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Found 4 results

  1. Hi, i have made a video where i show a coople of eyepieces that i use and like, and also show how a barlow works, compared to the "Powermate" from Televue. And how is the FOV affected when changing the magnification with different eyepieces? Feel free to comment and give me feedback - I hope you like the video! /Daniel
  2. I am intending to buy a Sky-Watcher Skymax-127 Maksutov-Cassegrain, primarily to make a number of daytime photographic terrestrial observations ranging from 1-5.5 miles, the most distant of which is of a target at 5.5 miles that is about 5 feet in size. The distant target would have an angular size of about 0.01 degrees. To get an object of an angular size of 0.01 degrees to appear as 100pixels vertically on the APS-C sensor (22.3mm x 14.9 mm) of my Canon EOS 250D camera, I would need a telescope/lens with a focal length of about 2.1m. The Skymax 127 with a focal length of 1500mm is pretty close to the required value and if used in combination with a Barlow lens I would be able to get more than the focal length I require. The FOV of the Skymax-127 with its 1500mm focal length in combination with the vertical sensor size of 14.9mm gives a field of view that can be derived from the following attached diagram which gives a FOV of 0.569°. But I would also like to make some lunar and solar observations with my system and capture images of the complete objects, and the maximum angular size of the Moon at perigee is 0.568°, so in theory the FOV of my system at 0.569 should be sufficient to gather the whole of the Moon at its largest angular size. But I have seen a video in which someone is using a Celestron SE6, which also has a focal length of 1500mm and when using it with a Canon 60da camera with an ASP-C sensor https://www.dpreview.com/products/canon/slrs/canon_eos60da which has the same size sensor as mine, then the camera does not seem able to capture a full view of the Moon. The video is at https://www.all-startelescope.com/video/scope-setup/nexstar-6se-dslr . The section where the Canon 60da is used is just before the end of the video and a clipped moon is shown, and then a full sized sensor camera is used to capture the whole Moon. Does this video not suggest that my camera with its APS-C sensor coupled with the Skymax 127 with its 1500mm focal length would not be able to capture the whole Moon, despite my calculations suggesting that it would? I understand that a reducer does not work well with a Muksatov-Cassegrain, so if I buy this telescope and use it in conjunction with my Canon250D, am I doomed never to be able to get an image of the complete moon? Thanks in advance for any suggestions/help etc
  3. I use a 200p/F5 for visual only. My longest FL EP is a 25mm Plossl with a 22mm field stop which gives a FOV of about 1.25 deg. I would like a much larger field to view the larger open clusters & similar size targets. Expensive SWA EP's & all 2" EP's are beyond my meagre budget. I could find say a 32mm (1.25") EP with a 27mm field stop which would deliver about 1.5 deg; better, but not enough. But (unbranded) focal reducers are not too expensive. If I used say a 0.5 reducer with my 25mm EP this would give me about about 2.5 deg which is just what I'm looking for. Is my analysis correct? Would this give pleasing views? Or would the shadow of the secondary be too noticeable at a Mag of 20X ? Any advice would be welcome. Thanks for reading.
  4. I have a Canon 5D Mark III with a 400mm lens and wanted to see if I captured a photo of the Andromeda Galaxy if I could print it as a 24"x18" print @ 200 PPI (pixels per inch). Since Andromeda is about 3.17 degrees x 1.03 degrees that translates into 11,412 ArcSeconds x 3,708 ArcSeconds. The Canon full frame sensor is 5784 x 3861 pixels with each pixel being 6.25 uM in size. For best-case atmospheric seeing of 2 ArcSeconds and a 2:1 Nyquist sampling, I think we want 1 ArcSecond of sky to map to one pixel of size 6.25 uM. Using the Image scale equation of Image Scale = (206.265 X pixel size)/ focal length I get Image Size = (206.265 x 6.25)/400 = 3.2 Arc Seconds per pixel. 11,412: (11,414 Arc Seconds) x (pixels/3.2 Arc Seconds) x (1 inch/200 pixels) = 17" (so the wide part of the Andromeda galaxy would print at 200 PPI for 17" 3,708: (3,708 Arc Seconds) x (pixels/3.2 Arc Seconds) x ( 1 inch/200 pixels) = 5.8" (so the narrow part of the Andromeda galaxy would print at 200 PPI for 5.8" Can anyone confirm that these calculations are correct or give some suggestions on how to correct? Thanks, Lloyd Linnell
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