A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: David Pike
Date: 2015 Feb 9, 14:05 -0800
The question asked:
How would early inland explorers/navigators determined a noon latitude when the sun was too far south ,for instance, making the use of an artificial horizon impractical.
I currently use a fence at home during the summer months but carrying a fence around the countryside is also impractical, plus the neighbors might object.
Excuse the ignorance but my knowledge of celestial nav is rather limited.
Hello Frank. I was going to sit on the fence on this one, but I can’t resist a good problem. What did you mean the sun is so low it makes the use of an artificial horizon impractical? Did you mean you couldn’t see over the edges, too many ghosts, a distorted sun, or you have to sit so close to the ground you risk getting ants in your pants? I had all of these problems at Christmas, except for me it was a wet bum off the snow. I got around that by raising the artificial horizon onto a table as Amundsen appears to have done in the classic photo in Antarctica. If you’re just practicing, I can’t see anything wrong with using a fence. You know were you are, so you could get an idea of the accuracy. I’ve used the roof of a house about a quarter of a mile away just for practice. What sort of accuracy do you think the travellers would be looking for? About 40 years ago, we were visited at the RAF School of Navigation by a chap who was planning to push a wheelbarrow across the Sahara wanting to be shown how to use a Mk IX bubble sextant he managed to get hold of. That should have given him about three miles, if the works didn’t get jammed with sand. The ancient astronomers like Tycho Brahe made everything very big and used a plumb bob for verticality. How about using a stick about 1.1m long with a mark at 1.0m? Push it into flat ground down to the mark and check verticality with a plumb bob. Then measure the length of the shortest shadow. If the shadow is 4000mm, tan-1 height of Sun is 1000/4000 = 0.25 = 14.036°. If height of stick was actually 1010mm and length of shadow was only 3990mm height of Sun would actually be tan-1 1010/3990 = 0.2531 = 14.205° . Difference = 0.169° = 10.14minutes = 10.14nm. It’s worth experimenting with. There were of course all the usual things you see in books: Astrolabes, plumb bob quadrants, astronomical rings etc, all of which relied upon gravity for a vertical and worked just as well on land as at sea. Dave