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PaperclipPerfector  1mo ago (text post)   2593 thread views

Small-Scale Question Sunday for March 17, 2024

Do you have a dumb question that you're kind of embarrassed to ask in the main thread? Is there something you're just not sure about?

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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!

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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