Solar Panel Angle Chart for Panel Tracker

JeepDaddy

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Does anyone have a resource which shows solar panel output by angle to the sun? For example 90 degrees = 100%, 60 degrees = 70%, 30 degrees = 20%, etc.

For the curious: This is for a single axis solar panel tracker. It is driven by two 3 watt panels in an inverted V configuration with the apex pointed directly towards the sun. The panels are connected + to - and - to +. This combines the current and determines the net polarity and amount of current to the motor. The more sun one panel gets, the more power applied to the motor in that polarity. If the other side gets more sun, the polarity is reversed and the motor turns the other way. The tracker will always move to have both panels at the same voltage. I am trying to determine the best angle of the inverted V of the panels for this to happen.

I am NOT asking about optimal panel angle by latitude, time of day or year. We all (mostly) understand that.
 

svetz

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Like this one from the FAQ entries? I believe it's for a dual-axis tracker; that is the losses are from atmospheric absorbtion.

Intensity_small-psmfb4.png

 
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svetz

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Thanks for that, but I'm looking at more along the lines of how well a solar panel generates power at angles to the sun.
Still not clear what you're looking for then.

For example, from the chart when the sun is at 90° you get 100% power, at 60° you get 87%, at 30° 50%.
The sun moves roughly 15° per hour (360° / 24 hrs), it's a bit more complicated because of the sun's elevation. So, at two hours from solar noon you are (2x15) 30° or 60° sun elevation. So, it you had a 200 watt panel on a perfect STC sort of day when solar noon was actual noon, then at 10:00 AM and 2:00 PM you could expect (200 x .87) = 174 watts from them.
 

svetz

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Are you by chance looking for the same chart type as #2 with data for a fixed panel configuration?
 

JeepDaddy

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Still not clear what you're looking for then
That chart show solar radiation intensity as a function of sun altitude. So it shows available radiation to be harvested at a different altitudes. It most likely takes into account the passage of solar radiation through the atmosphere. That is useful to know.

My question is related to the angle at which that solar radiation hits the solar panel, regardless of the sun's altitude. The power output chart from a panel at 90 degrees (100%) to 0 degrees (closer to 0%) to the sun. It's more of a function of the solar panel.
 

curiouscarbon

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That chart show solar radiation intensity as a function of sun altitude. So it shows available radiation to be harvested at a different altitudes. It most likely takes into account the passage of solar radiation through the atmosphere. That is useful to know.

My question is related to the angle at which that solar radiation hits the solar panel, regardless of the sun's altitude. The power output chart from a panel at 90 degrees (100%) to 0 degrees (closer to 0%) to the sun. It's more of a function of the solar panel.
cosine

this is the graph you are describing

1 means 100% power
0 means 0% power

multiply the watt rating of the panel by this number to get what you want.

1617741248950.jpeg


i believe you are talking about lamberts cosine law if i understand you correctly..

 
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curiouscarbon

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Possibly, agreed, that refers to the amount of solar radiation available to harvest. Does that translate directly into the panels ability to harvest it?
It’s not perfectly linear

That’s this graph maybe

1617741587331.jpeg

If i find or make a graph that combines these two functions, i’ll be sure to post it here.
 

svetz

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...My question is related to the angle at which that solar radiation hits the solar panel [vs. the panel's ability to absorb it at that angle]
Gotcha, getting that data independent of atmospheric absorption is something I've always wanted to find too; never seen it in a datasheet. It's probably dependent on the manufacturer and the local conditions (e.g., we have a lot of light-scattering here). I've heard a general rule of thumb that power isn't affected "much" if within 15°, then it starts to drop off.
 
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JeepDaddy

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i believe you are talking about lamberts cosine law if i understand you correctly.
I think that is probably the majority of it along with svetz's comments here:
Gotcha, getting that data independent of atmospheric absorption is something I've always wanted to find too; never seen it in a datasheet. It's probably dependent on the manufacturer and the local conditions (e.g., we have a lot of light-scattering here from the humidity/latitude, dawn/twilight are pretty short). I've heard a general rule of thumb that power isn't affected "much" if within 15°, then it starts to drop off.
At some point the radiation starts reflecting off the surface of the panel in a meaningful way and is not absorbed to be converted to power. Or perhaps the construction of the cells/panel plays a part in it.
 

svetz

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I suppose it would be easy enough to set up a test-bed. Put a panel on a rig that could be rotated from 0° to 90° and measure the output at high noon at various angles (high noon to eliminate atmospheric effects, the panel's output when perpendicular to the sun would give you the ratio to STC rating).
 

svetz

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From my raw data last summer solstice (sun at 88°, essentially flat panels) I can see at solar noon I'm at 100%, and at 30° sun altitude, I'm at 50% power. The chart in #2 suggests with atmospheric effects the available power should be 50%; so that says my LG Neons are freaking awesome with light hitting them at a 30° angle? Or, possibly, perhaps it was a bit hazy at solar noon that day ;-)1617743061732.png
 
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curiouscarbon

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At some point the radiation starts reflecting off the surface of the panel in a meaningful way and is not absorbed to be converted to power.

“Power (intensity) reflection and transmission coefficients”

if you wanna sweat the equations, i think those are relevant to the skimming angle condition. the reflected light will of course not be converted to power.

at this point i want to build a test setup and attach a meter lol
 

TomC4306

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I'm watching with interest. I'm thinking about a pair of 30 watt panels and a linear actuator. If the inverted v doesn't prove adequate then a 10ah battery with a little controller, charged by one of those 30w but in the same plane as the array. I look forward to your tests.
 

poorman137

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Use your latitude as a guide. If you live at 30° latitude then set your panels at 30°. Summer & winter may be a little less or more angle.
 

eXodus

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At some point the radiation starts reflecting off the surface of the panel in a meaningful way and is not absorbed to be converted to power. Or perhaps the construction of the cells/panel plays a part in it.
modern panels are far better at capturing the sun at 90 angles. And it will get better.

While trackers are a nice idea - panels are so cheap these days, Many just add panels for East and West orientations and over panel their inverters accordingly.

So you basically build 3 panel arrays - one east facing, one south (or the southern hemisphere) one west - just stationary. This allows maximum sun catching for the money. Not space efficient - but cost effective.
 

JeepDaddy

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I would agree for a stationary installation and you have space for more panels. I have an RV application with the roof being mostly shaded. I run my portable 160W panel in the sun about 25 feet from the RV. Adding more panels isn't really an option for me. I did end up building a tracker for under $50. https://diysolarforum.com/threads/diy-single-axis-solar-panel-tracker.21417/
 
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