Re: [APML] New All Sky Camera Images

[Aplanatic@aol.com]

Gene,

Oops.  I made the mistake of only doing half the problem.  You are right, the 
focal length of the system shortens by the factor

(1 + 2d/R)

where R is the radius of the sphere and d is the distance from the sphere to 
the lens.

As I stated previously, the aperture is reduced by the factor

(1+2d/R)

and the effective f/ratio does not change.

So, for R= 1 meter and d = 2.5 meters, a 50 mm f/2.8 lens will look like an 
8.33 mm lens with the same f/ratio.

Sorry for the error.

Dave Rowe








I wrote:

> To keep the argument simple, consider the on-axis case.  To find the 
>  effective f/ratio of the system, we need to find the diameter of the 
> parallel 
>  bundle of rays -- from the star that's directly overhead -- that makes it 
>  into the aperture of the camera lens.  There is a virtual focus for these 
>  rays behind the sphere.  In other words, it's as if the star is located a 
>  distance R/2 behind the apex of the reflecting sphere.  So far, so good?  
>  Draw a cone from this virtual focus to the aperture of the camera lens.  
> This 
>  cone intersects the sphere at the effective aperture circle of the system. 
 
>  Parallel light rays from the star that lie outside of this circle do not 
> make 
>  it into the lens.  
>  
>  If you make a quick drawing of this situation you will see that the 
> effective 
>  aperture is much smaller than the camera lens aperture.

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