"No, not spaceman. Specimin." - von Braun, in "The Right Stuff", speaking of the Mercury astronauts.

... What's ridiculous is your constant repetition of this bullshit idea. Yes, the cooler walls radiate inward but they have no effect whatsoever on the heat source. ALL of that radiation is reflected or scattered by the heat source. (It is not transmitted because we're dealing with diffuse gray bodies of significant mass.) ... [Jane Q. Public, 2014-09-21]

It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).

... You took a badly-worded sentence or two and jumped on them as though Latour made a mistake. But his only mistake was wording a couple of sentences badly. He does in fact NOT suggest that warmer objects absorb no radiation, and he has written as much many times. ... You have refuted NOTHING but a couple of unfortunately-worded sentences, which Latour himself publicly corrected shortly after that post appeared. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly object to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.

... shortly after Latour published that blog post, it became clear that the language he used implied that no radiation at all was absorbed by the warmer body. So a reader could not reasonably be blamed for inferring that. But Latour quickly apologized for the unfortunate wording and corrected himself to make it very clear he was referring to net, not absolute, heat transfer. ... [Jane Q. Public, 2014-07-27]

Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.

Or the right wing social network..

Or Ricochet wireless networks.

Besides, most Tor exit nodes are monitored. Using Tor is like screaming "I'm hiding".

That lasted until WWII.

Until the 1980s, anyone could enter the Pentagon and wander around the corridors. (George C. Marshall, Army Chief of Staff, decided during WWII that there was no way a building with as many people as the Pentagon could keep spies out, and requiring badges would give a false sense of security.) In the 1960s, anyone could enter most Federal buildings in Washington, including the Capitol and all the House/Senate office buildings, without passing any security checkpoints.

Isolated event, and the guy was brought down. There'll always be a risk as long as their are fanatics or loonies who don't give any though to their own personal safety, but there comes a point of diminishing returns.

Suppose they hired 10 times as many Secret Service agents? That just increases the odds of one of them going bad and offing the President himself. (Not a likely event, but having 10x as many agents also means more chances of confusion in a crisis, etc, etc.)

Security is never perfect (wasn't there an incident some years back where an intruder wandered into the Queen's living quarters at Buckingham Palace?) That's one reason we have a line of succession -- it's not like the government collapses in the case of an untimely death.

Mind, given the choices of VP over the past few presidencies, that line of succession might actually be helping lower the odds of someone trying to assassinate the Prez.

It was a Friday evening. The President had left for Camp David earlier, and his main protective detail went with him. Most staffers had gone home. The guy got just inside the outer doors, where there is a security checkpoint, before he was tackled.

The Secret Service made the right choice not shooting the intruder dead on the lawn. They certainly had the capability to kill him. They would have been heavily criticized, with pictures of the dead body on national TV.

On September 12, a man wearing a Pokemon hat and carrying a stuffed animal jumped the White House fence. He was tackled and arrested. Should he have been killed?

I agree and am happy to see this fork. As unpopular as it may make me, I actually like the initd functionality of systemd. I'm fine with using and writing the old init scripts, but systemd unit files are simple, concise, and powerful enough for my needs.

On the other hand, I find the kitchen-sink feature creep of systemd absolutely repulsive. Cramming all of that functionality into PID 1 as a unwieldy monolith seems like such a deeply flawed exercise. Uselessd seems like a perfect replacement for systemd: all of the benefits and none/less of the cruft.

If that were correct how would the Docklands Light Railway operate above ground without any drivers at all? The sad reason that drivers are needed is because of the unions. They automated the Victoria line years ago (1960s) but the unions threatened action and the resulting chaos that a drivers strike would have caused on the lines which were not automated forced them to keep drivers on each train even though they are completed unnecessary.

... Repeat: this ASSUMPTION of yours that the chamber walls must be accounted for in the power requirement of the heat source is a direct violation of the Stefan-Boltzmann law. There are no 2 ways around it. Established physics (the Stefan-Boltzmann law) says that the radiative power out (and therefore power in) of a gray body is dependent ONLY on emissivity and thermodynamic temperature. It is completely unrelated to any nearby cooler bodies. ... [Jane Q. Public, 2014-09-21]

Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary. But "power in" through a boundary around the heat source looks like this:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Jane refuses to account for the chamber wall radiative "power in" which would only be true if the source didn't absorb any of that radiation. Zero.

If you are sincere (you certainly haven't been acting like you are), then you must be postulating some kind of "tractor beam" effect that allows the chamber wall to "suck" power out of the heat source from a distance. I assure you that at least at out current level of technology, we have not managed to build such a sucking device. The heat source radiates out what it radiates out, and nothing around it is "sucking" any power from it. Although you seem to be doing your very best at "sucking" my time away over stupid bullshit. [Jane Q. Public, 2014-09-21]

That's ridiculous, Jane. I'm just noting that the chamber walls are hotter than 0K, so they emit radiation into a boundary around the heat source. Therefore Jane's wrong to ignore that radiation when applying the principle of conservation of energy:

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.

So the only "heat source" where we could validly ignore the radiation from the chamber walls would be a perfectly reflective "bobble" from Vernor Vinge's Marooned in Realtime. I assure you that at our current level of technology, we haven't managed to build such a device. And even if we could, it wouldn't be a heat source.

... No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer. ... [Jane Q. Public, 2014-09-20]

Jane's equation claims "none at all":

electrical power per square meter = (s)*(e)*Ta^4

Since Jane's equation for required electrical power doesn't even include a term for radiation from the chamber walls, Jane's equation wrongly says that no radiation at all is absorbed by the source. None. Zero.

It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source. Slayer "physics" are incoherent nonsense.

You are confusing "state capitalism" with "socialism".