X(lambda) has no filter of it's own, but is produced by scaling
> z(lambda). With broadband sources, this will probably not be a
> problem, but with narrowband sources, you may see a difference in
> cases.
>
> Best regards
>
> Dieter Michel

--
Today, Medgar Evers was buried from the bullet he caught.
They lowered him down as a king.
But when the shadowy sun sets on the one
That fired the gun
He'll see by his grave
On the stone that remains
Carved next to his name
His epitaph plain:
Only a pawn in their game.

[Dieter Michel]

Hi Alessandro,

>>> I think you can argue that, since this type of instrument is
>>> calibrated using an illiminant A type of source, the problem
>>> remains, the chromaticity is not truly independent of some
>>> illuminant.

>> as I unterstand it, the XYZ (and derived Yxy) chromaticities shoud be
>> independent of of any reference illuminant.

> In which sense are you saying that they are independent of any reference
> illuminant? As far as I know the X coordinate is computed as
> > X = K integral S(lambda)R(lambda) xbar(lambda) d(lambda)
>
> with S(lambda)spectrum pf the source, R(lambda) reflectance of the sample

^^^^^^^^^^^^^^^^^^^^^^^^^^
Proabably a small misunderstanding her: I was referencing
to Roger's posting who wrote about pointing the CS-100 into
the blue sky, i.e. performing a measurement of light directly
coming from a light source, not being reflected from a colored object.

In case of directly measuring a light source, the result
depends on the source (obviously) but there is no additional
reference illuminant.

Of course, talking about reflectance measurements, you are
perfectly right - the meter reading depends on both the color
of the object and the illuminant.

Best regards

Dieter Michel

[Gernot Hoffmann]

Dieter Michel schrieb:

Yes, the sky is a self luminous light source (no reflectance).
If we can measure the spectrum for the light as emitted
by the blue sky by a spectroradiometer, then we can
calculate the XYZ values.

In image processing we are using Lab coordinates.
D50 is the reference white, simply because print products
are checked under D50. This defines as well the chromatic
adaptation transform.
Therefore it's really correct to define something like average
blue skies in images by Lab.

Best regards --Gernot Hoffmann
np

[inv...@invalid.invalid]

>> Gernot Hoffmann wrote:
>>> Which color (Lab) has a nice blue sky at noon ?

Gerhard Fuernkranz wrote:
> Wouldn't it be better, to specify just the [x,y] chromaticity?

I think so too.

Figure 2 in
<http://www.usna.edu/Users/oceano/raylee/papers/RLee_JOSAA_skylight_paper.pdf>
shows CIE 1931 chromaticities of 1567 clear skylight spectra measured in
Granada / Spain.

Klaus

--
echo '4b6c617573204b6172636865722c206d61696c746f3a6c6973
7473406469676974616c70726f6f662e696e666f0a' | xxd -r -ps

[Gernot Hoffmann]

inva...@invalid.invalid schrieb:

> >> Gernot Hoffmann wrote:
> >>> Which color (Lab) has a nice blue sky at noon ?
>
> Gerhard Fuernkranz wrote:
> > Wouldn't it be better, to specify just the [x,y] chromaticity?
>
> I think so too.
>
> Figure 2 in
> <http://www.usna.edu/Users/oceano/raylee/papers/RLee_JOSAA_skylight_paper.pdf>
> shows CIE 1931 chromaticities of 1567 clear skylight spectra measured in
> Granada / Spain.
>
> Klaus

IMO you're missing the point. I'm really not interested in the
chromaticity coordinates of clear skylight. These can be found
already in W & S.
I'm interested in Lab coordinates for sky blue. For Lab, one
needs the Y level of the brightest diffuse white in the scene.
One solution could be this: the white level is defined by clouds
in sunshine. If we assign here L*=95 or L*=100 for the brightest
parts, then it should be possible to express the surrounding
blue by Lab (as already suggested by Mike Russell - quite clear
that there isn't just one blue).

In other words: what's the range of sky blue, expressed in Lab,
if white clouds have for instance Lab = 100 / 0 / 0 ?
Note: in image processing, a neutral white or gray has a=b=0
for D50. This is so even if the real sky should have a different
CCT.

Best regards --Gernot Hoffmann

[Alessandro]

Dieter Michel wrote:
> Hi Alessandro,
>
>>>> I think you can argue that, since this type of instrument is
>>>> calibrated using an illiminant A type of source, the problem
>>>> remains, the chromaticity is not truly independent of some
>>>> illuminant.
>
>>> as I unterstand it, the XYZ (and derived Yxy) chromaticities shoud be
>>> independent of of any reference illuminant.
>
>> In which sense are you saying that they are independent of any reference
>> illuminant? As far as I know the X coordinate is computed as
>> > X = K integral S(lambda)R(lambda) xbar(lambda) d(lambda)
>>
>> with S(lambda)spectrum pf the source, R(lambda) reflectance of the sample
>
> ^^^^^^^^^^^^^^^^^^^^^^^^^^
> Proabably a small misunderstanding her: I was referencing
> to Roger's posting who wrote about pointing the CS-100 into
> the blue sky, i.e. performing a measurement of light directly
> coming from a light source, not being reflected from a colored object.

You're right, I was too fast. Probably because I am used to put R=1 for
computing the XYZ coordinates of the few sources I am interested in

Alessandro

[Gernot Hoffmann]

Some practical advice for handling Sky Blue in images:
http://www.fho-emden.de/~hoffmann/skyblue14072008.pdf

Best regards --Gernot Hoffmann

[dea...@msn.com]

Do you want the color of the sky at the horizon or at the zenith? The
color at the horizon is much less saturated any more cyan than at the
zenith where it is more saturated and a darker blue.

If you want to create a realistic sky, you need to mimic the real
world transition from horizon to zenith...