Help:Shading

Soulbringer

Hello,
I want to call everyone from the forum who knows how to calculate the distance between to rows of modules to prevent shading. Do you use any PV software or formulas which can help.It`s about ground-mounted installation.
Thank you

crewzer

Re: Help:Shading

SB,

I’m not aware of any available software that will meet your need. However, it seems to me that some straightforward geometry and/or an Excel spreadsheet would do the job. However, calculations for the Arctic, Antarctic, and tropical areas might get a bit complicated.


Basic inter-row spacing (distance from bottom of row n to the bottom of row n+1) is dependent on two parameters: Module length, and latitude of the array location. Here’s a way to make the calculation for my location (39 degrees N) using a module that’s 60 inches long.

Step 1: Assume the module is south-facing and tilted up from horizontal at an angle equal to local latitude. In this case, the local latitude and therefore the tilt angle is 39 degrees.

Step 2: Assume the side view of the tilted module is the hypotenuse of a right triangle. One leg of the triangle is the vertical line segment “h” from the top of the module to the ground, and the other leg is the horizontal line segment “s” from the base of the module to the intersection with leg “h”. The intersection of legs “h” and “s” is a right angle.

Step 3: The cosine of 39 degrees is equal to “s” / module length. Therefore, “s” = 0.78 x 60 inches = 46.8 inches. This is part of the inter-row distance.

Step 4: Leg length “h” = sin 39 x 60 inches = 0.63 x 60 inches = 37.8 inches.

Step 5: The balance of the inter-row distance “d” is found by using a second right triangle. This triangle consists of previous leg “h”, a hypotenuse formed by the Sun’s ray from the top of row n to the bottom of row n+1, and by “d”.

Step 6: The horizontal angle of incidence of sunlight striking the Earth’s surface on the Winter solstice is 90 - 39 - 23.5 = 27.5 degrees.

Step 7: Tan 27.5 degrees = 37.8 inches / “d”, so “d” = 37.8 inches / 0.52 = 72.7 inches.

Step 8: The inter-row distance for 60 inch long south-facing modules tilted up at 39 degrees at my location is 37.8 inches + 72.7 inches = 110.5 inches.

HTH,
Jim / crewzer

Solar Guppy

Re: Help:Shading

If your math challenged, a general rule is you need 4x the height of the modules for spacing between the rows. So if your modules are 4ft off the ground at the peak, you would need 16ft row to row

There is software one can buy that can do dusk to dawn and any day of the year at any lattitude, but is a few hundred dollars

System2

Re: Help:Shading

To calculate without having to dig out my moldy old trig book that I haven't seen in a couple of decades, I would:

1) Decide the lowest sun angle I want all panels to be completely illuminated
2) Use a sun angle calculator to find the date and time that the sun is at that angle http://susdesign.com/sunangle/
3) Determine the height from horizontal of the lower and upper edges of an array, maybe I need that trig book after all.
4) Subtract the lower edge height from the upper edge height
5) Cut a 1" X 2" piece of lumber to the length determined in step 4
6) When the sun is at the angle you selected in step 1), plumb the board with the ground
7) Measure the length of the shadow. That should be the horizontal distance between the upper edge of the first array to the lower edge of the second array and so on.

This assumes level ground and each array has the same dimensions. Otherwise, add or subtract those diffences between the first and second array, the second and third, etc.

For example, using the solar angle calculator, if I want full illumination with the sun at 20° altitude, I would put in my location's coordinates (95w 30n), today's date (Dec. 6), time zone (-6). Change the time setting until I get the altitude angle output to equal 20°. In this case, about 8:59AM. At 8:59AM on Dec. 6th, I would go outside with my precut 1" X 2" and a tape measure. With one end of the 1 X 2 on the ground, plumb the board and measure the length of its shadow. Use the result as the distance between trailing and leading edges of the arrays. Seems like that should do it.


Cheers,

Bad Apple