Yeah, that "new law" one is f'ing annoying. It's been a "new law" since, uhm, for as long as I've been buying insurance? News flash: You pay different insurance rates based on your driving profile! WOW. They've started changing up the wording with other non sequiturs, but it's the same crap ad. (Why would the DMV give two shits about how much I'm paying for insurance?)

Unless it allows them to raise their per-click rates on the premise of "quality clickthroughs."

80-bit floats are not available on any platform other than x86

The 80-bit long double is also available on the 68881, 68882 coprocessors and later 68K family members that incorporate the FPU. The Itanium also supports the 80-bit format.

But yeah... those aren't particularly common these days.

Ok, it allows a few, but it definitely strips mu. I tried it. Here's 15 mu characters:

FTFA:

Any complaints can be submitted to the provided box in the kitchen to be used in our weekly Negate the Negativity Bonfire.

I said as much above.

In an AC system, that current is continuously changing, so those transmission lines are continuously radiating away some amount energy. But that's not all. If there are any conductors nearby, those E-M waves can induce a current in those conductors, and the resulting E-M waves from that induced current can drag on the AC line further. This mutual induction is how transformers work. But, along an AC transmission line, unwanted coupling results in transmission losses. So, an AC system has a built in, inherent source of losses in the alternating current itself.

...and...

In a DC system, with a fixed, perfectly resistive load, the current doesn't change, so there's no radiative losses. In the real world, though, the loading on the system is continually changing, so the actual current demand on the DC system will vary over time, and some energy will be radiated away. To some extent that can be filtered, but that's limited by the amount of storage you can put near the ends of the transmission.

The submitter probably forgot to use the HTML escapes for the less than and greater than (ie. < and >), and those got eaten (along with the mu character before m in micrometers.)

The Greek mu was probably there when it was copy/pasted. Slashdot silently eats characters outside the English alphabet though.

Capacitors store energy, they don't dissipate it. Likewise with inductors.

Transmission lines represent both capacitive and inductive loads simultaneously. The capacitance, inductance, resistance of the transmission line together combine to form the characteristic impedance of the line. (Ok, there's one additional term: the conductance of the dielectric between the conductors. But, for high voltage transmission lines that are widely separated, this term is effectively 0.)

The characteristic impedance of a transmission line is of primary importance for determining the ideal load impedance for the line. In an impedance matched system, the maximum power will be transmitted to the load with no reflections.

Reflections can cause a phase shift between voltage and current, making a transmission line effectively look reactive or inductive. (See surge impedance loading.) This can be corrected for in the same ways as reactive or inductive loads by adding capacitance or inductance elsewhere.

If the load itself is reactive or inductive then you can get reactive power transfer. Reactive vs. inductive is in some sense a matter of sign; in one, current leads voltage, in the other current lags voltage. In both cases, current is out of phase with voltage and that's the problem to be solved.

Reactive power doesn't transmit any actual power to the load, but it still sends current through the system. Current is subject to ohmic losses (thanks to our friend I*I*R). Sending current without delivering real power subjects you to losses without any benefits.

In general, the capacitance of the transmission line itself isn't the culprit on its own. Rather, if you have a reactive load (either capacitive or inductive), or you have imperfect impedance matching between the load and the transmission line, you can get current flowing through your wires that isn't driving a load. That excess current incurs plain ol' resistive losses.

There is one way high capacitance can cause real problems for transmission line management, though. The rate of propagation of waves through a conductor slows in proportion to the square root of the product of the inductance and the capacitance. So, for a highly capacitive line, reflections move slowly through the system, and it becomes more difficult to compensate for transients. That seems to be the real bugbear for buried high-capacitance lines. Again, you're not losing to the capacitance directly, but rather to the knock on effects that lead to poorly compensated reflections and reactive power transfer in the system.

(Dr. Jetton, if you're reading this... EE305 may have been 20 years ago for me, but I haven't completely forgotten it. And Dr. Schertz... I didn't completely forget my T-line theory either. I wouldn't be surprised if either of you would point out flaws in my summary above.)

I used 5v as an example as the linked article spoke specifically of running 5V and 12V everywhere. I agree that you really want a higher voltage for distribution. 48V goes a long way, although it still requires quite a lot more copper than 110V or 240V for the same power carrying capacity. (About 5x if I did my math correctly.)

Now, if those in-wall adaptors could store some charge locally (small capacitor bank), and you didn't have to wire for peak current, only sustained current, maybe you could get away with smaller wiring that way. I'm skeptical.

I see what you mean. Let's put some numbers to that for everyone's benefit.

According to the table I linked previously, the OOOO gauge wire is 0.16072 ohms per 1000m. So, for a 20m run, that's about 0.00321 ohms. The voltage drop incurred by 330A across that resistance would be just over 1.06 volts.

For a 5V run, that's pretty significant, really. And you'd be dissipating over 350W in that wire alone. Yow! At 330A, you'd be burning 20% of your power just in that cable if you used OOOO gauge cabling.

Now the same numbers for 10 gauge wire, 15A, 110V, 20m. That's 3.276392 ohms per 1000m, or 0.0655 ohms for 20m. Voltage drop at 15A is 0.983V. Peak power dissipated in the wire is 15A * 0.983V = 14.7W. (RMS power is only ~10W.)

I'd hate to think of just how much extra copper I'd need in my walls to distribute ample power throughout my house and not suffer resistive losses. A 15A 110V circuit would become a 330A 5V circuit. The wire that carries the 15A @ 110V circuit is about the thickness of a pencil lead (about 1/10"). The wire required to carry 330A would have conductors the size of a garden hose (about 1/2"). (Note: The table linked above only goes up to 302A with a thickness of 0.46"; 330A would be larger still.)

There's two main sets of losses, as I understand: Resistive losses and radiative losses. You can get into other issues, such as power factor and phase error related losses. The two biggies that hit you almost before you get started are resistive and radiative losses, though, if you just consider a single transmission line driving a resistive load.

You combat resistive losses by going up in voltage, so you can send more power with less current. Since resistive losses are proportional to the square of current, each doubling of voltage reduces your resistive losses by a factor of 4. That's why long haul transmission lines are high voltage.

Radiative losses are different. Whenever you accelerate a charged particle, you generate an electromagnetic wave. With respect to wires carrying current, that corresponds to changing the amount of current. (Current measures the rate at which electrons flow, so changing current means accelerating or decelerating electrons.) That's how radio transmitters works, for example.

In an AC system, that current is continuously changing, so those transmission lines are continuously radiating away some amount energy. But that's not all. If there are any conductors nearby, those E-M waves can induce a current in those conductors, and the resulting E-M waves from that induced current can drag on the AC line further. This mutual induction is how transformers work. But, along an AC transmission line, unwanted coupling results in transmission losses. So, an AC system has a built in, inherent source of losses in the alternating current itself.

In a DC system, with a fixed, perfectly resistive load, the current doesn't change, so there's no radiative losses. In the real world, though, the loading on the system is continually changing, so the actual current demand on the DC system will vary over time, and some energy will be radiated away. To some extent that can be filtered, but that's limited by the amount of storage you can put near the ends of the transmission.

The reason AC won out over DC in the early days is that we didn't have practical means to step DC voltages up and down. But, we had just invented the first practical transformers, and those can step AC to higher and lower voltages trivially.

HVDC is practical now since we've had 100 years to develop better technology for converting DC voltages on the grand scales required.