NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Theoretical Question - Height of Eye from Shore
From: Greg R_
Date: 2007 Nov 16, 21:24 -0800
From: Greg R_
Date: 2007 Nov 16, 21:24 -0800
--- frankreed@HistoricalAtlas.net wrote: > Myself, I usually just walk over to the water and look. That's the easy (and obvious) solution, but in this case the water's edge is a couple hundred feet from the parking lot, and I'm loath to drag all of the celnav gear that far over the sand (and risk dropping it), not to mention needing to also carry a chair, table, blanket, or something else to set it all down on while reading the time and recording the sextant reading. > I could imagine this issue could cause some confusion if you're > taking sights once a week on your lunch hour... this Tuesday, it's > high tide, next Tuesday it's low tide, the following Tuesday, high > tide again... Yabbut, I thought we'd all agreed that the "height" of the visible horizon should be the same as that at the shoreline. > You could easily get a "mysterious" fortnightly "error" of 2 > or 3 minutes of arc. Yep, that's what prompted the question in the first place. Last time I was out there taking sights I seemed to get some sort of "systemic error" (referenced to a known GPS position) that I couldn't account for. Index error is always dialed to zero on both the Davis and the Astra sextants, and I couldn't believe that my technique had suddenly gotten that "sloppy". The only thing that I could think of that might have affected the accuracy that one time (and consistently, too) would be either a different He than I usually used or some sort of anomalous refraction condition, and simple logic would tend to point to the former rather than the latter... ;-) > When you don't have a table or reference with the right formula > handy, the best short rule to remember is "dip is square root of > height of eye in feet" (example: height of eye = 25 feet, dip = 5 > minutes of arc, nearly). And, not by coincidence, the distance to > the horizon follows a similar rule, "distance to horizon in nautical > miles is square root of height of eye in feet, nearly." Thanks for that info - seems like I remember seeing a formula for it (probably the same one) years ago when I was first learning this stuff, but it's been long enough (and something that never got used in the real world) that I don't remember it today. -- GregR --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To , send email to NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---