How to calculate azimuth and elevation angle?

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RegGuheert

Well-known member
Joined
Mar 19, 2012
Messages
6,419
Location
Northern VA
I'm stumped.

I'm helping a friend install a PV array on his workshop. He wants to mount the array on a 4:12-pitch roof which faces about 30 degrees north of West. In order to get the panels to face South, the intent is to mount the rails at *some* oblique angle to the roof and then use an elevation kit to raise the backs of the panels.

But I've come to the conclusion that as you raise the elevation of the panels relative to the roof, both the true azimuth and the elevation of the panels also change. So the question becomes: If I want the solar panels to face true south with a particular elevation, at would angle relative to the roof do I run the rails and at what elevation angle relative to the roof do I raise the panels?

I've tried to find an online calculator to provide the results, but I haven't found anything. Does anyone know of one? If not, is there a simple transformation that I can perform to calculate the true azimuth and elevation? TIA!
 
So if I understand correctly you are changing the angle of the panels and also running the rails NOT parallel to the building? That seems like some fun math.

So you know the buildings azimuth and you know that the rails are going to go up from the edge of the roof at X angle from parallel. So it seems like the geometry of this is that you need to create a triangle where the rail ends. The roof being one side, the other side being the height above the start of the rail at the roof and then the third side being how far into the building you are, level at the end of the rail. Then with that last number you can calculate the angle change when flat to give you the azimuth modifier.

As for the end result of the angle of the panels, I have no idea how to think about that.
 
Thanks, QueenBee!

Unfortunately, I'm having a bit of trouble with this part:
QueenBee said:
Then with that last number you can calculate the angle change when flat to give you the azimuth modifier.
I don't think there can be a single number to modify the azimuth since I believe it changes continuously as you increase the tilt angle WRT the roof. For instance, the azimuth is always 30 degrees north of West when mounted flat, regardless of the angle chosen for the mounting rails.
 
RegGuheert said:
Thanks, QueenBee!

Unfortunately, I'm having a bit of trouble with this part:
QueenBee said:
Then with that last number you can calculate the angle change when flat to give you the azimuth modifier.
I don't think there can be a single number to modify the azimuth since I believe it changes continuously as you increase the tilt angle WRT the roof. For instance, the azimuth is always 30 degrees north of West when mounted flat, regardless of the angle chosen for the mounting rails.

Yeah yeah, agreed, the first part only gets you what the azimuth of the rails is then somehow gets modified by the additional angle which I have no idea how to even start thinking about. But as you said if the panels were installed flat to the roof but not parrallel the azimuth and angle are still the same as if they were parrallel.
 
RegGuheert said:
I'm helping a friend install a PV array on his workshop. He wants to mount the array on a 4:12-pitch roof which faces about 30 degrees north of West. In order to get the panels to face South, the intent is to mount the rails at *some* oblique angle to the roof and then use an elevation kit to raise the backs of the panels.
OK, I'm going to work this out as I type, so bear with me.

I think the way to approach this is to use vector math on the vector orthogonal to the face of the panels. Vector length will be arbitrary, as we are only interested in direction.

So let's have the x-axis point east, the y-axis point north, and the z-axis point up.

If the 4:12 pitch roof faced due south, the orthogonal vector would be (0,-4,12) or (0,-1,3).

Since the building is turned 120 degree CCW from above, this become (-sin 120, -cos 120, 3) = (-sqrt(3)/2 , 0.5, 3)

Say you want the panels to face south at an elevation of 40 degrees. That vector would be (0,-sin 40, cos 40).

The cosine of the angle between the vectors is their dot product divided by their lengths. The angle is

arccos((0 + cos120 sin40 + 3 cos 40)/sqrt(10)) = arccos(0.625) = 51 degrees.

So you would need to an elevation kit that would rotate your panels 51 degrees, not sure if that is practical. We still need to find what direction to run the rails.

The rails would need to run perpendicular to both the pre and post rotation orthogonal vectors, i.e. perpendicular to both (-sin120, -cos120, 3) and (0, -sin 40, cos 40). That would be in the direction of their cross product. So the direction is given by

(-cos 120 * cos 40 - 3 * (-sin 40), 3 * 0 - (-sin 120) * cos 40, -sin 120* (-sin 40) - (-cos 120) * 0) = (2.31, 0.766, 0.557)

If the rails ran horizontally across the roof, their direction would be (cos 120, -sin 120, 0). So the angle between this and the desired rail direction is

arccos( (2.31 * cos 120 - 0.766 * sin 120 + 0) / sqrt (2.31^2 + 0.766^2 + .557^2) ) = 137 degrees.

Since the rail directions are only defined upto +-1, this is really 180 - 137 = 43 degrees. I haven't calculated whether that's CW or CCW, but from the roof it should be obvious which way to go.

The upshot is that for the inputs of a 1:3 slope, 120 degrees off south, and a desired south elevation of 40 degrees, you need to rotate the rails 43 degrees from the horizontal on the roof, and you need to elevate the panels 51 degrees relative to the roof. If I didn't make any calculation errors.

You could easily check this with a very small model. For other values of the inputs, repeat the calculation above substituting the new inputs.

If you have more than one row of panels, you will have to figure out how far apart to space the rows to avoid shading.

Cheers, Wayne
 
wwhitney said:
RegGuheert said:
I'm helping a friend install a PV array on his workshop. He wants to mount the array on a 4:12-pitch roof which faces about 30 degrees north of West. In order to get the panels to face South, the intent is to mount the rails at *some* oblique angle to the roof and then use an elevation kit to raise the backs of the panels.
OK, I'm going to work this out as I type, so bear with me.

I think the way to approach this is to use vector math on the vector orthogonal to the face of the panels. Vector length will be arbitrary, as we are only interested in direction.

So let's have the x-axis point east, the y-axis point north, and the z-axis point up.

If the 4:12 pitch roof faced due south, the orthogonal vector would be (0,-4,12) or (0,-1,3).

Since the building is turned 120 degree CCW from above, this become (-sin 120, -cos 120, 3) = (-sqrt(3)/2 , 0.5, 3)

Say you want the panels to face south at an elevation of 40 degrees. That vector would be (0,-sin 40, cos 40).

The cosine of the angle between the vectors is their dot product divided by their lengths. The angle is

arccos((0 + cos120 sin40 + 3 cos 40)/sqrt(10)) = arccos(0.625) = 51 degrees.

So you would need to an elevation kit that would rotate your panels 51 degrees, not sure if that is practical. We still need to find what direction to run the rails.

The rails would need to run perpendicular to both the pre and post rotation orthogonal vectors, i.e. perpendicular to both (-sin120, -cos120, 3) and (0, -sin 40, cos 40). That would be in the direction of their cross product. So the direction is given by

(-cos 120 * cos 40 - 3 * (-sin 40), 3 * 0 - (-sin 120) * cos 40, -sin 120* (-sin 40) - (-cos 120) * 0) = (2.31, 0.766, 0.557)

If the rails ran horizontally across the roof, their direction would be (cos 120, -sin 120, 0). So the angle between this and the desired rail direction is

arccos( (2.31 * cos 120 - 0.766 * sin 120 + 0) / sqrt (2.31^2 + 0.766^2 + .557^2) ) = 137 degrees.

Since the rail directions are only defined upto +-1, this is really 180 - 137 = 43 degrees. I haven't calculated whether that's CW or CCW, but from the roof it should be obvious which way to go.

The upshot is that for the inputs of a 1:3 slope, 120 degrees off south, and a desired south elevation of 40 degrees, you need to rotate the rails 43 degrees from the horizontal on the roof, and you need to elevate the panels 51 degrees relative to the roof. If I didn't make any calculation errors.

You could easily check this with a very small model. For other values of the inputs, repeat the calculation above substituting the new inputs.

If you have more than one row of panels, you will have to figure out how far apart to space the rows to avoid shading.

Cheers, Wayne
Thanks, Wayne! I'm impressed!

That's exactly what I'm looking for! Right now we have a fixed-tilt kit (and I'm not sure of the exact tilt, since I believe it depends on the distance between the rails), so let me see how much South-pointing elevation we can achieve and which way the rails need to run. I'll post back once I figure out how that all works out. Perhaps a simple computer program is in order!
 
RegGuheert said:
wwhitney said:
(-cos 120 * cos 40 - 3 * (-sin 40), 3 * 0 - (-sin 120) * cos 40, -sin 120* (-sin 40) - (-cos 120) * 0) = (2.31, 0.766, 0.557)

If the rails ran horizontally across the roof, their direction would be (cos 120, -sin 120, 0). So the angle between this and the desired rail direction is

arccos( (2.31 * cos 120 - 0.766 * sin 120 + 0) / sqrt (2.31^2 + 0.766^2 + .557^2) ) = 137 degrees.

Since the rail directions are only defined upto +-1, this is really 180 - 137 = 43 degrees. I haven't calculated whether that's CW or CCW, but from the roof it should be obvious which way to go.
Thanks, Wayne! I'm impressed!

That's exactly what I'm looking for! Right now we have a fixed-tilt kit (and I'm not sure of the exact tilt, since I believe it depends on the distance between the rails), so let me see how much South-pointing elevation we can achieve and which way the rails need to run. I'll post back once I figure out how that all works out. Perhaps a simple computer program is in order!

And that's why I'm in IT instead of computer engineering/science. I'm thoroughly impressed.
 
O.K. It looks like we will be installing these panels on Friday so I need to finalize my calculations pretty soon. Since this system will be used in an off-grid application, I want to get the tilt as steep as possible. My goal is 50 degrees (actual elevation), but I may need to settle for a lower angle if that is unachievable with the tilt kit which was purchased.

That kit does not appear to be a true 4-bar linkage, but rather the two different-length legs appear to be perpendicular to the PV module when mounted. That means my only degree of freedom when it comes to angle is how close together I mount the rails and therefore how much of the PV module is cantilevered above the top connection point.

So, that brings me to my two questions:

Q1) What is the maximum amount of PV panel which you would allow to cantilever above the top rail? For reference, here is a datasheet for the modules which we are mounting: LG265S1C-B3. Unlike panels which I have used before, these panels are thin and light. Thickness is only 1.38".

Q2) What is the minimum spacing you would allow between the two rails when mounting PV modules which will have an elevation angle of about 50 degrees from horizontal?

Since it may impact wind loading slightly, I will point out that there are quite a few deciduous trees around the building where we are mounting the panels.

Thanks in advance for any thoughts!
 
RegGuheert said:
Since this system will be used in an off-grid application, I want to get the tilt as steep as possible.
I was going to ask you why you were going through so much trouble to do this goofy looking install but that definitely makes sense.
RegGuheert said:
Q1) What is the maximum amount of PV panel which you would allow to cantilever above the top rail? For reference, here is a datasheet for the modules which we are mounting: LG265S1C-B3. Unlike panels which I have used before, these panels are thin and light. Thickness is only 1.38".

Q2) What is the minimum spacing you would allow between the two rails when mounting PV modules which will have an elevation angle of about 50 degrees from horizontal?

I've always tried to follow the manufacturer's guidelines on where to install the rails. For example looking in the install manual they show either using the predrilled bolt holes or with clamps a range that starts at 270mm and ends at 400 mm from the top/bottom. Page 8 of the install manual.
 
RegGuheert said:
He wants to mount the array on a 4:12-pitch roof which faces about 30 degrees north of West.
I take it either this is a shed roof, so there is no other side, or there is some other reason not to use the other side of the roof, that faces 30 degrees south of East?

How big is the roof, and how many panels are you putting up?

RegGuheert said:
My goal is 50 degrees (actual elevation), but I may need to settle for a lower angle if that is unachievable with the tilt kit which was purchased.
The web info you linked to doesn't give the length of the long leg and the short leg. Do you have that info?

RegGuheert said:
Q1) What is the maximum amount of PV panel which you would allow to cantilever above the top rail?
The tilt kit manual says to limit the cantilever to a maximum of 15% of the panel length. Without knowing anything else or doing a wind load calculation, I would think that would be OK. Your panels are 65 inches long, so that would be 9.75 inches.

Cheers, Wayne
 
Let me reprise my calculation in pseudo-code.

inv_slope: the roof slope is 1:inv_slope
house_angle: CW angle in plan view the roof direction deviates from due south
target_elevation: desired elevation of the panels in degrees from straight up

roof_normal = (-sin(house_angle), -cos(house_angle), inv_slope)
panel_normal = (0, -sin(target_elevation), cos(target_elevation))
tilt_angle = arccos( roof_normal *dot* panel_normal / length(panel_normal) )

rail_vector = roof_normal *cross* panel_normal
roof_horizontal = (cos(house_angle), -sin(house_angle), 0)
rail_angle = arccos( rail_vector *dot* roof_horizontal / length(rail_vector) )

For inv_slope = 3, house_angle = 120 degrees, and target_elevation = 50 degrees, I get

roof_normal = ( -0.866, 0.5, 3)
panel_normal = (0, -0.766, 0.643)
tilt_angle = 60.7 degrees

rail_vector = (2.620, 0.557, 0.663)
roof_horizontal = (-0.5, -0.866, 0)
rail_angle = 130.5 degrees (or 49.5 degrees)

Cheers, Wayne
 
QueenBee said:
I've always tried to follow the manufacturer's guidelines on where to install the rails. For example looking in the install manual they show either using the predrilled bolt holes or with clamps a range that starts at 270mm and ends at 400 mm from the top/bottom. Page 8 of the install manual.
Good info! I found this installation manual which seems to extend the clamp region from 270mm to 500mm (page 7). Frankly, that seems like a pretty wide range, since it would allow the clamps to be only 650mm apart with 500mm overhangs.
wwhitney said:
The tilt kit manual says to limit the cantilever to a maximum of 15% of the panel length. Without knowing anything else or doing a wind load calculation, I would think that would be OK. Your panels are 65 inches long, so that would be 9.75 inches.
Interestingly, the range specified by IronRidge (0 to 246mm) does NOT overlap the range provided by LG (270mm to 500mm). Since it is not possible to comply with both specifications, I will have to choose what to do.

I have no choice except to clamp (or bolt) close to the end on the lower side of the panel in order to prevent the bottom of the panel from hitting the roof surface. But on the upper end, I am going to make the first calculations with the clamps (or bolts) 500mm from the top of the panel and see what I get.

Since I can access the backs of the panels, I'm going to see if I can mount by using the bolt holes in the panels (although that requires almost twice as many bolts). I think that might give the best support to the rails (which are cantilevered off the ends of the tilt kits).
wwhitney said:
RegGuheert said:
He wants to mount the array on a 4:12-pitch roof which faces about 30 degrees north of West.
I take it either this is a shed roof, so there is no other side, or there is some other reason not to use the other side of the roof, that faces 30 degrees south of East?
There is an eastern roof, but it is much shorter than the western side. The result is that the sunniest portion of the roof is on the western portion near the peak due to trees being on both sides.
wwhitney said:
How big is the roof, and how many panels are you putting up?
I don't recall the exact dimensions of the roof. We are mounting two rows of five panels each on a set of four 18' rails. I'm pretty sure the long rails will not fit on the eastern roof, even if we cut them down to 17'.
wwhitney said:
The web info you linked to doesn't give the length of the long leg and the short leg. Do you have that info?
It's a bit hard to come by, but I'm pretty sure the short leg is 6" while the long leg is 40". This kit is discontinued, but even the old catalog doesn't have the details.
wwhitney said:
Let me reprise my calculation in pseudo-code.

inv_slope: the roof slope is 1:inv_slope
house_angle: CW angle in plan view the roof direction deviates from due south
target_elevation: desired elevation of the panels in degrees from straight up

roof_normal = (-sin(house_angle), -cos(house_angle), inv_slope)
panel_normal = (0, -sin(target_elevation), cos(target_elevation))
tilt_angle = arccos( roof_normal *dot* panel_normal / length(panel_normal) )

rail_vector = roof_normal *cross* panel_normal
roof_horizontal = (cos(house_angle), -sin(house_angle), 0)
rail_angle = arccos( rail_vector *dot* roof_horizontal / length(rail_vector) )

For inv_slope = 3, house_angle = 120 degrees, and target_elevation = 50 degrees, I get

roof_normal = ( -0.866, 0.5, 3)
panel_normal = (0, -0.766, 0.643)
tilt_angle = 60.7 degrees

rail_vector = (2.620, 0.557, 0.663)
roof_horizontal = (-0.5, -0.866, 0)
rail_angle = 130.5 degrees (or 49.5 degrees)

Cheers, Wayne
Cool! Thanks again, Wayne!
 
RegGuheert said:
QueenBee said:
I've always tried to follow the manufacturer's guidelines on where to install the rails. For example looking in the install manual they show either using the predrilled bolt holes or with clamps a range that starts at 270mm and ends at 400 mm from the top/bottom. Page 8 of the install manual.
Good info! I found this installation manual which seems to extend the clamp region from 270mm to 500mm (page 7). Frankly, that seems like a pretty wide range, since it would allow the clamps to be only 650mm apart with 500mm overhangs.

Oh, sorry, math is hard. Both your direct link and the one that I found on the LG datasheet site you linked to say the same thing. 270+230=500mm not 400mm ;) That does seem like a really wide range as that's only a couple inches away from putting the rails at the third mark
 
RegGuheert said:
It's a bit hard to come by, but I'm pretty sure the short leg is 6" while the long leg is 40".
With the legs perpendicular to the roof and a 34" height differential, to achieve a 60 degree roof-relative elevation you'd need to space the supports only 34/tan 60 = 19.6 inches apart. With a 6" lower leg, you could cantilever the bottom a bit less than 6/sin 60 = 6.9"; call it 6". So you'd have to cantilever the top 40" which is crazy.

Any chance you can turn the panels and fit them in "landscape" orientation? They're only 39" in the dimension, so you'd only have to cantilever the top 14" which seems perhaps doable.

RegGuheert said:
Thanks again, Wayne!
Better check my results, although they look plausible to me. You could do a physical check easily with some small squares of plywood, some small squares of plywood, and a hot glue gun or a brad gun.

Cheers,
Wayne
 
P.S. Have you considered whether during the morning the roof will shade the array, or the first row will shade the second row? I can't quite visualize it. If morning shading is unavoidable, then the optimal orientation will be somewhat west of south, which might reduce your roof-relative elevation requirements.

Cheers, Wayne
 
Coming into this late, and fortunately Wayne's already done the nasty calcs (I'd have had to look up the equation), but I have a question. Since this is off-grid, is there some reason why you can't use what would appear to be a much simpler ground or pole mount or two? I don't know what the issues are in the area regarding space or vandalism, but if I had to go through all those gyrations to roof-mount the panels, I'd first look for a simpler approach. You're going to be dealing with strong wind-loading anyway with that steepness, so there doesn't seem to be any advantage on that score with roof-mounting the panels.
 
GRA said:
Coming into this late, and fortunately Wayne's already done the nasty calcs (I'd have had to look up the equation), but I have a question. Since this is off-grid, is there some reason why you can't use what would appear to be a much simpler ground or pole mount or two? I don't know what the issues are in the area regarding space or vandalism, but if I had to go through all those gyrations to roof-mount the panels, I'd first look for a simpler approach. You're going to be dealing with strong wind-loading anyway with that steepness, so there doesn't seem to be any advantage on that score with roof-mounting the panels.

Doing some math on a computer is a whole lot easier and cheaper than digging post holes and filling them with concrete, etc.

Roof mount is pretty simple and all he's doing different than normal is running the rails off from parrallel and installing tilt legs. Nothing complicated...
 
QueenBee said:
GRA said:
Coming into this late, and fortunately Wayne's already done the nasty calcs (I'd have had to look up the equation), but I have a question. Since this is off-grid, is there some reason why you can't use what would appear to be a much simpler ground or pole mount or two? I don't know what the issues are in the area regarding space or vandalism, but if I had to go through all those gyrations to roof-mount the panels, I'd first look for a simpler approach. You're going to be dealing with strong wind-loading anyway with that steepness, so there doesn't seem to be any advantage on that score with roof-mounting the panels.

Doing some math on a computer is a whole lot easier and cheaper than digging post holes and filling them with concrete, etc.

Roof mount is pretty simple and all he's doing different than normal is running the rails off from parallel and installing tilt legs. Nothing complicated...
You're forgetting wind loading and potential roof leakage issues, as well as maintenance/snow clearance. Given the option, I'd almost always opt for an installation that doesn't penetrate the roof. That's rarely possible in an on-grid application, but off-grid there's often the ability to do so. It may be they've already considered and rejected it for good reasons, but in case they fastened on roof-mounting without considering the options, I thought I'd mention it. I prefer the simplest approach that does the job.
 
I didn't get a chance to write a program to make the calculations, so I followed Wayne's "paper" calculations. When I arrived on-site, the owner had decided he wanted to mount the panels on the South-of-East-facing roof, even though the rails were too long for that roof, so I redid the calculations for that case. A Google Earth plot gave me an azimuth of 28 degrees South of East versus the 30 I had previously guesstimated (not that it really makes a difference versus 30 degrees).

Here are the calculations and their somewhat-surprising results:
wwhitney said:
I think the way to approach this is to use vector math on the vector orthogonal to the face of the panels. Vector length will be arbitrary, as we are only interested in direction.

So let's have the x-axis point east, the y-axis point north, and the z-axis point up.

If the 4:12 pitch roof faced due south, the orthogonal vector would be (0,-4,12) or (0,-1,3).

Since the building is turned 120 degree CCW from above, this become (-sin 120, -cos 120, 3) = (-sqrt(3)/2 , 0.5, 3)
Actually, looking at this, I think the building was 150 degrees CCW from East. No matter, here is what I calculated for the South-of-East-facing roof:

Since the building is turned 28 degrees CW from above, this become (cos 28, -sin 28, 3) = (0.88 , -0.46, 3)
wwhitney said:
Say you want the panels to face south at an elevation of 40 degrees. That vector would be (0,-sin 40, cos 40).
We decided to shoot for 45 degrees elevation angle, so this is what are target was:

That vector would be (0,-sin 45, cos 45) = (0, -0.707, 0.707).
wwhitney said:
The cosine of the angle between the vectors is their dot product divided by their lengths. The angle is

arccos((0 + cos120 sin40 + 3 cos 40)/sqrt(10)) = arccos(0.625) = 51 degrees.

So you would need to an elevation kit that would rotate your panels 51 degrees, not sure if that is practical. We still need to find what direction to run the rails.
We did some experimentation with the tilt kit on the ground and determined that with a spacing of 53" we could hit the bottom of the panel close enough from the bottom to keep the panels above the ground and 1/3 of the distance from the top. That gave an elevation angle right around 50 degrees. More on this later.

In any case, the I calculated for the tilt angle was as follows:

Tilt angle = arccos((0 + sin 45 sin 28) + 3 sin 45)/sqrt(10)) = 39 degrees.

That's good, since it is LOWER than the maximum we could achieve with the tilt kit which was purchased.
wwhitney said:
The rails would need to run perpendicular to both the pre and post rotation orthogonal vectors, i.e. perpendicular to both (-sin120, -cos120, 3) and (0, -sin 40, cos 40). That would be in the direction of their cross product. So the direction is given by

(-cos 120 * cos 40 - 3 * (-sin 40), 3 * 0 - (-sin 120) * cos 40, -sin 120* (-sin 40) - (-cos 120) * 0) = (2.31, 0.766, 0.557)
For my calculations, I found the direction of the rails as follows:

Direction of rails = (-sin 28 cos 45 - 3 (-sin 45), 3 * 0 - (cos 28 cos 45, (- cos 28)(-sin45) - (-sin 28)(0))
= (1.65, -0.624, 0.624)
wwhitney said:
If the rails ran horizontally across the roof, their direction would be (cos 120, -sin 120, 0).
In the chosen case: (-sin 28, -cos 28, 0)
wwhitney said:
So the angle between this and the desired rail direction is

arccos( (2.31 * cos 120 - 0.766 * sin 120 + 0) / sqrt (2.31^2 + 0.766^2 + .557^2) ) = 137 degrees.

Since the rail directions are only defined up to +-1, this is really 180 - 137 = 43 degrees.
Final calculation:

Rail rotation angle = arccos((1.165*(-sin28) + 0.628*cos 28 + 0)/sqrt(1.65^2 + 0.624^2 + 0.624^2)) = 83 degrees.
wwhitney said:
I haven't calculated whether that's CW or CCW, but from the roof it should be obvious which way to go.
I wasn't quite sure which way this angle went, but we discussed things and decided that this was close enough to perpendicular that we would take advantage of that fact and simply go with the perpendicular for the installation. Further, we decided to go with a 48-inch spacing between the feet of the tilt kit rather than the 53 inches we had previously determined. Together these two decisions allowed us to mount the feet directly into the rafters and therefore minimize the amount of time and effort spent in the attic (particularly close to the bottom of the roof where space is very limited).

The resulting installation achieved a result very close to what was desired. I wasn't really sure how to accurately measure either azimuth or elevation, but the azimuth appeared to be very close to South and my attempt to measure the elevation with my iPad yielded 40 degrees. That elevation result was a bit of a surprise to me, since we had been targeting 45 degrees and had INCREASED the tilt angle on the kit by spacing the two feet closer than needed to get 50 degrees tilt. (I'm wondering if I misinterpreted the measurement and the real number is 50 degrees. I'm just not sure.)

Some thoughts on the resulting installation:

1) We couldn't be happier with the resulting pointing angle. I was more than a bit surprised that running the rails vertically up the roof was sufficient to achieve the desired pointing angle.

2) Because we moved the feet of the tilt kit to be 48 inches apart, we cantilevered more of the PV module off the top of the panel than is specified by LG. The final result is that nearly one meter of the panels is above the top rail. That will mean additional stresses on both the PV module, the roof and the tilt kit which we used to mount it.

3) I really don't think the tilt kit is intended to be used in any application other than on a flat roof or with rails running horizontally along the roof. Hopefully the PV modules are able to take most of the odd diagonal stresses which were created via this approach. The owner is also considering adding some cross-bracing between the tilt kits to give additional rigidity.

4) We had there people working when mounting the panels on the roof. We concluded that this could not have been reasonably been done with only two people. This coming from the owner and his son, who are both builders who have done major projects with just two people. The reason is that there were simply two many moving parts and strange forces all acting at one time until the parts could be tightly connected together.

5) One interesting result of mounting the modules in this orientation is that they cast REALLY long shadows in the afternoon. This is because the top of the western-most module is very high and casts a shadow which hits to bottom of the eastern-most panel even when they are very far apart. This would require about 20 feet of spacing between rows to achieve a decent amount of shadow-free operation around the winter solstice. This array was mounted on the far North end of the roof, so there IS room for another row, but the owner wants to keep the rest of the roof shadow-free for a possible addition of a grid-tied array. As a result, we decided that the second array for this off-grid application will go on the western roof and will not point directly South, but will be more easterly to better catch the rays of the morning Sun. The owner will purchase some square aluminum bars to allow us to extend the remaining tilt kilts significantly. We may also need to purchase another tilt kit (or two) to accommodate the longer rails.

6) Since the rails were about four feet longer than the Eastern roof, we cantilevered about four feet off the top of the roof. At the end of the day, we decided to cut off these rails and NOT mount a fifth panel up there. There were two main reasons for this: A) Since the second array will point in a different direction, we need to group the PV modules in pairs to go with the two charge controllers that are being used. That necessitates having one array of four panels and one array of six panels. (They were already going to be grouped that way ELECTRICALLY, but the different angles require them to also be grouped that way PHYSICALLY.) We all felt like the four tilt kits ALREADY have their work cut out for them to support the four modules up there. Adding another one cantilevered off the end would likely result in the entire structure being ripped from the roof in high winds. The owner intends to cut off the ends of the rails and we will add them to the other rails to be able to accommodate all six panels in the other array.

7) It will be interesting to see how long this system lasts. The equipment which we installed is OLD. The charge controllers are RV Power Products Solar Boost 3048s which were built in 1999 (operated about four years total) and the inverter is a Trace Engineering SW5548 built in 2002 (operated about nine years total). I stepped up the voltage on the battery side slowly in 12V steps to anneal the electrolytic capacitors before exposing them to the full 48 Volts. I really like both of these products and hope the owner gets a long time out of them. We'll see.
wwhitney said:
Cheers, Wayne
Thanks again, Wayne! I was stuck without your assistance.

I forgot to take photographs of the final result, but I will try to do that and post them here when I have a chance.
 
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