Article 30421 of alt.solar.thermal: Path: news.misty.com!not-for-mail From: nicksanspam@ece.villanova.edu Newsgroups: alt.solar.thermal Subject: Re: South window shading in Texas Date: 6 May 2008 15:58:09 -0400 Organization: Villanova University Lines: 66 Message-ID: References: NNTP-Posting-Host: acadia.ece.villanova.edu X-Trace: max.inside.misty.com 1210100264 1915 153.104.44.130 (6 May 2008 18:57:44 GMT) X-Complaints-To: abuse@misty.com NNTP-Posting-Date: Tue, 6 May 2008 18:57:44 +0000 (UTC) Xref: news.misty.com alt.solar.thermal:30421 >NREL recommends As = 108-lat and Aw = 71-lat, based on no shading at solar >noon at 11/17 and 1/25 and complete shading at solar noon on 5/12 and 8/2... Maybe they used Cooper's 1969 equation (line 60 below): DS=23.45*SIN(2*PI*(N+284)/365) A more accurate equation from Spencer (1971) and Iqbal (1983): d = 0.006918 -0.399912cosB +0.070257sinB -0.006758cos2B +0.000907sin2B -0.002697cos3B +0.00148sin3B in radians, 20 PI=4*ATN(1) 30 DATA 25,319,320,321,322,323 40 FOR I=1 TO 6 50 READ N 60 DS=23.45*SIN(2*PI*(N+284)/365) 70 B=(N-1)*2*PI/365 80 D=.006918-.399912*COS(B)+.070257*SIN(B) 90 D=D-.006758*COS(2*B)+.000907*SIN(2*B) 100 D=D-.002697*COS(3*B)+.00148*SIN(3*B) 110 D=180*D/PI 120 IF N=25 THEN DS25=DS:D25=D 130 PRINT N,D,DS,DS-DS25,D-D25 140 NEXT I 25 -19.16716 -19.26362 0 0 319 -18.30314 -19.14784 .1157837 .8640156 320 -18.55995 -19.37802 -.114399 .6072045 321 -18.81123 -19.60246 -.3388405 .3559265 322 -19.05686 -19.8211 -.5574723 .1102982 <-- 323 -19.29675 -20.03387 -.7702427 -.129591 says day 322 (11/18) would be a better match to 1/25... Iqbal includes the equation of time, but it doesn't look easy to find a matching date for a given day's declination. And Cooper's equation can be improved. It's fine for lots of purposes, but there's no fall day that perfectly matches a given spring day's declination, and how can we average in leap years? If sin(360(355+C)/365) = -1 on 12/21 and 360(355+C)/365 = 270, C = -81.25, equivalent to 365-C = 283.75. If the peak happens 365/2-10 = 172.5 days sooner, spring day N = 25 will match fall day 2x172.5-N = 345-N = 320: 20 PI=4*ATN(1) 30 DATA 25,319,320,321 40 FOR I=1 TO 4 50 READ N 60 DS=23.45*SIN(2*PI*(N+283.75)/365) 70 IF N=25 THEN DS25=DS 80 PRINT N,DS,DS-DS25 90 NEXT I 25 -19.32099 0 319 -19.0894 .2315941 320 -19.32102 -2.861023E-05 :-) 321 -19.54689 -.2258988 Ah, that feels much better. Now how about leap years? Nick