How can I estimate local solar time from direct solar radiation, knowing only latitude and longitude?
If I know the location (latitude and longitude) of a panel that measures solar radiation (in watts/m$^2$) and that it is at at 0 degrees to the horizon (i.e. facing skyward), can I estimate the local solar time that a particular level will be achieved?
From the reading I have done, I think this is possible in principle if I am happy to ignore local weather and transient atmospheric conditions like air pressure changes.
I think I understand some of the pieces:
* I have been able to calculate a zenith angle from a local solar time, Julian day and latitude...
* ...and use that as an input to the Air Mass formula to get the reduction in power...
* ...and then use $R = 1.353 \times ({0.7} ^ A) ^ {0.678}$ to get a value in kW/m$^2$ based on that distance.
This came out with some sensible values which match some worked examples I found online.
* Separately, I have been able to calculate elevation from a local solar time and latitude and verify the calculation by looking at a sunrise/sunset calculator.
I think my approach should be:
1. Determine what reduction in power will give the level in kW/m$^2$ that I want to achieve.
2. Determine the zenith angle that would cause that reduction in power.
3. Determine the elevation angle from the zenith angle (trivial: $elevation = 90° - zenith$)
4. 371
Determine the local solar time from the elevation and latitude. I think this is just a rearranging of my elevation from LST and latitude formula from the point above.
Unfortunately for me, I am an idiot. When I try to plug these things together I just get nonsense values. After several hours of getting nowhere I thought I might do well to validate my approach with the smart people on here; people for whom humanity will never be grateful enough!
It may be I have a simple problem of mathematics and was torn between posting here and on the Math StackExchange, but my intuition tells me I may be overlooking some approach that makes sense because of the physical characteristics of the problem...so I thought I'd try here first.
Thanks in advance!
-- UPDATES --
* Re. point #1 of my approach, I am using... $$ A =- {log({{10000 \times 10^{48/113} R^{500/339} \over 1353 \times 1353^{161/339}})}\over log(\frac{10}{7})} $$ ...to get a reduction in power (Air Mass) from a kW/m$^2$ figure. 565