I am very used to working with seeing in arcseconds, but sometimes I come across the seeing in cm, e.g. here.

I have looked online for a way to convert from one to the other, but I can't find any. How would this be done?

  • 1
    $\begingroup$ Do you have an example you could point to for seeing stated in cm? The reason I ask is because you can't specify seeing in physical distance units like cm in a general way. You could possibly do it for a specific instrument if you know that instruments plate scale but really that's just converting the angular size of the seeing to the physical size for that instrument only. I've never seen anyone do such a thing though. $\endgroup$
    – zephyr
    Oct 16 '17 at 15:59
  • $\begingroup$ Here for example: royac.iac.es/seeing.html $\endgroup$
    – Coolcrab
    Oct 16 '17 at 16:01
  • $\begingroup$ And I agree with what you are saying, this is why it is confusing for me. They say that the 'true' seeing which doesn't help $\endgroup$
    – Coolcrab
    Oct 16 '17 at 16:04

Seeing is typically measured as the FWHM of the seeing disk, but can also be expressed though the Fried parameter $r_0$, which measures the size, or strength, of the parcels of gas that cause the turbulence in the atmosphere. The relation between the two is (e.g. Vernin & Munoz-Tunon 1995) $$ \text{seeing} = 0.98\frac{\lambda}{r_0}, $$ where $\lambda$ is the wavelength of the light.

$r_0$ can by calculated by integrating the "turbulence strength" along the line of sight$^\dagger\!\!$, and scales as $\lambda^{6/5}\!$. Because of this dependence, the "usual" seeing has only a weak ($\lambda^{-1/5}$) dependence on wavelength.

From the Wikipedia article on seeing:

At visible wavelengths, $r_0$ varies from 20 cm at the best locations to 5 cm at typical sea-level sites

$\dagger$Actually, $r_0$ refers to a distance corresponding to the line of sight towars zenith; observing at an angle $\zeta$ from zenith introduces a factor $\cos^{3/5}\!\zeta$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.