Here is my understanding of this issue.
Perhaps the simplest way to visualize this is to consider two spots
on a surface. One of the spots contains a uniform distribution of
stain, while the other has a center with a high amount of stain,
which then tapers off towards the edge of the spot. Let's assume
that you want to know the total amount of the stain in each spot.
For the first, you could measure the density of a single pixel and
multiply by the number of pixels in the spot. However, if you did
this with the second spot, you would be unable to select a single
pixel that was typical of all of the pixels in that image. The
solution is to measure the density of each of the pixels that
comprise the spot and add them up. This will give you an "integrated
density".
You could do the same thing with the first spot, and it should be
obvious that the integrated density will equal the density of one
pixel multiplied by the numer of pixels. The number of pixels is, of
course, the area of the spot.
Now, when ImageJ computes the density (not the integrated density)
of a spot, regardless of whether the spot is uniform or not, it adds
up
the density of each pixel and divides by the number of pixels (i.e.)
the
area.
So, to convert this to integrated density, you would take the density
measurement and multiply it by the area.
Am I making any sense?
J
--
Joel B. Sheffield, Ph.D.
Biology Department, Temple University
1900 North 12th Street
Philadelphia, PA 19122
[hidden email]
(215) 204 8839, fax (215) 204 0486
http://astro.temple.edu/~jbs
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