In sci.physics, BradGuth
>
> wrote
> on Mon, 01 Oct 2007 16:19:59 -0700
> :
>> >>>
>>> Where? Oh, you must be seeing things again. In any event, anyone
>>> who knows chemistry can figure out the above fact. I'm still curious
>>> as to how you get 40 kW/m^3 from variants of solar energy.
>>
>> A 100~125 meter tall tower will take up roughly 100 m2 worth of
>> surface footprint at it's base, that which can't easily be utilized
>> for all that much other than a fluid storage tank or perhaps on behalf
>> whatever fluid processing that could rather easily be contained within
>> the somewhat less than 100 m2 interior. However, on top of this
>> sucker is a good 3.5~4.5 MW wind turbine, and well enough below the
>> blade sweep is a very large DVD like disk of those 25% efficient PV
>> cells (we're talking better than twice that amount if using William
>> Mook's multi-band and special lens enhanced PV cells) that'll track
>> sunrise to sunset, as well enough elevated above the local terrain and
>> whatever trees so that full solar benefit is easily maintained.
>
> The only problem is that those solar cells cast a shadow. Is there
> anything nearby? Oh, another tower? Won't do that other tower
> much good, will it?
>
Or worse, modules around the one tower. The idea of one 8000 m2 PV disk tracking the daily sun is pretty daunting. But if you break it up into several smaller modules, say 100 of 80m2 each, then when the sun isn't directly overhead, the modules closest to the sun are casting shadows on the ones directly behind them. Space them out far enough that you can get full sun on all of them for about six hours a day and you just about double the distance between them. Any more than that and simple trigonometry expands the spacing needed very quickly.
Oops.
Yep, it's a pretty big 'oops'.
daestrom