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Comment Re:Huh? (Score 1) 106

Not much energy, on inspection:
the math is about high-school physics level, so here goes:

For one square meter of square module, inclined at your latitude angle L, at 2.54 cm of precipitation per hour, the power from the water sliding across the module is good for about

cos L * 1 sin L * 1 * 25400 / 2 * 0.000098 watts.

For 45 degrees latitude, that's about .622 watts, at 100% efficiency (which it is not). That's the sweet spot.

That's about 1/350 the energy available falling on a 22% efficient module on June 21 in the northern hemisphere at that latitude.

(that's for a square meter of square module, whose effective area for rain capture is proportional to the cosine of its inclination; with the average potential energy of a parcel of water on the module equal to the average of the highest elevation of the parcel and the lowest elevation on the parcel, and whose elevation is proportional to the sine of its inclination, in Earth's gravity).

Comment Re:Fake Fake Fake! (Score 1) 276

I'm here to say the same thing.

A 45 ACP round in a Colt 45 Automatic pistol is 600 foot-lbs, or 450 joules, and it'll do more to steel-backed plywood than just dent it.

No way the projectile is ending up with 1800 joules in kinetic energy. Perhaps that is what the capacitor bank starts with, but that's not what the projectile ends up with.

Try to be charitable about the 3 million joule thing. I work on a NASA campus that has a regular old utility transformer outside labelled 23,000 Kilovolts (and it's not).

Mistakes are made :-)

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