Andrew:
I think your numbers make sense.
You didn’t mention the date, but I suspect your calculation was for July 30,
2011. For this date at 11:18 a.m. EDT, and the lat/long of Boston, I find the
following values ---
Sun’s declination: +18.5 degrees
Sun’s azimuth: 133.4 degrees (from north through east)
When you say the wall “declination” is 20 degrees east (from due south), this
is the same as saying that the normal to the wall faces azimuth 160 degrees
(measured from north through east).
Then, you measured the Sun’s azimuth to be 26.8 degrees (east) with respect to
the wall, implying that the Sun’s azimuth is 160 – 26.8 = 133.2 degrees. This
is rather close to the 133.4 that I got independently above.
Some reference books define azimuths as the angle from due south, while others
define it as the angle from true north, so it is important to keep these
straight.
Roger
From: sundial-boun...@uni-koeln.de [mailto:sundial-boun...@uni-koeln.de] On
Behalf Of Andrew Theokas
Sent: Sunday, July 31, 2011 9:57 AM
To: sundial@uni-koeln.de
Subject: Azimuth calculation
Fellow dialists:
I am using the following well known formula to calculate the sun’s azimuth for
a particular time and location:
Azimuth= tan-1 (sin H/(sin φ*cos H – cos φ*tanδ)
where
H= Sun’s hour angle
φ= the latitude - 42.3 degrees
δ is the sun’s declination - 18.62 degrees
The location is in Boston, USA or 42.3 degrees N and 71.04 degrees west
I am using the azimuth-azimuth approach to find the declination of a wall found
here:
http://www.mysundial.ca/tsp/wall_declination.html
the time the measurement was made was 11:18 am (daylight savings time is in
effect)
I can easily calculate that the azimuth with respect to the wall is 26.8
degrees.
Here is the problem: using two other independent methods I find that the wall’s
declination is 20 degrees East.
So 26.8 degrees – Sun’s Azimuth should equal about twenty degrees.
But, using the above equation I cannot get an Azimuth value to work. One place
where I might be in error is the value of the Hour angle which I compute to be
about –16 degrees.
But you can also find the Hour Angle on line here at
http://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-calculator
Where might I be going wrong?
Many thanks for a reply!
Andrew Theokas
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