@power.saver looking at your spreadsheet, for month 1 you list 22.21. This seems like a very low optimal angle considering most of your sun producing times are above that. What am I missing?
Panel angle
- Thread starter zƬesla
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OK, it would be interesting to see the formulas you are using. Here's the answer I came up with (because the dot product is bilinear):
For a given day, consider the sun's path across the sky. Express the path as points on the unit sphere, and integrate each coordinate over the course of the day separately. [For the x coordinate (E-W) the integral will obviously be zero.] The resulting vector will no longer be a unit vector, but the optimal panel direction is pointed along that direction.
For a longer time period, you can just add up those integrated sun coordinate vectors to get the optimal panel direction. Also, if you have a shading forecast as a function of time, you can multiply the sun coordinates by the shading factor before integrating. And if you are trying to maximize economic production according to a time of use plan, you can multiply the sun coordinates by the value of electricity at each time before integrating.
Cheers, Wayne
power.saver
Supporting Member
The angle reported by my program is measured from vertical. I do this because the program calculates everything in angles of elevation (of the sun). If you measure your panel angle (tilt) from horizontal, subtract those values from 90 degrees. So January has a tilt of 67.79 degrees.
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OK, I have a simple answer to this question now (at least for the northern hemisphere above the tropic and excluding the arctic winter):
On any day, the sun path through the sky (if we ignore the earth's orbital motion) will be a planar cut of the sky hemisphere (an arc of a circle). If we call O our location (the origin), E the east compass point (90 degree azimuth, 0 degree elevation) and W the west compass point, and S the location of the sun in the sky at solar noon (180 degree azimuth, some non-zero elevation), then O-E-S (equivalently W-E-S) defines a plane P.
On the equinoxes, the sun rises at E, hits S, and sets on W, so the sun's entire path is in the plane P. That means S is the average direction of the sun, and the solar panel should be pointed at S. During the spring/summer, the sun rises north of E, hits S, and sets north of W. That means the sun's entire path is above the plane P (except at the point S of incidence), and the optimal panel elevation will also be above the point S, shifted towards the sun's path relative to P. Conversely, in the winter, the sun rises south of E, hits S, and sets south of W. So now the sun's entire path is below the plane P, and the optimal panel elevation is below the point S.
Still working on quantifying the above.
Cheers, Wayne
power.saver
Supporting Member
Years ago I searched for such a site but didn't find any. So I wrote a program to do the calculation for myself.
If you would like me to run it for your specific location, send me the following information:
1. Location (lat/lon or nearby city)
2. If your array is fixed, the orientation (azimuth) of the panels
3. The minimum elevation (in degrees) when the sun is on the array. This removes the time when the sun in blocked by hills, trees or buildings from the angle calculation.
I'll post a table of the tilt angle for each month.
Appreciate the offer @power.saver but would like to run these on my own so I can play with params and see change factors, especially as I have winter time tree shadows that are not an issue rest of year.
I've been thinking about this question some from a purely geometric point of view, ignoring atmospheric effects. I may have an exact answer, but I don't think it's actually useful in reality, because of those atmospheric effects. It likely overvalues the contribution from times when the sun is low in the sky.
So here's a tedious way to get what should be a very accurate answer: use PVWatts. For the web application, you have to run it one tilt angle at a time, and it gives you an estimate by month of production (or you can download by hour if you like). If you're a programmer, maybe they have an API so that you could write a simple program to query it for all the tilt angles of interest.
Otherwise, you can run it, say, 10 times, for your latitude +/- 20 degrees at multiples of, say, 4 degrees. Then for each month you can look at the modeled kWh generated as a function of tilt angle. If you see an increasing trend that then reverses, you know the maximum is within that 4 degree interval, and you can run PVWatts a few more times to get the maximum down to a degree. If you don't see the trend reverse, then +/- 20 degrees wasn't broad enough, and you can extend the range on the appropriate end.
Cheers, Wayne
So here's a tedious way to get what should be a very accurate answer: use PVWatts. For the web application, you have to run it one tilt angle at a time, and it gives you an estimate by month of production (or you can download by hour if you like). If you're a programmer, maybe they have an API so that you could write a simple program to query it for all the tilt angles of interest.
Otherwise, you can run it, say, 10 times, for your latitude +/- 20 degrees at multiples of, say, 4 degrees. Then for each month you can look at the modeled kWh generated as a function of tilt angle. If you see an increasing trend that then reverses, you know the maximum is within that 4 degree interval, and you can run PVWatts a few more times to get the maximum down to a degree. If you don't see the trend reverse, then +/- 20 degrees wasn't broad enough, and you can extend the range on the appropriate end.
Cheers, Wayne
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