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Reflector CO math?


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I hate maths and might regret asking this...

But is there a simple formula for working out the light loss percentage due to the telescopes CO? For example a newt with

  • 750mm focal length
  • F5
  • 150mm aperture
  • 50mm secondary mirror

 Thanks for info!

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Ratios of areas will get you that:

Light collecting area: 75mm ^ 2 * PI

(half diameter squared multiplied by PI)

Light blocking area: 25mm ^ 2 * PI

(same but this time take diameter of secondary)

Divide the two and multiply by 100 to get percent of light blocked by secondary.

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Should simply be:: Take the secondary size (50) divide by the mirror size (150) =0.33333, then take the square of that =0.1111.

Percentage then is simply x100 so 11.1%

It assumes tha the 50mm secondary is treated as a circle of 50mm diameter which it should obstruct light as since it is an elipse at a 45 degree angle. Guess it will always be a bit of an approximation.

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1 hour ago, vlaiv said:

Ratios of areas will get you that:

Light collecting area: 75mm ^ 2 * PI

(half diameter squared multiplied by PI)

Light blocking area: 25mm ^ 2 * PI

(same but this time take diameter of secondary)

Divide the two and multiply by 100 to get percent of light blocked by secondary.

Yep I did regret it! Er...what`s the answer?!

1 hour ago, John said:

This piece might go a little deeper than you are interested in but it might be worth a read:

http://www.cloudynights.com/page/articles/cat/articles/optical-theory/aperture-comparision-r1901

Thanks John, interesting article, so not just a numbers game if trying to evaluate overall performance.

32 minutes ago, ronin said:

Should simply be:: Take the secondary size (50) divide by the mirror size (150) =0.33333, then take the square of that =0.1111.

Percentage then is simply x100 so 11.1%

It assumes tha the 50mm secondary is treated as a circle of 50mm diameter which it should obstruct light as since it is an elipse at a 45 degree angle. Guess it will always be a bit of an approximation.

That kind of tallies with the CN article. Thanks that looks a lot clearer to understand.

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