sunrise bearings mid summer, equinox, mid winter.

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Shearwater

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I feel I ought be able to work this out but I can't; I have no knowledge of solar nav. I live at a point 41 degs North and 0 degs 40 mins East. I saw and recorded with a stick sunrise at June 21. I ditto'd sunrise at Sept 21. I would not be surprised if sunrise at Dec 21 as the same angle duplicated. But I haven't been able to watch sunsets at these dates. Can anyone tell me where to expect sunset on Dec 21 and where it would be last and next equinox and summer solstice. Is it reasonable I couldn't work it out for myself? What I'm sure I couldn't do is calculate the height of the sun above the horizon on these dates, You might guess this is slightly non-boatie but has something to do with where we position the sun awning and pergola. Well, it could be boatie !!
 
I don't know whether this will do what you want

http://www.suncalc.net/

Put in your postcode et voila.

You can go back & forth using the calendar & see the sunrise/set angles relative to your location.

HTH

Works for Ampolla

ampolla.jpg
 
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Sunrise at 41°00' N, 0°40'E

- Jun 21st: 57°T at 0425 GMT
- Sept 21st: 89°T at 0544 GMT
- Dec 21st: 121°T at 0719 GMT

Sunset
- Dec 21st: 239°T at 1632 GMT.

For local time (Spain) add 2 hours in Summer, 1 hour in Winter. These are the times of apparent sunrise/set allowing for atmospheric refraction - i.e. as it appears to you. They are taken as the point when the upper edge of the sun touches the horizon at sea, for someone near sea-level. If you are in, or looking at the mountains, the times you first/last see the sun will obviously be a little different*.

Your supposition about doubling the angle is approximately correct. The position and time on the same date changes very little from one year to the next, certainly less than will matter for positioning your sun awning!

*PS: From Ampolla, you are looking at mountains that rise to about 1200m to the west. You will last see the sun roughly 7 minutes before actual sunset.
 
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1st define sunrise!

It's reasonable for it not to be trivial!

First step is to define sunrise: I define it as the sun's upper limb being at 0 degrees observed altitude, for an eye height of 0m. This may be mathematically ok but not what you want: for instance if you're in a pergola you may want to chose the centre of the sun to be on the horizon. The land is probably not flat either (nor your eye at zero).

Then find the true altitude when the upper limb is at zero. In principle, this needs the height of eye, the sun's semi-diameter (which varies with time of year), and the refraction (which is a weak function of air pressure and tempertaure). To cut a long story short, it is about -51' for my definition of sunrise.

Now you can find the time at which the true altitude is at -51'. In your location (41 00'N, 00 40'E) that is 07h 18' 48" on 21st December 2012. The azimuth at that time is 120.9 degrees True.

You probably aren't after quite this resolution and maybe can find a faster way, but setting True Altitude = 0 will not give the right result.

There are various JavaScript or spreadsheet calculators on the web which can be used to calculate the True Alt given a time and location, and you just have to iterate the time (eg a binary chop algorithm) to get T.A. = -51'.
 
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