Before anything starts, I want to explain AA. AA is my own abbreviation, it’s Amateur Astronomer in full.
What’s the common requirement of an AA’s scope? Unless you are a serious AA with deep pocket, you won’t have an observatory that allows you to mount a gigantic telescope inside. Some AA travels to the dark site with his scope or just move his scope from his house to the backyard. It can’t be too heavy. So, portability comes into one of the major considerations. We want to get the largest aperture within the portability.
8″ SCT has it all. For example, my fork mounted LX90 has an aperture of 203.2mm, 2000mm focal length and weights about 24kg. I can carry it around easily. (Not for a marathorn though)
Left: 8″ LX90 of Edwin Kats from Netherlands.
Continue reading ‘8″ SCT The Best For AA?’
The exit pupil is the image of the objective that is formed by the eyepiece. It’s where you place your eye to see the full field of view. You can calculate the diameter of the exit pupil by dividing the focal length of the eyepiece by your scope’s focal ratio or you can divide aperture by magnification.
Exit pupil (mm)= Fe ÷ f/# (N) | D ÷ magnification
Fe: Focal length of eyepiece
f/# (N): Focal ratio of telescope
D: Diameter of aperture (mm)
Continue reading ‘Exit Pupil’
Astronomers use a system of magnitudes to indicate how bright a stellar object is. An object is said to have a certain numerical magnitude. The larger the magnitude number, the fainter the object. Each object with an increased single number is approximately 2.5 times fainter. The faintest star you can see with your naked unaided eye is about 6th magnitude whereas the brightest stars are negative number magnitude.
The skies and any instrument mentioned are at the best condition.
I would like to know the field of view I can archieve with my scope if I am going to buy an eyepiece, especially a wide angle eyepiece. Apply the formula below, you can get a rough figure of the true field of view (TFOV). Why is it just a rough figure? There are still several factors to be taken account but I think this is good enough. There is a more precise formula to do the same work.
TFOV = AFOV of Eyepiece / Magnification
The apparent FOV (AFOV) of an eyepiece is provided by the manufacturer. To know the magnification, please refer to Calculate Eyepiece Magnification.
Apparent Field of View of Several Eyepiece Designs
Orthoscopic - 43º
Plossl - 50º
Radian - 60º
Panaoptic - 68º
Nagler - 82º
We determine how powerful a telescope is mostly by resolution and the light gathering power. So, what is light gathering power?
Left is a Celestron CGE-800 which has a light gathering power of 843.
The light gathering power of a telescope is the theoretical ability of a telescope to collect light compared to your fully dilated eye. It takes both aperture and your fully dilated eye into factors. The larger the aperture, the higher the light grasp. It is directly proportional to the square of the aperture. To calculate this, first, you have to divide the diameter of aperture (in mm) by the diameter of fully dilated eye (7mm for a normal young man) and then square the result. For instance, an 8″ telescope has a light gathering power of 843 [(203.2/7)² = 843].
The focal ratio is the ratio of the focal length to the aperture. So, a telescope with a 2500 mm focal length, and 250 mm aperture would be an f/10 scope. A scope with a smaller focal ratio always benefits the users in imaging as it shortens the exposure time. Some people will call it photographic speed.
Why it takes more time to have the same result using the same imager but a slower telescope? Let’s say I have a f/5 and f/10 telescopes and both of them have the same aperture. Let’s make them 8″. Now, I move from f/5 to f/10. Since the aperture is fixed, focal length is the only increasing factor. In this case, it is doubled. As the image you see is in 2 dimensions, it needs a least 2 rays of lights to form an area. Therefore, the lights are spreaded into a factor of 4. In conclution, you take 4X more the exposure time to get the same S/N.
*S/N refers to Signal/Noise ratio
Below is the formula to calculate the exposure factor,
Exposure Factor = (Focal ratio 1)^2 / (Focal ratio 2)^2
Once you figured out the exposure factor, you know how much is A scope faster than B scope or vice versa.
To master your imaging platform, you’ll have to know the resolution of your setup. Why should we bother about this? Let’s say, if your setup’s resolution is sub arcsec/pixel (measurement unit for resolution), you need a very stable mount, usually a high end German Equatorial Mount (GEM), for this. As the higher the resolution is (the lower the value of the resolution), any minor defect is easily seen. However, you are able to take high resolution images which are always honoured.
To calculate the resolution, follow the formula:
Resolution (arcsecond) = [CCD Pixel Size (micron) ÷ FL (mm)] * 206
Credit to Webfoot at Cloudy Nights forum for providing this formula. 
Isn’t it a GoTo telescope is capable of tracking celestial objects? Why do I need the autoguiding? ![:-\](http://www.astronomynotes.net/smilies/yahoo_question.gif ":-\")
At first, tracking is just similiar to guiding. For instance in the Autostar Suite from Meade, tracking allows you to define a bright spot (centroid) that it (Autostar Suite) will use to anchor images when combining (stacking) them into one image.
Continue reading ‘Autoguiding By Imager’
What’s the common requirement of an AA’s scope? Unless you are a serious AA with deep pocket, you won’t have an observatory that allows you to mount a gigantic telescope inside. Some AA travels to the dark site with his scope or just move his scope from his house to the backyard. It can’t be too heavy. So, portability comes into one of the major considerations. We want to get the largest aperture within the portability.
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