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Comment Shut up and lawyer up .. (Score 1) 656

If it's a wrongful dismissal, then it's better to lawyer up before you start talking about it in public forums. There's a protocol to follow. Might even get crowdfunding from other Gors.

It looks to me like somebody outed this guy, and then he got fired. It didn't sound like he was bringing it to work with him. I wouldn't consider it a frivolous lawsuit.

Comment Everybody benefits... (Score 5, Informative) 79

Student saw an in-band indication that the detector was in a non-radiation reporting state, and asked NASA about it.

NASA says, huh, that's weird. It's not supposed to happen that often. Hey kid, wanna do us a solid (in more ways than one)?

Hell yes, says kid.

BBC, realizing that story is too complicated, bowdlerizes title to get people to read it.

Slashdot talks about something else entirely.

Comment A sufficient number of refractory projectiles .. (Score 1) 130

.. will overwhelm the power output of a single truck-mounted laser, even under ideal conditions.

For example, a 300 gram tungsten projectile will require a full second at 58 KW to be melted, assuming no reflection. An alumina projectile of 42 grams will require the same full second at that power.

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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