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From: george huxtable (no email)
Date: Fri Jul 15 2005 - 07:17:14 EDT
>Peter Fogg wrote:
>
> > Have found this one: (Azimuth formula)
> >
> > Chuck Pettis' Azimuth equation:
> > AZ = acos ((sin D - (sin L * sin H) / (COs L * COs H))
> > where H = horizon height (degrees)
> >
> > (The H has me puzzled. Perhaps it refers to Dip, or could it refer to
> > altitude?)
No, it's simply the (corrected) height of the body above the horizon, in
degrees, same as h in the expression given below..
> > Here's another:
> >
> > Z = cos^-1 * [sin Dec - sin Lat * sin h / cos Lat * cos h]
> > where h = vertical angle to the sun corrected for parallax and refraction
> > (h = altitude?)
No, it's not another, it's exactly the same expression, just expressed in a
slightly different way.
>I think they are the same sine method.
No, they both derive azimuth from its cosine, not its sine. That expression
(unlike the expression that calculates azimuth from its sine) has no
difficult ambiguities about it. However, it becomes imprecise for azimuths
near North and South, and (in that case) VERY dependent on the precise
value of the body's measured altitude. Indeed, in some circumstances, near
North and South, a small error in altitude can result in a value of cos az
that exceeds 1, for which there's no solution. The great advantage of the
tangent formula is that it doesn't even call for a measurement of altitude,
at all. Also, it maintains its precision for all azimuths.
>The first version may be for terrestrial navigators, with its added factor
>of 'horizon height'.
No.
>Along the way, I found a formula for the calculation of the Sun's
>declination:
>
>Dec = 23.45 sin (360/365.25)
>Its such a simple formula even I can understand it. Its the maximum
>declination of the sun expressed as a proportion of its change.
Peter may claim to understand it, but I don't. Nor do I understand his
description of what it does. As it stands, that formula calculates the
Sun's declination approximately one whole day after the moment of the
Vernal equinox, to arrive at an answer of 0.403 degrees North. Something
is missing. The quantity 360/365.25 should be multiplied by the number of
whole days elapsed since the vernal equinox, to get an approximate answer
for the Sun's declination, expressed in degrees (not as a proportion of its
change).
>Here's another version:
>
>Dec/23.45 = sin(0.985*t)
>
>0.985 is a truncated version of (360/365.25)
>and t = the number of days from the vernal equinox
>or
>t = (inv sin(Dec/23.45))/(360/365.25)
>
>These formulae come from:
>www.geomancy/org./sunfinder
Those formulae are defective, as Fred Hebard has pointed out, because they
ignore the elliptical nature of the Earth's orbit round the Sun. Just
because information can be found on a website does not imply that it is
accurate or reliable, or even true.
George.
===============================================================
Contact George at ,or by phone +44 1865 820222,
or from within UK 01865 820222.
Or by post- George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13
5HX, UK.
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