## RE: [atm_free] zone one challenge

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• Jerry, ... I will respond to this right away. I hope to respond later today to all the other comments people have made. I didn t lose track of the right one,
Message 1 of 14 , Apr 1, 2005
Jerry,

> I chose "I" as my favorite. In my defense on this error, I find it
> interesting that when I look at your intensity plots across the openings,
> that at least to my eye window "I" is more symmetrical. Symmetrical both
> in left and right intensity curve shape and the cross-over is in the
> center.
>
> ...........Are you sure you didn't loose track of which is the right one?

I will respond to this right away. I hope to respond later today to all the

I didn't lose track of the right one, but which one is right depends on how you
define right. I chose to use the root-mean-square method for determining the
zone center, which puts the best null a little out from the middle of the
aperture. I then used this null radius to compute the stage position for the
zero point of the simulation. If you look at the intensity ratio in the "answer"
diagram, J is .97 and I is 1.05, if I remember correctly. Thus, the location of
perfectly balanced L/R intensity is about 1/3 of the way from J to I.

-- Steve Koehler
steve_koehler@...
• So what you are saying is that the image I is more correct when you take the center of the zone (Rsub2 + Rsub1 / 2) than what J is? I ve always assumed that
Message 2 of 14 , Apr 1, 2005
So what you are saying is that the image I is more correct when you take the
center of the zone (Rsub2 + Rsub1 / 2) than what J is?
I've always assumed that the midpoint between the two radii was the proper
location for the zone center.

Bob May
bobmay@...
http://nav.to/bobmay
http://bobmay.astronomy.net
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