Optimal Panel Elevation Angle For a Single Day

[GeorgeB]

That previous post wasn't worded very well, but what I meant was one could approximate a given day ideal angle by looking at which quarter that day falls in (and where in that quarter), and using the formulas on the solar panel tilt website. For my latitude of 43 degrees, I get summer: 15 degrees, spring and fall: 40 degrees, winter: 62 degrees. The ideal fixed annual tilt is 36 degrees (which is cool since it is constructible with compass and straight edge).

Although as a math guy, I would like to see the exact formula.

I tried to find these details when I had a small array installed 4 years ago, more for academic reasons (my roof is not adjustable, and for a small array, custom mounts would not be realistic) than practical.

Also a math guy (son has a degree in the Operations Research branch of maths), what I found is that the azimuth and elevation are readily determined. It is generally assumed that normal incidence maximizes production (energy conversion). What are not easily calculable are atmospheric effects. Shade for those with some shading "things" around is very hard to calculate.

I have a "small" gas flue from my furnace. It is ABOUT 3" in diameter and protrudes about 2' above the roof. I can watch dramatic production losses from the 2 nearer panels vs sun angle (well, really time) over especially in the morning and afternoon. Noon, +/- an hour has little if any shade.

The OP seems to want to charge a "small" battery, IIRC. What ___I___ would do if I really cared is to pick an angle from that reference. Then, rather than a fixed South direction, I would point it (NOT SCIENTIFIC) SSE from daybreak until about 1.5 hours before solar noon, S from 1.5 hours before until 1.5 hours after, then SSW from 1.5 hours after until sunset.

I'm in upstate South Carolina where clouds are common; real measurements without nature introduced variation are darned near impossible to even imagine. I paid for the enhanced reporting from my Enphase Envoy ... if anyone would like to see any detail, let me know. I'll share coordinates so you can see the layout on Google maps or earth. I'll certainly share graphical production (kW for several days, kWh over a longer period) with anyone. Nothing is confidential.

I bought it to POSSIBLY save a little money; payback should come in about 8 years. But really, it was to tell our kids we are environmentally "good". My problem is a wife who has moved every 7 or 8 years, and we'll probably not realize anything except the respect of our children. And I'm an unhealthy 70 in a 3 story house ... how dumb is that. But we're close to the grandkids.

 

[LarryFine]

The OP seems to want to charge a "small" battery, IIRC. What ___I___ would do if I really cared is to pick an angle from that reference. Then, rather than a fixed South direction, I would point it (NOT SCIENTIFIC) SSE from daybreak until about 1.5 hours before solar noon, S from 1.5 hours before until 1.5 hours after, then SSW from 1.5 hours after until sunset.

That's what I've been trying to say, except with numbers.

But we're close to the grandkids.

That's paramount.

 

[Carultch]

Although as a math guy, I would like to see the exact formula.

Unfortunately, it is one of those math problems that has no exact formula as a solution.

You'd first have to neglect cloud cover, and assume a uniform clear sky throughout the year, in order to get anything close to a problem that can be generalized with an exact formula as the answer. Even with that concession, you end up at an integral that isn't possible to solve in closed form, and ultimately requires a numeric method for approximation. Which is what we do, with an hour-by-hour simulation in the production modeling software. We also include historical models of cloud cover (known as TMY, for typical meteorological year), to get a more realistic estimate of production.

It is like integrating e^(-x^2) dx, which is of interest to probability and statistics. You can't do it with elementary functions alone. We defined the error function known as erf(x), to solve integrals of this family of functions. It is ultimately evaluated with the numeric method of an infinite series behind the scene.

 

[electrofelon]

Unfortunately, it is one of those math problems that has no exact formula as a solution.

You'd first have to neglect cloud cover, and assume a uniform clear sky throughout the year, in order to get anything close to a problem that can be generalized with an exact formula as the answer. Even with that concession, you end up at an integral that isn't possible to solve in closed form, and ultimately requires a numeric method for approximation. Which is what we do, with an hour-by-hour simulation in the production modeling software. We also include historical models of cloud cover (known as TMY, for typical meteorological year), to get a more realistic estimate of production.

It is like integrating e^(-x^2) dx, which is of interest to probability and statistics. You can't do it with elementary functions alone. We defined the error function known as erf(x), to solve integrals of this family of functions. It is ultimately evaluated with the numeric method of an infinite series behind the scene.

Sure, but I was assuming more of the simplified version, assuming no cloud cover, simplify atmospheric effects to a simple fixed attenuation per unit distance, no (or simple) cell temperature correction.......in that case, it seems like a relatively simple problem of a fixed plane on a rotating sphere and an integration of the light concentration on that plane throughout the rotation. I am not immediately seeing that integral as not being solvable in closed form.

 

[wwhitney]

OK, I came back to this question after a while. And I have a handle on when it is optimal not to try to capture the full day's sun at a glancing angle, but instead capture part of the day's sun at a more direct angle. It only occurs above 60 degrees latitude, so for simplicity this post sticks to 60 degrees or less (each column is one latitude, as labeled on the top row).

Here's what I come up with for optimal local elevation angle for a given sun angle (declination angle \delta) and latitude (\phi). Again, this is just for the geometric effects and ignores atmospheric effects completely. Negative angles mean point towards the north, rather than the south. For a clear view to the horizon in all directions:

Date (Approx)​	Delta​	Phi=0​	10​	20​	30​	40​	50​	60​
December 20​	-23.5​	34.3​	43.1​	51.9​	60.6​	69.2​	77.5​	85.4​
January 20​	-20​	29.8​	38.8​	47.8​	56.8​	65.7​	74.4​	82.6​
February 8​	-15​	22.8​	32.2​	41.6​	51.0​	60.3​	69.4​	78.2​
February 23​	-10​	15.5​	25.2​	34.9​	44.6​	54.2​	63.8​	73.1​
March 8​	-5​	7.8​	17.7​	27.7​	37.6​	47.5​	57.4​	67.2​
March 21​	0​	0.0​	10.0​	20.0​	30.0​	40.0​	50.0​	60.0​
April 3​	5​	-7.8​	2.1​	12.0​	21.9​	31.8​	41.6​	51.3​
April 16​	10​	-15.5​	-5.8​	3.9​	13.5​	23.0​	32.2​	40.8​
May 1​	15​	-22.8​	-13.5​	-4.2​	4.9​	13.7​	21.8​	28.2​
May 20​	20​	-29.8​	-20.8​	-12.0​	-3.6​	4.3​	10.8​	13.4​
June 20​	23.5​	-34.3​	-25.7​	-17.3​	-9.3​	-2.2​	2.9​	2.0​

And if we ignore the sun until it is at least 5 degrees above the horizon (say there are hills, or as a very basic attempt to model atmospheric effects):

Date (Approx)​	Delta​	Phi=0​	10​	20​	30​	40​	50​	60​
December 20​	-23.5​	32.8​	41.7​	50.5​	59.3​	67.8​	76.1​	83.9​
January 20​	-20​	28.4​	37.5​	46.6​	55.6​	64.5​	73.1​	81.2​
February 8​	-15​	21.7​	31.2​	40.6​	50.0​	59.2​	68.3​	77.0​
February 23​	-10​	14.7​	24.4​	34.2​	43.8​	53.4​	62.9​	72.1​
March 8​	-5​	7.4​	17.4​	27.3​	37.2​	47.0​	56.8​	66.5​
March 21​	0​	0.0​	10.0​	20.0​	30.0​	40.0​	50.0​	60.0​
April 3​	5​	-7.4​	2.5​	12.5​	22.4​	32.4​	42.3​	52.3​
April 16​	10​	-14.7​	-5.0​	4.8​	14.5​	24.2​	33.8​	43.2​
May 1​	15​	-21.7​	-12.3​	-2.8​	6.5​	15.7​	24.5​	32.6​
May 20​	20​	-28.4​	-19.3​	-10.3​	-1.5​	7.0​	14.7​	20.5​
June 20​	23.5​	-32.8​	-24.0​	-15.3​	-7.0​	0.9​	7.6​	11.2​

Cheers,
Wayne

 

[GeorgeB]

(I trimmed for space issues; your full post will allow anyone the full data. I left 30 and 40 since I'd interpolate for my 35 degree latitude.

OK, I came back to this question after a while.

Here's what I come up with for optimal local elevation angle for a given sun angle (declination angle \delta) and latitude (\phi). Again, this is just for the geometric effects and ignores atmospheric effects completely. Negative angles mean point towards the north, rather than the south. For a clear view to the horizon in all directions:

And if we ignore the sun until it is at least 5 degrees above the horizon (say there are hills, or as a very basic attempt to model atmospheric effects):

Date (Approx)​	Delta​	Phi​			30​	40​
December 20​	-23.5​				59.3​	67.8​
January 20​	-20​				55.6​	64.5​
February 8​	-15​				50.0​	59.2​
February 23​	-10​				43.8​	53.4​
March 8​	-5​				37.2​	47.0​
March 21​	0​				30.0​	40.0​
April 3​	5​				22.4​	32.4​
April 16​	10​				14.5​	24.2​
May 1​	15​				6.5​	15.7​
May 20​	20​				-1.5​	7.0​
June 20​	23.5​				-7.0​	0.9​

Cheers,
Wayne

2 questions ...

what about the other 6 months, just correspond November with January, October with February, September with March, August with April, July with May?

Negative declination angles are East of South?

Oh, a 3rd question ... care to share the formulae?

George

 

[wwhitney]

A few notes. Latitude \phi can be -90 to 90 degrees, and declination angle \delta can be -23.5 to 23.5 degrees, but if you multiply everything by -1, it just flips the problem upside down, so I only considered positive \phi (northern hemisphere).

Under the circular approximation to the Earth's orbit, each \delta occurs twice a year at equal times before or after the solstice (except the solstice itself). So \delta = 20 degrees occurs 31 days before and after the June 20 solstice, i.e. May 20 and July 21. I did the math in terms of \delta and used an approximation to convert that to date; you can look up formulas of different accuracy for converting between declination angle and date.

Negative \delta means the sun is "below" the equatorial plane (northern hemisphere fall/winter).

Ignoring the case that you're better off losing some exposure time to get more direct exposure (high latitudes only), the formulas are these:

\beta is the absolute panel angle, it's the latitude at which the panel would be straight up. \phi is the latitude. So usual panel elevation (as in the table) is \phi- \beta. [The math is easier in a non-latitude dependent coordinate system, basically the coordinates at the equator.]

The hour angle \omega is a measure of time in radians where 2\pi = 1 day and \omega = 0 at solar noon. The value A of \omega at which the setting sun drops below \gamma degrees above the horizon is given by:

A = arccos((sin \gamma - sin \delta sin \phi) / (cos \delta cos \phi)) if defined; otherwise 0 or 180 degrees, as appropriate (a polar phenomenon).

Then the optimal panel angle is given by pointing the panel at the center of mass of the sun's path in the sky as omega varies from -A to A. This is given by

\beta = arctan(A tan \delta / sin A)

where it is important that A be in radians, not degrees.

Now you can find an accurate formula for converting date into \delta, choose the \gamma you want, and for each day, find the corresponding \delta to compute A, \beta, and the unusual elevation angle.

This is just for a single day. If you want the optimal geometric angle for a number of days, the simple average of the angles is only an approximation of the correct answer. If you make it this far and want to do the proper average, go ahead and ask me. : - )

Cheers, Wayne

 

[Carultch]

Now you can find an accurate formula for converting date into \delta.

If the year started on the March equinox, it would roughly be a sine wave with a period equal to 1 year, and an amplitude equal to 23.5 degrees. This is what it would be if you assumed a circular orbit. The elliptical orbit prolongs summer (94 days) and shortens winter (89 days), but this model does not consider that detail. This model considers uniform time between the solstices and equinoxes. The March equinox is generally the 80th day of the year, so we start by subtracting 80 days. Then translate the date to a fraction of the year and multiply by "1 cycle" in the units applicable to the sine function.

In radian mode:
delta = (23.5*pi/180)*sin((date - 80)*2*pi/365.25)

Or in degree mode:
delta = 23.5*sin((date - 80)*360/365.25)

 

[wwhitney]

Yeah, that's the model I used for my approximate dates. But I think it's worth using an elliptical model if actually applying anything I posted. I'm mainly interested in the geometry, so \delta is my starting point, not the date.

Cheers, Wayne