>> Solar insolation / day is about 1.3 kWh/m^2 x disc area in m^2 x 24 >> hrs/day kWh > Errr - hang on, that fails dimensional analysis! You can't have kWh > at both ends and also multipy by 24hrs... > Do you mean 1.3kW/m^2 ? Well, I was being sloppy - I meant 1.3 kWh/h/m^2, which of course is 1.3 kW, but one is power and the other energy :-). > ... and if so, I don't think anywhere on Earth gets this much for > 24hrs/day :-). Oh, it does. It's falls on projected stationary earth sized disc I used in my calculation. Peak solar insolation in 1.3 kW/m^2 more or less. The sun always shines on somewhere on earth. And the sun "sees" a disk of an area equal to the projected circular outline of the earth. It doesn't care that there's a turning planet inside the projection. Some of that will have very long atmospheric paths (near the poles) and some a short path (neat the equator) but the sun doesn't care - it just throws its 1.3 kW/m^2 onto the disk. End result is as I said, more or less. > And it's "renewable". There's just the old problem of collecting > the energy where it's available and > transporting it to where it's needed. A lot is already used where it's needed. makes plants growm pumps surface water back into the sky, drives the wind , waves, ... (apart form the aforesaid parts which are driven by the earth's rotation :-) ), and much more. But there's plenty 'left over' for other things. We just haven't got the technology fully sorted yet. RM -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist