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    Re: sextant index error measurement
    From: James R. Van Zandt
    Date: 2006 Nov 11, 21:45 -0500

    I hope you'll forgive me for falling a bit behind in studying these
    postings.  Back on Nov 5, George Huxtable wrote:
    > I wrote, in NavList 1588, about estimating the offset between the
    > two views from a sextant, when it's set to zero on the arc.
    > | Or you can do it by geometry. For that, you will need the
    > | effective distance d between the parallel reflecting surfaces of
    > | the two mirrors. You can measure this between the upper part of
    > | the horizon glass and the lower part of the index glass, where
    > | there will usually be some overlap between those parallels. If
    > | they are front-silvered, that's easy. Otherwise, if you want to be
    > | precise, you will have to add two-thirds of the combined thickness
    > | of the two glasses; the two-thirds factor allowing for the
    > | refractive index of glass.
    > |
    > | And you need the angle A by which the horizon mirror is tilted
    > | from the plane that's at right angles to the collimation line of
    > | the telescope. You should be able to estimate this using some sort
    > | of protractor (or course plotter). Then the spacing between the
    > | sightlines is then d x cos 2A.
    > But it isn't, I'm afraid. It's d x sin 2A, if A is defined in the
    > way I had suggested.
    Actually I think it's 2 d sin A.
    If d = distance between the two parallel front-silvered reflecting
       p = distance between the spots on the two mirrors where an
           initially horizontal ray is reflected
    then the offset after the two reflections is
       s = p sin(2A)
         = (d/cos A)(2 sin A cos A)
         = 2 d sin A
    I've posted a diagram at
    Note, it is not necessary to determine the offset very accurately.  I
    like George's earlier suggestion:
    > Put some sort of boldly-marked ruler at a convenient (short; a few
    > inches perhaps) distance in view of the sextant, so that two views of
    > it can be seen, one through the index mirror, one through the horizon
    > glass, and record the offset between them. ...It's dead easy to get the
    > answer to a millimetre or so.
    Suppose we take a sighting of a horizontal line at a distance l, and
    we read a value Hs off the sextant.  Then the index error is
       IE = Hs + s/l
    where s is the offset, and both IE and Hs are in radians.  Now suppose
    our estimation of s has uncertainty ds = 1 mm and we want to estimate
    IE with an uncertainty dIE < 10".  Then
        dIE < ds/l
             ds    (0.001 m)(60"/')(60'/deg)(180 deg)
        l > ---- = ---------------------------------- = 20 m
            dIE       (10")(pi rad)
    Actually we would want the target a little farther away, to allow a
    few percent uncertainty in l as well.
                        - Jim Van Zandt
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