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College Students: Want To Earn More? Take a COBOL Class

Jane Q. Public Re:COBOL: Why the hate? (122 comments)

However, since if it is still being used, then it still has some capability that is not available in other solutions.

No, no, no, no!

COBOL is still in use because because mid-to-large corporations spend many millions of dollars on systems that WORKED, and now it's far cheaper to keep them working, the same old way, than it is to do it all over again with modern equipment and languages.

This is called "installed base" and it's a particular problem for COBOL because that was one of the first business languages, and has one of the largest, large-corporation "installed bases".

COBOL has nothing to offer that newer languages don't do better. Not. One. Thing.

5 hours ago
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College Students: Want To Earn More? Take a COBOL Class

Jane Q. Public Re:The UK Cobol Climate Is Very Different (122 comments)

Every professional workplace has an expectation of a formal atire.

No, they don't. This is a statement made by someone about ready to REtire.

Most high-paying tech jobs today do not require a suit and many not even an office to go into. Often you can work at home in your pajamas, if you like.

Yes, really.

5 hours ago
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Micron Releases 16nm-Process SSDs With Dynamic Flash Programming

Jane Q. Public Re:Lifetime at 16nm? (48 comments)

seems like the average life expectancy of SSDs are well beyond the needs of most people at the moment, unless you're doing some serious content creation with massive amounts of read/writes.

The lifetime has been exaggerated from Day 1. Further, multiplying this problem manyfold, is that when an SSD fails, it tends to fail totally. In contrast, when a hard drive i failing, you tend to get a few bad sectors which flag an impending problem, and you main lose a file or two. Bad SSD usually means "everything gone with no warning".

If you use SSD you should have a good HDD backup.

6 hours ago
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

And, last comment here: you have confirmed that you have not abandoned your incorrect (and actually quite ludicrous) version of heat transfer, which violates the Stefan-Boltzmann radiation law on its very face.

That was all I needed. I am now done. Have a nice day. You can have the last word all you like; it won't make you any more correct.

yesterday
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

And no, I don't have to ask myself that, because it doesn't happen.

I have already found the solution to a reasonable degree of precision. Your solution, as stated (approximately 241 degrees F for the central heat source) does not check out, even using your own equations.

yesterday
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.

That is neither correct, or an answer to my question.

yesterday
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

In other words, the electrical heating power is determined by drawing 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:

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

Right?

No. Not right. Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. Because the only power transfer taking place here is heat transfer, which is a function of (emissivity) * (S-B constant) * (Ta^4 - Tb^4).

You DO know what a minus sign is, yes?

Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler than the source, NONE of the power comes from the chamber walls, because of that minus sign in the equation above. Nothing has changed in that respect, and that's what the Stefan-Boltzmann law requires.

The only time that changes is if the walls are at an equal temperature, in which case heat transfer is 0 and you can begin to use "ambient" temperature as input. You are still supplying the same input power, you are just supplying it a different way.

If the chamber walls were hotter than the central source, then heat transfer would be in the other direction (because the sign of the solution to the equation above changes), and only THEN are you getting net heat transfer TO the central sphere.

And BOTH of those situations are a violation of Spencer's conditions.

yesterday
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

But wait. I take that back. Before I declare that I am done and go away, I just want to ask you: do you still maintain that after the enclosing passive sphere is inserted, the central heat source raises in temperature to approximately 241 degrees F? You haven't said anything about that in a while, so I'm just checking.

yesterday
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.

Nonsense. It would take power to bring the chamber walls up to 150F (338.71K). How else do you expect them to get to that temperature? Where are you getting that power from? This is so utterly obvious that I honestly don't believe you don't get it.

For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.

You could, but we haven't. Regardless, it still remains the same. Power output at that temperature remains constant because P = (emissivity) * (S-B constant) * T^4 says it has to.

The only thing you are doing is ADDING energy to the system by putting it in an ambient environment of 150F. That's not irrelevant at all, because if you're at thermal equilibrium, there is no heat transfer. Since this is all about heat transfer, how could it be irrelevant?

I have finally concluded that you are just a very good troll. I honestly -- and I mean that: honestly -- don't believe you could be this stupid and possess a degree in physics.

The ONLY time the power output changes is if you change the temperature. You can do that by making the walls HOTTER than the "heat source", thereby causing a net heat transfer TO it from the walls, OR you can input more electrical power to the heat source, thereby making it hotter, but that would be a violation of the conditions Spencer stipulated.

Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.

That's not our disagreement at all. Not even frigging close. Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees. I just got done saying that. But it still does have power input. It' just that it comes from the environment in this case rather than an electrical element.

Because its radiant output power remains constant according to the Stefan-Boltzmann law. All you have done is raise the environment's output power to match, and raised the input to that environment enough to achieve that temperature. Big deal. That takes energy of its own, and proves exactly nothing. You haven't proved that it needs no power, you just changed the source of that power. And used up even more power in the process, because the environment is larger than the central sphere.

You're just wrong about how this works. And not just a little bit wrong, but completely out there in lala-land wrong.

And you have made it perfectly obvious that I am wasting my time talking to you. You are either crazy, or stupid, or a very talented troll. Based on my experience, I vote for that last one, but I think that necessarily implies a little bit of the first, too.

So we're done. I'm going to write this up as it stands here. I don't need anything else, and you've made it very clear that anything else would be further waste of my time. You refuse to change your tune, so fine. I'll just write it up that way. Don't worry: I am going to include your exact words.

yesterday
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.

Are you REALLY the moron you make yourself out to be? NET radiation from a cooler surface that passes the boundary is reflected, transmitted, or scattered and passes right back out through the boundary. This is a corollary of the Stefan-Boltzmann radiation law, which states that NET heat transfer is always from hotter to cooler.

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.

Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.

And again: by that same law, it just passes right back out again because the same NET amount of radiative power that crosses the boundary and intercepts the smaller sphere is either reflected, transmitted, or scattered. (Since we are discussing diffuse gray bodies here, we can consider it all reflected or scattered because there is no transmissivity.) The radiation that crosses the boundary that does not strike the smaller sphere due to view factor also just passes right back out. You are ignoring (e*s) * (Ta^4 - Tb^4). Anything other than what I described does not add up.

Once again, no. Draw a boundary around the heat source: power in = electrical heating power + radiative power in from the chamber walls

Just NO. Net heat transfer is ALL from hotter to colder, by (e*s) * (Ta^4 - Tb^4).

Let me put it another way: we can easily show how you have gotten your thermodynamics backward by referring to a question you asked earlier. You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F.

The answer is YES, and here is why:

You are proposing to bring the whole system up to a level of higher thermodynamic energy, rather than just the heat source. And you are somehow proposing that it doesn't take more energy to do that. But of course it does.

The power required to bring the heat source up to 150F remains the same, because the Stefan-Boltzmann law says it has to be. But NOW, you are ALSO bringing the walls up to that higher temperature, and THAT would require even more power (because of the slightly larger surface area).

This clearly illustrates your ass-backward thermodynamic thinking. The radiative power output of the heat source does not change due to the temperature of the walls. At all. The only thing that changes as the wall temperature changes is the heat transfer, which would lessen as you brought up the temperature of the walls. But that isn't because the heat source is using less power, it is because you are putting more power into raising the wall temperature. You are creating a more thermodynamically energetic environment, and that requires power.

Just like your other arguments: you invent power in out of thin air, and claim you can do that because it's "moving" in the opposite direction in which heat transfer is actually taking place.

You are giving physicists a bad name, and I repeat that I am going to show this to all the world to see.

yesterday
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Extent of Antarctic Sea Ice Reaches Record Levels

Jane Q. Public Re:Time for new terminology (586 comments)

GISS is precisely the dataset that has been accused of the the most egregious "adjustments".

Further, it was recently found that GISS was improperly averaging in "missing" data over a period of years, which they admitted to about 2 months ago.

It is interesting that the historical HCN data disagree quite a bit with the modern versions of the data sets.

yesterday
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Extent of Antarctic Sea Ice Reaches Record Levels

Jane Q. Public Re:Time for new terminology (586 comments)

Actually NASA (or was it NOAA?) changed their tune again and are saying it was 1937.

Gotta keep up with this stuff, man.

The raw, unadjusted temperature records always have said 1937. It's the adjustments that are questionable, not the historical record.

2 days ago
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Comcast Allegedly Asking Customers to Stop Using Tor

Jane Q. Public Re:This may be the way to escape from Comcast (405 comments)

In the end, you signed a contract and are legally bound to continue to pay for almost any type of service inturruption.

Except that I didn't. When my cable was installed I signed a small receipt acknowledging that the tech had been there. I signed no contract.

That might have been an oversight on their part, but that doesn't matter.

Further, the KIND of contract that Comcast has customers sign is known in the legal industry as a "contract of adhesion". What that means is that it was a non-negotiable, take-it-or-leave-it "contract". The problem being that contract law assumes that every party is free to negotiate before signing.

So in many genuine, legal senses of the term, it's not a "real" contract anyway, and honest judges are required in principle to view them "with a jaundiced eye", and lean toward the customer when a dispute arises.

I'm not saying all judges are honest enough to do that, but they're supposed to.

2 days ago
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

And one more thing I would like to make very clear:

The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0).

It is NOT, as you assert, because the heat source would be using less power. That's false, by the S-B equation. Its power output remains the same because (Spencer's stipulation) the power input remains the same.

The reason my solution does not violate conservation of energy, is that the power consumption of the chamber wall is allowed to vary. THAT is where the change takes place, not at the heat source. Again, this is a stipulation of Spencer's challenge.

Once again: power out of heat source remains constant, because P = (emissivity) * (S-B constant) * T^4. There is nothing in these conditions that changes this at all. Therefore, BECAUSE the power out and power in at the heat source remain constant, so does the temperature. It's all in that one little equation.

2 days ago
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

In fact let's just face this directly, with no mincing of words:

It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

We are not AT thermal equilibrium, so that is a ridiculous straw-man argument.

One question only: do you agree with the Stefan-Boltzmann relation: power out P = (emissivity) * (S-B constant) * T^4 ??

No more bullshit. "Yes" if you agree that equation is valid, or "No" if you deny that it is valid. Just that and no more.

I'm not asking your permission. I'm just trying to find out whether you're actually crazy or just bullshitting.

2 days ago
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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

Jane Q. Public Re:Jane/Lonny Eachus goes Sky Dragon Slayer (155 comments)

You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!

Nonsense. This is textbook heat transfer physics. We have a fixed emissivity. Therefore, according to the Stefan-Botlzmann radiation law, the ONLY remaining variable which determines radiative power out is temperature. NOTHING else. That's what the law says: (emissivity) * (S-B constant) * T^4. That's all. Nothing more. This makes it stupidly easy to calculate the radiative power out, and therefore the necessary power in.

YOU are disputing the Stefan-Boltzmann law. But it is a known physical law, and this is a textbook demonstration of it. You lose.

It's so adorable that Jane keeps insisting that Jane kept the power constant, even after I showed that Jane's calculation was only able to hold the source temperature constant after the enclosing shell was added by halving the actual electrical heating power.

You showed no such thing. Your calculations contradict themselves, and your methodology contradicts itself.

EVEN IF we accepted your idea that the "electrical" power required to be input to the heat source is dependent on the temperature difference between the heat source and chamber wall (a violation of the S-B law), you still contradict yourself because your answer of a hotter heat source would still then require MORE power, because the difference is greater. But that is not allowed by the stated conditions of the experiment, and you keep glossing over that simple check of your own work which proves it wrong.

So no matter how you cut it, your answer is wrong, by your own rules.

It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

This is a simple requirement of the Stefan-Boltzmann law. The radiative power output of a given body does not depend on other nearby bodies. It's inherent in the law itself. And this is precisely where you are getting it wrong.

I find it highly amusing that you derive your own calculations from the Stefan-Boltzmann law, then deny that it is valid. Every time you try to squirm out of this you just contradict yourself again.

I am further amused that you find it "adorable" that you've been proven wrong. Be a man for a change and admit it. Or show us your own replacement for the Stefan-Boltmann law. You don't get to have it both ways.

2 days ago
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High School Student Builds Gun That Unlocks With Your Fingerprint

Jane Q. Public Re:Great one more fail (581 comments)

Besides, 99.99% is not nearly reliable enough. (And besides, this number is misleading... probably outright false.)

According to calculations I did a year or two ago, in order for a "smart firearm" to be worthwhile and actually solved the problem for which is supposed to be designed, for modern arms, it needs to have AT LEAST three 9s behind the decimal point for true positives: 99.999%, and probably actually 4.

And that's assuming the stats are correct. What does that 99.99% represent? True positives? What is its rate at rejecting true negatives? After all, that's the entire purpose it was designed for.

Further yet: how long does the battery last? What is its success rate with a dead battery? Current battery tech is not capable of delivering 99.99% reliability because batteries go bad even on the shelf.

2 days ago
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Why Atheists Need Captain Kirk

Jane Q. Public Re:illogical captain (877 comments)

Well, I didn't interpret the article the same way you did. I thought the article was saying that you can be logical and still feel wonder. It wasn't saying that science-oriented people need to be religious, but rather that religious people should stop seeing them as somehow inhuman and unfeeling without a belief in their God.

3 days ago
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Why Atheists Need Captain Kirk

Jane Q. Public Re:illogical captain (877 comments)

Atheists will be in for a rude awakening when they die as they will realize that their belief was incomplete. Regardless, they can be just as good, (or as bad) as theists if they practice the golden rule.

Why would they be in for a "rude" awakening, when one would think that any awakening at all should be a pleasant surprise?

Further, as Sam Harris argues quite well, one need not be a theist to have moral values. Science + secular society are perfectly capable of agreeing upon ethical and moral rules, without resorting to theism.

3 days ago
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The Future According To Stanislaw Lem

Jane Q. Public Re:Maybe... (195 comments)

No, you don't. You just have to give enough people cheap energy that they're no way to put the genie back in the bottle.

Once you do that, the Gateses and Kochs of the world can be safely ignored. They won't have any power anymore.

3 days ago

Submissions

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Global Warming Researchers Trapped In Antarctic Ice

Jane Q. Public Jane Q. Public writes  |  about 9 months ago

Jane Q. Public (1010737) writes "Christ Turney, a climate researcher at University of New South Wales, and some other researchers chartered a ship to go to Antarctica to further their Anthropogenic Global Warming ("climate change") research.

The expedition, consisting of 74 researchers and crew, radioed for help on Christmas day, stating that they are trapped in the ice.

A chinese ice breaker called "Snow Dragon" came within a few miles of the stuck ship but had to turn back. The researchers and crew are now hoping that the ice breaker Aurora Australis, out of Australia, will be able to reach them."
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Airport Announcement Threatens Arrest For TSA Jokes

Jane Q. Public Jane Q. Public writes  |  about a year ago

Jane Q. Public (1010737) writes "In this YouTube video posted just 2 days ago, the PA system in the Houston airport tells passengers that "... inappropriate remarks OR JOKES concerning security may result in your arrest".

Even under GWB, this would have been unthinkable. And the timing is — for lack of a better way to put it — very interesting."
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Slashdot Drastically Throttles Submission Frequency

Jane Q. Public Jane Q. Public writes  |  about a year ago

Jane Q. Public (1010737) writes "Remember when you could submit a comment in one thread, then submit a comment in another thread after 1 minute?

Slashdot has now limited your submissions to once every 5 minutes.

I don't know about you, but there have been rare occasions in which I found even 1 minute to be stifling. 5 minutes is ridiculous. Sometimes it's possible to browse through 3 whole new topics in less than 5 minutes."
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Obama Administration Asks Supreme Court To Not Hear Jammie Thomas Case

Jane Q. Public Jane Q. Public writes  |  about a year and a half ago

Jane Q. Public (1010737) writes "The Jammie Thomas-Rasset case has been in the news for years now. As of the last court ruling, she has been ordered to pay $222,000 for sharing 24 songs. Her attorney argues that you can buy the same songs on iTunes for $24, and imposing a penalty of almost 10,000 times as much is "excessive and oppressive". The case has been appealed to the Supreme Court.

The Obama Administration has asked the Supreme Court to not review the case. Is this another example of this administration pandering to the copyright tro... I mean corporations, rather than The People they are supposed to represent?"

Link to Original Source
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The Best Dennis Ritchie Quote

Jane Q. Public Jane Q. Public writes  |  more than 2 years ago

Jane Q. Public writes "

"Dennis Ritchie (1941-2011). His pointer has been cast to void *; his process has terminated with exit code 0."

Thus spake James Grimmelmann (@grimmelm), on Twitter"

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MIT Prof. Says Power From Water is Near.

Jane Q. Public Jane Q. Public writes  |  more than 5 years ago

Jane Q. Public writes "At the Aspen Environmental Forum yesterday, MIT Professor Daniel Nocera claimed that MIT research has found a more efficient way to hydrolyze water at room temperature with the use of cobalt and potassium phosphate, and that tomorrow's home will get its power from feul cells charged with hydrogen from plain water and a bank of inexpensive solar cells. If true, this is a major breakthrough in energy distribution and could solve many of our global energy needs."
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Apologies!

Jane Q. Public Jane Q. Public writes  |  more than 5 years ago

Jane Q. Public writes "I admit that I was a bit less than diplomatic; frankly I did not think I would get your attention, and it really was the kind of error that can cause bad feelings.

I will do better next time."
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Hey, Editors!

Jane Q. Public Jane Q. Public writes  |  more than 5 years ago

Jane Q. Public writes "Hey! Re: my article that you just posted, "FTC Warns Against Deceptive DRM"... webcasts are NOT available, and you should have checked before you changed the article to say that they were. Live streaming webcasts were available when the talks were going on, but they don't work now.

So now, you are going to get lots of readers trying to download webcasts, and blaming me when they can't. Thanks a shitload."
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FTC says "We'll 'come calling' about deceptive

Jane Q. Public Jane Q. Public writes  |  more than 5 years ago

Jane Q. Public writes "At the FTC's Seattle conference on DRM, FTC Director Engle started off by referencing the Sony rootkit debacle, and said that companies are going to have to get serious about disclosing DRM that may affect the usability of products. She also said that the fine print in a EULA is not good enough, and "If your advertising giveth and your EULA taketh away, don't be surprised if the FTC comes calling."
The conference was webcast live from the FTC website."
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Scotty's Final Mission

Jane Q. Public Jane Q. Public writes  |  more than 6 years ago

Jane Q. Public writes "According to a recent article at Ars Technica, the ashes of James Doohan, who played "Scotty" in the original Star Trek series and several movies, were aboard the SpaceX III launch yesterday and were lost when the launch vehicle failed.

A fitting epitaph might be: "The engines are not meeting specification, Captain! She kinna hold out much longer!""

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