A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Position-Finding
From: David Pike
Date: 2018 Oct 8, 15:52 -0700
I wrote earlier:
I was rather hoping that someone would tear my geometry to pieces, because I know roughly where I held the camera, and the ratio seemed to reach 45% long before I got down to the 2700km predicted by my maths. DaveP
I realised this morning that I’d overlooked the fact that the photo is of a hemisphere, so the USA is nearer the camera than the Earth’s periphery. I roughed something out, but had to leave it, because I was committed elsewhere until this evening. Below is my revised solution. Fig I assumes SF to Miami is 45% if disc diameter (as measured by me off my computer) and shows what the camera would see neglecting Earth curvature. The line from N through MI gives the position of the lens. Fig II then adjusts for Earth curvature. The line from Q to the disc’s edge defined B. The line from U through MI gives the revised position of the lens T.
Ideally, more iteration is required, but we’re close enough. The camera is approximately 14,000km above the USA. This ties in better with my earlier photos of my globe. I hope you can read the 2B pencil. It’s a bit dim, but it’s easy to rub out. DaveP