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kon1966
14-03-2020, 01:21 PM
This is interesting, rough 5 sec photo of orion ed80 with celestron SCT f6.3 reducer and eos 1100D camera compared to the FOV calculator using same settings.
Wonder what spacing they use to calculate the image square.
I expect that the field of view would be nominally calculated from the full-frame equivalent field of view of that optical setup, without regard to any changes spacing may result in.
1. ED80 is an 80mm aperture f/7.5= 600mm focal length
2. The SCT f6.3 reducer may be equivalent to a 0.63x reducer (assuming it doesn't vignette the frame)
3. The Canon EOS 1100D has a 1.6x FOV crop factor.
Taking that in to account the field of view of your setup should be equivalent to the field of view of a 600 x 0.63 x 1.6 = 605mm focal length on a Full frame senor, which has a ~3.4 x 2.3 degree field of view.
Or not....:D
Best
JA
Merlin66
14-03-2020, 02:23 PM
Nothing to do with crop factor....
Most (all?) FOV calculators work on pixel size x number of pixels.
Depending on the reducer spacing the magnification factor may vary.
Do a plate solve on your image to confirm the size (or just estimate using say CdC with the image sitting next to it)
The attached CCDCalc FOV shows your camera with/ without reducer.
Merlin66
14-03-2020, 03:10 PM
Solved your image
See display against CdC etc.
FOV 2.2 x 3.27 deg (132 x 196 arc min)
Comparing this with an ED80 - no reducer (84.5 x 126.8 arcmin), gives an effective reduction of x0.64.
Hope this helps.
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