Posted by
Michael Schmid on
Oct 18, 2010; 1:34pm
URL: http://imagej.273.s1.nabble.com/Raw-versus-calibrated-integrated-density-tp3686613p3686614.html
Hi anonymous,
your question is not very clear to me.
There is spatial calibration and grayscale (pixel value) calibration,
you have to differentiate between them.
Remember that an integral has the unit of the integrand times the
unit of the variable of integration (in our case, area).
E.g., you have a camera that measures the beam profile of a laser:
Assuming that your spatial unit (x, y) is cm, and your grayscale unit
(after calibration) is watts/cm^2, then your integrated intensity has
the unit cm * cm * W/cm^2 = watts (the total power of laser light
falling on the area of interest).
The mean intensity over an area should be independent of spatial
calibration, but, of course, it is sensitive to grayscale
calibration. If you have no grayscale calibration, the calibrated and
uncalibrated mean are the same.
E.g., if all your pixels, have a value of 10, the mean should be 10,
irrespective of spatial calibration.
I have done a quick test with the 'Measure' command of ImageJ on a
calibrated 16-bit image and did not find any inconsistency for Mean,
IntDen and RawIntDen; it behaves as expected.
Michael
________________________________________________________________
On 18 Oct 2010, at 15:03, IJperson wrote:
> Hi everyone.
>
> This could well be a misunderstanding on my part, but why are the
> raw and
> calibrated integrated densities different? I understand the per
> micron or
> per pixel means being different, but isn't the total intensity unit-
> free?
>
> Here is a quick calculation, where as I calculate it, the total
> integrated
> intensity should be the same. Where have I gone wrong? Is it my
> understanding of what 'density' means?
>
> Pix count Calibration Cal area
> 100000 2 200000
>
> Total Intensity over image
> 555555
>
>
> Cal mean Pix mean
> 2.777775 5.55555
> um-2 pix-2
> mean umean
>
>
> IntDensity (mean*cal area) RawIntDensity (umean*num pixels)
> 555555 555555
>
> Many thanks for the clarification.