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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
       surfaces
   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
http://jrv.oddones.org/path-offset-calculation.png


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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