How To Understand an Imaging Setup - Pixel Size, Arc-Seconds/pixel, Sampling and More
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I have looked up my CCD's QE (quantum efficiency) which was 60%@green (not a nice deep sky imaging color as most people simply remove this - so its most important is 50%@red and blue (ie H-alpha) for most deep sky objects. This is not great, but probably better than most DSLRs and low-end ccds. Green is often removed or decreased in most pro image processing steps so its a ploy to try to boost QE values by companies.
Getting An Optimal Imaging Setup Together - Star Size, and Sampling - Pixel Size / Array
I inputted all my setup details and got the following:
Image Scale: 1.79 ARCSEC/Pixel (at 1X1 Bin)
Field of View: 60X90 arcmin (I can fit the entire horsehead but its not wide enough to get all the details of the nebula.)
Chip Size: 15.7mm X 23.6 mm
Pixel Size: 7.8X7.8uM
What does this mean?
Now from here we know that there is some consensus that the optimal pixel size is 2-3 arcsec/pixel for deep sky imaging - some say <0.5 arcsec per pixel. At 1.79 arcsec/pixel with 1X1 binning I am close to the optimal 2-3, but far from the less than 0.5.
However from Stan Moore's page above, the ideal is actually around a pixel size of 0.5 to 1.5 arcsec. Being at 1.7 arcsec/pixel my setup is a bit outside this range.
More on this
Since we want nice stars, we should calculate what a star's size will be on our chip - this is dependent on several factors but the most important are seeing (in arc-seconds) and your telescopes focal length in mm.
Star Size (µm) = (Seeing in arcsec * Focal Length mm)/206.3
From the book Digital astrophotography: the state of the art By David Ratledge, he mentions that in a suburban location the resolution is 3-5 arc seconds.
Assuming a seeing a 3, lets calculate the star size in microM for my scope (900mm).
Star Size (µm) = ( 3 X 900 ) / 206.3 = 13.09 µm
so the star size on the ccd chip will be 13.09 µm
Being the Nyquist criterion is FWHM = 3.3 pixels, we would need pixels of 13.09 / 3.3 = 3.97µm
Looking at my CCD's pixel size of 7.8 µm, the ideal pixel size would be 3.97 µm
When a CCD's pixels are too big (ie 7.8 to about 4) compared to the optimal pixel size this is called undersampling, and can lead to blocky square shaped stars, or even a loss of details such as stars that are missing completely.
When a CCD's pixels are too small, compared to the optimal pixel size calculated this is termed oversampling. In this case, you are not missing out on any stars or details however you are not capturing any extra details as well.
Calculate your CCD's Optimal Focal Length (FL)
FL = 8 / (P/S)
where P is the image scale per pixel in arc seconds (Nyquist is 0.33)
and S is the size of your CCD pixels in microns
FL = 8 / ( 0.33 / 7.8 )
FL = 189 inches or 4 802 mm
so to get an image scale of 0.33 arc seconds per pixel (of 7.8 microM of my CCD) to correctly sample 1 arc second detail in the object therefore I would at least need a 5X barlow (900 mm X 5 = 4500mm)
Benefits of Large Pixelsor Undersampling
So as I am using my setup without a barlow my camera setup is Undersampling, meaning my pixels are too big (7.8 to 3.97 microM) but there are benefits of this.
One of the benefits of undersampling or having larger pixel size is improved sensitivity. This is similar to collecting rain - the bigger the buckets (ie pixel size) the more rain drops (or photons) your chip's pixels will capture compared to a ccd with identical QE but smaller pixels.
You can therefore image with shorter exposure times, as you can get brighter details than an equivalent QE but smaller pixel chip - so your larger pixel chip helps with exposures at the expense of some lost resolution.
Name: Gus OTAs: ED 100 PRO refractor, Orion ST80 (not the CF), 8" Dob stuck in Canada Mounts:HEQ5PRO Synscan mount, Manfrotto Tripod CAMS: QHY8L not recommended, Guidecam Philips SPC900 webcams (4), Canon unmodded-450D DSLR
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