Back to the original question:
"I was wondering if anyone has a link to something explaining arcseconds, particularly how these apply to seeing and pixel resolution?"
Hi Guys
Well, as the thread hasn’t been added to, I’ll continue.
Just like a telescope/eyepiece combination has a field of view (FOV) that can be expressed in deg, mins & secs, a telescope/sensor combination can be described in the same way. Here are some FOV examples:
- Celestron C9.25 at F10 + Baader 21mm Hyperion = 36.5 arcmins.
- Celestron C9.25 at F10 + Vixen LV 5mm = 5.74 arcmins.
- Celestron C9.25 at F10 + Philips ToUcam = 5.27x3.95 arcmins, or 5.27’x3.95’.
- Celestron C9.25 at F10 + Pentax *istDS = 34.38’x22.97’.
- Vixen 102mm at F9 + Pentax *istDS = 80.77’x53.97’.
So, let’s take the C9.25 at F10 with a ToUcam and a FOV of 5.27’x3.95’.
We know the following about our ToUcam:
- Chip size = 3.6mm x 2.7mm.
- Resolution is 640x480 pixels.
- Pixels are 5.6um square.
A FOV of 5.27’x3.95’ is the same as 316”x237” which would fit in approx 9 Jupiter’s with the current angular size of Jupiter being around 32”.
So, we can image a field of 316”x237” with our 640x480 pixel chip.
316/640=0.49
237/480=0.49
We are just dividing the FOV by the number of pixels on our chip to find out what each pixel can "see" or record. This means that in theory, the smallest detail on Jupiter’s disc that each pixel can resolve would be 0.49”, or ½ arcsec. So, you would record a feature on the disc if it appeared to be say, 2” in size and this feature would occupy 4 pixels in your image. You would not be able to record any detail that is say, 0.25” in size.
I’ll leave it there for the moment as I’m concerned I’m hogging the post and of course, as always, I could be wrong or over complicating things, so I always welcome corrections, better explanations and constructive feedback.
Cheers
Dennis