A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Position-Finding
From: David Pike
Date: 2018 May 9, 01:29 -0700
Fred you wrote: My trig is very rusty and i’m having a hard time solving your problem, yet it is occupying my mind. In the interest of helping me get back to an important project, would you please provide a solution.
For the distance off I was simply using the 1 in 60 rule (or 1 in 57 for the purist). One foot on the circumference of a 57 ft radius circle subtends one degree of arc at the centre. As far as I remember a nautical mile is still 6080 ft, and I guessed a typical warship would be about 300ft long (poor guess, Bismarck was 820ft long!). Therefore, using more realistic values, at ten miles range Bismarck would subtend a horizontal sextant angle of 820 in 10x6080 or 0.82 in 60.8 which is near enough 0.82 in 57 or 0.82 degrees or 49 minutes of arc (or the classroom answer without a tight flying helmet on in a cold noisy aeroplane: angle subtended = 820/10x6080 radians x 57 x 60 = 46 minutes of arc). It’s just like distance-off from lighthouses during yacht pilotage.
For the R34 Airship I feel on less secure ground. I only came across it because I was once asked to read another speaker’s talk for him and had to seriously modify it before I was even moderately happy with it. The quote I had, and I don’t know from where it originates, was Using the sextant they
measured the angle subtended by the length of the shadow; knowing the length of
the ship was 640ft they calculated their height to be 2,100ft – the aneroid
barometer indicated 1,200ft an apparent error of some 900ft, fortunately on this
occasion on the safe side. When they eventually obtained an accurate pressure
setting from a ship some 50-60 miles away they found that the actual error was
1000ft. Thus proving the accuracy of their improvised method. My slides rely upon the fact that, because light rays from an object as far away as the Sun are effectively parallel, the R34’s shadow on the sea will always be 640 ft long, the same length as the airship itself. The problem I soon realised is that over the North Atlantic the shadow is not going to be directly below the aircraft, and it’s not going to be dead abeam, so how’s that going to affect the angle you measure? I couldn’t really get my mind around that, so I decided to try for a graphical approach. I’m still not sure if it solves the problem, fortunately no one in the audience asked for further explanation. If anyone had, I was going to say “Well you’d have to get the Captain to turn the airship until its shadow was dead abeam”. I was happier about that. A note on dip, because dip varies with the square root of the height of the observer the effect of a 10ft error in calculating dip reduces as the height of the observer increases, so an airship height didn't have to be known quite so exactly as the height of an observer on a ship's bridge. Please return to your important project. DaveP