Re: (ATM) Coma and Telescope Design

[Chuck Grant <grant@aretha.llnl.gov>]

> (unless I get a bigger trunk.)  Chuck's comment "Not with a 20% 
> obstruction diagonal and NGF-3 focuser" with respect to my desire to 
> achieve 70% illumnination of an approximately 1.4 degree true field  
> has me a bit worried that I am not using the NEWT/Vignette/Diagonal 
> software correctly.  I'm figuring on a 9.5" inside diameter tube 
> that is .25" thick with a .25" velvet lining for controlling light 
> reflection (barring input that says I need additional/different 
> baffels).  I think I can put the focal plane 2" above the tube with 
> the 2" diameter JMI focuser and still have a little in travel (no 
> coma corrector to worry about anymore.)  Am I incorrect here?  With 
> NEWT I get a 1.25 degree diameter field with 75% illumination with a 
> 1.6" diagonal, and with Vignette I get a 1.4 degree diameter field 
> with 70% illumination.  Diagonal gives results that are consistent 
> with these.  I'm working with a pupil diameter of 6mm which would 
> allow me to get one of the 25-35 mm eyepieces with wide (~68 degree) 
> apparent fields.  Any thoughts, or was Chuck referring to the 
> infeasability of putting the heavy 14mm UWA on the focuser with the 
> (now unneeded) coma corrector?  Or am I incorrect on the height of 
> the racked in JMI focuser above the tube?  I thought I had read it 
> was about 1.5" - 1.75".
> 

You may have trouble with tube currents with only 0.5" between
the edge of the mirror and the velvet lining.  I would suggest at
least 1.5 inch. With a 52" tube with a 9" id. (incuding the velvet)
and 1.25 degree field of view, you are already vignetting 
against the end of the tube.

avaliable space = ( 9" id - 8" mirror ) / 2 = 0.5 on each side.

expansion of light cone = 52" tube * sin ( 1.25 field / 2 )
                        = 0.567" on each side.

I did this back of the envelope calculation from first
principles.  You are welcome to check my math.  I ignore
the central obstruction, which would only make the calculated
70% illuminated field smaller if considered.

You are placing the image plane 7" from the optical axis.
The focal length is 52" and we are at f/6.5
The minor axis of the diagonal will be 1.6" (20% of 8")

A point on the image plane receives light from an elipise on the
diagonal with a minor axis of 7" / 6.5 = 1.077"

So you have ( 1.6" - 1.077" ) / 2 = 0.2615" on both sides with
which to play for full illumination.

0.2615" at the diagonal (52 - 7 inches from the primary) gives 
2 * arcsin ( 0.2615 / (52 - 7) ) = 0.666 degrees, so you have a
fully illuminated field of 0.666 degrees.

I figure that when the 1.077" circle is offset from the center of
the 1.6" circle by about (mumble, algebra, mumble...) 0.3879" you
get 70% coverage.  Ok so this step is a little hard to check.
Check your CRC Standard Mathematical Tables for the formulae for
the area of a sector of a circle.  
Area = area(1.077"circle) - area(sector of 1.077circle) 
                          + area(sector of 1.6"circle)

2 * arcsin ( 0.3879 / ( 52 - 7 ) ) = 0.988 degrees

So did I make a mistake here somewhere?

Chuck