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  1. 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
  2. Wow! Those mounts are gorgeous. Really gorgeous! Beautifully elegant yet robust. I don't think any of them are motorised, is that a conscious decision? I appreciate that motorised altazimuth mounts aren't terrible useful in astrophotgraphy because of field rotation.
  3. Thank you so much for your answer. Yes I hadn't fully appreciate the need for a SLOW motion to manipulate the telescope, because of course I haven't handled high magnification optics before. But I now appreciate it after having watched some videos of hi-magnification telescopes in action. Wow! So you can do 10k terrestrial observations in good conditions, fabulous. Sometimes the atmosphere can cooperate. That does sound like lots of fun. Presumably doing birdwatching at 10k for flying birds might be a bit tricky, but then I suppose if they don't move across the field of view, then you could.
  4. Comsic Geoff, Thank you so much for your reply. I am familiar with the issues of looking over long distances such as 5.5 miles, or 8km, horizontally, and you are absolutely correct that the atmosphere will do terrible things for much of the time, and perhaps I hadn't made clear that I was willing to put up with lots of horrible distortion of the image at the extreme distances, I just needed to be able to identify the target. I suspect that your low altitude observations of planets would be much worse because you would be looking through far more atmosphere than 5.5 miles and you would be doing so through turbulence etc. I think you are right about the mount, you have made me realise that fiddling with an equatorial mount for terrestrial observations would be difficult. I had vaguely thought about the issues, but as I am not familiar with telescopes etc and eq mounts I hadn't considered all the issues and I can see your points. I think I will go for a Celestron SE6 SCT with a 1500mm focal length with an altazimuth mount to achieve what I want, and use the adaptor, the Barlow you mentioned, to get near the 4m I was aiming for. So thank you so much for enlightening me.
  5. Sorry to come straight to the point but action on my part is required immediately. I would like to have a telescope which could be used for astrophotography, particularly of the planets, and hopefully of nebulae as well, but my urgent requirement is to have a very powerful lens for a camera to make a terrestrial observation of targets at about 1km, 2 km, 4km, and 8km distances, each of which will have an angular size of about 0.01 degrees. As I explain below I believe that if I want to have the targets form a 100 pixel high image in a photograph this requires a telescope/camera combination with a focal length of about 3.8m. I believe I can get a camera/telescope combination to achieve this focal length with a Meade 8" SCT as shown below in the pictures and here https://www.ebay.co.uk/itm/8-Meade-SCT-Telescope-Motorised-EQ5-Mount/303633495209?hash=item46b1f774a9:g:cw0AAOSwMmRfHGE2#viTabs_0 The telescope, which has a focal length of 2m is not sufficient by itself, and would be acting as a prime lens i.e. the camera replaces the eyepiece of telescope and would be at the primary focus of the telescope, so I would need a 2x adaptor between the scope and the camera to get the focal length up to 4m. Is this a viable option? I would like to make a number of terrestrial observations with his telescope in addition to the one described, perhaps nature observations, and I think I read on some website that this sort of telescope would NOT be suitable of nature observation, could anyone tell me why? Is it 1) The default 2000mm focal length is too high? So everything is magnified too much? Can one not put a reducing eyepiece on it to halve the focal length? 2) The f/10 focal ratio is too small? 3) It is too bulky/heavy/ difficult to handle 4) Is there a focusing issue? Can this telescope be focused at the distances I described above and could I focus on objects as close as 100m? Additionally, the telescope I am considering comes with a motorised equatorial mount, does this make it difficult to use for nature observations where the telescope would be near horizontal and where I would not be using the motor? Is there a better way to achieve the terrestrial observation with a different method, perhaps just using a horrible Canon P1000 camera which has a focal length of 3000mm? Thanks in advance for any assistance, and sorry for the urgency in my questions. ---------------------------------------------------------------------------------------------------------------------- Here is my calculation for determining a required focal length of the telescope for target of 5 feet high at 5.5 miles and I wish the target to form at least 100 pixels in an image taken by a camera. Specifically there will be a 5 feet high target at 5.5 miles (8800m), so size of target in metres is 5*12*2.54/100 which needs to be divided by distance to the target to get the angular size in radians, and finally multiply by 180/Pi to get degrees, so that gives angular size = 60*2.54/100 / 8800 * 180 / Pi = 0.01 degrees The above calculation does not feature in my calculation below but the angular size I am dealing with is useful to consider. I wish the target to form at least 100 pixels pixel height in an image, and so if using a full size 24mm high sensor in a Canon DSLR camera with the ability to output images of 3700 pixels in height, I believe that gives a magnification (the ratio of image height to target height) of magnification mag = 100 / SensorPixels * SensorY / TargetHeight , where 100 = desired pixel height of target in final image SensorPixels = 3700, the number of pixels vertically in the sensor SensorY = 24mm, the size of sensor in mm vertically TargetHeight = 1.5m This gives a magnification mag = 0.0004 or 1/2312 Using the lens maker equation 1/f= 1/u+1/v, where f = focal length u = image distance, v = object distance, and using that magnification is the ratio of the u and v distances, and v = 8800m, therefore knowing that the magnification is mag = u/v, so u = v * mag 1/f = 1/(v * mag)+1/v that gives the focal length f f = 1/ ( 1/(v * mag)+1/v) putting in the numbers = 1/ ( 1/(8800 * 0.0004)+1/8800) = 3.8m
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