## Solar angle OT

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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
Message 1 of 10 , Jul 7 4:05 AM
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.

Approx. 55ºN, 2ºW (Northumberland, UK)
• 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
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.

Approx. 55ºN, 2ºW (Northumberland, UK)

_____

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[Non-text portions of this message have been removed]
• 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
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.

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.
>
> 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
>
>
>
>
• 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
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>
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.

Approx. 55ºN, 2ºW (Northumberland, UK)

[Non-text portions of this message have been removed]
• 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

Cheers, Peter.

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.
>
> 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
>
>
>
>
• 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
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.

Approx. 55ºN, 2ºW (Northumberland, UK)

On 07/07/2011 23:12, Peter Vasey wrote:
> See my reply to Joe!
>
> Cheers, Peter.
>
> 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.
>>
>> Approx. 55ºN, 2ºW (Northumberland, UK)
• ... Please share it (so that us lurkers will learn) thanks --dick
Message 7 of 10 , Jul 7 6:09 PM
--- 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
• 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
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
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.

Approx. 55ºN, 2ºW (Northumberland, UK)

On 07/07/2011 23:12, Peter Vasey wrote:
> See my reply to Joe!
>
> Cheers, Peter.
>
> 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.
>>
>> Approx. 55ºN, 2ºW (Northumberland, UK)

[Non-text portions of this message have been removed]
• 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
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.

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.
>
> Approx. 55ºN, 2ºW (Northumberland, UK)
>
> On 07/07/2011 23:12, Peter Vasey wrote:
>> See my reply to Joe!
>>
>> Cheers, Peter.
>>
>> 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.
>>>
>>> Approx. 55ºN, 2ºW (Northumberland, UK)
• 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
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.