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How about the amount of solar energy that is reflected? That amount should be lower at a higher angle right, thus allowing more solar energy to reach the ground closer to ecuator? —Preceding unsigned comment added by 83.255.186.55 (talk) 17:13, 10 October 2009 (UTC)[reply]
This effect is included on some websites, but not in the most reliable sources. It is unclear how significant this effect actually is. On average, 19% of insolation is absorbed by cloud, but this energy is not lost. Only 6% of insolation on average is reflected back into space by the atmosphere, and perhaps the lower angle of incidence reduces the 4% that is normally reflected back into space by the earth's surface. I haven't found a website that discusses this in detail in relation to the seasons. PhysicalGeography.net has some detail, but doesn't discuss seasonal effects. The claim of a seasonal atmosphere effect has been removed. Dbfirs12:12, 3 May 2016 (UTC)[reply]
An observation for User:Michael Hardy who originated this article. You talk of a sunbeam one mile wide. Note that this width is a first order (linear) measure. You next say that "One sunbeam one mile wide falls upon the ground at a 90°-angle, and another at a 30°-angle. The one at a shallower angle covers twice as much [second order measure] area with the same amount of light."
Now, I presume that your sunbeams are things which exist within three-dimensional space and, as such, have a cross-sectional area, else you'd not have used the term "area" in your comment. Then, in fact, the one at the shallower angle of 30° will cover the same area with 1/4 the light, not one half. As you probably know, rigorously, this is not even true. It would be more correct to say that the beam at a 30° angle will cover an area 4 times as large with the same amount of light. Of course, I stand to be corrected.
Finally, are you linking from another article to this one? I sincerely hope so since I think it's exceedingly unlikely that someone will hit upon this title in a search. — Dave00:00, 7 February 2006 (UTC)[reply]
You're mistaken about 1/4 versus 1/2. If we imagine the sun shining from the south at noon, the north-south width doubles; the east-west width does not.
Not sure if it is applicable, but sunbeams are much larger then 1 mile wide, so wouldn't the angle not matter since sunbeam would be wider then earth anyways? 65.167.146.130 (talk) 19:39, 21 November 2008 (UTC)[reply]
I'm wondering if someone has suitable numbers, or a graph, showing the relationship between J/level m^2 and peak W/level m^2 against latitude. Naively, we would expect it to just be cos(lat), but things like day length and atmospheric scattering might make a significant difference?
At the very least it should be "Effect of sunlight angle on climate"...After all, it's not the sun that is angled, but the light that comes from it that is angled. BootitooB (talk) 08:40, 13 May 2019 (UTC)[reply]