Andrew M.A. Cater wrote: > On Mon, Apr 27, 2009 at 11:56:24AM -0700, Lux, James P wrote: >> Urban legend. >> Do the math and calculate how many turns it takes, and don't forget that >> those wires have some non-zero resistance. So, unless you happen to have a >> large vat of liquid helium and lots of superconductors, the "stealing power >> by capacitive/inductive coupling" is probably bogus. >> >> Don't forget that "real" power lines tend to be fairly well balanced, so the >> net field at any reasonable distance is quite low. >> >> Detect the field, certainly. >> Get some trivial amount of mechanical work out of the system (spin a low >> mass dial or something), probably. >> Power your house/barn/dairy shed, not a chance. >> >> Let's use real numbers: 1000A, unbalanced, 20m away (extremely unlikely >> that 1000A unbalanced, but we'll go ahead anyway) >> >> Ampere's law >> B = mu0 * I/(2*pi*r) >> = 4*pi*1E-7 * 1000 /(2*pi*20) >> = 2E-7*1000/20 = 2E-4/2E1 = 1E-5 Tesla (note Earth field is 0.5E-4 T) >> >> >> >> >> Faraday's law of induction >> V = -dPhi/dt = -N*Area*dB/dt >> dB/dt = omega*cos(omega*t)*Bpk (for sinusoid) >> = 2*pi*f*Bpk = 377*Bpk (for 60Hz) >> >> Let's assume that Bpk = 1.414Brms (because the above Ampere's law >> calculation assumed 1000A rms) >> dB/dt = 377*1.414*1E-5 T/s = 0.005 T/s >> >> OK. We're going to put that big coil along our roof. Let's say 2m by 10m, >> so the area is 20 square meters. The induced voltage per turn is 20*.005 = >> 0.1 Volt. That's pretty low, so let's use 100 turns, so the voltage is now >> 10 Volts. But, the length of the wire in that coil is 22 meters/turn, so >> we've got about 2200 meters of wire. Assume it's AWG20 (fairly small, but >> not too small that it will easily break), which has a resistance of about 10 >> ohms/1000ft (sorry for the US units) or, 72 ohms total for our 100 turn >> coil. That works out to about 130mA, so we actually are going to dissipate >> about 1.4Watts in the coil. The optimum load impedance is also 72 ohms, so >> if we do that, we'll have half the current, and we'll capture a whopping >> 350mW in our load. >> >> Let's see.. At $0.34/kWh (the peak rate I pay at home), assuming I do this >> big coil thing for a year (8000 hours), I will have scammed the power >> company for 8000*.35 = 2.8kWh or about a dollar's worth of electricity. >> >> Leaving aside the substantial labor, I also had to go out and buy about 10kg >> of copper wire, which is about $5-10/kg, depending on the copper market. So >> I spent $50-100 on wire. The payback period is 50-100 years. >> > > Beautifully done. The version I'd heard - which is fractionally more > plausible - is that the person concerned lived next to one of the Royal > Navy's submarine communication transmitters operating at lower than > 16KHz. These things run megawatts at piddling efficiency to get watts > output because the antenna matching is so poor. Said anonymous bloke - > they're always anonymous - allegedly tuned his heating system to > resonance therefore acting as a lossy dummy load and, incidentally, > heating his water. Found only because there was an arc of the N. > Atlantic where there was no signal AT ALL and the subsequent > investigation turned up what he'd done. He wasn't receiving the signal > content per se - so no unauthorised interception - and he couldn't be > prosecuted readily so they agreed to forget about it. >
My favorite characteristic of these urban legends is that anyone smart enough to pull off a trick like this is probably already making enough money that they don't need to steal electricity. -- Prentice _______________________________________________ Beowulf mailing list, Beowulf@beowulf.org sponsored by Penguin Computing To change your subscription (digest mode or unsubscribe) visit http://www.beowulf.org/mailman/listinfo/beowulf
