Hank,
Yes, thanks indeed. Your explanation has really helped me visualize
what's going on.
It seems that trying to find South or North from the moon isn't a great
method, but perhaps Bill's modification would help? Anyway, as a last
resort any compass is better than none. I have twice in my life become
severely disorientated when hiking. These days I keep a magnetic compass
permanently attached my knapsack so that I can't forget to bring it, but
on both those occasions I had to resort to using whatever I could see in
the sky to try to get my bearings.
It seems that the time measurement idea I set out in my earlier email is
a dud. A bearing error of 20+ degrees could result in up to a couple of
hours error in the time estimate. Actually, as well, for the sun and
moon to both be above the horizon at the same time doesn't happen that
much overall.
Steve
On 12/05/2014 12:31 PM, Bill Gottesman wrote:
Thanks, Hank. Very helpful observations. The ecliptic north pole
lies in the curve of Draco's (The Serpent) neck. From your
explanation, it seems that the line connecting the crescents of the
moon should always point approximately to that location, and this
should be something easy to test in the night sky. Is there a way to
use this information (perhaps the time of year) to help refine finding
south by the moon's crescent? -Bill
On Mon, May 12, 2014 at 3:07 AM, Hank de Wit <h.de...@bom.gov.au
<mailto:h.de...@bom.gov.au>> wrote:
HI Steve,
I think I can answer this one approximately. The maths is also
beyond me, but we can get an intuitive answer without causing too
much brain strain.
The first point to remember is that both the Sun and the Moon
travel on paths nearly along the Ecliptic. The Sun sits exactly on
the Ecliptic, and the Moon, deviates plus or minus 5 degrees,
because it's orbit is inclined by 5 degrees to the Ecliptic. This
means that that shadow of the terminator between light and dark on
the moon must be aligned nearly perpendicular to the path of the
Ecliptic in the sky - they are in the same plane. So the problem
reduces to the angle that the path of the Ecliptic makes in the sky.
To reduce variables even more, let's just think about the Moon
when it is highest in the sky, along the meridian through South
(North in the SH).
We need a planisphere to visualise, and I found a nice online one
here:
http://drifted.in/planisphere-app/app/index.xhtml
This planisphere has the Ecliptic marked as a blue line in the
sky. If you rotate the outer disk to move through the months, and
imagine the Moon along the Ecliptic and sitting on the north-south
meridian you can clearly see the tilt of the Ecliptic line, and
therefore the line through the horns of the Moon if it were
located at that point. You can see that this line is not directly
through north for most of the year, and can be either side. The
biggest deviations are at the two equinoxes. It is pointing south
(north) at the solstices. I wonder if the amount of maximum
deviation from due south (north) is plus and minus 23.5 degrees.
Many regards
Hank
-----Original Message-----
From: sundial [mailto:sundial-boun...@uni-koeln.de
<mailto:sundial-boun...@uni-koeln.de>] On Behalf Of Steve Lelievre
Sent: Sunday, 11 May 2014 12:22 AM
To: sundial@uni-koeln.de <mailto:sundial@uni-koeln.de>
Subject: Using the moon to find south
Hi folks,
Only loosely related to my question just posted, I'm interested to
know more about a primative navigation method I've read of. The
idea is that if one projects an imaginary line through the cusps
of a crescent moon down to the horizon, that gives the approximate
position of South (or perhaps North depending on your hemisphere).
How accurate is this position compared to true south? I'm guessing
it depends on the time of year, phase of moon and latitude - can
any one supply formulae? Working it out from first principles is
beyond my math ability.
I'm thinking that if I can use the moon to find south, I can then
measure the azimuth of the sun and use that to get time of day...
Thanks,
Steve
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