It's still faster than Ruby. :D
PS: I love both anyway.

I'm not a native speaker and I still often make mistakes in English, but I cannot understand how people can mix those up : then/than, your/you're, its/it's, there/their/they're.
http://theoatmeal.com/comics/m...

My kingdom for a mod point!

I also think many Python programmers know only the bare minimum.
But it's not a bug, it's a feature : there are so many libraries that help you make cool stuff with audio/video/websites/graphs in a few lines of code.
All you need to know is Google "Python + stuff you want to do", use a few assignments and make a few method calls.

Wonderful!
This service has even been certified : http://2.bp.blogspot.com/-x__K...

No problem, apologies accepted.
And yes, MathML would be nice, so would be m**2.
The conversion efficiency eta is actually included/hidden in nominal_power :
nominal_power = STC_irradiance*area*eta
To calculate the expected yield, you need either nominal_power or both area and eta.
Most of the time, you know the nominal_power but not the exact area. Almost every PV module is named according to its nominal_power.

And the efficiency isn't included in performance_ratio (which is typically between 0.7 and 0.9, 0.9 being state of the art installations) or STC_irradiance (set at 1000W/m**2 by definition).

^YES. Check your brain and your glasses.
This formula is correct, I use it every day and double checked it again tonight.
Slashdot ate away m**2 from STC_irradiance. Units are correct otherwise.

It's
W rated * irradiance / irradiance
not
W rated * irradiance * irradiance

Directly from Qalc :
6.8kWh/(m**2*day)*(3600W)/(1000W/m**2)*0.9 to kWh/day
->
22.032((kW*h) / d)

As you said, units never lie.
And yes, the 1000W/m2 is a "lab test value for standardizing comparisons", but so is nominal_power.

So, this is my last attempt to explain my formula. I still expect a "You're correct and I was wrong" from you.

performance_ratio = effective_yield/theoretic_yield
= effective_yield/(yearly_irradiation*area*eta)
= (specific_yield * nominal_power)/(yearly_irradiation*area*eta)
= (specific_yield * STC_irradiance*area*eta)/(yearly_irradiation*area*eta)
= specific_yield*STC_irradiance/yearly_irradiation

So you get
specific_yield = yearly_irradiation*performance_ratio/STC_irradiance
and
effective_yield = nominal_power*yearly_irradiation*performance_ratio/STC_irradiance

*) First of all, it's not because you don't understand something that it's "purely out of luck".

*) I agree I didn't do a good job explaining it, though.

*) The reason I calculated a yearly specific yield is because it's a very common value, it's easy to compare from one location to another (1000kWh/kWp.y in south Germany, almost 2000kWh/kWp.y in Spain, ...) and daily values are less common, except in off-grid systems. You might consider this value useless, but it surely isn't erroneous. Also, the performance ratio is defined for yearly values, and the "daily performance ratio" fluctuates over the year.

*) I used one, and only one method :

daily_yield = daily_irradiance*(nominal_power/STC_irradiance)*performance_ratio

daily_yield in [kWh/day]
daily_irradiance in [kWh/(m**2*day)]
nominal_power in [W], often written [Wp]
STC_irradiance in [W/m], defined as 1000W/m**2
performance_ratio is unitless

6.8kWh/(m**2*day)*(3600W)/(1000W/m)*0.9
->
22 kWh/day

Why is it ridiulous to consider nominal power and not area?
You pay your installation per kWp, not per m**2.
You choose your inverter depending on kWp, not m**2.
You choose your cable section depending on kWp, not m**2.
As long as a pv installation fits on a roof, carport or satellite, nobody cares about its area.

You assumed a wrong carport area (40m**2 instead of 27m**2) and a wrong efficiency (20% instead of ~15%) instead of just googling and finding out that the carport nominal power is 3.6kWp.

And once again, I do use the 6.8kWh/m2.day irradiance information, but no, I don't need to know the cell efficiency to calculate the maximum yield of this carport : about 22kWh/day in Arizona.

From http://www.photon.info/photon_...

The theoretical yield possible is thus calculated by multiplying the accumulated solar irradiation per area with the STC power of the module, divided by the STC irradiation power of 1kW/m. The module area does not contribute to the calculation, since it appears in the numerator and denominator of the fraction and is thus omitted by cancellation

Let's ignore the losses (i.e. performance ratio = 100%) and compare two installations.

PV Installation #1, 20% cell efficiency, 1kWp :
With 6.8kWh/m**2.day of irradiance and 20% efficiency , you get 1.36kWh/m**2.day of electricity.
The installation covers 5m**2, so you get 1.36kWh/m**2.day*5m**2/kWp = 6.8kWh/kWp.day of electricity

PV Installation #2, 10% cell efficiency, 1kWp :
With 6.8kWh/m**2.day of irradiance and 10% efficiency , you get 0.68kWh/m**2.day of electricity.
The installation covers 10m**2, so you get 0.68kWh/m**2.day*10m**2/kWp = 6.8kWh/kWp.day of electricity

You see? The results don't depend on the cell efficiency.

Sorry if I wasn't clear.
I don't need to assume *any* cell efficiency, since this information is irrelevant.
The only needed information to calculate the energy yield are :
*) Solar irradiance
*) Nominal power
*) Performance ratio

Two 3.6kWp pv installations will produce the same energy yield for a given performance ratio, independently of their cell efficiency (i.e. size).

Take a look at the units for solar irradiance and specific yield.
They are *not* the same.
The 20% cell efficiency (or anything between 0.05 and 0.44) is included in the conversion between m2 and kWp : e.g. you need 5m2 of PV modules at 20% for 1kWp.
Performance ratio and cell efficiency are different notions. Performance ratio can theoretically be higher than 100%.
http://www.photon.info/photon_...

My method is just fine, thank you very much. I happen to work at a german research center on solar energy.

The performance ratio takes all losses in consideration (cable, MPP, inverter, shadowing,...) and isn't dependent on either the area (which is 27m2 for this carport, BTW) or the efficiency (about 15% for this carport).

With the performance ratio, you can convert solar irradiance (in kWh/m2.year) directly into specific yield (in kWh/kWp.year).
This carport has 3.6kWp capacity, and seems to be developed by Solarwatt.

22kWh/day really is your best case scenario.

http://www.nrel.gov/gis/images...
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y)
With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
With 3.6kWp (see http://www.solarwatt.de/en/pro...), you get 8100kWh/y, which is about 22kWh/day.

But this is only in the sunniest place in the US, with a tilted roof and a very good performance ratio.
You'll get close to 10kWh in Europe and many other places in the US.

It doesn't go both ways.
I suppose the camera will "inadvertently" be turned off anytime the cop feels like doing something unjustified.