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Solar angle OT

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  • Peter Vasey
    Off Topic but maybe useful. I ve had solar panels installed, and I m trying to calculate the angle of incidence of the sunlight given the altitude and azimuth
    Message 1 of 10 , Jul 7 4:05 AM
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      Off Topic but maybe useful.

      I've had solar panels installed, and I'm trying to calculate the angle
      of incidence of the sunlight given the altitude and azimuth of the Sun
      at different times of day - I can get those values from my planetarium
      program. My panel is tilted 30 degrees from the horizontal and faces
      due South.

      So there are four angles to consider. Solar Altitude, Solar Azimuth,
      Panel tilt and Panel orientation.

      If the panel was flat on the ground, no problem, only the altitude value
      would be used. But the tilt and orientation muddies the water. I was
      always comfortable with standard trigonometry, but this one has me beat.

      Any mathematicians out there able to come up with a formula?

      Cheers, Peter.

      http://www.madpc.co.uk/~peterv
      Approx. 55ºN, 2ºW (Northumberland, UK)
    • Simon Morgan
      Try these sites: http://pvcdrom.pveducation.org/SUNLIGHT/MODTILT.HTM and a site I use to calculate the “standard” insolation for my solar PV panels
      Message 2 of 10 , Jul 7 4:47 AM
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        Try these sites:



        http://pvcdrom.pveducation.org/SUNLIGHT/MODTILT.HTM



        and a site I use to calculate the “standard” insolation for my solar PV
        panels



        http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php?lang=en
        <http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php?lang=en&map=europe>
        &map=europe



        I am not sure if these are exactly what you are after. You could try, of
        course, using a Nautical Almanac which gives you Latitude for sun’s
        elevation so use in reverse so to speak.





        From: lx90@yahoogroups.com [mailto:lx90@yahoogroups.com] On Behalf Of Peter
        Vasey
        Sent: 07 July 2011 12:06
        To: SBIG@yahoogroups.com; lx90@yahoogroups.com; QCUIAG@yahoogroups.com
        Subject: [lx90] Solar angle OT





        Off Topic but maybe useful.

        I've had solar panels installed, and I'm trying to calculate the angle
        of incidence of the sunlight given the altitude and azimuth of the Sun
        at different times of day - I can get those values from my planetarium
        program. My panel is tilted 30 degrees from the horizontal and faces
        due South.

        So there are four angles to consider. Solar Altitude, Solar Azimuth,
        Panel tilt and Panel orientation.

        If the panel was flat on the ground, no problem, only the altitude value
        would be used. But the tilt and orientation muddies the water. I was
        always comfortable with standard trigonometry, but this one has me beat.

        Any mathematicians out there able to come up with a formula?

        Cheers, Peter.

        http://www.madpc.co.uk/~peterv
        Approx. 55ºN, 2ºW (Northumberland, UK)



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      • Peter Vasey
        Thanks, Simon, but that s not what I m after. I want to collect historical data from day to day, and correlate the power generated with the Sun s position at
        Message 3 of 10 , Jul 7 6:05 AM
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          Thanks, Simon, but that's not what I'm after. I want to collect
          historical data from day to day, and correlate the power generated with
          the Sun's position at that time. And the best figure for correlation is
          the actual angle of the Sun with respect to the panel.


          http://www.madpc.co.uk/~peterv
          Approx. 55ºN, 2ºW (Northumberland, UK)

          On 07/07/2011 12:47, Simon Morgan wrote:
          > Try these sites:
          >
          >
          >
          > http://pvcdrom.pveducation.org/SUNLIGHT/MODTILT.HTM
          >
          >
          >
          > and a site I use to calculate the “standard” insolation for my solar PV
          > panels
          >
          >
          >
          > http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php?lang=en
          > <http://re.jrc.ec.europa.eu/pvgis/apps4/pvest.php?lang=en&map=europe>
          > &map=europe
          >
          >
          >
          > I am not sure if these are exactly what you are after. You could try, of
          > course, using a Nautical Almanac which gives you Latitude for sun’s
          > elevation so use in reverse so to speak.
          >
          >
          >
          >
          >
          > From: lx90@yahoogroups.com [mailto:lx90@yahoogroups.com] On Behalf Of Peter
          > Vasey
          > Sent: 07 July 2011 12:06
          > To: SBIG@yahoogroups.com; lx90@yahoogroups.com; QCUIAG@yahoogroups.com
          > Subject: [lx90] Solar angle OT
          >
          >
          >
          >
          >
          > Off Topic but maybe useful.
          >
          > I've had solar panels installed, and I'm trying to calculate the angle
          > of incidence of the sunlight given the altitude and azimuth of the Sun
          > at different times of day - I can get those values from my planetarium
          > program. My panel is tilted 30 degrees from the horizontal and faces
          > due South.
          >
          > So there are four angles to consider. Solar Altitude, Solar Azimuth,
          > Panel tilt and Panel orientation.
          >
          > If the panel was flat on the ground, no problem, only the altitude value
          > would be used. But the tilt and orientation muddies the water. I was
          > always comfortable with standard trigonometry, but this one has me beat.
          >
          > Any mathematicians out there able to come up with a formula?
          >
          > Cheers, Peter.
          >
          > http://www.madpc.co.uk/~peterv
          > Approx. 55ºN, 2ºW (Northumberland, UK)
          >
          >
          >
          > _____
          >
          > No virus found in this message.
          > Checked by AVG - www.avg.com
          > Version: 10.0.1382 / Virus Database: 1513/3738 - Release Date: 07/01/11
          >
          >
          > ----------
          >
          >
          > No virus found in this outgoing message.
          > Checked by AVG - www.avg.com
          > Version: 9.0.901 / Virus Database: 271.1.1/3749 - Release Date: 07/07/11 07:34:00
          >
          >
          > [Non-text portions of this message have been removed]
          >
          >
          >
          > ------------------------------------
          >
          > To unsubscribe from this group mailto lx90-unsubscribe@yahoogroups.comYahoo! Groups Links
          >
          >
          >
          >
        • Fernando Campuzano
          Have you considered entering into the planetarium program the site coordinates 30 degrees toward the equator? (your longitude & your latitude minus 30)
          Message 4 of 10 , Jul 7 10:26 AM
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            Have you considered entering into the planetarium program the site coordinates
            30 degrees toward the equator? (your longitude & your latitude minus 30)

            Fernando




            ________________________________
            De: Peter Vasey <petevasey@...>
            Para: SBIG@yahoogroups.com; "lx90@yahoogroups.com" <lx90@yahoogroups.com>;
            "QCUIAG@yahoogroups.com" <QCUIAG@yahoogroups.com>
            Enviado: jue,7 julio, 2011 13:05
            Asunto: [lx90] Solar angle OT

             
            Off Topic but maybe useful.

            I've had solar panels installed, and I'm trying to calculate the angle
            of incidence of the sunlight given the altitude and azimuth of the Sun
            at different times of day - I can get those values from my planetarium
            program. My panel is tilted 30 degrees from the horizontal and faces
            due South.

            So there are four angles to consider. Solar Altitude, Solar Azimuth,
            Panel tilt and Panel orientation.

            If the panel was flat on the ground, no problem, only the altitude value
            would be used. But the tilt and orientation muddies the water. I was
            always comfortable with standard trigonometry, but this one has me beat.

            Any mathematicians out there able to come up with a formula?

            Cheers, Peter.

            http://www.madpc.co.uk/~peterv
            Approx. 55ºN, 2ºW (Northumberland, UK)




            [Non-text portions of this message have been removed]
          • Peter Vasey
            See my reply to Joe! Cheers, Peter. http://www.madpc.co.uk/~peterv Approx. 55ºN, 2ºW (Northumberland, UK)
            Message 5 of 10 , Jul 7 3:12 PM
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              See my reply to Joe!

              Cheers, Peter.

              http://www.madpc.co.uk/~peterv
              Approx. 55ºN, 2ºW (Northumberland, UK)

              On 07/07/2011 18:26, Fernando Campuzano wrote:
              > Have you considered entering into the planetarium program the site coordinates
              > 30 degrees toward the equator? (your longitude& your latitude minus 30)
              >
              > Fernando
              >
              >
              >
              >
              > ________________________________
              > De: Peter Vasey<petevasey@...>
              > Para: SBIG@yahoogroups.com; "lx90@yahoogroups.com"<lx90@yahoogroups.com>;
              > "QCUIAG@yahoogroups.com"<QCUIAG@yahoogroups.com>
              > Enviado: jue,7 julio, 2011 13:05
              > Asunto: [lx90] Solar angle OT
              >
              >
              > Off Topic but maybe useful.
              >
              > I've had solar panels installed, and I'm trying to calculate the angle
              > of incidence of the sunlight given the altitude and azimuth of the Sun
              > at different times of day - I can get those values from my planetarium
              > program. My panel is tilted 30 degrees from the horizontal and faces
              > due South.
              >
              > So there are four angles to consider. Solar Altitude, Solar Azimuth,
              > Panel tilt and Panel orientation.
              >
              > If the panel was flat on the ground, no problem, only the altitude value
              > would be used. But the tilt and orientation muddies the water. I was
              > always comfortable with standard trigonometry, but this one has me beat.
              >
              > Any mathematicians out there able to come up with a formula?
              >
              > Cheers, Peter.
              >
              > http://www.madpc.co.uk/~peterv
              > Approx. 55ºN, 2ºW (Northumberland, UK)
              >
              >
              >
              >
              > [Non-text portions of this message have been removed]
              >
              >
              >
              > ------------------------------------
              >
              > To unsubscribe from this group mailto lx90-unsubscribe@yahoogroups.comYahoo! Groups Links
              >
              >
              >
              >
            • Peter Vasey
              Oops, Joe was on the SBIG group! I said to him Good lateral thinking, Joe. And I thought of that also, and played with it. But it would be nice to have
              Message 6 of 10 , Jul 7 4:00 PM
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                Oops, Joe was on the SBIG group! I said to him "Good lateral thinking,
                Joe. And I thought of that also, and played with it. But it would be
                nice to have confirmation from trigonometry at my location."

                Well, I now have a formula (from William Hamblen on the QCUIAG group),
                and it agrees within a degree or so.

                :-)

                Cheers, Peter.

                http://www.madpc.co.uk/~peterv
                Approx. 55ºN, 2ºW (Northumberland, UK)

                On 07/07/2011 23:12, Peter Vasey wrote:
                > See my reply to Joe!
                >
                > Cheers, Peter.
                >
                > http://www.madpc.co.uk/~peterv
                > Approx. 55ºN, 2ºW (Northumberland, UK)
                >
                > On 07/07/2011 18:26, Fernando Campuzano wrote:
                >> Have you considered entering into the planetarium program the site coordinates
                >> 30 degrees toward the equator? (your longitude& your latitude minus 30)
                >>
                >> Fernando
                >>
                >>
                >>
                >>
                >> ________________________________
                >> De: Peter Vasey<petevasey@...>
                >> Para: SBIG@yahoogroups.com; "lx90@yahoogroups.com"<lx90@yahoogroups.com>;
                >> "QCUIAG@yahoogroups.com"<QCUIAG@yahoogroups.com>
                >> Enviado: jue,7 julio, 2011 13:05
                >> Asunto: [lx90] Solar angle OT
                >>
                >>
                >> Off Topic but maybe useful.
                >>
                >> I've had solar panels installed, and I'm trying to calculate the angle
                >> of incidence of the sunlight given the altitude and azimuth of the Sun
                >> at different times of day - I can get those values from my planetarium
                >> program. My panel is tilted 30 degrees from the horizontal and faces
                >> due South.
                >>
                >> So there are four angles to consider. Solar Altitude, Solar Azimuth,
                >> Panel tilt and Panel orientation.
                >>
                >> If the panel was flat on the ground, no problem, only the altitude value
                >> would be used. But the tilt and orientation muddies the water. I was
                >> always comfortable with standard trigonometry, but this one has me beat.
                >>
                >> Any mathematicians out there able to come up with a formula?
                >>
                >> Cheers, Peter.
                >>
                >> http://www.madpc.co.uk/~peterv
                >> Approx. 55ºN, 2ºW (Northumberland, UK)
              • autostaretx
                ... Please share it (so that us lurkers will learn) thanks --dick
                Message 7 of 10 , Jul 7 6:09 PM
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                  --- In lx90@yahoogroups.com, Peter Vasey <petevasey@...> wrote:
                  ...
                  > Well, I now have a formula (from William Hamblen on the QCUIAG
                  > group), and it agrees within a degree or so.

                  Please share it (so that us lurkers will learn)

                  thanks
                  --dick
                • Fernando Campuzano
                  I love the concept of lateral thinking . As we say in Spain: Sometimes the trees don t allow to see the forest ... ________________________________ De:
                  Message 8 of 10 , Jul 8 8:25 AM
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                    I love the concept of "lateral thinking". As we say in Spain: "Sometimes the
                    trees don't allow to see the forest"
                    :-)




                    ________________________________
                    De: Peter Vasey <petevasey@...>
                    Para: lx90@yahoogroups.com
                    Enviado: vie,8 julio, 2011 01:00
                    Asunto: Re: [lx90] Solar angle OT

                     
                    Oops, Joe was on the SBIG group! I said to him "Good lateral thinking,
                    Joe. And I thought of that also, and played with it. But it would be
                    nice to have confirmation from trigonometry at my location."

                    Well, I now have a formula (from William Hamblen on the QCUIAG group),
                    and it agrees within a degree or so.

                    :-)

                    Cheers, Peter.

                    http://www.madpc.co.uk/~peterv
                    Approx. 55ºN, 2ºW (Northumberland, UK)

                    On 07/07/2011 23:12, Peter Vasey wrote:
                    > See my reply to Joe!
                    >
                    > Cheers, Peter.
                    >
                    > http://www.madpc.co.uk/~peterv
                    > Approx. 55ºN, 2ºW (Northumberland, UK)
                    >
                    > On 07/07/2011 18:26, Fernando Campuzano wrote:
                    >> Have you considered entering into the planetarium program the site
                    coordinates
                    >> 30 degrees toward the equator? (your longitude& your latitude minus 30)
                    >>
                    >> Fernando
                    >>
                    >>
                    >>
                    >>
                    >> ________________________________
                    >> De: Peter Vasey<petevasey@...>
                    >> Para: SBIG@yahoogroups.com; "lx90@yahoogroups.com"<lx90@yahoogroups.com>;
                    >> "QCUIAG@yahoogroups.com"<QCUIAG@yahoogroups.com>
                    >> Enviado: jue,7 julio, 2011 13:05
                    >> Asunto: [lx90] Solar angle OT
                    >>
                    >>
                    >> Off Topic but maybe useful.
                    >>
                    >> I've had solar panels installed, and I'm trying to calculate the angle
                    >> of incidence of the sunlight given the altitude and azimuth of the Sun
                    >> at different times of day - I can get those values from my planetarium
                    >> program. My panel is tilted 30 degrees from the horizontal and faces
                    >> due South.
                    >>
                    >> So there are four angles to consider. Solar Altitude, Solar Azimuth,
                    >> Panel tilt and Panel orientation.
                    >>
                    >> If the panel was flat on the ground, no problem, only the altitude value
                    >> would be used. But the tilt and orientation muddies the water. I was
                    >> always comfortable with standard trigonometry, but this one has me beat.
                    >>
                    >> Any mathematicians out there able to come up with a formula?
                    >>
                    >> Cheers, Peter.
                    >>
                    >> http://www.madpc.co.uk/~peterv
                    >> Approx. 55ºN, 2ºW (Northumberland, UK)




                    [Non-text portions of this message have been removed]
                  • Peter Vasey
                    Changing latitude alone worked well because my panels face due South. If they were angled away from South, then I would have to alter my longitude as well.
                    Message 9 of 10 , Jul 8 1:32 PM
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                      Changing latitude alone worked well because my panels face due South.
                      If they were angled away from South, then I would have to alter my
                      longitude as well. But I've incorporated Bud's formula in a
                      spreadsheet, so just have to input my local parameters.

                      Cheers, Peter.

                      http://www.madpc.co.uk/~peterv
                      Approx. 55ºN, 2ºW (Northumberland, UK)

                      On 08/07/2011 16:25, Fernando Campuzano wrote:
                      > I love the concept of "lateral thinking". As we say in Spain: "Sometimes the
                      > trees don't allow to see the forest"
                      > :-)
                      >
                      > ________________________________
                      > De: Peter Vasey<petevasey@...>
                      > Para: lx90@yahoogroups.com
                      > Enviado: vie,8 julio, 2011 01:00
                      > Asunto: Re: [lx90] Solar angle OT
                      >
                      >
                      > Oops, Joe was on the SBIG group! I said to him "Good lateral thinking,
                      > Joe. And I thought of that also, and played with it. But it would be
                      > nice to have confirmation from trigonometry at my location."
                      >
                      > Well, I now have a formula (from William Hamblen on the QCUIAG group),
                      > and it agrees within a degree or so.
                      >
                      > :-)
                      >
                      > Cheers, Peter.
                      >
                      > http://www.madpc.co.uk/~peterv
                      > Approx. 55ºN, 2ºW (Northumberland, UK)
                      >
                      > On 07/07/2011 23:12, Peter Vasey wrote:
                      >> See my reply to Joe!
                      >>
                      >> Cheers, Peter.
                      >>
                      >> http://www.madpc.co.uk/~peterv
                      >> Approx. 55ºN, 2ºW (Northumberland, UK)
                      >>
                      >> On 07/07/2011 18:26, Fernando Campuzano wrote:
                      >>> Have you considered entering into the planetarium program the site
                      > coordinates
                      >>> 30 degrees toward the equator? (your longitude& your latitude minus 30)
                      >>>
                      >>> Fernando
                      >>>
                      >>>
                      >>>
                      >>>
                      >>> ________________________________
                      >>> De: Peter Vasey<petevasey@...>
                      >>> Para: SBIG@yahoogroups.com; "lx90@yahoogroups.com"<lx90@yahoogroups.com>;
                      >>> "QCUIAG@yahoogroups.com"<QCUIAG@yahoogroups.com>
                      >>> Enviado: jue,7 julio, 2011 13:05
                      >>> Asunto: [lx90] Solar angle OT
                      >>>
                      >>>
                      >>> Off Topic but maybe useful.
                      >>>
                      >>> I've had solar panels installed, and I'm trying to calculate the angle
                      >>> of incidence of the sunlight given the altitude and azimuth of the Sun
                      >>> at different times of day - I can get those values from my planetarium
                      >>> program. My panel is tilted 30 degrees from the horizontal and faces
                      >>> due South.
                      >>>
                      >>> So there are four angles to consider. Solar Altitude, Solar Azimuth,
                      >>> Panel tilt and Panel orientation.
                      >>>
                      >>> If the panel was flat on the ground, no problem, only the altitude value
                      >>> would be used. But the tilt and orientation muddies the water. I was
                      >>> always comfortable with standard trigonometry, but this one has me beat.
                      >>>
                      >>> Any mathematicians out there able to come up with a formula?
                      >>>
                      >>> Cheers, Peter.
                      >>>
                      >>> http://www.madpc.co.uk/~peterv
                      >>> Approx. 55ºN, 2ºW (Northumberland, UK)
                    • Peter Vasey
                      Hi, Dick, Here it is: cos(d) = sin(h1)*sin(h2)+cos(h1)*cos(h2)*cos(A1-A2) Where: h1 and A1 are the altitude and azimuth of the Sun h2 and A2 are the altitude
                      Message 10 of 10 , Jul 8 1:49 PM
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                        Hi, Dick,

                        Here it is:

                        cos(d) = sin(h1)*sin(h2)+cos(h1)*cos(h2)*cos(A1-A2)

                        Where:

                        h1 and A1 are the altitude and azimuth of the Sun
                        h2 and A2 are the altitude and azimuth of the *perpendicular* from the
                        solar panel

                        d is the angle between the perpendicular to the panel and the Sun.

                        So the actual angle of incidence is 90-d or in full:

                        90-arccos(sin(h1)*sin(h2)+cos(h1)*cos(h2)*cos(A1-A2))

                        My panel faces due South and the roof angle is 30 degrees from the
                        horizontal. So h2 is 60 degrees and A2 is 180 degrees. I just then
                        need the Altitude and Azimuth of the Sun for the time I'm checking, and
                        I get that from Skymap. Put the formula in a spreadsheet and it's
                        easy-peasy ;-)

                        Cheers, Peter.

                        http://www.madpc.co.uk/~peterv
                        Approx. 55ºN, 2ºW (Northumberland, UK)

                        On 08/07/2011 02:09, autostaretx wrote:
                        > --- In lx90@yahoogroups.com, Peter Vasey<petevasey@...> wrote:
                        > ...
                        >> Well, I now have a formula (from William Hamblen on the QCUIAG
                        >> group), and it agrees within a degree or so.
                        >
                        > Please share it (so that us lurkers will learn)
                        >
                        > thanks
                        > --dick
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