From: Jerry Hudson 
To: atm{at}shore.net
Reply-To: Jerry Hudson 


Oops -

The derivation in my Apendix contained a couple of typos; will repeat here,
hopefully correctly:

Appendix:  Derivation of L.A. for conic section under test

From Herzberger's formula for a conic section:

    c K z^2 + 2 z - c r^2 = 0

where
    c = Curvature = 1/R, R the vertex radius of curvature
    K = Conic constant = e^2 - 1
    z = Sagitta
    r = Zone radius

Note that this can represent any conic section that is a surface of
revolution about the axis, with the vertex at r=0, z=0.

Differentiate the above with respect to r:

    dz/dr = c r / (1 + cKz)    [was erroneously c r / (R + cKz)]

Note that this is also the slope of the surface normal at r, which is equal to

    slope = r / ( R - z + L.A. ).

It helps to draw a diagram, which I can't submit to the List,
unfortunately.  Here's an attempt:

         (Point r,z)
       *-      Surface normal:
      /  \-    Slope = r / (R - z + L.A.)
     | |   \-
     /       \-
    |  |       \-
   -A--B------C--D

where the points
    A = Vertex of curve
    B = Point directly under the zone
    C = Vertex center of curvature
    D = Normal (ray) intercept
Note that AB = sag = z
          AC = radius of curvature, R
          CD = L.A.
          B* = r
          slope = B* / BD

    Equating the two slopes, we obtain

    r / (R - z + L.A.) = c r / (1 + cKz),
                    [above orig. had R + cKz in denom.]

[I managed to get it right here:]
    L.A. = -R + z + (1 + cKz)/c

        = (1 + K) z = e^2 z (Q.E.D.)

- Jerry Hudson

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