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I should be able to calculate this... but what's the resistance of that cord then? That's quite a voltage drop.

EDIT: Wow, apparently this is pretty typical and expected as per this calculator: http://www.csgnetwork.com/voltagedropcalc.html


Did you try measuring the power at the wall end of the cord while the heater was running? Most space heaters aren't very smart, and it wouldn't surprise me if the draw varied wildly depending on the power supply and how warmed up the heater is (I know mine will go between 1 and 1.5kW plugged straight into the wall as it warms up).

I'd be surprised if a 100-foot extension always consumed 850 watts.


Elsewhere in the thread I calculate the life support draw at 29 watts from available data (53kWh battery, 11 weeks to idle to empty). I think you can afford the $100 if you have a Tesla.


This is based on a wrong premise, that 900 watts is needed to keep the thing from losing charge, derived from vague anecdotal evidence and not supported in any way by technical fundamentals. If it was true, the 85 kwhr battery pack would last less than 3 days when the car was unplugged.

The actual number is probably more than an order of magnitude smaller.


"The article indicates that a 100 foot cord connected to a standard outlet does not provide enough power to prevent a Tesla car from discharging..."

Pretty sure they say that to make sure people don't just use any old cord. A 24 AWG cord with 66% voltage drop as the low end would change your cost analysis radically.


You probably meant 4 AWG, a 100' of 24 AWG cord has about 2.5 ohms of resistance in it. Since power dissipation is exponential with current pulling even 2 amps through it would have the wire trying to dissipate 10 watts and that would represent 100mW/foot, easily enough to raise the temperature of the copper to the point where it would be too soft and break.

[1] http://www.cirris.com/testing/resistance/wire.html


Since power dissipation is exponential with current

Minor nitpick: power is proportional to the square of current (e.g. P=I^2), not exponentially proportional (which would be something like P=e^I).


I was just following an online calculator, will plead ignorance on the physics but I think the basic point holds for some gauge wire.




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