[Loadstone] solar calculations

Paul Woffenden paul.woffenden at talktalk.net
Thu Oct 25 14:47:27 BST 2007


Hello Shawn and All,
Most of the information from your program looks very useful. Not sure what 
units the heliocentric (sun-to-earth) distance is in though; this is 
presently just less than one Astronomical Unit which is 93 million miles.
As regards a solar azimuth, the 24-hour day is the time taken for the Earth 
to rotate once; which results in the Sun completing a circle of 360 degrees. 
If we divide 360 by 24, we of course get 15 degrees per hour; which would 
give you an approximation for solar azimuth -- 360 degrees at midnight and 
180 degrees at noon. However there are many factors which pull and push the 
azimuth to either side of this average value of 15 degrees and I am 
currently trying to find formulae which take these into account. The best 
program on a computer for really exact solar/lunar information is RedShift; 
which has the advantage that it outputs information numerically as well as 
pictorally so it works with a screen reader such as JAWS.
I am tweaking my Pascal code to try to get my values to agree with those 
output by Redshift and will let you know as soon as I have any more 
information.
Really precise astronomical calculations use matrices; but my mathmatical 
knowledge isn't really up to this - I'm a lawyer not a mathematician hehe.
The best book on these formulae is called Positional Astronomy; but my 
scanner won't read the symbols so its rather a slow process.
Hope this helps.
Best regards.
Paul Woffenden
----- Original Message ----- 
From: "Shawn Kirkpatrick" <shawn at odyssey.cm.nu>
To: <loadstone at loadstone-gps.com>
Sent: Thursday, October 25, 2007 9:14 AM
Subject: [Loadstone] solar calculations


> This message is for those people that want Loadstone to have a sunrise,
> sunset, moon phase, etc. calculator.
> I have some code on the pc that outputs a lot of information about these
> things. Not knowing much about this stuff I'm not sure what we'd want as
> output or the best way to represent it.
> The output from my program is as follows:
> local time: wed oct 24 23:52
> timezone: -700
> latitude: 49.2251716
> longitude: -122.9915750
> moon age: 14.08
> moon visible percentage: 98.70
> moon cycle percentage: 46.37
> moon is visible
> today's moonrise: 17:08
> today's moonset: 05:47
> moon azimuth: -3.42
> moon altitude: 51.15
> moon distance: 56.16
> moon right ascention: 01:02:34
> moon declination: 10.41
> sun distance: 23350.68
> sun right ascention: 13:57:20
> sun declination: -11.99
> Eccentricity: 0.02
> Obliquity: 0.41
> Ecliptic Long: 3.69
> today's sunrise: 07:47
> today's sunset: 18:05
>
> Some of the information is very obvious but some just isn't. There's a 
> moon
> asimuth but no sun azimuth. Is it possible to calculate this from some of
> this other information?
> Any help or suggestions on how to use this information would be welcome.
>
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