From: "Dwight K. Elvey" 
To: atm{at}shore.net
Reply-To: "Dwight K. Elvey" 


Hi Jack
 I only get a 8mm 100% spot. Here is my math:

What I want to determine first what the triangle's ratio is that would be
formed by using your secondary, primary and placement. I form a similar
triangle at one side of the mirror that has a vertex right at the
secondary's edge and the inside edge is parallel with the line from the
opposite side of the primary to the opposite side of the secondary.

 ( 413 - 160 ) / ( 106 - 46 ) = 4.2166...
 This is the ratio of the triangle formed by the edges of the
primary and secondary. We can now use this to see how wide the 100%
illuminate spot size is.

  413 / 4.2166... = 97.9

 This is the base of the triangle that would be formed from
the edge of the 100% field to the primary. Since we used a similar
triangle, the inside edge is parallel to the opposite edge of the larger
triangle. This means that the 100% spot size is the difference between this
value and the size of the primary.

 106 - 97.9 = 8.1mm

 So, your secondary is actually a little small, rather than
too large. It helps to draw this out so you can see what I've done. I hope
my descriptions are useful enough. Dwight


>From: "Jack Schmidling" 
>
>I would appreciate it if someone would go over my numbers for the diagonal
>in my 4.25" astrograph.
>
>I do this by laying out a scaled drawing and measuring what I come up with
>but it would be nice if someone could crunch the numbers in one of the
>equations that I don't understand for a second opinion.
>
>Here are the numbers in mm's:
>
>Aperture................................................106
>Focal Length..........................................413
>Focal Plane to center of tube................. 160
>Fully Illuninated Field (diameter).............  12
>
>I came up with............  Diagonal Minor Axis =  46
>
>Actually, that was a bit too big but a readily avaiable size.
>
>How did I do?
>
>
>
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