NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: sextant index error measurement
From: James R. Van Zandt
Date: 2006 Nov 11, 21:45 -0500
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 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 --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To , send email to NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---