Calculating the position of the sun.

I was looking for information regarding the sun's elevation versus the
amount of available light in a photographic situation. In other words, what
would be the range of exposure values at different times of the day? I found
an answer but in my search, I got sidetracked into another interesting
question. Can you figure out where the sun is given your latitude,
longitude, time of day, and day of the year? I found an answer to that as
well over he http://answers.google.com/answers/threadview?id=782886

There are some equations that can be plugged into a spreadsheet and voila,
you come up with the sun's elevation and azimuth. So I created a spreadsheet
with those equations, plugged in the sample numbers given from the above
link, and came up with the exact same answer as the author did. Then I
changed the latitude and longitude to the Los Angeles area, entered a time
of 10:43 (AM) and came up with the sun's elevation as a negative 50.93
degrees and an azimuth of a positive 64.35 degrees. I double checked my
answers with a calculator and came up with the same result. It doesn't make
sense to me that the sun's elevation would be negative at 10:43 in the
morning nor does it make sense that the sun's azimuth would be at 64.35
degrees on a winter morning. (By the way, I would rate my astronomy
knowledge as entry level as I have just completed reading H. A. Rey's book,
"The Stars, A New Way To See Them." It set a good foundation for a beginner
like myself.) Has anyone else every tried to create a spreadsheet like this
and put in random locations to check the accuracy of the equations? I can
e-mail you a copy of the spreadsheet if you request.

Thanks for your reply.

--
David Farber
L.A., CA

"David Farber" wrote in message
...
I was looking for information regarding the sun's elevation versus the
amount of available light in a photographic situation. In other words,
what
would be the range of exposure values at different times of the day? I
found
an answer but in my search, I got sidetracked into another interesting
question. Can you figure out where the sun is given your latitude,
longitude, time of day, and day of the year? I found an answer to that as
well over he http://answers.google.com/answers/threadview?id=782886

There are some equations that can be plugged into a spreadsheet and voila,
you come up with the sun's elevation and azimuth. So I created a
spreadsheet
with those equations, plugged in the sample numbers given from the above
link, and came up with the exact same answer as the author did. Then I
changed the latitude and longitude to the Los Angeles area, entered a time
of 10:43 (AM) and came up with the sun's elevation as a negative 50.93
degrees and an azimuth of a positive 64.35 degrees. I double checked my
answers with a calculator and came up with the same result. It doesn't
make
sense to me that the sun's elevation would be negative at 10:43 in the
morning nor does it make sense that the sun's azimuth would be at 64.35
degrees on a winter morning. (By the way, I would rate my astronomy
knowledge as entry level as I have just completed reading H. A. Rey's
book,
"The Stars, A New Way To See Them." It set a good foundation for a
beginner
like myself.) Has anyone else every tried to create a spreadsheet like
this
and put in random locations to check the accuracy of the equations? I can
e-mail you a copy of the spreadsheet if you request.

Thanks for your reply.


Just a quick question whether you assumed that 10.43 is local time for LA?
It could be that the spreadsheets assume you enter UTC time, in which case
you need to add 8 hours.

"David Farber" wrote in message

I was looking for information regarding the sun's elevation versus the
amount of available light in a photographic situation. In other
words, what would be the range of exposure values at different times
of the day? I found an answer but in my search, I got sidetracked
into another interesting question. Can you figure out where the sun
is given your latitude, longitude, time of day, and day of the year?
I found an answer to that as well over he
http://answers.google.com/answers/threadview?id=782886

There are some equations that can be plugged into a spreadsheet and
voila, you come up with the sun's elevation and azimuth. So I created
a spreadsheet with those equations, plugged in the sample numbers
given from the above link, and came up with the exact same answer as
the author did. Then I changed the latitude and longitude to the Los
Angeles area, entered a time of 10:43 (AM) and came up with the sun's
elevation as a negative 50.93 degrees and an azimuth of a positive
64.35 degrees. I double checked my answers with a calculator and came
up with the same result. It doesn't make sense to me that the sun's
elevation would be negative at 10:43 in the morning nor does it make
sense that the sun's azimuth would be at 64.35 degrees on a winter
morning. (By the way, I would rate my astronomy knowledge as entry
level as I have just completed reading H. A. Rey's book, "The Stars,
A New Way To See Them." It set a good foundation for a beginner like
myself.) Has anyone else every tried to create a spreadsheet like
this and put in random locations to check the accuracy of the
equations? I can e-mail you a copy of the spreadsheet if you request.

Thanks for your reply.

I haven't looked too closely at the calculations on
the web page, just a quick scan. But I suspect that
they are lacking a correction for the time zone
difference between UT and local time, something that
would have been not needed for the example observer
at just 2 degrees East longitude.

Try adding eight hours for your Los Angeles location
to the hour variable in the SHA equation:

SHA = (hour - 12 + 8)*15 + Long + TC

You might want to compare results with those of the
JPL Horizons system for your given location, date,
and time (Trimmed output):

************************************************** *****************************
Start time : A.D. 2006-Nov-15 10:30:00.0000 UT-08:00
Stop time : A.D. 2006-Nov-15 10:40:00.0000 UT-08:00
Step-size : 5 minutes
************************************************** *****************************

************************************************** *****************************
Date_(ZONE)_HR:MN Azi_(a-appr)_Elev S-O-T /r S-T-O
************************************************** ***********
$$SOE
2006-Nov-15 10:30 *m 160.3611 34.9515 0.0000 /? 0.0000
2006-Nov-15 10:35 *m 161.7502 35.2869 0.0000 /? 0.0000
2006-Nov-15 10:40 *m 163.1532 35.5983 0.0000 /? 0.0000
$$EOE
************************************************** *****************************

I used the Horizons web-interface to generate the above
starting at page:

http://ssd.jpl.nasa.gov/?horizons

I specified the time as local by entering the timezone
offset in the starting time: 2006-11-15 10:30 UT-8

On Tue, 19 Feb 2008 11:57:19 -0800, "David Farber"
wrote:

There are some equations that can be plugged into a spreadsheet and voila,
you come up with the sun's elevation and azimuth. So I created a spreadsheet
with those equations, plugged in the sample numbers given from the above
link, and came up with the exact same answer as the author did. Then I
changed the latitude and longitude to the Los Angeles area, entered a time
of 10:43 (AM) and came up with the sun's elevation as a negative 50.93
degrees and an azimuth of a positive 64.35 degrees.

In addition to what has already been suggested about local time versus
UT, make sure as well that the longitude was entered correctly. Most
formulas use +/- 180 degrees, with west longitudes negative (Los Angeles
is about -118). Occasionally, you'll also see longitude expressed as
0-360 degrees (LA at 242).

Meeus, used by many as a reference, does things completely ass
backwards, so it all depends on the details of the formulas used in your
spreadsheet.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

"OG" wrote in message
...

"David Farber" wrote in message
...
I was looking for information regarding the sun's elevation versus the
amount of available light in a photographic situation. In other words,
what
would be the range of exposure values at different times of the day? I
found
an answer but in my search, I got sidetracked into another interesting
question. Can you figure out where the sun is given your latitude,
longitude, time of day, and day of the year? I found an answer to that
as
well over he http://answers.google.com/answers/threadview?id=782886

There are some equations that can be plugged into a spreadsheet and
voila,
you come up with the sun's elevation and azimuth. So I created a
spreadsheet
with those equations, plugged in the sample numbers given from the above
link, and came up with the exact same answer as the author did. Then I
changed the latitude and longitude to the Los Angeles area, entered a
time
of 10:43 (AM) and came up with the sun's elevation as a negative 50.93
degrees and an azimuth of a positive 64.35 degrees. I double checked my
answers with a calculator and came up with the same result. It doesn't
make
sense to me that the sun's elevation would be negative at 10:43 in the
morning nor does it make sense that the sun's azimuth would be at 64.35
degrees on a winter morning. (By the way, I would rate my astronomy
knowledge as entry level as I have just completed reading H. A. Rey's
book,
"The Stars, A New Way To See Them." It set a good foundation for a
beginner
like myself.) Has anyone else every tried to create a spreadsheet like
this
and put in random locations to check the accuracy of the equations? I
can
e-mail you a copy of the spreadsheet if you request.

Thanks for your reply.


Just a quick question whether you assumed that 10.43 is local time for LA?
It could be that the spreadsheets assume you enter UTC time, in which case
you need to add 8 hours.



By adding the 8 hours, the new numbers a

Elevation angle = 37.68 degrees
Azimuth = 150.95 degrees
(I forgot to mention the date I used in my calculation which was Feb. 12,
2008)

Those numbers make more sense and if the sun is out tomorrow, I'll compare
those numbers with the date and time adjusted accordingly to see if they
check out.

Also, http://answers.google.com/answers/threadview?id=782886 says,

"Note that hour is the local hour and it is expressed in fractions of
hours..." So is that statement incorrect or does local hour mean UTC time?

Thanks for your reply.
--
David Farber
L.A., CA

"Greg Neill" wrote in message
m...
"David Farber" wrote in message

I was looking for information regarding the sun's elevation versus the
amount of available light in a photographic situation. In other
words, what would be the range of exposure values at different times
of the day? I found an answer but in my search, I got sidetracked
into another interesting question. Can you figure out where the sun
is given your latitude, longitude, time of day, and day of the year?
I found an answer to that as well over he
http://answers.google.com/answers/threadview?id=782886

There are some equations that can be plugged into a spreadsheet and
voila, you come up with the sun's elevation and azimuth. So I created
a spreadsheet with those equations, plugged in the sample numbers
given from the above link, and came up with the exact same answer as
the author did. Then I changed the latitude and longitude to the Los
Angeles area, entered a time of 10:43 (AM) and came up with the sun's
elevation as a negative 50.93 degrees and an azimuth of a positive
64.35 degrees. I double checked my answers with a calculator and came
up with the same result. It doesn't make sense to me that the sun's
elevation would be negative at 10:43 in the morning nor does it make
sense that the sun's azimuth would be at 64.35 degrees on a winter
morning. (By the way, I would rate my astronomy knowledge as entry
level as I have just completed reading H. A. Rey's book, "The Stars,
A New Way To See Them." It set a good foundation for a beginner like
myself.) Has anyone else every tried to create a spreadsheet like
this and put in random locations to check the accuracy of the
equations? I can e-mail you a copy of the spreadsheet if you request.

Thanks for your reply.

I haven't looked too closely at the calculations on
the web page, just a quick scan. But I suspect that
they are lacking a correction for the time zone
difference between UT and local time, something that
would have been not needed for the example observer
at just 2 degrees East longitude.

Try adding eight hours for your Los Angeles location
to the hour variable in the SHA equation:

SHA = (hour - 12 + 8)*15 + Long + TC

You might want to compare results with those of the
JPL Horizons system for your given location, date,
and time (Trimmed output):

************************************************** **************************
***
Start time : A.D. 2006-Nov-15 10:30:00.0000 UT-08:00
Stop time : A.D. 2006-Nov-15 10:40:00.0000 UT-08:00
Step-size : 5 minutes
************************************************** **************************
***

************************************************** **************************
***
Date_(ZONE)_HR:MN Azi_(a-appr)_Elev S-O-T /r S-T-O
************************************************** ***********
$$SOE
2006-Nov-15 10:30 *m 160.3611 34.9515 0.0000 /? 0.0000

My spreadsheet cranks out with the UTC time adjustment:
160.01 34.77

2006-Nov-15 10:35 *m 161.7502 35.2869 0.0000 /? 0.0000

For these numbers I get:

161.40 35.12

2006-Nov-15 10:40 *m 163.1532 35.5983 0.0000 /? 0.0000
$$EOE

And finally:

162.79 35.43
************************************************** **************************
***

I used the Horizons web-interface to generate the above
starting at page:

http://ssd.jpl.nasa.gov/?horizons

I specified the time as local by entering the timezone
offset in the starting time: 2006-11-15 10:30 UT-8


That telnet horizons interface program looks quite interesting. It will take
me a while to figure out how it use the interface.

Thanks for your reply.
--
David Farber
L.A., CA

"Chris L Peterson" wrote in message
...
On Tue, 19 Feb 2008 11:57:19 -0800, "David Farber"
wrote:

There are some equations that can be plugged into a spreadsheet and
voila,
you come up with the sun's elevation and azimuth. So I created a
spreadsheet
with those equations, plugged in the sample numbers given from the above
link, and came up with the exact same answer as the author did. Then I
changed the latitude and longitude to the Los Angeles area, entered a
time
of 10:43 (AM) and came up with the sun's elevation as a negative 50.93
degrees and an azimuth of a positive 64.35 degrees.

In addition to what has already been suggested about local time versus
UT, make sure as well that the longitude was entered correctly. Most
formulas use +/- 180 degrees, with west longitudes negative (Los Angeles
is about -118). Occasionally, you'll also see longitude expressed as
0-360 degrees (LA at 242).

Meeus, used by many as a reference, does things completely ass
backwards, so it all depends on the details of the formulas used in your
spreadsheet.

_________________________________________________

Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com

I followed the instructions on the site which are in agreement with your
first statement. (Use negative numbers if the location is a western
longitude.)

Thanks for your reply.
--
David Farber
L.A., CA