[Beowulf] Re: ECC Memory and Job Failures (Huw Lynes)

Lux, James P james.p.lux at jpl.nasa.gov
Mon Apr 27 11:56:24 PDT 2009

> -----Original Message-----
> From: beowulf-bounces at beowulf.org 
> [mailto:beowulf-bounces at beowulf.org] On Behalf Of John Hearns
> Sent: Monday, April 27, 2009 8:09 AM
> To: Beowulf Mailing List
> Subject: Re: [Beowulf] Re: ECC Memory and Job Failures (Huw Lynes)
> 2009/4/27 Vincent Diepeveen <diep at xs4all.nl>:
> >
> >>
> > Would this measurement prove that ionising particles, which 
> cause of 
> > course cancer, sometimes travel further away from these cables?
> It probably shows that an oscillating 50Hz magnetic field 
> travels further away from these cables.
> As I remember, there is a case on record of someone in the UK 
> living under a high tension line - he had an enormous 
> transformer coil in his attic, and was powering his house via 
> the induced current. He was successfully prosecuted for 
> stealing electricity.
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.

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