Re: Measuring liquids
Re: Measuring liquids
- Subject: Re: Measuring liquids
- From: email@hidden
- Date: Thu, 18 Oct 2001 10:15:50 +0100
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>I have a client that needs to measure colored liquids and output them to
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>a
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>printer. The way I though of doing this was to sandwich the liquid between
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>two sheets of acetate or thin glass on top of bright white paper and
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>measuring with the spectrolino. I am having doubts with this technique
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>as
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>the density of the liquid will change as it is sandwiched.
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>
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>Anyone have any ideas?
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This will vary with the depth of liquid viewed, lighting color, angle etc...
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might as well shoot with a digital camera, correct on screen by eye, to match
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the liquids in a light box or booth, and call it close enough.
I agree that Lab or XYZ measurements will vary (see below). But depending on
the precision needed, spectral measurements wouldn't be bad: I'd use a vial
(a cuvette) and take a measurement with the solvent alone and then with
solvent plus colorant. If you manage not to change the experimental setup
in-between, and if you are measuring (almost) prependicular to the surface of
the rectangular cuvette, you can zero out any glass influence and the measuring
color, since in most cases the colored liquid is a perfect diffusor (and let's
hope it's not flourescent).
But you need to take spectral readings, and you need to know the concentration
of the colorant. Normalize the spectrum with colorant by dividing by the
remission values without colorant. If you later need the color at double
concentration, take the square of the normalized remission values, and for
other concentrations and/or optical pathlengths, apply Beer's power law.
Then either calculate XYZ (and then Lab) from the spectra yourself, or export
the changed spectra to some software that can handle it (ColorShop?).
The point is that XYZ relates linearly to the spectral intensity, but spectral
intensity decays exponentially with concentration. Assume that some dye
transmits 90% in a cyan spectral band but only 50% in a yellow one. At doubled
concentration you still have 81% in cyan (= 0.9 * 0.9) but only 25% in yellow.
Your color becomes more saturated, the X:Y:Z ratios disbalance, and likewise
Lab values will change hue and saturation. For the same reason, cyan ink looks
quite blue in the ink pot, and magenta looks red.
The calculations can quickly be done e. g. in Excel. So if your client changes
his mind about the concentration, you do not need to remeasure. You can even
calculate mixed colorants. (If they don't react with each other -oops)
But you see that simulating colored liquids in circular tubes or
"Frankensteiny" glass cylinders (with varying optical pathlength) would be a
nice little exercise...
(Of course using a digital camera and re-shooting for a new concentration is
probably FAR quicker. American pragmatism meets German perfectionism ;-)
Cheers,
Hanno