Name: Catherine Jean C.
Date: Sunday, April 21, 2002
I asked this before, but did not give enough info. One of
my students is doing a science fair project on fading construction paper.
She cut strips of each color and exposed them to sunlight for 1 hour
intervals for a period of 6 hours. At the end, the green paper appeared
to have faded the most. She took the strips to a photo shop and the
photographer let them use a densitometer. The readings (abbreviated
version)she took measured the density of blue, green, red, and just plain
density of each color of paper for each time interval. The density of
blue for the green construction paper was .78 at the beginning (unexposed
paper) and .71 at the end (paper had been exposed the full 6 hours). The
density of green stayed the same
.38; the density of red was .77 (beginning) and .52 (end). The density of
the black paper (which appeared to have faded the least) was:
(before and after) Blue density 1.29/ 1.04 Red density 1.20/ 1.02;
green density 1.26/1.08 and plain density 1.25/1.09. Her conclusion is
that if she subtracts the numbers of the before and after density, the
black shows a greater loss, so therefore it must have faded more.
I do not think that is correct. I think it has to do with the density of
blue number. I do not want to tell her the wrong thing, but I am over my
head here. Can you help? Thanks.
From what you describe, I do not think you are using the proper measuring
tool. A densitometer (as usually defined) measures the lightness or darkness
of the sample being tested. The reading that the densitometer provides does
not take into account color change per se, but only the change in the
lightness or darkness, from white through gray to black without regard to
the color. That is not what the eye/brain perceives. The change in a deep
blue will be more noticeable than a change in a pale yellow for example, and
so the color change needs to have a way of taking this into account.
Color is defined in three coordinates: hue (which is what we usually refer
to as "the color", meaning blue, red, yellow, green, etc.), the saturation
(which is how "pure" the color is (a laser pointer is very saturated, as
opposed to a pastel of the same hue), and density (which is how white, gray,
or black the color is regardless of its hue or saturation). Navy blue is
more dense than lemon yellow for example.
These coordinates are measured by a number of instruments, of which a
tri-stimulus color meter is the most common. The outputs of such a meter are
the parameters: L, a, and b. The L-axis is the white/black axis, the a-axis
is the red/green axis, and the b-axis is the blue/yellow axis. (I may have
the "a" and "b" axes reversed.). So every sample has three components
(L,a,b). These can also be expressed in cylindrical coordinates where, then,
the angle measures hue; the "r" measures the saturation; and the "Z" axis
measures the density. The change in the color of a sample can then be
expressed by a single parameter which is essentially the length of the
vector from the origin to the point (L,a,b) but then of course just "how"
the color changed is lost.
Copy centers may have instrumentation that can provide you with these data
points, I am not sure. However, any paint store outlet -- Sherwin Williams,
for example -- will have this instrumentation, because it is used (along
with the (L,a,b) coordinates) for matching paint colors.
I do not think you would have any trouble getting them to measure and collect
this data for you. Then you can decide whether and how you want to massage
the numbers to illustrate the changes most clearly. I personally prefer the
cylindrical coordinate representation because it isolates changes in hue,
saturation, and density most clearly, but in that respect I am a minority.
The largest U.S. manufacturer of tri-stimulus meters is Hunter Laboratories.
They may be of assistance in obtaining background info on the instruments
and the theory behind their operation and interpretation.
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