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RE: [atm_free] zone one challenge

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  • Koehler, Steve
    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
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      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
      other comments people have made.

      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@...
    • Bob May
      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
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        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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