"Would you take a real Rolls Royce to an unauthorized dealer for repairs?"

you bet your ass I would. I take my BMW to a non BMW mechanic. in fact I refuse to let the thieves at the BMW dealership TOUCH my BMW as they are no talent hacks.

Only a complete and utter fool believes that authorized means anything other than scam.

Oh look, here come the same "social engineers" that brought us soaring male suicide rates and burgeoning single motherhood with it's associated social outcomes,

The only problem with this statement, is while these theories on raising children are a relatively new thing (last 100 years or so) - it can not be demonstrated that any kids are ACTUALLY being raised this way. Maybe a few, here and there, but by and large, most parents still raise their kids using traditional violence-based methods.

So to blame these new science-backed techniques for the "decline of modern civilization" is just a bunch of bullshit; to justify frustrated parents whose first tool in their parenting toolbox is the paddle.

Their MSP430 is great tech but their libraries suck horribly. They need to embrace the public community and less of their in house engineers on making libraries that are worth using AND stop using a compile/IDE that is complete crap and crippled.

the GCC toolchain is decent but still clunky as all hell.

Not healing like we used to.... Damn it's annoying for healing to take so frigging long.

Oh and GET OFF MY LAWN!

Most people are too stupid to use them. There is a reason that US market cars are dumbed down compared to Japan Market and Europe Market.

Standard resistive Touchscreen tech is dirt freaking cheap. I can get 7" resistive types for $9.00 each all day long at single quantities. If I was a car maker I could get them at less than $1.00 each in 1000+ quantities.

Honestly this IR system is a rehash of really old tech that is just not needed.

What is needed is the important buttons existing as REAL HARD BUTTONS. the systems that are 100% touch are complete crap. Yes I do want my hard buttons back on android, the on screen home button is really 100% crap.

If anyone reading this is curious what a troll looks like, find thia dude's "energy conservation" post in that thread. And i'll write your next comment for you to save you from having to consult your one-line script yet again: "y u ask me kill myslef" [Rujiel, 2014-12-14]

Do you mean this post where I explained that Jane Q. Public's climate science denial violates conservation of energy? Again, why did that prompt you to accuse me of being a paid oil troll?

Are they hiring you losers while still in high school these days? The bar for paid oil trolls sure is a low one--any stupid thing to prevent the discussion of the oil cartel's impunity. Do the world a favor and kill yourself. [Rujiel, 2014-11-20]

Why would a paid oil troll defend mainstream climate science? This is one reason why I think you might be mistaking me for someone else. Why would the oil industry pay me to debunk the same baseless accusations they're helping to spread?

Another reason I think you might be mistaking me for someone else is that in that post I quoted Jane Q. Public to respond to his baseless accusation:

.. Ever since I challenged his incorrect answer to a question of physics several years ago, he has been rude and insulting.. [Jane Q. Public, 2014-11-20]

... seriously, "rude and insulting"? Here are just a few of Jane's most recent charming statements to me. If Jane was telling the truth about my comments, Jane should be able to produce quotes of similar length which are just as "rude and insulting" as Jane's. Jane can't do that because he's just projecting his own rude, cuss-filled insults onto me.

".. Jesus, you're a dumbshit. .. your adolescent, antisocial behavior .. keep making a fool of yourself. .. you're being such a dumbass .. your analysis of it is a total clusterfuck. .. you're so damned arrogant you think I'm the one being stupid. .. you were too goddamned stupid .." [Jane Q. Public]

As you can tell by clicking those links, all those insulting comments were actually quotes from Jane Q. Public, directed at me. As you can tell, Jane Q. Public has been cussing at me for months, and I never responded in kind. That's why I found it bewildering that he accused me of being "rude and insulting". So I quoted some of Jane's bizarre insults to show that Jane's baseless accusation was textbook psychological projection.

It's still not clear why this caused you to hate me so much that you've suggested I kill myself three times.

Jane's "interest" in that NAS report evaporated after I showed that Jane had been fooled by "Steven Goddard" once again. So let's return to Jane's confusion about basic thermodynamics.

But net radiative power out of a boundary around the source = "radiative power out" minus "radiative power in", so the equation Jane just described also says:

NO!!!!! As I have explained to you innumerable times now, you can also consider your heat source, by itself, that "sphere". The only NET radiative power out comes from the electrical power in. Further, the cooler walls do not contribute any of that NET power out. That's what net means. [Jane Q. Public, 2014-12-16]

I've already pointed out that Jane's hopelessly confused about the word "net", but that's just one of the mistakes Jane packed into these few sentences.

Jane's also wrong to imply that energy conservation across one choice of boundary could somehow contradict energy conservation across another boundary choice. That's impossible. Many boundary choices are inconvenient but they all have to be consistent. Otherwise, how could we possibly tell which boundary choice was correct?

So Jane can't object to the simple energy conservation equation I derived by claiming that some other boundary choice would somehow contradict my equation. That's completely impossible, and if Jane doesn't understand that point then he should learn about conservation of energy: example (backup), example (backup), example (backup).

As you can tell after reading those introductions, here's how to apply conservation of energy. Draw a boundary around the heat source:

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 through any boundary where nothing inside is changing:

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

I put the boundary around the heat source so the boundary is in vacuum. That's because radiation can't travel through opaque solids like the heat source. So the only way to obtain an energy conservation equation with radiative terms is to place the boundary around the heat source.

For example, I calculated the enclosing shell's inner temperature by drawing the boundary within the enclosing shell. This boundary was inside aluminum, so heat transfer through it was by thermal conduction, not radiation. Notice that even this boundary choice leads to a conduction equation where electrical heating power depends on the cooler chamber wall temperature. That's because all boundary choices have to be consistent. They can't contradict each other unless one of them is wrong.

After I asked Jane to explain exactly where his boundary would be drawn, Jane replied:

... You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency. ... [Jane Q. Public, 2014-09-15]

Nonsense. I've repeatedly explained that my boundary is drawn around the heat source, so it's in vacuum and therefore contains radiative terms both for radiation going out and radiation going in.

Choosing to put the boundary somewhere else, like inside the heat source, leads to an energy conservation equation with conduction rather than radiative terms. But even those conduction equations agree that electrical heating power depends on the cooler chamber wall temperature. They can't contradict each other. Putting the boundary somewhere else might be inconvenient, but it couldn't possibly contradict the fact that electrical heating power depends on the cooler chamber wall temperature.

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Once again, Jane's wrong. There is literally no choice of boundary which will lead to his absurd equation. Once again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power".

At least, that's the most charitable explanation. Once again, I'm trying to rule out less charitable explanations like the disturbing possibility that Jane isn't honestly confused about basic thermodynamics. Maybe Jane/Lonny Eachus has simply betrayed humanity by deliberately spreading civilization-paralyzing misinformation.

Jane/Lonny Eachus could help convince posterity that he was just honestly confused by thinking carefully about conservation of energy, explaining exactly where his boundary lies, and carefully listing all the power going in and out of that boundary.

Or Jane/Lonny Eachus could help convince posterity that he's betrayed humanity by continuing to spread civilization-paralyzing misinformation.

Jane's "interest" in that NAS report evaporated after I showed that Jane had been fooled by "Steven Goddard" once again. So let's return to Jane's confusion about basic thermodynamics.

But net radiative power out of a boundary around the source = "radiative power out" minus "radiative power in", so the equation Jane just described also says:

NO!!!!! As I have explained to you innumerable times now, you can also consider your heat source, by itself, that "sphere". The only NET radiative power out comes from the electrical power in. Further, the cooler walls do not contribute any of that NET power out. That's what net means. [Jane Q. Public, 2014-12-16]

I've already pointed out that Jane's hopelessly confused about the word "net", but that's just one of the mistakes Jane packed into these few sentences.

Jane's also wrong to imply that energy conservation across one choice of boundary could somehow contradict energy conservation across another boundary choice. That's impossible. Many boundary choices are inconvenient but they all have to be consistent. Otherwise, how could we possibly tell which boundary choice was correct?

So Jane can't object to the simple energy conservation equation I derived by claiming that some other boundary choice would somehow contradict my equation. That's completely impossible, and if Jane doesn't understand that point then he should learn about conservation of energy: example (backup), example (backup), example (backup).

As you can tell after reading those introductions, here's how to apply conservation of energy. Draw a boundary around the heat source:

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 through any boundary where nothing inside is changing:

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

I put the boundary around the heat source so the boundary is in vacuum. That's because radiation can't travel through opaque solids like the heat source. So the only way to obtain an energy conservation equation with radiative terms is to place the boundary around the heat source.

For example, I calculated the enclosing shell's inner temperature by drawing the boundary within the enclosing shell. This boundary was inside aluminum, so heat transfer through it was by thermal conduction, not radiation. Notice that even this boundary choice leads to a conduction equation where electrical heating power depends on the cooler chamber wall temperature. That's because all boundary choices have to be consistent. They can't contradict each other unless one of them is wrong.

After I asked Jane to explain exactly where his boundary would be drawn, Jane replied:

... You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency. ... [Jane Q. Public, 2014-09-15]

Nonsense. I've repeatedly explained that my boundary is drawn around the heat source, so it's in vacuum and therefore contains radiative terms both for radiation going out and radiation going in.

Choosing to put the boundary somewhere else, like inside the heat source, leads to an energy conservation equation with conduction rather than radiative terms. But even those conduction equations agree that electrical heating power depends on the cooler chamber wall temperature. They can't contradict each other. Putting the boundary somewhere else might be inconvenient, but it couldn't possibly contradict the fact that electrical heating power depends on the cooler chamber wall temperature.

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Once again, Jane's wrong. There is literally no choice of boundary which will lead to his absurd equation. Once again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power".

At least, that's the most charitable explanation. Once again, I'm trying to rule out less charitable explanations like the disturbing possibility that Jane isn't honestly confused about basic thermodynamics. Maybe Jane/Lonny Eachus has simply betrayed humanity by deliberately spreading civilization-paralyzing misinformation.

Jane/Lonny Eachus could help convince posterity that he was just honestly confused by thinking carefully about conservation of energy, explaining exactly where his boundary lies, and carefully listing all the power going in and out of that boundary.

Or Jane/Lonny Eachus could help convince posterity that he's betrayed humanity by continuing to spread civilization-paralyzing misinformation.

It has been well known that all Corporate IT security is a complete joke. CIO refuses to spend the money on it, COO refuses to make users actually follow real security procedures, and the CFO loves the "it wont happen to us" line that means they will not have to actually spend money on real IT security.

This is not new, I'm just glad that it's happening in a very public way so that maybe the worthless executives out there will actually listen to their IT experts about the fact that we NEED to spend the money to try and keep the bad guys out.

The old guy there was trying to get internet to the island and they threw him in jail. Let's start with forcing the Cuban Government to take bi-polar meds first?

I find it stupid. My problem is, I can't quite figure out who is being stupid here, except that it doesn't seem to have much to do with N Korea, if I'm any judge. They may be idiots, but they have been clever enough to hold on to power for decades in the most astonishingly ridiculous circumstances - a bit like the goings-on in the declining Roman empire. But it clearly doesn't add up, the idea that North Korea are somehow able to threaten the US into submission. I would be very interested in knowing the actual truth of the matter.

Not saying there is evidence it has happened. Saying it is a great idea and should have happened.

Have you ever retired a riddle by mistake?

The Federal numbers are an average for cars that cost $500,000 to $25,000 my 2007 civic will lose less than $3.00 for the 3000 miles added to it, it's already at the bottom of the curve and even adding 10,000 miles will not change it's "resale value" that has no real meaning as I dont intend to sell it.

And "major repairs" don't come from miles, they come from abuse and lack of proper maintenance.

Now my Ferrari F40, that would have a much higher depreciation for those miles.