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PaperclipPerfector  8mo ago (text post)   3499 thread views

Small-Scale Question Sunday for March 17, 2024

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I'm trying to calculate the strength of solar radiation for a given latitude as a simple factor from 1 (equatorial) to 0 (polar), assuming no axial tilt or any other complications. My trigonometry is weak. How do I do this?

Arccos curve looks good. Is it arccos? Arccos(latitude / 90) * (2 / pi) looks pretty decent.

Edit: It was Cos all along!

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.

arccos is gonna give you too sharp a result near the equator (i.e. predict that the last few degrees as you get closer matter the most). What you want is just cos(latitude/90 * pi/2).

edit: the way you visualise this is by holding a square piece of paper in front of you, and tilting it until you're looking at it edge-wise. The "visual area" of the piece of paper in your field-of-view is what will give you the proportionality factor.

Yep, that does the trick. Thanks!

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