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the trigonometry of solar power |
I am not flagging this as "off topic" because it is math/engineering...sort of. Heartland wife and I are talking about putting our money where are liberal mouths are and installing solar panels. It turns out you can get solar shingles...look like regular shingles except a little shinier and blacker, and also keep the rain out. But, you are stuck with the orientation of your roof, while with regular solar panels you can install them at an angle (which might be ugly). We have sections of roof that face to the east-southeast, to the south, and to the west-northwest. The slope (for geology, dip) of the roof is 14 deg everywhere. Normally, you would put solar on the S-facing roof. But, that part of the house is shaded after 1 PM. The ESE-facing part of the roof is shaded until about noon (an hour or so before solar noon). The WNW-facing part of the roof is sunny from mid morning until late afternoon. Heartland wife derived an equation that gives the inclination of the sun at noon relative to a plane (roof) of any orientation. You need to know that at the equinoxes the sun inclination at solar noon = (90-your latitude) so for us at latitude 39, the inclination is 51. For our WNW-facing (300 deg azimuth) roof, the maximum inclination of sun relative to roof is 20 deg at winter solstice, 43 at the equinoxes, and 64 deg at the summer solstice. We would get about half the watts per meter squared at a relative inclination of 30 deg. So, we would get little power production from, say, late October to early February from this roof. But, the south-facing roof would be less shaded in winter (no leaves), so we could wire up both. Now have to look at cost. Follow Ups: ● Re: the trigonometry of solar power - Canie 15:29:36 - 10/7/2007 (72742) (1) ● Re: the trigonometry of solar power - heartland chris 06:11:47 - 10/9/2007 (72748) (0) |
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