# How can we calculate the visibility in CMAQv5.3.1?

Hello,

I have seen a few discussions in the forum but the method is not clear.
I know there is no visibility output anymore and in the CCTM_PHOTDIAG3 output these variables are given:

• total extinction from Rayleigh scattering at 294, 303, 310, 316, 333, 381, 607 nm
• aerosol extinction at 294, 303, 310, 316, 333, 381, 607, 550 nm
From these variables, how can I calculate the visibility? alternately, is there another method to get the visibility?

Regards,
Matthieu

Hello Matthieu,

quoting from @bhutzell (who is the expert here) in this thread, I believe the calculation equivalent to earlier versions of CMAQ would be

`visibility = 10.0 log( (aerosol_extinction_550nm+Rayleigh_Extinction_550nm)/Rayleigh_Extinction_550nm)`

where `aerosol_extinction_550nm` is EXT_AERO_W550 from CCTM_PHOTDIAG3 and `Rayleigh_Extinction_550nm` is 0.014 1/km

@bhutzell , please chime in if I’m misrepresenting your previous answer.

ok, thank you.

Does it mean the visibility is only calculated with this simple formula: 10*log((EXT_AERO_W550+0.014)/0.014) ?

regards
Matthieu

The formula works assuming (1) no extinction from clouds or fog and (2) the air number density equals 2.54E+19 molecules/cm3. To correct for the latter assumption, see the closing post in the Forum discussion.

ok.
In other words, and if I understand correctly, the formula should be:

10*log( (4.499E-22 x AIR_DENS +0.014) /0.014).

is it correct?

No, the formula is

10*log( (aerosol_extinction_w550+4.499E-22*air_number_density)/(4.499E-22*air_number_density) )

where air number density is in molecules per cm**3.

ok, thank you!

I better understand.

Is there an ouptut with this air number density in molec/cm3 or should I convert the AIR_DENS which is in kg/m3 ?

You need to convert AIR_DENS in the combine extracts or DENS in METCRO3D files.

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