RE: ATM Spider Design

["Alan Levin" <ailevin@netcom.com>]

> From:          andydtg@alpha1.phoenix.net
> Date:          Mon, 19 Aug 1996 16:55:35 -0500
> To:            ailevin@netcom.com
> Subject:       RE: ATM Spider Design

> Hi Alan,
> I agree with the above except for the assertion that the thickness (area)
> of objects in the aperture somehow affects the amount of diffraction. There
> may be some
> second order effect of varying the location of the diffracting edges relative
> to each other or to the other existing edges of the imaging aperture but I'd
> like
> to see a quantitative analysis before I'll believe that a 'thin' spider has
> less diffraction than a 'thick' one.
> 
> > The area determines how much light goes there.
> 
> I really don't understand why this would be so. If I'm missing
> something I hope you can explain; one thing I've learned in my
> 40+ years in this hobby is that there's always someone who
> knows more than you in any particular area :>)

Andy,

Here is the way I figure.  If you had a clear circular aperture, you
would get a diffraction pattern that is the Fourier Transform (FT)
of that that aperture.  (It turns out to be a Bessel function.)  If
you looked at this diffraction pattern it would have a peak in the
center and then come down on either side of that center  into a
trough (negative) that was much less deep than the central peak and
then go positive to a peak much less high than the first trough was
deep and so on as the peaks and troughs get smaller and smaller in
amplitude toward the edge.  The peaks and troughs are evenly spaced
according to the ratio of the wavelength of the light to the
diameter of the circle.  The heights of the peaks and troughs would
be related to the area of the circle.  The basic principle here is 
that a large diameter gives closely space peaks and troughs of large 
amplitude, and a small diameter gives broadly space peaks and troughs 
of small amplitude.  Our measure of close and small is the wavelength 
of light.  If you squared that diffraction pattern you would see the 
Airy disc pattern for that aperture and wavelength.

The way you deal with a blockage, either your secondary mirror, or a 
spider, or an apodizing mask of some sort, is to take the FT of the 
blockage and subtract it from the FT of the clear aperture.  
Thankfully these things are linear.

If you put a circular blockage in the center of your clear aperture,
you calculated the FT of the blockage and subtract it from the FT of
the clear aperture.  Because the blockage is smaller in diameter than
the clear aperture, the first peak will be broader, as will the
spacings of the other peaks and troughs.  Because the blockage is
smaller in area, the heights of the peaks and troughs (really the
area under the peaks and troughs if you sqaure them) will be smaller
as well.  This is where the area factor for a blockage comes in.

Now imagine a .333 by diameter blockage.  The diffraction pattern of
the obstruction will have a  main peak three times as wide as the
clear aperture pattern.  Although this obstructed pattern will be
reduced due to the smaller area of the obstruction, when you subtact
it, you will be subtracting from the first peak of the clear
aperture, and making the first trough of the clear aperture more
negative.  According to how the peaks and troughs of the obstruction
line up with the clear aperture, you will make various peaks and
troughs either stronger or weaker.  Since you get the intensity by
squaring the resulting pattern, you see more light in the first
diffraction ring of an obstructed telescope and less light in the
central spot.

For a spider vane you have to calculate the FT of a rectangle of
length equal to the length of the vane and width equal to the
thickness of the vane.  The FT of this thing is going to be made up
of things that look like [sin(x)/x]*[sin(y)/y].  They will once again
have that peaky and troughy charater with the oscillations damping
out.  Again, a wide rectangle will have narrowly spaced peaks and
troughs and a narrow rectangle will have broadly spaced peaks and
troughs.  Since most spider vanes are thin compared to the aperture,
when you subtract the diffraction pattern of the vane, the vane's
broad main peak will be quite wide compared to the Airy disc.  That
is why we see a wide diffraction spike compared to the Airy disc.

Sorry this is so long winded, but you see that the thicker vane will
have slightly narrower spikes (they will still be long compared to
the Airy disc) but there will be more energy in the spikes since the
spikes block more area.
'
Hope this helps,

Alan

P.S.  I left out why the diffraction pattern is the FT of the 
aperture, as that would take another lengthy post.