What is the optimal tilt and azimuth angle for solar panels?

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The discussion revolves around finding the optimal tilt (Tm) and azimuth angle (AZm) for solar panels to minimize the angle of incidence (AOI). The user presents a complex equation and seeks assistance in rearranging it to isolate AZm. Several participants suggest using calculus, specifically derivatives, to identify minimum values, emphasizing that setting the derivative to zero is crucial for finding maxima or minima. The conversation highlights the challenges of manipulating trigonometric functions and derivatives, with a focus on simplifying the equations involved. The user expresses a need for further guidance and considers using advanced math software for assistance.
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I have a math problem that I cannot wrap my head around. Maybe there is no easy answer but I thought I would ask for some help.
So first the equation:

AOI = cos-1 {cos(Tm) cos(Zs)+sin(Tm) sin(Zs) cos(AZs-AZm)}

Where:
AOI = solar angle of incidence (deg)
Tm = tilt angle of module (deg) (0deg is horizontal)
Zs = zenith angle of sun (deg)
AZm = azimuth angle of module (0deg = North, 90deg= East)
AZs = azimuth angle of sun (0deg= North, 90deg= East)


*Note that the -1 beside the cos means "ACOS" not "COS to the power of -1"

So my problem is that for a given sun postion (Zs, AZs), I want to know the best combination of Tm and AZm to give me the lowest possible value for AOI.
I calculate Tm based on AZm with the formula:

Tm = AZm x .5

I know that I need to substitute AZm x .5 for all the Tm’s in the original equation but I cannot seem to rearrange it to isolate AZm onto one side. I fear my lack of trig identity knowledge is what is preventing me from solving. I would appreciate any help or guidance anyone could give. Even just a hint.
Thanks!
 
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welcome to pf!

hi drogerssolar! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
drogerssolar said:
AOI = cos-1 {cos(Tm) cos(Zs)+sin(Tm) sin(Zs) cos(AZs-AZm)}

Tm = AZm x .5

I know that I need to substitute AZm x .5 for all the Tm’s in the original equation but I cannot seem to rearrange it to isolate AZm onto one side.

is the aim to minimise cos-1{cos(Tm) cos(Zs)+sin(Tm) sin(Zs) cos(AZs-AZm)} ?

(btw, that's the same as maximising what's inside the {})

can't you do it by finding the derivative?​
 
Well my calculus is a bit rusty I was able to find the derivative (I think). So first I made AZm a function of Tm (Tm/.5). I also am looking just inside the brackets of the ACOS.
So my starting equation looked like this:

(cos(Tm)*cos(Zs))+(sin(Tm)*sin(Zs)*cos(AZs-(Tm/.5)))

The derivative I ended with was:
f'x = 2*sin(Tm)*sin(Zs)*sin(AZs-2*Tm)+cos(Tm)*sin(Zs)*cos(AZs-2*Tm)-sin(Tm)*cos(Zs)

So with this derivative I assume I want to make f'x = 1 in order to obtain the minimum value of AOI?
 
hi drogerssolar! :smile:
drogerssolar said:
The derivative I ended with was:
f'x = 2*sin(Tm)*sin(Zs)*sin(AZs-2*Tm)+cos(Tm)*sin(Zs)*cos(AZs-2*Tm)-sin(Tm)*cos(Zs)

So with this derivative I assume I want to make f'x = 1 in order to obtain the minimum value of AOI?

what's "x" ? :confused:

yes that's the correct value for ∂f/∂(Tm) :smile:

but don't you also need ∂f/∂(Zs)?

and no, you need f' = zero for a maximum or minimum, don't you? :wink:
 
Thanks for the help Tim. This is a good refresher for my brain. Once I figure this all out I will get write back with the correct answer. You have pointed me in the right direction. Thank you :)
 
So if I make ∂f/∂(Tm)=0
and I make ∂f/∂(Zs)=0
That means ∂f/∂(Tm)=∂f/∂(Zs)?
and then I just rearrange the get Tm on one side plug in my value for Zs and solve.
Am I on the right track here?
Thanks again Tim!
 
hi drogerssolar! :wink:
drogerssolar said:
So if I make ∂f/∂(Tm)=0
and I make ∂f/∂(Zs)=0 …

these are two separate equations,

you can solve them quite separately (which is easier) :smile:
 
Tim,

I may require a little more guidance if you don't mind. I have a slightly different equation than I had in the previous posts as my initial equation was not correct. I have determined that:

y=(cos(35)*cos(x)*cos(25))+(sin(acos(cos(35)*cos(x)))*sin(25)*cos(180-(acos(atan(sin(x)/tan(35))*cos(x)))))

Where:

Y = Angle of incidence
X= Degrees of Rotation (around panel axis)

I was not able to take the derivative myself; however, with the help of a derivative solving web page I ended up with a rather large derivative. So large in fact that I cannot possible type it out so I will save it as an image and attach it to this post.

Now I know I have to set the derivative equal to zero and solve for X.
My question:
Is there anyway to shrink this equation down to something a little more manageable? Is it even solvable?
Thanks again for any help you can provide.

Regards,
Dan
 

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  • Axis Tilt Derivitive.jpg
    Axis Tilt Derivitive.jpg
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erm …

thumbnail totally unreadable :redface:
 
  • #10
Sorry Tim,

I just realized when I uploaded, the file size was shrunk way too much. Let me see if I can ZIP the original and attach it. The only other way would be to go to:
http://www.numberempire.com/derivatives.php
and copy and paste the following into the form:
(cos(35)*cos(x)*cos(25))+(sin(acos(cos(35)*cos(x )))*sin(25)*cos(180-(acos(atan(sin(x)/tan(35))*cos(x)))))

I have a feeling I might need to help of some math software like Mathematica or MatLab to rearrange such a large derivative.
 

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