The US is the largest power consumer in the world - using about 102 ExaJoules/Year - from the Lawrence Livermore study for 1999. In the more familiar KiloWatt Hours, this is 102 x 2.78 x 10⁸ KWH/year = 2.8356 x 10¹⁰KWH/year.

1. The average terrestrial photovoltaic modules have at least 10% efficiency.
2. Sunlight reaches the Earth with power of about 1kilowatt/square meter
3. 1KW x 10% = 100W/square meter
4. Hours per day for the southern half of the US on average, is 5 peak solar hours, therefore each square usable meter could produce 100W/m² x 5H = 0.5KWH/m² per day.
5. (2.8356 x 10¹⁰KWH/year)x (1/0.9) (inverter loss) x (1/0.89) (temperature derating x (1/0.93) (dust and dirt derating x (1/0.95) (panel mismatch and wiring loss) x (1year/364 days))/ 500WH/m² = (2.20 x 108 m²)(1mile/(1.61 x 10³m)²) = 84.9 square miles - is the amount of area that is needed to power the entire country with solar power!!