Difficult rearranging problem

  • #1
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!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
welcome to pf!

hi drogerssolar! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
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?​
 
  • #3
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?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,836
251
hi drogerssolar! :smile:
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:
 
  • #5
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 :)
 
  • #6
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!
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,836
251
hi drogerssolar! :wink:
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:
 
  • #8
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
 

Attachments

  • Axis Tilt Derivitive.jpg
    Axis Tilt Derivitive.jpg
    2.8 KB · Views: 265
  • #9
tiny-tim
Science Advisor
Homework Helper
25,836
251
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.
 

Attachments

  • Axis Tilt Derivitive.zip
    26.6 KB · Views: 76

Related Threads on Difficult rearranging problem

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
9
Views
9K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
15
Views
2K
Top