Raw versus calibrated integrated density

[IJperson]

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.



                       

-- View this message in context: http://imagej.588099.n2.nabble.com/Raw-versus-calibrated-integrated-density-tp5646703p5646703.html Sent from the ImageJ mailing list archive at Nabble.com.

[Michael Schmid]

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.

[Norbert Vischer-2]

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?

Integrated density in ImageJ relates indeed to scaled area units. However if one really was interested in the number of photons per micron, he needed to know a lot of other factors as well. Personally I think it would have been better not to include the image scale - in practice it may lead to misunderstandings, which often happened in our lab. Users often set the scale to their images only if spatial measurements are needed; sometimes images are silently scaled (or inadvertedly re-scaled) because the "Global" flag in the scale dialog is switched on. I always remove the scale in such cases and divide the result by 1000, so I am independent of scaling, and I get smaller numbers.

Norbert

[IJperson]

Hi Michael.

Many thanks for the reply.
To clarify, I am not referring to greyscale calibration. Lets assume I am just interested in the intensity of the pixels as they appear, with no extra scaling.  My confusion lies with different integrated intensities (IntDen and RawIntDen) over spatial calibration.

The example you present with Watts is helpful.  So the integrated density has units:
 cm * cm * W/cm^2 = watts.  I.e. The spatial units disappear.

So the calibrated (IntDen) has units:
<realworld unit> * <realworld unit> * <intensity unit>  / <realworld unit>^2  =<intensity unit>

and the uncalibrated (RawIntDen):
<pixel unit> * <pixel unit> * <intensity unit>  / <pixel unit>^2                    =<intensity unit>

I.e. both have the same unit, which appears to be independent of a spatial dimension. So should both therefore have the same value?

I expect I am still misunderstanding something, but this is a very helpful discussion.

Many thanks
Andrew

[Michael Schmid]

Hi Andrew,

maybe my example has been deceiving - watts/cm2 is the intensity  
(grayscale) unit in my example.

The IntDen unit is:
  <realworld unit> * <realworld unit> * <intensity unit>

Maybe there is a better example:

Assume, you take an image of a bright spot, 10um*10um with 1 um pixel  
spacing, and all pixels having intensity = 100.

Then you get integrated density = 10 um * 10 um * 100 = 10000 um^2.

Now, scale the image so that you have 2 um pixel size. The bright  
spot will be just 5*5=25 pixels, and raw integrated density will be  
2500.
Nevertheless, scaled integrated density will still be 5*2 um * 5*2 um  
* 100 = 10000.

You can easily try this with an image of your choice and the  
"Image>Adjust>Size" function of ImageJ.

Is this explanation better?

Michael
________________________________________________________________

On 18 Oct 2010, at 16:42, IJperson wrote:

> Hi Michael.
>
> Many thanks for the reply.
> To clarify, I am not referring to greyscale calibration. Lets  
> assume I am
> just interested in the intensity of the pixels as they appear, with  
> no extra
> scaling.  My confusion lies with different integrated intensities  
> (IntDen
> and RawIntDen) over spatial calibration.
>
> The example you present with Watts is helpful.  So the integrated  
> density
> has units:
>  cm * cm * W/cm^2 = watts.  I.e. The spatial units disappear.
>
> So the calibrated (IntDen) has units:
> <realworld unit> * <realworld unit> * <intensity unit>  /  
> <realworld unit>^2
> =<intensity unit>
>
> and the uncalibrated (RawIntDen):
> <pixel unit> * <pixel unit> * <intensity unit>  / <pixel unit>^2
> =<intensity unit>
>
> I.e. both have the same unit, which appears to be independent of a  
> spatial
> dimension. So should both therefore have the same value?
>
> I expect I am still misunderstanding something, but this is a very  
> helpful
> discussion.
>
> Many thanks
> Andrew

[Gabriel Landini]

On Monday 18 Oct 2010  16:21:12 IJperson wrote:
> Many thanks for the reply.
> To clarify, I am not referring to greyscale calibration. Lets assume I am
> just interested in the intensity of the pixels as they appear, with no
> extra scaling.  My confusion lies with different integrated intensities
> (IntDen and RawIntDen) over spatial calibration.
 
This has been discussed many times in this mailing list.
I suggest that you search for "Integrated density" and see that the values do
change if you take into consideration spatial calibration in the computation,
but you do not gain any more information.

I personally think it is more intuitive to calculate the integrated density in
pixels (i.e. samples, not calibrated units) and provide the interpixel spacing
for reference of the resolution of the measurement. The measurement is then
just the sum of the greyscale values of the blob.

Regards

Gabriel

[Herbert M. Geller]

  Friends,

I would like to have a plug-in to identify the relative maxima of an
image across a horizontal line selection.  Presumably, i would need some
threshold to eliminate noise.  Any suggestions would be appreciated.

--
--------------------------------------
Herbert M. Geller, Ph.D.
Developmental Neurobiology Section
National Heart Lung and Blood Institute, NIH
10 Center Drive MSC 1754
Bldg 10, Room 6D18
Bethesda, MD  20892-1754
Tel: [phone]; Fax: [phone]
e-mail: [hidden email]
Web: http://dir.nhlbi.nih.gov/labs/ldn/index.asp
---------------------------------------

[Michael Schmid]

Hi Herbert,

the easiest would be selecting a rectangle of height 1 pixel,  
duplicate it and run 'Find Maxima' on the 1-pixel-high image.

Michael
________________________________________________________________

On 18 Oct 2010, at 17:41, Herbert M. Geller wrote:

>  Friends,
>
> I would like to have a plug-in to identify the relative maxima of  
> an image across a horizontal line selection.  Presumably, i would  
> need some threshold to eliminate noise.  Any suggestions would be  
> appreciated.

[vischer]

Andrew, I think your critics is correct. I guess that RawIntegratedDensity (introduced in june 2010) is the only meaningful quantity, and IntegratedDensity survived for backward compatibility. Michael, when talking about light quanity and enlarging the image without making it darker, you introduce already an "error" here, which then is automatically corrected by Integrated Density. It is one of those automatic corrections that I always try to avoid.

Norbert

[IJperson]

Michael,

Many thanks. Its the units of the integrated density that has been confusing me.  But it seems it has unit of <spatial unit>^2

So in your new example, with the um calibration the result is 10000 um^2, and with the raw value it is 2500 pixel^2
And with one pixel equalling 4um^2 (2x2) I can see how you can convert between 2500 pixel^2 and 10000 um^2

I still don't quite understand how the units of the integrated density is solely a spatial area, but there you go, at least I understand the maths now! Thanks to everyone for pointing out this IS confusing, and its not just me!

Andrew

[Michael Schmid]

Hi Andrew,

in the general case, the unit of integrated density is <grayscale
unit>*<spatial unit>^2.
You get <spatial unit>^2 only if you don't have a grayscale unit.

Michael
__________________________________________________________________________

On Tue, October 19, 2010 09:53, IJperson wrote:

> Michael,
>
> Many thanks. Its the units of the integrated density that has been
> confusing
> me.  But it seems it has unit of <spatial unit>^2
>
> So in your new example, with the um calibration the result is 10000 um^2,
> and with the raw value it is 2500 pixel^2
> And with one pixel equalling 4um^2 (2x2) I can see how you can convert
> between 2500 pixel^2 and 10000 um^2
>
> I still don't quite understand how the units of the integrated density is
> solely a spatial area, but there you go, at least I understand the maths
> now! Thanks to everyone for pointing out this IS confusing, and its not
> just
> me!
>
> Andrew
> --
> View this message in context:
> http://imagej.588099.n2.nabble.com/Raw-versus-calibrated-integrated-density-tp5646918p5649894.html
> Sent from the ImageJ mailing list archive at Nabble.com.
>