what is the unit of light intensity values in a grayscale image ?

asked 2018-05-31 02:13:23 -0500

ak1 gravatar image

updated 2018-05-31 02:14:19 -0500

I want to calculate average luminance of RGB image.
My approach - I am converting RGB image into grayscale image and further taking average all the intensities values of that grayscale.

I am confused about the units of above result. Is it right way to calculate variation of light intensity over images ? Is this is a relative luminance or absolute luminance ?

As grayscale represents light intensity values at every pixels. Hence is it right to say intensity values at pixels will have unit candella/ m^2 ?

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Comments

Doesn't look correct to me since that value is the result of 2 processes: first conversion of the amount of radiation hitting a specific pixel on the camera sensor (which gives an analog signal), second the digitalization of the analog signal. Here different levels of precision in the conversion can be chosen, thus leading to different grayscale values. So I don't think there's any correspondence to the physical unit. Not to mention that for an RGB sensor we have not considered the Bayer filter influence so far...

David_86 gravatar imageDavid_86 ( 2018-05-31 02:29:59 -0500 )edit

As grayscale represents light intensity values at every pixels

not exactly. not every image comes from a camera. and this is not a lighting problem.

if you look at the opencv logo above, your question does not make sense at all.

berak gravatar imageberak ( 2018-05-31 02:35:17 -0500 )edit

@berak and @David_86 I have series of images taken at different lightning conditions in which planar marker detection is done. I don't have any proper instrument to measure light intensity. So I am looking for a parameter for comparing marker detection in different lightning conditions, like the opencv algo which I have mentioned above in the question. Can I go with this parameter? What is the suitable way to acheive this?

ak1 gravatar imageak1 ( 2018-05-31 03:24:51 -0500 )edit
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LBerger gravatar imageLBerger ( 2018-05-31 03:40:46 -0500 )edit

@LBerger Thanks. I think this will solve my problem.

ak1 gravatar imageak1 ( 2018-05-31 04:06:29 -0500 )edit

What do you mean by "different lightning conditions"? You'll be measuring the amount of light reflected from the target surface, not the light intensity. If the position of the light source changes the results won't have any sense.

David_86 gravatar imageDavid_86 ( 2018-05-31 08:17:04 -0500 )edit