Keep in mind that HFR is a radius, FWHM is a diameter. So you would need to double HFR to compare it to FWHM. Even then, though, you can't compare even twice the HFR to FWHM. They are calculated completely differently. HFR is the radius at which half of the light is concentrated within the radius and the other half is outside of the radius. For a Gaussian distribution, twice the HFR encompasses approximately 1.2 standard deviations whereas FWHM encompasses approximately 2.4 standard deviations. So HFR (actually twice the HFR) is quite a bit narrower. or the same Gaussian distribution, an HFR of 1" corresponds to a FWHM of 4.8".
I'm willing to believe that my FWHM is closer to the 4 or 5 range!
Do I have this totally wrong in my head:
- with a longer FL (2800mm) the FWHM of a star has to be larger than with a short FL (200mm) because the star is magnified more and therefore covers more pixels. Is that correct?
Generally not. It's complex, but two things that figure in are the higher resolution of a large aperture scope, and the quality of the optics. The "magnification of the star" doesn't really figure in to it, stars are (without optics) extremely tiny points of light that cannot be "magnified" even by a large telescope. For example, Betelgeuse is 0.05 arc sec, about one-tenth of a pixel at your image scale.
What you're seeing are things like imperfections (such as optical aberrations) and noise, not a magnified star. This chart is fun, can't guarantee the accuracy, but the general picture is correct.
And, of course, blur due to atmospheric turbulance, aka "seeing" trumps everything. Rough idea (not intended to be precise). If your FWHM is 4 to 5, it's likely that your seeing is 3 to 4. Note the sig, please.
Edited by bobzeq25, 21 April 2017 - 05:50 PM.