A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Position-Finding
From: Fred Hebard
Date: 2018 May 10, 21:03 -0400
On May 10, 2018, at 11:01 AM, David Pike <NoReply_DavidPike@fer3.com> wrote:
Your slides 2 and 3 had a central ray originating at the location of the observer, defining the height of a right triangle with two dashed-line sides; that is not there. Assume the observer is in the center of the ship. His position defines the angle between the ends of the shadow. The right triangles are defined by his two observation lines of sight.FredI made the slides about ten years ago for more effect than accuracy. Looking at slide 2, I realise I drew it for an into-Sun course so that I could show a side view of the R34 rather than a cross section. In fact, downloading EM Maitland’s ‘The log of the R34’ from Kindle, which is more a diary than a navigational log, he says the shadow was on the starboard side almost immediately underneath. The actual position of the shadow can be worked out from the course, time, date, and approximate position. First thoughts are that it couldn’t have been more than just over 60 degrees down, but I’ll work this out when I next get chance. Then, assuming the shadow was almost beam-on (they were heading west and it was around noon), it shouldn’t be too hard to calculate airship height using geometrical drawing from schooldays many decades ago. The fact that the observation cabin was nearer the front than the rear of the airship will complicate the drawing slightly. DaveP
The following assumes a flat earth and blimp parallel to that. The flat earth assumption would not affect the results, much like its lack of effect on lines of position. If the blimp were pointing in the direction of the sun on departure of the sun from zenith, the shadow of the blimp would decrease in length until it reached zero at sun set, assuming the blimp is a straight, one-dimensional line. Of course the shadow would not be cast upon the earth at sunset, but it would be zero. I would think the length would be decreased by the cosine of the zenith angle. If the blimp were right angles to the sun, the shadow would not decrease in length except that the distance to the shadow would increase with departure of the sun from zenith. i expect the increased distance to the shadow would scale as the sine of the zenith angle. Now combine these two guys and you’ve got it.
Regarding the flat earth and parallel blimp, both of those factors would decrease the length by the cosine of the angle. One minus cosine is quite small for small angles, less than 0.02 for 10 degrees, so these corrections can probably be ignored except in more extreme cases. I haven’t thought about the effect of a two or three dimensional blimp.