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Notes -
Why arccos? Like ignoring diffuse horizontal irradiance, .......
The insolation is just proportional to the incident cross-section, or sin(90° - ϕ) = cos(ϕ), where ϕ is the latitude.
you can check for yourself though. The instantaneous insolation is given by
Q = S_0 * (d_bar / d)^2 * (sin ϕ sin δ + cos ϕ cos δ cos h)
In your case S_0, d_bar, and d are constant (the solar constant, mean distance, actual distance)
δ is the declination angle, 0 in your case, so sin δ -> 0, cos δ -> 1.
The hour angle h, or deviation from local solar noon you can consider for h = 0, since you can scale to the max if everyone has a 12 hour day. Or cos h -> 1
giving Q ∝ cos ϕ.
Edit: I see @JhanicManifold already answered and got the same thing. Obviously I mean like cos of degrees latitude and he was working in cos of radians and converting the degrees with the factor of 1/90 * pi/2
Thanks, that will come in handy when I decide to complicate matters.
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