[Beowulf] Re: ECC Memory and Job Failures (Huw Lynes)
Prentice Bisbal
prentice at ias.edu
Tue Apr 28 05:14:59 PDT 2009
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
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