How do you think that's even a fair comparison? I could easily wipe the hard drive and put a Windows install that is finely tuned and put it in the box as well, and there's a good chance my install will have a better user experience than yours.
It's also about fame and power. Being noticed, having people rely on you, and being able to directly affect the lives of so many... these are like a drug.
And even if it starts out unselfishly, I've seen a few politicians get a taste of the drug and change ways. It sucks when I voted for them and feel the need to apologize for my support.
Ballmer's best-before date was before he became CEO. He rode the monopoly train well past the station. The fact that he allowed the company to finally take risks with new product is pretty much the only twinkling of worthiness he had during his tenure.
And they should lie in it. Microsoft's monopoly in IE was one of the principal causes of stagnation in the industry during the mid 2000s.
Then again, that stagnation arguably led to some great innovations by others in the industry, which is why we've witnessed the mobile revolution and downfall of IE since.
Email allows me to respond according to my schedule. Call me if something's really important.
Actually this brings up a side point... I would really like to have a good/standardized way to leave a voice message that works as you describe email, allowing the recipient to respond when you can get around to it. The problem with voicemail is that I have to call you to leave the message.
Calling interrupts you and you might answer, since you can't tell whether it needs immediate attention. So to leave an unimportant message, I need to type a text message or email... both of which are suboptimal in many situations (walking, driving, etc.).
I was thinking of this thread again (the tug-of-war definition) and thought of another interesting thing to calculate: What if I swap Pluto and Charon in the equation? Would it indicate that Pluto is also more influenced by Charon than the sun at times?
The answer is yes, meaning Pluto does not have a concave orbit either. The tug-of-war value focusing on Pluto as the primary body is 337.3 at perihelion, but focusing on Charon as the primary body it is 39.3. Since both are greater than 1, this means Pluto and Charon cause each other to have an occasionally convex orbit.
So perhaps this definition would still find Pluto/Charon to be a binary planet, but in a different way than the Earth/Moon are. Compare the numbers above with Earth/Moon: Earth-centric has a value of 0.45, Moon-centric has a value of 0.006. Both being less than 1, Earth and Moon always have concave orbits around the sun.
For clarification, the system requirements for Windows 9 developer preview are already known
There is no such thing as "Windows 9 developer preview". There's speculation that it will be called "Windows 9". There's speculation that it will have a developer preview. There's speculation about the system requirements. But that's all it is, speculation. No announcements and to my knowledge no leaks about any of those specific points have been made.
Until we see leaks or at least strong rumors by sources that have been correct in the past, nothing is "known". (Remember how many rumors we've had about Apple's iWatch and for how many years? Some even by credible sources? Yet we still don't have an iWatch.)
They could let the host venue determine whether to blackout. Some might, others wouldn't, and eventually the better outcome would become the most popular.
Thus, all we need to do is figure out at what new distance R2 the new orbital velocity V2 = V * (1 / square root of 2).
Change it to the following:
Thus, all we need to do is figure out at what new distance R2 the original orbital velocity V = V2 * (square root of 2), where V2 is the velocity of a circular orbit at distance R2.
Another reason I don't care for the barycentric approach is because it depends so highly on the radius of the larger body. What if the barycenter of the moon were right above the surface but well within the atmosphere? What about a gas giant where the definition of the radius is a bit fuzzier?
Agreed, that one is a bit far fetched. It's a many-body problem and all it takes is a bit of eccentricity or pull from other moons, planets, and the sun to destabilize.
But back to the first question, what if the barycenter moves in and out of the planet due to multiple moons? This would be akin to the solar system, where the barycenter moves in and out of the sun. I don't know if we could easily call it a ternary planet, quaternary planet, etc.
I think I prefer Isaac Asimov's tug-of-war definition of a binary planet. It would be considered a binary planet if the smaller body has a concave orbit around the sun; in other words, the two are both primarily orbiting the sun and just happen to be close to each other. This would, however, define the earth/moon system as a binary planet and Pluto/Charon would be a planet/moon.
Windows 8 has been optimized to run faster in most benchmarks than Windows 7. I am sure this trend will reverse itself eventually but for now I'll take your bet.
a couple Qs if you don't mind, because you obv know a lot about this.
I'm just a guy who was interested enough to Google and throw together some calculations.
so I guess if the barycenter is not in the middle of earth, then the earth wobbles as the moon goes around. Is this what causes tides, it's essentially the sloshing of the ocean as the earth wobbles? I always knew that "the moon causes tides", but I never understood the mechanism.
The gravitational forces between the earth and moon are major components of tides. However the barycenter doesn't seem to contribute directly. Moving the earth and moon farther apart (and thus moving the barycenter further away from the center of the earth) actually causes the tides to become weaker. This actually happens regularly as the moon gets closer and then farther from the earth in its orbit (the moon's orbit is not perfectly circular, but slightly elliptical). When the moon is at its closest, the tides are barely higher, and at its furthest the tides are barely lower.
I guess a second question would be, is there a certain distance at which the moon would escape earth's gravity? I wonder what it is, esp compared to the current distance away? would it be 2x, or 10% or 10x?
Gravity accelerates two objects toward one another, based on their mass and their distance. It works the same whether the two objects are initially moving toward each other or away from each other (away from each other, we usually call "decelerating", but there's no difference in the math).
Escape technically occurs when the two objects are moving away from each other, but the deceleration due to gravity will never be enough to overcome their initial velocity at their initial distance from each other. Gravity diminishes as the objects move farther apart, which results in less deceleration over time. In the case of escape, the velocity will never reach zero.
The answer is "yes"... assuming the moon magically appeared at that new farther distance but traveling at the same velocity as it is currently. According to the Wikipedia link on escape velocity:
The escape velocity at a given height is (square root of 2) times the speed in a circular orbit at the same height
Also, the orbital velocity of an object decreases as its distance increases. So increasing the distance of the moon would decrease how fast it would need to be going to stay in orbit.
But remember that the orbital distance suddenly increased but the orbital velocity did not change...
Let's say the moon is orbiting at distance R with orbital velocity V. Thus, all we need to do is figure out at what new distance R2 the new orbital velocity V2 = V * (1 / square root of 2).
This page contains the formula we need. Solving for r, r is proportional to 1/(v^2). So R2 is proportional to 1 / (V2^2), and substituting the equation above we find that R2 is proportional to 2 / (V^2), which equals 2 * (1/V^2), which equals 2 * R. R2 = 2*R.
Thus the answer is "2 times the original orbital distance, 478,000 miles".
That's a bit unfair to compare Mac computers with Windows phones. Comparing computers, Microsoft has pretty much an unbeatable record when it comes to supporting old versions.
But iPhones are updated for somewhat longer than Windows Phones (my 3GS was supported from iOS 3 through iOS 6, so 4 years worth).
My 2008 Mac Pro has the latest version of OSX and the Apple applications.
My 2006 Macbook hasn't been eligible for an OS upgrade since Lion (3 versions ago). To upgrade to the latest OS means buying a new Macbook (Air) at or around $1000 with tax.
Meanwhile I can still purchase the full latest version of Windows for that computer for $120-200 depending on edition. I'd bet that any Intel version of Windows Microsoft releases in the next 10, possibly 15 years will still run on it.
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And you are an idiot if you think people will not pay for good/familiar software. They have been doing it for decades.
But please, continue to tell me about how THIS is the year of Linux on the desktop.
PEBKAC.
How do you think that's even a fair comparison? I could easily wipe the hard drive and put a Windows install that is finely tuned and put it in the box as well, and there's a good chance my install will have a better user experience than yours.
It's also about fame and power. Being noticed, having people rely on you, and being able to directly affect the lives of so many... these are like a drug.
And even if it starts out unselfishly, I've seen a few politicians get a taste of the drug and change ways. It sucks when I voted for them and feel the need to apologize for my support.
A solution to this problem is to organize decisions and action items into a centralized repository that can be viewed and tracked by everyone.
Email is best used for quick communication, not for project tracking.
Ballmer's best-before date was before he became CEO. He rode the monopoly train well past the station. The fact that he allowed the company to finally take risks with new product is pretty much the only twinkling of worthiness he had during his tenure.
And they should lie in it. Microsoft's monopoly in IE was one of the principal causes of stagnation in the industry during the mid 2000s.
Then again, that stagnation arguably led to some great innovations by others in the industry, which is why we've witnessed the mobile revolution and downfall of IE since.
Email allows me to respond according to my schedule. Call me if something's really important.
Actually this brings up a side point... I would really like to have a good/standardized way to leave a voice message that works as you describe email, allowing the recipient to respond when you can get around to it. The problem with voicemail is that I have to call you to leave the message.
Calling interrupts you and you might answer, since you can't tell whether it needs immediate attention. So to leave an unimportant message, I need to type a text message or email... both of which are suboptimal in many situations (walking, driving, etc.).
Sure the web is decentralized, except for IP address assignment and DNS.
Watching one second unfold over 4650.77 years might just be unfathomable to me.
No, I'm pretty sure it is exactly unfathomable to any of us.
This is like saying Google is just a massive number of bits with a diverse range of activation functions finely tuned by evolution and experience.
While the statement is true, it bears little practical importance.
Whose reality? This is reality for many people.
Besides, if you take the "per GiB" out and just talk about a 500 GiB drive for $250, that is exactly how it works.
It's called money. Learn it.
I suppose, once you know when the bot is going to say that line, just preemptively ask it:
Can I record this call?
Then when it says
This call may be recorded for quality and training purposes.
They almost certainly wouldn't have a leg to stand on.
I was thinking of this thread again (the tug-of-war definition) and thought of another interesting thing to calculate: What if I swap Pluto and Charon in the equation? Would it indicate that Pluto is also more influenced by Charon than the sun at times?
The answer is yes, meaning Pluto does not have a concave orbit either. The tug-of-war value focusing on Pluto as the primary body is 337.3 at perihelion, but focusing on Charon as the primary body it is 39.3. Since both are greater than 1, this means Pluto and Charon cause each other to have an occasionally convex orbit.
So perhaps this definition would still find Pluto/Charon to be a binary planet, but in a different way than the Earth/Moon are. Compare the numbers above with Earth/Moon: Earth-centric has a value of 0.45, Moon-centric has a value of 0.006. Both being less than 1, Earth and Moon always have concave orbits around the sun.
Unless they have a copy of the password hash
Add proper salting and an adequate number of iterations over the hashing function, and the attacker's computer will be necessarily slowed as well.
That definition would reduce the problem to the relationship of their masses.
For clarification, the system requirements for Windows 9 developer preview are already known
There is no such thing as "Windows 9 developer preview". There's speculation that it will be called "Windows 9". There's speculation that it will have a developer preview. There's speculation about the system requirements. But that's all it is, speculation. No announcements and to my knowledge no leaks about any of those specific points have been made.
Until we see leaks or at least strong rumors by sources that have been correct in the past, nothing is "known". (Remember how many rumors we've had about Apple's iWatch and for how many years? Some even by credible sources? Yet we still don't have an iWatch.)
They could let the host venue determine whether to blackout. Some might, others wouldn't, and eventually the better outcome would become the most popular.
In retrospect I don't like how I phrased this:
Thus, all we need to do is figure out at what new distance R2 the new orbital velocity V2 = V * (1 / square root of 2).
Change it to the following:
Thus, all we need to do is figure out at what new distance R2 the original orbital velocity V = V2 * (square root of 2), where V2 is the velocity of a circular orbit at distance R2.
Another reason I don't care for the barycentric approach is because it depends so highly on the radius of the larger body. What if the barycenter of the moon were right above the surface but well within the atmosphere? What about a gas giant where the definition of the radius is a bit fuzzier?
Agreed, that one is a bit far fetched. It's a many-body problem and all it takes is a bit of eccentricity or pull from other moons, planets, and the sun to destabilize.
But back to the first question, what if the barycenter moves in and out of the planet due to multiple moons? This would be akin to the solar system, where the barycenter moves in and out of the sun. I don't know if we could easily call it a ternary planet, quaternary planet, etc.
I think I prefer Isaac Asimov's tug-of-war definition of a binary planet. It would be considered a binary planet if the smaller body has a concave orbit around the sun; in other words, the two are both primarily orbiting the sun and just happen to be close to each other. This would, however, define the earth/moon system as a binary planet and Pluto/Charon would be a planet/moon.
Windows 8 has been optimized to run faster in most benchmarks than Windows 7. I am sure this trend will reverse itself eventually but for now I'll take your bet.
a couple Qs if you don't mind, because you obv know a lot about this.
I'm just a guy who was interested enough to Google and throw together some calculations.
so I guess if the barycenter is not in the middle of earth, then the earth wobbles as the moon goes around. Is this what causes tides, it's essentially the sloshing of the ocean as the earth wobbles? I always knew that "the moon causes tides", but I never understood the mechanism.
The gravitational forces between the earth and moon are major components of tides. However the barycenter doesn't seem to contribute directly. Moving the earth and moon farther apart (and thus moving the barycenter further away from the center of the earth) actually causes the tides to become weaker. This actually happens regularly as the moon gets closer and then farther from the earth in its orbit (the moon's orbit is not perfectly circular, but slightly elliptical). When the moon is at its closest, the tides are barely higher, and at its furthest the tides are barely lower.
I guess a second question would be, is there a certain distance at which the moon would escape earth's gravity? I wonder what it is, esp compared to the current distance away? would it be 2x, or 10% or 10x?
Gravity accelerates two objects toward one another, based on their mass and their distance. It works the same whether the two objects are initially moving toward each other or away from each other (away from each other, we usually call "decelerating", but there's no difference in the math).
Escape technically occurs when the two objects are moving away from each other, but the deceleration due to gravity will never be enough to overcome their initial velocity at their initial distance from each other. Gravity diminishes as the objects move farther apart, which results in less deceleration over time. In the case of escape, the velocity will never reach zero.
The answer is "yes"... assuming the moon magically appeared at that new farther distance but traveling at the same velocity as it is currently. According to the Wikipedia link on escape velocity:
The escape velocity at a given height is (square root of 2) times the speed in a circular orbit at the same height
Also, the orbital velocity of an object decreases as its distance increases. So increasing the distance of the moon would decrease how fast it would need to be going to stay in orbit.
But remember that the orbital distance suddenly increased but the orbital velocity did not change...
Let's say the moon is orbiting at distance R with orbital velocity V. Thus, all we need to do is figure out at what new distance R2 the new orbital velocity V2 = V * (1 / square root of 2).
This page contains the formula we need. Solving for r, r is proportional to 1/(v^2). So R2 is proportional to 1 / (V2^2), and substituting the equation above we find that R2 is proportional to 2 / (V^2), which equals 2 * (1/V^2), which equals 2 * R. R2 = 2*R.
Thus the answer is "2 times the original orbital distance, 478,000 miles".
That's a bit unfair to compare Mac computers with Windows phones. Comparing computers, Microsoft has pretty much an unbeatable record when it comes to supporting old versions.
But iPhones are updated for somewhat longer than Windows Phones (my 3GS was supported from iOS 3 through iOS 6, so 4 years worth).
My 2008 Mac Pro has the latest version of OSX and the Apple applications.
My 2006 Macbook hasn't been eligible for an OS upgrade since Lion (3 versions ago). To upgrade to the latest OS means buying a new Macbook (Air) at or around $1000 with tax.
Meanwhile I can still purchase the full latest version of Windows for that computer for $120-200 depending on edition. I'd bet that any Intel version of Windows Microsoft releases in the next 10, possibly 15 years will still run on it.