Definitely not a hijack, as this is quite relevant, and hard data is quite appreciated.
So, feel-free to double-check my numbers on this, but let's look at energy per square mile from this map:
For Arizona (erring toward the conservative side of average) 6,500 Wh / sq m * 2,589,988 sq m / sq mi = 16,834,922,000 Wh / sq mi / day (sq meter to sq mile conversion provided by Google calculator).
16.8 Gigawatt hours per day from one square mile of solar collectors? I'm thinking I've done something wrong here in my numbers. I checked them at least 3 times.
While not all of that land will be used for just solar panels, and you'll certainly lose some energy in the conversion and transmission processes, you're probably still operating on the order of several GWh/day of actual usable energy if you can get a total of 1 square mile worth of solar panels from a number of sources. Holy s***.
Also, for sake of comparison and context, 1 gigawatt hour is 3.6 trillion joules (per Google calculator), and coal has an energy density of 24 megajoules per kilogram (per wikipedia). So it theoretically takes 150,000 kg of coal to produce 1 GWh of energy, which is ~330,000 lbs, or 165 tons.
Just for kicks, what would a home in Wisconsin generate (theoretically, keep in mind potential transmission and conversion losses)? I'm going to assume a "typical" semi-detached home, with a roof area of 72 sq m (http://www.resurgence.org/energy/heac/index.htm
). I'll also assume that due to conversion and transmission losses, as well as the fact that a roof can't follow the sun and not all of the roof will be exposed, that you'll only get 50% of the theoretical maximum.
3,500 Wh / sq m (low side of average) * 72 sq m * .5 = 126,000 Wh / day = 126 KWh / day. That would help take quite a chunk out of my electric bill. Even at 1/3 total efficiency you're still talking about 84 KWh / day. Yikes.