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My first Good Jupiter Sept 18.


Atlas

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After 4 months of collecting dust in the garage, I finally pulled out the 254mm Reflector. Glad I did...cirtainly my best to date.

5x powermate/3x digital zoom on the 600D. Seeing was fair. Image is 880 frames. I have a few more with 4000+ frames but having onion ring issues I'm working through.

Top image is 100% size. Definatly much better detail than my little 6" :cool:

Edit* The images are taken from a single, color .AVI.

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That's a size of an orange. :eek: What is the total focal length?

I honestly have no idea. If anyone knows of a caculator for that, I would be grateful.

You can't beat extra aperture! Very good image with that camera setup! I prefer the smaller image. :smiley:

Couldn't agree more...looks like the 152mm will be retired and stored for the littles ones to use when they feel like viewing when they get older. (And saved as a portable vacation scope).

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I honestly have no idea. If anyone knows of a caculator for that, I would be grateful.

1200 telescope focal length x 5 Powermate would be 6 meters of focal length but not sure how 3x digital zoom applies to that. 18 meters would be too much I think, it must be something else than just multiplying it three times.

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with a dslr ! great result.

Now,:-

Measure the equatorial diameter of the planet (rotate it first so it's equator is horizontal) in pixels. Then use a program like WinJupos which will give you the apparent diameter of the planet in arcseconds at the time of the observation.

Divide this figure( arcseconds) by the measure number of pixels in the diameter of the planet to give you your image scale in arcseconds/pixel - call this value A

To calculate the effective focal length, use this formula

F = 206 x Dp /A , where Dp= the size of your camera's pixels in microns.

so as an example Jupiter measures 360 pixels wide. The diameter of the planet was 49.1" , so this would give an image scale of (A) of 0.136"/pixel.

the size of the camera's pixels was 7.4microns, so the calculation is .........

F= 206 x 7.4 / 0.136

F= 11209

If the scope diameter is 350mm, this gives an effective focal ratio of f/32

Simples !!

courtesy of Pete Lawrence . many thanks Pete don't know where you are

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