Ohms tells you how much power you lose to resistance given the voltage drop from one end to the other.

The voltage you choose to put across it is a free parameter in the range +/- 1e6 V or so.

As most people would consider "mildly heating the Pacific" to be pointless, you add the resistance of the line to the effective resistance of all the load on the grid and then use P=IV=I^2R=V^2/R to work out how many volts/amps/watts are going though the line.

You actually have to have a number for grid load to turn this into "what percentage are we losing to the transmission line".

(Also: I'm simplifying by assuming DC, because AC resistance depends on the environment and the frequency and that length is long enough to just be an antenna at normal grid frequencies).

Right, I understand now, you are talking about transmission losses.

And using units of resistance is a way of talking about the loss independently of the amount of power transmitted.

Although, I think your original statement I quoted, is omitting some information you are relying on – e.g. how much energy is "$265 billion of aluminium", and so how much energy are we talking about such that 1 ohm transmission loss would amount to that

> how much energy is "$265 billion of aluminium", and so how much energy are we talking about such that 1 ohm transmission loss would amount to that

I don't understand the question.

$265 bn of aluminium is a square metre cross section going around the equator and back to the start, or any equivalent network.

How much energy can be conveyed through that depends how long it lasts, which depends on the specifics, and how much power on average it is used to convey over that lifetime.

Many grids are showing their age at 70-100 years. Perhaps it would last that long.

Perhaps we'd choose to put the entire world electricity supply through it continuously so that Anchorage nighttime in winter is powered by the simultaneous New Zealand daytime summer. Or perhaps we won't.

But these kinds of question are the domain of geopolitics and global economics, so the best I can do is give the numbers that show a very small electrical resistance is objectively quite cheap — significantly less than the US alone is planning on spending to modernise its grids.

I misunderstood what you were originally saying.

I thought you were talking about how much electricity was consumed in smelting that much aluminium. Not actually using that much aluminium to transmit power