Bearing for Sunrise

cagey

cagey

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I know I am going to go doh at the answer but where or how do I find the bearing for sunrise for any day for Falmouth, cos I am watching the dawn rise and it is not where I expected it to be, also linking to my previous post re sunrise /sunset now I am seeing at 180deg opposition two dawns one approx. SE and other NW. I know the NW light is not the dawn, but still for the last month it has been there.
K
 
"sunrise azimuth".

That Solar calculator from Dylan seems to do the job, otherwise the Admiralty Nautical Almanac at £30 odd prints it out every 3 days.
 
Table 3(5) within Reeds Nautical Almanac shows "True bearing of sun at sunrise and sunset"

Another way, probably, would be to refer to "sunrise" time from the daily page of a Nautical Almanac (the Hydrographic Office version) and follow the sight reduction process. Find the GHA and Dec for the time of sunrise; assume a Longitude of Falmouth to give whole number LHA; use AP3270 to determine Zn.
 
cagey said:
I know I am going to go doh at the answer...
...re sunrise /sunset now I am seeing at 180deg opposition two dawns one approx. SE and other NW. I know the NW light is not the dawn, but still for the last month it has been there.
K
Click to expand...

At a guess I'd say you are looking at the loom of the street lights in Redruth.
 
The formula for the azimuth of geometric rise (northern hemisphere) is:

azimuth = acos(sin(dec)/cos(lat))

so with lat +50 and dec +26.43 we have az = acos(0.3981 / 0.6428) = acos(0.6193) = 51.7
and with lat +50 and dec -26.43 we have az = acos(0.3981 / -0.6428) = acos(-0.6193) = 128.3

The azimuth for apparent sunrise (rather than for geometric sunrise) was 50.5 in the first case and 127.0 in the second.

The difference is that apparent sunrise is when the upper limb appears to touch the horizon. This allows for the Sun's semidiameter (about 16') and refraction (about 34'). From an astro navigation point of view the difference is significant. From as observational point of view it may not be. Calculating the apparent azimuth needs the full formula: azimuth = acos((sin(dec) - sin(H)sin(lat)) / (cos(H)cos(lat))) where H is the altitude (about -50' for the Sun)

The formula works for any body - you just need to know its declination. However, for the Moon horizontal parallax is a significant factor and makes the difference between geometric and apparent azimuths quite large.

You can get the Sun's declination from online sources or from a nautical almanac. If you want a very approximate value you can use:
dec = asin(0.3978(sin(280.5+360Y+1.9sin(357.5+360Y)))) where Y is the year number (including decimals) - 1/7/13 would be 13.500
This gives a value for 12:00 each day that's accurate to around half a degree (the rate of change is up to about half a degree a day).
 
Please just stay with this conversation. my posn 050.23.5 N 005.46 W by GPS Newquay Cornwall UK. Local Time now is 0025 360 deg horizon is dark.
Every morning at approx. 0500 BST I am watching at approx 040 deg the sunrise as expected but at approx. 150 deg is another display in the sky, too powerfull and colourfull to be a loom and at a reciprocall at about 330 deg is another display. Simple answers please.
 
cagey said:
Please just stay with this conversation. my posn 050.23.5 N 005.46 W by GPS Newquay Cornwall UK. Local Time now is 0025 360 deg horizon is dark.
Every morning at approx. 0500 BST I am watching at approx 040 deg the sunrise as expected but at approx. 150 deg is another display in the sky, too powerfull and colourfull to be a loom and at a reciprocall at about 330 deg is another display. Simple answers please.
Click to expand...

Did you look at the website www.sollumis.com ?

Sunrise 51, sunset 310deg

But puzzled what you could be seeing at 150

Whether you are in falmouth or newquay
 
cagey said:
Please just stay with this conversation. my posn 050.23.5 N 005.46 W by GPS Newquay Cornwall UK. Local Time now is 0025 360 deg horizon is dark.
Every morning at approx. 0500 BST I am watching at approx 040 deg the sunrise as expected but at approx. 150 deg is another display in the sky, too powerfull and colourfull to be a loom and at a reciprocall at about 330 deg is another display. Simple answers please.
Click to expand...

Here are a couple of diagrams of the sky at 04:00UT this morning from the position you gave. The first one has bodies labeled, the second one has more detail. The horizontal line across the middle represents the horizon, the vertical line down the middle repesents South.

I can't see anything celestial that would light up the horizon in those positions at that time. Perhaps there's a terrestrial explanation.

View attachment 33105View attachment 33106
 
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