Because different wavelengths of light are scattered differently in air (there are actually two distinct types of scattering involved) light on the red end of the spectrum is scattered far more than light on the blue end, so things in the distance appear blue and somewhat washed out — aerial perspective.

I have a large number of photos. I know the distance to the subject in every photo. It is large enough that their is some aerial perspective, which they would look better without. I want to correct it automatically based on the distance involved. I don't want to correct them by hand.

  • Does anyone know, or know a source for, the degree of loss in different wavelengths of light over a distance D?

  • Does anyone know, or know a source for, the degree of ambient background light of a given wavelength accumulated washing out an object at distance D?

Yes, I know these would depend on all sorts of factors to get exact just like the color of the sky does, but I'm sure someone has studied reasonable averages for middle of the day photos.

The cheap version

The above is what I am really after, but if nobody can answer that, the simple version is this: If I know the distance D in a photo to a subject, what percentages would I use to mix the subject photographed close up and the sky to the get the same aerial perspective effect. This really isn't the same, but if it is the best I can get, I will take it.

  • \$\begingroup\$ Can you post an example photo? \$\endgroup\$ Mar 6, 2015 at 14:09
  • \$\begingroup\$ I think this is verey complex, becouse atmospheric conditions, humidity, dust particles, wind, altitude... \$\endgroup\$
    – Rafael
    Mar 6, 2015 at 14:50
  • \$\begingroup\$ I would start by looking at the science of color-correcting underwater photos -- same idea, but much more extreme. \$\endgroup\$
    – feetwet
    Mar 6, 2015 at 17:18
  • \$\begingroup\$ If you know a detail that should be white, select that to set automatic white-balance. Free <a href="irfanview.com">IrfanView</a>, for example has that feature: press Ctrl-G to open the dialog and click on a white pixel in the left pane -- the right pane shows the adjusted image. \$\endgroup\$ Mar 6, 2015 at 18:43
  • \$\begingroup\$ @mattdm Aerial perspective and linear perspective both show distance. Geometric perspective shows it by distortion, aerial shows it by color change (mountains in the distance are blue and hazy, i.e. a change in color shows distance) \$\endgroup\$ Mar 6, 2015 at 21:12

3 Answers 3


Of note to photographers: It should be pointed out that the cause of aerial perspective is the same as what causes sky color and (and this may interest photographers) the sky is also the light source for all shadowed ares in outdoor photography. Thus the light spectrum below is actually also the light spectrum for your average light source for all areas in shadow in an outdoor photograph, i.e. this is why shadows appear blue in outdoor photographs (not from aerial perspective, but because any area not under direct sunlight is being lit by the secondary light source which is the sky, and this gives you a typical color spectrum for a blue sky.)

Answer to the question OK, I haven't worked out all the details yet but this might make for a quick cheap answer. I am still interested in a real answer if anybody know one.

Calculate the weight of the air above you by using the fact that at sea level the atmospheric pressure is 1 atm. Then use the color spectrum of the sky and assume the effect is linear (obviously the color spectrum of the sky varies with things like temperature and sun angle, so use an average)

Then assume the effect is linear in the amount of air (by mass, i.e. use the density of air at sea level to figure how much air you are seeing something through at distance D)

Then you can assume your object has been averaged with that much average sky photo, so subtract off that much sky spectrum and renormalize to make up for the amount of stuff you subtracted off.

Here is an average blue sky color spectrum from wikipedia.

From wikipedia licensed under GNU Free Documentation License and Creative Commons Attribution-Share Alike 4.0 International, 3.0 Unported, 2.5 Generic, 2.0 Generic and 1.0 Generic.

P.S. It might be interesting to apply this "undoing aerial perspective" to rainbow pictures to see what you get, since rainbows are obviously typically more blue and washed out than they should be due to aerial perspective. Not sure what distance to use off the cuff since they are an accumulated effect over a bunch of rain (i.e. not at fixed distance) but probably estimating half the distance from yourself to the edge of the storm would be a good start.


There has been work on image processing of this sort. The typical terminology would be something like "Blind haze removal", where "blind" would refer to the processor having no special information about the image, just the pixels.

There is an example and link to the relevant scientific paper on this Github project. It uses Matlab ( a mathematics software package ! ). I'm not aware of a specific general plug-in for any graphics application, although there may be one. Be cautious with those as most will probably be based on quite simplistic approaches which are not generally effective.


If you want to correct the image by haze or mist, you should read this thread at Rational painting forum :


You should forget wavelengths and extravagances such as Mie scattering and Rayleigh scattering, because the law governing the artist model is more basic : exponential

The formula was derived from Beer-Lambert-Koschmeider equation You should use RGB[0,255] color notations, gamma-corrected I'm a painter and use it extensively, is very easy to use when you grasp the variables, I built an excel worksheet that make the job for me ...

1) Set the Visual Range (D) from observer where the constrast decays to 2% 2) Set the RGB "true" (local) color for the object to project into the mist 3) Set the distance (d) where you want to put the object in the mist 4) Set the Gamma correction factor (i.e. 2.4) 5) Apply formulae explained in attached files

The rest of the explanation you could seen it in those files, the last one is a test showing how it works in a real case

enter image description here enter image description here enter image description here enter image description here

  • \$\begingroup\$ This link is not accessible to me. Could you summarise the information in you answer . \$\endgroup\$ Dec 15, 2015 at 3:03
  • \$\begingroup\$ Yes, in a while I'll add it here ... \$\endgroup\$ Dec 15, 2015 at 13:43

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