grid parity, page-37

The one third sun measure is being used for real world conditions in northern Europe, Japan and US.

In those locations, you don't get 1000W/m2. I believe they are allowing for latitude and average weather/atmospheric conditions (i.e. overcast/haze). They are allowing for what on average really hits the panels in the real world.

So the calculation should be 333w/m2 from the sun * 8.5% effeciency (assuming near term improvements) to get Wh/ms2.

We also need to know an average number of hours per day we should expect the 1/3 sun. So while dawn to dusk numbers may vary from 6-18 hours (only a guess), the average might be 9 (again a guess).

Now assuming the near term improvements are made (8.5%) and the Austock 85kWh/m2 we have:

85000 / (333 * 0.085 * 365) = 8.2 hours

That would be the average number of hours per day you get 1/3 sun conditions. The number of hours may already be derated to allow for system losses.

To sumarise, you get 28.3Wh/m2 in 1/3 sun conditions. If you get that for (on average) 8.2 hours a day, you have 85kWh per year.

For an equivalent setup with traditional solar you apparently get 30% less power.

That's my current understanding. It may or may not be correct.