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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Distance off with Chicago buildings-2
From: Bill B
Date: 2005 Dec 1, 20:15 -0500
From: Bill B
Date: 2005 Dec 1, 20:15 -0500
Frank Appreciated the URL dealing with refraction. Using his formula of distance in meters^2 x 2.55 x 10^-8, Sears would be lifted by 153 feet and Hancock by 158 ft Recalculating the angles with that refraction correction I obtain: Sears 0d 30' 19.26" Hancock 0d 21 38.36" Diff 0d 08 40.90 That leaves calculated targets 0.5' below observed. So many unknowns and fuzzy variables. Correction to pressure of 0.983 at lake level and 0.941 at Sears roof line, heat from the mills and Chicago, exact height of base above lake level., (I did find a topo map for Chicago, and on the face of things Sears *may* be at 595 above sea level, and Hancock slightly lower--who knows?) Plus the accuracy of d = 1.17 sqr rt (H-h) in calculating visible height. Distances calculated with Bowditch with above angles and height adjustments (nm): Sears: 23.05 nm Hancock: 23.49 nm Out of curiosity I calculated the amount of building (Sears) which would be hidden by curvature with no refraction. Height of eye 0.0, distance 23.08 nm. If I did my homework correctly 470 feet would be hidden. If refraction lifts the image 153 feet, anything over 317 feet could appear above the horizon. Compare that to Bowditch where h is 0: 23.08 nm = 1.17 sqr rt (H-h)= 389 feet. If my thinking is straight, 1454 ft (Sears) + 30 ft (base above water) + 153 ft (lift from refraction) =1637 ft. Subtract the hidden part (471 ft) and the observer could *see* 1166 ft of the structure. Using 1166 ft and a distance-to-base of 23.08 nm, the calculated angle (height of eye 0) would 0d 28' 35". Plug that into the Bowditch formula and calculated distance would be 22.2 nm. Ignoring the predicted lift from refraction the visible height would be 1013 ft. with an angle of 0d 24' 50" The distance, calculate with Bowditch, would be 24.1 nm. Adding dip back into your 30.8' figure to correct it to 0 ft height of eye, we have 31.1', compared to calculated (with generous refraction lift as the base is approx. 595 ft above sea level) of 28.6', we find your observation 2.5' high. Once again your concerns over Bowditch's refraction adjustments seem warranted. This is the point where the hog runs out of acorns. It would also appear that in addition to problems with Bowditch, refraction is above predicted levels? Thanks for an interesting exercise and learning experience. Bill