permaculture

[Calum Ewing <calumewing@eastlink.ca>]

Hello Declan,

Greetings from Nova Scotia.  Your "dodgy" math is essentially correct 
and the reverse of how mariners have calculated their latitude from the 
sun's position and noon.

Fortunately, there are a number of good online calculators that will do 
all the math for you (and use some complicated formulas to correct for 
some of the assumptions you point out).  Many of them are designed to 
give the sun's exact position in the sky for a given place at a specific 
date and time.

There is a really good one at the University of Oregon's Solar Radiation 
Monitoring Lab.  It will generate a nice graph of sun angles for a 6 
month period )between solstices) for the whole day. It's at:
http://solardat.uoregon.edu/SunChartProgram.html

Enjoy!

Calum

Calum Ewing
QuackaDoodle Farm
Seaforth, Nova Scotia
quackadoodle.wordpress.com


On 2016-07-16 03:44 AM, scrooge McDuck wrote:
> Has anyone done the research on this? One of the basic inputs to any 
> permaculture design is the angle of the Sun's maximum at summer and 
> winter solstices.
>
> If the earth's position is exactly level with the sun, , it should be 
> possible to calculate the Sun's max and Min points for any site from 
> one's latitude. Have any of you tried this?
> I am hoping to get a better approach than I am taking, but let me try. 
> All assumptions are from general knowledge and imprecise thus far. The 
> next paragraph contains the ugly math.
>
> /Begin dodgy math calculations
>
> Take my location @ 53.30095 North, -6.3296 West. We know the earth 
> wobbles by 23 degrees. Let's presume that the sun sits exactly over 
> the equator on each equinox. So an imaginary right angled triangle can 
> be constructed between the sun, my site, and a perpendicular to an 
> imaginary line between the equator at 6.3296 West, and earth's centre. 
> Another imaginary triangle with the same perpendicular site to 
> equator, and the Centre of the earth has an angle at the centre of 
> 55.30095 degrees at earth's centre and a hypotenuse of ~8000 miles, 
> which yields the length of that perpendicular from site to Zero 
> latitude. From This I can calculate the Opposite side length, = common 
> side length. In my original triangle, That side over the hypotenuse 
> (my distance from the sun) provides the angle at the sun between the 
> equator line and my site. Subtract that from 90 degrees to get the 
> Sun's angle on an equinox. Add/subtract 11.5 degrees for the max & Min 
> points.
>
> /End dodgy maths calculations.
>
> This presumes many simplifications like the earth being spherical 
> (which it isn't) and the exact distance to my site from the sun would 
> have to be an approximation which could skew the results by a degree 
> or two, but better can be done. Has this been done properly? Has 
> anyone got the URL?
>
> --
> Declan Moriarty
> Sent from my Android tablet with mail.com <http://mail.com> Mail. 
> Please excuse my brevity.
>
>
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