Others have mentioned the varying thickness of the atmosphere that the sunlight has to pass through, but cloud cover is the more important factor. The map you are looking at seems to include cloud cover in its calculations.
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You're probably right, but atmosphere depth isn't a minor impact. It's visually observable from almost anywhere in the world by watching a sunrise or sunset in a clear day: the sun, which can't be comfortably stared at at noon, is often watchable as it approaches and sinks behind the horizon. At even 0% occlusion, the ambient light levels are dramatically lower than at noon, and that's due almost entirely to just the atmosphere.
Granted the line between clouds and "pure" atmosphere is fuzzy. There's a lot of particulate matter, smog, pollen, non-cloud-formed water vapor.
The suns rays go through less atmosphere near the ~~poles~~ equator. Not to scale, but should show the overall concept.
"The sun rays go through less atmosphere near the poles"?
I think you meant to say sunlight goes through more atmosphere near the poles, but otherwise nice quick sketch diagram 👍
I had so much fun making the image I totally the words
I totally the words
The whole thing!
You can actually see this in action. At sunrise or sunset, it is possible to look directly toward the sun. That's because more light is scattered at that angle and so it is less direct. At noon, the same sun will sear your eyes.
Please note whilst the jist of this diagram is correct, it's not drawn properly. The sun is so far away and much larger than the Earth. This means sunlight is about as parallel as it can be once it gets to Earth. So the lines aren't going through the atmosphere at different angles. The angle is the same, but since the Earth is a sphere it will travel through more atmosphere before hitting the ground.
Helpful diagram. Thank you!
isn't the atmosphere transparent for most of the light, though? (except UV, but that gets filtered out even at the equator, so it's the same everywhere again)
The atmosphere is mostly transparent; see the table near the bottom of this Wikipedia page for some numbers. At an angle of 45 degrees from vertical you're getting 91% of the energy, and at 60 degrees you're getting 81%. A bigger problem is seasonal variation: during the winter at high latitudes you get very little energy. My city, at around 47 degrees latitude, sees the Sun peak at less than 30 degrees with under 10 hours of daylight for a quarter of the year. A solar array isn't as useful if it produces almost no power for much of the year, especially when people need a lot to keep warm.
It isn't perfectly transparent. It has dust, moisture, and other particles in the air that block or deflect a portion of the light.
Lie on the beach in winter: the sun is pathetically un-warm.
Adjusting the angle at which you recline won't make the sun any warmer.
So your intuition is off here.
Imagine someone puts an opaque shell around the earth at cruising altitude, and cuts a metre-square window in it.
Put that window directly over the equator, wait for noon. You will have a metre-square patch of sunlight on the ground.
As the sun heads towards the horizon, the patch of sunlight stretches into a long east-west tail, just like a long shadow.
The same amount of energy is coming through the window, but it's spread over a much larger area on the surface, so there's fewer watts-per-square-cm hitting the ground.
Now move that window north 30 degrees. Wait until noon, and the patch is already smeared into a long tail north-south, and that's before applying any east-west smear. As the sun heads for the horizon, it's going to be even more spread out into a great big enormous oblong, extended in both directions.
Now, entirely replace that shell with windows, and through the raytracing gets more involved, the same principles are at work.
That's why mornings and evenings are cold, that's why winter is cold, that's why it gets colder towards the poles. You're getting a smaller and smaller share of the sunlight hitting the area where the earth is.
Really great visuals.
This covers much of it. Also inderect sunlight matters too. So any percentage non sky (groud/horizon) the panel "sees" wil have less indirect sunlight than the (blue or even cloudy) sky going towards the panel.
It's not just the cloud cover and going through more atmosphere, it's also the amount of energy per square meter hitting the ground. But a large part of it is most solar panels aren't tracking and even if they are it's usually in the horizontal and not in the vertical.
It's impractical to mount solar panels at such an extreme angle, but it also won't help very much since the sun is so close to the horizon shadows will be terrible. Imagine two rows of panels, once you set up the first one almost vertical, the second row won't get any sun. And that's if there is even a clear view to the horizon, in most places that's not true. It's also very hard to mount panels at such an extreme angle because the wind will catch it more easily. Mounting flush to the roof is usually preferred, or at a fixed angle with struts for flat roofs.
Because the panels don't track, higher latitudes are less efficient, as the sun varies more in angle during the year. From just peaking out over the horizon in winter, to high in the sky in summer.
There's several factors here.
The most important probably being the energy per square meter. Higher latitudes gets less energy per square meter than equatorial latitudes.
There's also the mentioned cloud cover and atmosphere density.
The climate it's also important. As higher latitudes tends to be more cloudy.
And sun hours, here is not about the total energy but how it's distributed. As sun hours are more estable near the equator (12 hours light 12 hours dark) while in higher latitudes you can get 4 hours light some times of the year that can't amount to nothing, and 20 hours of day other times of the year which are nice, but there's no way to store that energy for the winter lack of sunlight.
I see what you're getting at here. The solar constant is the solar constant. If you've got a perfect angle to the sun, you should be getting the same amount of power regardless of latitude. I mean I suppose it's possible there might be a slight attenuation with the sun at a lower angle due to there being more atmosphere to traverse? Otoh solar panels don't function as efficiently at high temperatures, so it's possible they may be more efficient in some cases.
But you have to consider that averaged out, you're looking at shorter daylight hours overall at high latitudes, even if there are periods in mid-summer when days can be super long, so that's a consideration. So yes, the panels should pull in similar amounts of power while the sun is up, but it's not up as much.
Solar panels are more efficient closer to the equator because of the most direct light from the sun. At higher or lower latitudes, there is more atmosphere for the light to pass through. The actual distance from the sun is basically irrelevant without an atmosphere. There might be a measurable difference based on distance alone but not much.
Efficiency does not generally equate to optimal power generation. There are probably hundreds of other variables that directly translate to maximum power production.
Assuming an ideal mounting angle...
Near the equator, sunlight has less distance of atmosphere to travel through than at higher or lower latitudes, generally speaking.
Then there's also the average of the weather patterns for different areas, like some areas get way more cloud cover and rain than others.
And then there's also the air pollution factor, and since significantly more people live in the northern hemisphere than the southern, of course the northern hemisphere has more air pollution.
I hope my insight on this has been helpful.
I'm no specialist, but (1) don't most solar panels have a fixed inclination and orientation? Depending on the location that will not be as optimal at each day of the year because of the tilt of the earth. (2) The weather/cliimate will have an effect too. Your 2nd image is apparently showing kWh/kWp - so the longer term yield (h=hours) divided by the theoreticall maximum (p=peak). So it shows longer term efficiency , vs the yield at a certain moment like you appear to be focusing on.
Since the kW part can cancel out, the resulting kWh/kWp value is basically measured in hours. There are 8,766 hours in a year and half of those are at night, so these numbers would make sense if you think of them as "this is how many hours of peak production equivalent you will actually get each year". You're in the Sahara, you get the equivalent of 2,400 hours of peak production. You're in Finland, you get the equivalent of 1,000 hours. If it actually magically ran at the peak production value all year 24/7, you get 8,766.