<html><head></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; ">Jim,<div>Obviously, your latitude will make a big difference. The further north you go the wider the summer to winter sun angle. Also, I'd want to know if that data was taken using just I/V curves or through a controller which had MPPT capabilities. In a real world situation with MPPT, I'm guessing that matching the array tilt to the sun angle would make more than a 6% difference at 40°+ latitudes.</div><div>Besides, what's wrong with an extra 6%??</div><div>Cheers, bob-O</div><div><br><div><div>On Feb 19, 2011, at 8:03 PM, North Texas Renewable Energy Inc wrote:</div><br class="Apple-interchange-newline">
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<div><span class="765552802-20022011"><font face="Arial">At least that's the
conclusion of Gottfried Wilhelm Leibniz. </font></span></div>
<div><span class="765552802-20022011">"The largest difference of the
[PV] plant yield was less than 6% for tilt angles between 0° and
70°."</span></div>
<div><span class="765552802-20022011"><font face="Arial">This begs the question,
where did the notion that tilt to latitude is critical come from. Surely
NREL or someone else has tested this concept before. </font></span><span class="765552802-20022011"><font face="Arial">Anyway if N-S angle energy
production loss is only 6% to +/-35° then E-W should be too,
right? But it's not. </font></span></div>
<div><span class="765552802-20022011"><font face="Arial">Here's why. If you measured
irradiance at 10°-70° only at noon over 12 months, the air mass would at
its minimum during the entire test and so irradiance
deviation would be too. AM would not be constant at +/- 35° E-W which has
been verified by NREL and others for a long time, AM increases the further from
solar noon the sun gets. </font></span></div>
<div><span class="765552802-20022011"><font face="Arial">But if the Earths tilt is
23.5 degrees and Gottfried measured to 35 degrees, the difference is 11.5
degrees at summer and winter solstice. And if your array angle is +/- 11.5
deg from true south, rule of thumb is that irradiance losses are minimal. Maybe
only 6% or so. </font></span></div>
<div><span class="765552802-20022011"><font face="Arial">This puts the significance
of array tilt in a whole new light. Pun intended... </font></span></div>
<div><span class="765552802-20022011"><font face="Arial"><span class="765552802-20022011">Of course there is a fee to download the entire
document but the abstract is here </span></font></span></div><font face="Arial"></font>
<div><span class="765552802-20022011"></span><strong><a href="http://tinyurl.com/4zf2syk">http://tinyurl.com/4zf2syk</a></strong></div>
<div><font face="Arial"><a href="http://www.scopus.com/record/display.url?eid=2-s2.0-78951495350&origin=inward&txGid=kX6CkwoH_w_VL01NbmaciIC:2">http://www.scopus.com/record/display.url?eid=2-s2.0-78951495350&origin=inward&txGid=kX6CkwoH_w_VL01NbmaciIC%3a2</a></font></div>
<div><strong><font face="Arial" size="2"></font></strong><font size="2"><font face="Arial"><strong></strong></font></font> </div>
<div><font size="2"><font face="Arial"><strong></strong></font></font> </div>
<div><font size="2"><font face="Arial"><strong>Jim
Duncan</strong></font></font></div>
<div><font face="Arial" size="2"><strong>North Texas Renewable
Energy</strong></font></div>
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