Can someone explain focal length calcs please

[KE400]

Hi

This is probably a silly question but I'm wondering what the calculations of focal lengths are and how to work out what people are using in kit to achieve magnifications etc.

I'll try and explain a bit more, I've been looking at some really nice planetary images which state they where taken by the same scope as mine, (C9.25) and that they where taken at F27 & F20 etc?? I think this must be to do with the focal ratios but not sure what they are using to get to those magnifications.

The focal length of the scope is 2350mm and the aperture is 234.95mm, how then do you calculate these F27 & F20 figures???

Any help would be much appreciated

[Cornelius Varley]

Focal ratio is the focal length of the telescope divided by the aperture. To work out the effective focal length of a telescope multiply the aperture diameter by the focal ratio. A C9.25 at F27 would have a focal length of 234.95x27 = 6343.65mm.

Peter

[KE400]

Thanks Peter for explaining that but how is the for example, the 6343.65mm then achieved? If you added a Barlow x2 or x3 to the scope with a 6mm EP say, is there a formula of calculating the required EP required to achieve the 6343,65mm??

[Cornelius Varley]

The effective focal length is achieved by using a barlow lens. A x2 barlow lens effectively doubles the focal length of the telescope, 2350 x 2 = 4700mm, 4700/235 = F20, a x3 barlow - 7050/235 = F31.

A SPC900 webcam gives an equivalent field of view to a 6mm eyepiece. A C9.25 imaging at F20 (FL=4600mm) gives a similar field of view to a 6mm eyepiece at the same focal length. This would give an equivalent eyepiece magnification of 4700/6 = 783x. This magnification would not be possible because the telescope's useful magnification is about x470 in good seeing conditions. At an effective focal length of 6343mm a 6mm eyepiece would be giving a magnification of 1057x.

Don't confuse the apparent "magification" of a webcam with the optical magnification of an eyepiece. When you increase the focal length of a telescope with a barlow lens you reduce the field of view of the webcam. It appears to give a greater "magnification" but, in reality, you are reducing the area of the sky/moon you are imaging.

Telescopes have a practical maximum magnification of 2x per mm or 50x per inch of aperture in good seeing conditions, it is usually nearly 30x per inch.

In imaging this rule does not apply because webcams don't magnify. So apparent magnifications in imaging can easily be higher than the practical maximum optical magnification.

Peter

[KE400]

Thanks again Peter, thats a really good explanation of it all.

Much appreciated