Posted by
Herbie on
Jun 22, 2015; 1:12pm
URL: http://imagej.273.s1.nabble.com/image-summing-details-tp5013231p5013236.html
Sorry Martin,
your questions have nothing to do with images but with logic and perhaps
a bit of mathematics.
What you may consider is a single pixel in your image stack.
Let's assume you have three images in the stack and we consider the
pixels at coordinate 0,0 in every image. If this pixel has a gray value
of 100 in every image, then the sum is 300 which is beyond the scope of
an 8bit image that is limited to 2^8 = 256 values (0...255). That is why
the summing result is represented as a 32bit image which can have deal
with very high values. Hence the resulting pixel in our example is
represented having the correct value of 300.
In short: nothing is scaled.
But...
In general, computer displays can show only 256 shades of gray and you
may ask how the value of 300 is displayed. In fact there is a scaling
for the display but the considered pixel in the result image still has
the value of 300. You may use the "Image > Adjust >
Brightness/Contrast..." to change display parameters (for a description
see the ImageJ-UserGuide at
<
http://rsb.info.nih.gov/ij/docs/guide/index.html>). (The values in the
32bit result image remain unaltered!)
If you want to keep the 8bit representation, you should use the
average-option. So in our example the resulting pixel from averaging
will show the value 100.
HTH
Herbie
:::::::::::::::::::::::::::::::::::::::::
Am 22.06.15 um 13:59 schrieb Martin Ward:
> Hi Herbie,
>
> As it may be quite clear by now, I am in no way an expert in image analysis.
> I understand that something is being scaled but I do not know what the
> scaling factor is; therefore, I cannot directly compare different stacks of
> images as I wish to. If it is possible I would like to establish if there
> is a way that I can find out what the scaling factor is. Also, if anyone
> has done something similar, it would be good to find out if they would
> recommend a different projection type to use when merging the images(perhaps
> average or median?).
>
>
> Regards,
> Martin.
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