File 15326598517.png - (21.74KB , 1115x717 , theta.png )

Here's part of what I'm doing. Assuming a panel can rotate side to side and track the sun, but cannot rotate up and down, these are the angles of inclination of the panel for a given latitude that will yield the maximum energy production throughout the year. The problem is that this assumes that as long as you point the panel directly at the sun, you will generate the same amount of power at noon as you would just before dusk. So for example at the equator the optimal angle is predicted to be 42°, rather than something flatter. At the same time, energy production is predicted to be higher in Antarctica (99% efficiency) than at the equator (92%), which is nonsense.

Ok with out much context about what the hell you're using to make that graph.

"efficiency" can mean a few different things.

efficiency of the solar panels increases with lower temperature.
There is also financial efficiency of longer daylight hours.

So because I don't really get what you're asking I'm just going to explain some things.

First the sky is blue because of diffraction,the sky filters out blue light more quickly than red light.

This is why the sun is more red looking as it sets and why clouds turn red during sun sets.

So can see when the sun is losing power.

Now how much power are you losing? well that's more of a question of your solar panels than and what spectrum of light they absorb. If your panels are tracking the sun and its not setting i doubt its very much at all.

ok after re-reading this, Im to wrecked to be explaining shit.

Just watch this
https://www.youtube.com/watch?v=ESAPg7w3wm8

Skip to 34:00
because the first part is about polarization.

https://www.pveducation.org/pvcdrom/properties-of-sunlight/calculation-of-solar-insolation

Also i think you're asking more about insolation at the end than efficiency.

Just so you know, you're replying to a question I asked almost a year ago.

I figured it out myself. I cobbled together a function f(h) that approximates atmospheric density (or attenuation. I can't remember which one) as a function of altitude, and another function that describes the non-linear curve for the altitude that a photon would follow when the sun is at a given azimuth. Then it was just a matter of integrating numerically.
That gave the correct result that power production is much greater at the equator than at the poles, as well as determining the optimal inclination for the panel.

And no, the word I was looking for was "efficiency". If you consider a solar panel always pointed directly at the sun as 100% efficient (there's no way to generate more energy than that), how much less efficient is a panel that has its rotation restricted on one of its axes? The answer is the integral of cos(angle_of_panel - angle_of_sun(t)) over the period that the sun is above the horizon.

Some of us start at the first page and work through responding to interesting posts without looking at when they were posted.
I looked too because solar is my main hobby.

No way you have the same working in Canada. We have sunny days only in summer, and we don't use it. Also it is very dark in our houses due to that. So I am sure it will not work.

>>18170
Sorry to hear that. Try moving to a less shitty latitude?

You could try to find a climatic compromise, for example. Suppose you get solar panels power supply in summer and traditional electricity in winter. How about that? As for house darkness you can get to newmarket window replacement https://thwindowsdoors.com/newmarket-window-replacement/ and change your small old windows to multiple new ones. This will attract some extra light, I assume.

Get more solar absorbent material...