> While the math may have not been understood back when it was developed
"It" being the concept of an octave?
The math behind it was pretty well understood (though not quite in the terms we use today) by about 2500 years ago. We certainly have music theory treatises from back then which are pretty clear on what's going on.
Then again, http://en.wikipedia.org/wiki/Pythagorean_tuning mentions that the concepts involved may date back to about 3800 years ago. The record is a bit scant as to which came first at that point.;)
What you said was "Given that I might spend $100 a month to power a house (conservatively, even if you went to 10 times the power consumption, the numbers are *still* not in favor of this device)".
I read that as meaning that $100 was your conservative number. I can see now that wasn't what you meant.
> Now, I am not in any way qualified to say that number is to large or to small.
Depends on where it comes from. 100mloc of hand-written stuff (any kind of language) is insane. 100mloc of mostly auto-generated boilerplate is ok as long as you're pretty sure your (presumably smaller) code-generator is correct.
Realistically, I would expect that the "code" in question here is effectively an intermediate representation compiled from some higher-level language. That intermediate representation might be C, but that doesn't make it the original source code any more than the assembly generated from your typical C program is the original source code, no matter whether one can hand-write assembly.
> Given that I might spend $100 a month to power a house
It really depends on where you are and what parts of your house run on electricity. DC area, with heat, A/C, hot water, cooking, lighting all on electricity, a 2300-sq-ft or so house (nothing ridiculous; basic suburban house) ends up averaging about $400 a month in electricity costs. If you're further south, it might run more; A/C is _expensive_.
If you're in the northeast and use gas/oil for all your heat/cooking needs, and don't have A/C, so all you run is lights, then it's possible to hit numbers on the order of $30 a month. Not sure where your $10 a month number comes from.
At low bitrates, Dirac produces lower-quality output than Theora. At high bitrates it's quite good. Something to do with Dirac being designed for high-quality archival video, not today's web video.
Apple is a member of the H264 patent pool. As in, they own some of the H264 patents. Which means they probably (can't say for sure; the exact agreement covering the patent pool is not public) have no licensing fees to use H264. They can just use it.
Sure. The original post in this sub-thread was explicitly talking about the energy consumption today and about the fact that we need to fundamentally change how we do energy generation to reasonably launch a mission to Alpha Centari that will get there in a matter of years. You're saying you think that sometime we'll get there, just like we got from 100AD to today. I'm pretty sure we will too, if we don't do ourselves in first.
> Your problem is lack of imagination.
I'm not sure what made you decide I have a problem. All I said is that with current technology, and even then only under the condition that one does not try to gather up interstellar hydrogen, travel to Alpha Centauri in several years elapsed time is not feasible.
> Not if you dump half of your spacecraft halfway down the road and have used up more than > half of your fuel already.
Ah, fair point.
OK, let's run the numbers a bit more carefully. Say you want to go to Alpha Centauri at 0.5c top speed (so the trip is about 10-15 years). You need to pick up kinetic energy equivalent to 15% of your final mass. Call your final mass M. Assume we're doing fission; uranium fission converts 0.1% of the mass of the fissioning nucleus to energy. Let's assume we fission all our uranium (probably unrealistic) and that we discard the spent fuel continuously so we don't have to cart it around (probably also unrealistic). Minimal amount of fuel is obviously about 150x the final mass of the spacecraft if it's all fissioned at once and doesn't pick up any velocity. But in practice, the amount of fuel depends on your acceleration profile.
It's not clear to me what an optimal profile is (or rather I'm too lazy to really sit down and crunch the numbers), but if you assume that fuel use per unit time is constant over the duration of the burn (totally not realistic for fission, but so is total fission anyway) then our dE/dt is basically constant. Let original fuel mass be F and final fuel mass be 0. Then initial spaceship mass if F + M. Fuel mass as a function of time if F(1-kt) for some constant k.
Then dE/dt = kFc^2*0.1%.
The fraction of energy that goes to our spaceship (as opposed to the remaining fuel) is M/(M+F(1-kt)). So for just the spaceship, dE/dt = kFc^2*0.1%*M/(M+F(1-kt)).
Integrating, delta E = kFc^2*0.1%*M*(ln(M+F(1-k(delta t))) - ln(M + F))*-1/kF
Since by assumption (delta t) is such that k(delta t) == 1, this simplifies to:
Since we want delta E to be 15%*Mc^2 or so (for speed 0.5c):
15% = 0.1% * ln(1+F/M)
150 = ln(1 + F/M)
F/M = 1.4e65 or so
That's not a good sign. Not enough mass in the universe to do that.
If we wanted just 0.1c, then we only need 0.5% instead of 15% of our final mass as energy, which gives F/M = 149 or thereabouts. That's at least physically possible, though we still made some very optimistic assumptions. We do need about 15000 tons of fissile material to accelerate a shuttle-size craft. That's about 1/3 of global yearly uranium production, so not totally ridiculous.
Different acceleration profiles might give better results that exponential growth of F/M in fractional energy gain, of course. Hard to say for me without crunching the numbers.
So you're arguing that the externalities of energy production are no priced into the cost of energy, right?
That may well be, which is why I said that a _locally_ rational decision may be to not go with a more energy-efficient device.
Of course the right solution to this is to price the externalities into the price of energy. That involves understanding the impact of the externalities; something that is sadly notoriously difficult.
The "dump the engine" approach works ok if you don't have to stop on the other end, or if the stopping is cheaper (e.g. moon landing, due to shallower gravity well) than the starting.
It works for getting out of a gravity well because the whole goal is to get up to speed and stay there.
But for a trip that involves actually stopping at AC, that won't work; you have to turn over and slow down and that needs as much engine as speeding up.
Clearly for an AC flyby a multi-stage design would be the way to go.
Yes, I'm well aware of what the formula for gamma is. If you note, in the above calculation gamma was about 8500 or so. If you drop down to 0.5c, gamma is closer to 1.2, and the top speed is about half, so the power draw drops by a factor of about 1.7e4. So instead of being 1e7 the total power production of humanity right now it's "only" 1e3.
So yes, ultra-relativistic speeds involve all sorts of problems, but even getting up to anything resembling relativistic speeds for a macroscopic object requires huge amounts of energy.
The other way to think of this is that getting up to a speed k*c requires you to completely convert a fraction of your mass equal to "1/sqrt(1-k^2) - 1" to energy. If k == 0.1, that's about 0.5%. Uranium fission converts about 0.1% of the mass each fissioned nucleus into energy, so you can't get to 0.1c using fission if you're carrying along all the fuel. Deuterium+tritium fusion converts about 0.3% of the mass of the original nucleons to energy for each event, so it's not quite good enough either.
If k == 0.5, then you need to convert about 15% of your mass into energy. Either you're carrying along a whole bunch of antimatter, or you're picking up mass along the way; no other way to do it. But really, even to get to 0.1c you need one of those two approaches.
Your k == 0.9 case involves generating kinetic equal to 1.3 times your rest mass to get you up to speed. If we go back to that 2e6 kg shuttle, that means you end up with 2.3e23 J of kinetic energy. Your trip time is about 4.6 years, half of that is 2.3 years. That's average power draw of 3.2e15 Watts while accelerating. Which is 200 times total human power production in 2008...
Even the k == 0.5 case gives you 25 times total human power production as average power draw. The k == 0.1 case gives 0.8 times total human power production. That's almost starting to look manageable, and only involves a 40-year trip time to Alpha Centauri.:)
1) Google does not hold a patent on PageRank. Stanford does. 2) If you read the patent, it's more than just the ranking system. At least as far as I can
tell. I am not a patent lawyer, of course.
That's the theory, and it might work out if the caps are set correctly and evaluated on a yearly basis and if the trading market in the caps is efficient and if there are no other externalities involved...
In practice, every single one of those assumptions is false.
The grandparent specifically said constant 4g boost to Alpha Centauri. That ends up near lightspeed.
But even if you restrict to just getting there in "years" (as great-grandparent did), you end up with an average speed of close to 0.5c if you want to stay under 10 years... Which means your top speed needs to be pretty close to c.
No, what's being discussed is not R&D funding but carbon cap&trade schemes. If your goal is R&D funding, then fund R&D instead of creating complex systems for people to game in the hopes that they spur R&D.
> I don't think that accelerating a shuttle-sized craft at 4 or 5g requires the current > total energy consumption of humanity
Let's call it 4g in the spacecraft's frame, or a force of 4g*m (rest mass). Doing a quick special-relativistic approximation to the process, I get:
v = c * tanh(4gt/c)
as the spacecraft's velocity in the earth's frame as a function of time. Then its position is:
s = c^2/(4g) * ln(cosh(4gt/c))
or in other words:
cosh(4gt/c) = exp(s*4g/c^2)
Plugging in s = 2.19 ly (half the distance to Alpha Centauri) we get:
t = 2.36 years
or so. At that point you turn over and start decelerating.
The peak power draw in this setup will be at this t=2.36 year mark, at which point we have v = 0.99999999c if my calculations are right. dE/dt = mva*gamma or so. Gamma is 8500 or so in this case. So we're looking at about 1e14 watts per kilogram. The space shuttle's mass is about 2e6 kilograms, so you're talking 2e20 watts.
World average power consumption in 2008 was 1.5e13 watts according to . So in fact peak power draw for your proposed constant-4g trip would be about 10,000,000 times more than the power consumption of all of humanity today.
Never underestimate the energy of macroscopic objects traveling at near-lightspeed.
> and why 100Mbps, it seems so excessive, I find my 5Mbps adequate.
I have 3MBps right now, and it's definitely a bottleneck day-to-day for me (e.g. I'm bandwidth-gated on simple things like binary searches on Gecko nightlies to find when a bug appeared, necessitating that I keep a large local cache of nightlies).
Slashdot Mirror
> While the math may have not been understood back when it was developed
"It" being the concept of an octave?
The math behind it was pretty well understood (though not quite in the terms we use today) by about 2500 years ago. We certainly have music theory treatises from back then which are pretty clear on what's going on.
Then again, http://en.wikipedia.org/wiki/Pythagorean_tuning mentions that the concepts involved may date back to about 3800 years ago. The record is a bit scant as to which came first at that point. ;)
> of course the one, where it's really hard to say no to such a bill.
Yep. http://en.wikipedia.org/wiki/First_they_came... comes to mind.
What you said was "Given that I might spend $100 a month to power a house (conservatively, even if you went to 10 times the power consumption, the numbers are *still* not in favor of this device)".
I read that as meaning that $100 was your conservative number. I can see now that wasn't what you meant.
> Now, I am not in any way qualified to say that number is to large or to small.
Depends on where it comes from. 100mloc of hand-written stuff (any kind of language) is insane. 100mloc of mostly auto-generated boilerplate is ok as long as you're pretty sure your (presumably smaller) code-generator is correct.
Realistically, I would expect that the "code" in question here is effectively an intermediate representation compiled from some higher-level language. That intermediate representation might be C, but that doesn't make it the original source code any more than the assembly generated from your typical C program is the original source code, no matter whether one can hand-write assembly.
OK, fair. I guess a 30-year time horizon is in fact what we're talking about here.
> Given that I might spend $100 a month to power a house
It really depends on where you are and what parts of your house run on electricity. DC area, with heat, A/C, hot water, cooking, lighting all on electricity, a 2300-sq-ft or so house (nothing ridiculous; basic suburban house) ends up averaging about $400 a month in electricity costs. If you're further south, it might run more; A/C is _expensive_.
If you're in the northeast and use gas/oil for all your heat/cooking needs, and don't have A/C, so all you run is lights, then it's possible to hit numbers on the order of $30 a month. Not sure where your $10 a month number comes from.
I dare you to find a safe bond or cd yielding 5% right about now.
At low bitrates, Dirac produces lower-quality output than Theora. At high bitrates it's quite good. Something to do with Dirac being designed for high-quality archival video, not today's web video.
> I don't see how this is true.
More or less by definition of "patent pool". See http://en.wikipedia.org/wiki/Patent_pool
> Start thinking about the successor to H.264, and better yet, start building it, write some
> code.
VP8 is in fact a plausible successor to H.264, last I checked. (Disclaimer: I'm not a video codec expert.)
Apple is a member of the H264 patent pool. As in, they own some of the H264 patents. Which means they probably (can't say for sure; the exact agreement covering the patent pool is not public) have no licensing fees to use H264. They can just use it.
> You can run for office yourself.
It is in fact quite rare that someone cannot vote but can run for office....
> Today is the keyword.
Sure. The original post in this sub-thread was explicitly talking about the energy consumption today and about the fact that we need to fundamentally change how we do energy generation to reasonably launch a mission to Alpha Centari that will get there in a matter of years. You're saying you think that sometime we'll get there, just like we got from 100AD to today. I'm pretty sure we will too, if we don't do ourselves in first.
> Your problem is lack of imagination.
I'm not sure what made you decide I have a problem. All I said is that with current technology, and even then only under the condition that one does not try to gather up interstellar hydrogen, travel to Alpha Centauri in several years elapsed time is not feasible.
> Not if you dump half of your spacecraft halfway down the road and have used up more than
> half of your fuel already.
Ah, fair point.
OK, let's run the numbers a bit more carefully. Say you want to go to Alpha Centauri at 0.5c top speed (so the trip is about 10-15 years). You need to pick up kinetic energy equivalent to 15% of your final mass. Call your final mass M. Assume we're doing fission; uranium fission converts 0.1% of the mass of the fissioning nucleus to energy. Let's assume we fission all our uranium (probably unrealistic) and that we discard the spent fuel continuously so we don't have to cart it around (probably also unrealistic). Minimal amount of fuel is obviously about 150x the final mass of the spacecraft if it's all fissioned at once and doesn't pick up any velocity. But in practice, the amount of fuel depends on your acceleration profile.
It's not clear to me what an optimal profile is (or rather I'm too lazy to really sit down and crunch the numbers), but if you assume that fuel use per unit time is constant over the duration of the burn (totally not realistic for fission, but so is total fission anyway) then our dE/dt is basically constant. Let original fuel mass be F and final fuel mass be 0. Then initial spaceship mass if F + M. Fuel mass as a function of time if F(1-kt) for some constant k.
Then dE/dt = kFc^2*0.1%.
The fraction of energy that goes to our spaceship (as opposed to the remaining fuel) is M/(M+F(1-kt)). So for just the spaceship, dE/dt = kFc^2*0.1%*M/(M+F(1-kt)).
Integrating, delta E = kFc^2*0.1%*M*(ln(M+F(1-k(delta t))) - ln(M + F))*-1/kF
Since by assumption (delta t) is such that k(delta t) == 1, this simplifies to:
delta E = c^2*0.1%*M * (ln(M+F) - ln(M))
= c^2*0.1%*M * ln(1+F/M)
Since we want delta E to be 15%*Mc^2 or so (for speed 0.5c):
15% = 0.1% * ln(1+F/M)
150 = ln(1 + F/M)
F/M = 1.4e65 or so
That's not a good sign. Not enough mass in the universe to do that.
If we wanted just 0.1c, then we only need 0.5% instead of 15% of our final mass as energy, which gives F/M = 149 or thereabouts. That's at least physically possible, though we still made some very optimistic assumptions. We do need about 15000 tons of fissile material to accelerate a shuttle-size craft. That's about 1/3 of global yearly uranium production, so not totally ridiculous.
Different acceleration profiles might give better results that exponential growth of F/M in fractional energy gain, of course. Hard to say for me without crunching the numbers.
So you're arguing that the externalities of energy production are no priced into the cost of energy, right?
That may well be, which is why I said that a _locally_ rational decision may be to not go with a more energy-efficient device.
Of course the right solution to this is to price the externalities into the price of energy. That involves understanding the impact of the externalities; something that is sadly notoriously difficult.
The "dump the engine" approach works ok if you don't have to stop on the other end, or if the stopping is cheaper (e.g. moon landing, due to shallower gravity well) than the starting.
It works for getting out of a gravity well because the whole goal is to get up to speed and stay there.
But for a trip that involves actually stopping at AC, that won't work; you have to turn over and slow down and that needs as much engine as speeding up.
Clearly for an AC flyby a multi-stage design would be the way to go.
Yes, I'm well aware of what the formula for gamma is. If you note, in the above calculation gamma was about 8500 or so. If you drop down to 0.5c, gamma is closer to 1.2, and the top speed is about half, so the power draw drops by a factor of about 1.7e4. So instead of being 1e7 the total power production of humanity right now it's "only" 1e3.
So yes, ultra-relativistic speeds involve all sorts of problems, but even getting up to anything resembling relativistic speeds for a macroscopic object requires huge amounts of energy.
The other way to think of this is that getting up to a speed k*c requires you to completely convert a fraction of your mass equal to "1/sqrt(1-k^2) - 1" to energy. If k == 0.1, that's about 0.5%. Uranium fission converts about 0.1% of the mass each fissioned nucleus into energy, so you can't get to 0.1c using fission if you're carrying along all the fuel. Deuterium+tritium fusion converts about 0.3% of the mass of the original nucleons to energy for each event, so it's not quite good enough either.
If k == 0.5, then you need to convert about 15% of your mass into energy. Either you're carrying along a whole bunch of antimatter, or you're picking up mass along the way; no other way to do it. But really, even to get to 0.1c you need one of those two approaches.
Your k == 0.9 case involves generating kinetic equal to 1.3 times your rest mass to get you up to speed. If we go back to that 2e6 kg shuttle, that means you end up with 2.3e23 J of kinetic energy. Your trip time is about 4.6 years, half of that is 2.3 years. That's average power draw of 3.2e15 Watts while accelerating. Which is 200 times total human power production in 2008...
Even the k == 0.5 case gives you 25 times total human power production as average power draw. The k == 0.1 case gives 0.8 times total human power production. That's almost starting to look manageable, and only involves a 40-year trip time to Alpha Centauri. :)
Two things:
1) Google does not hold a patent on PageRank. Stanford does.
2) If you read the patent, it's more than just the ranking system. At least as far as I can
tell. I am not a patent lawyer, of course.
That's the theory, and it might work out if the caps are set correctly and evaluated on a yearly basis and if the trading market in the caps is efficient and if there are no other externalities involved...
In practice, every single one of those assumptions is false.
> You don't need to go anywhere near lightspeed
The grandparent specifically said constant 4g boost to Alpha Centauri. That ends up near lightspeed.
But even if you restrict to just getting there in "years" (as great-grandparent did), you end up with an average speed of close to 0.5c if you want to stay under 10 years... Which means your top speed needs to be pretty close to c.
No, what's being discussed is not R&D funding but carbon cap&trade schemes. If your goal is R&D funding, then fund R&D instead of creating complex systems for people to game in the hopes that they spur R&D.
That should have been "according to http://en.wikipedia.org/wiki/World_energy_resources_and_consumption "
> I don't think that accelerating a shuttle-sized craft at 4 or 5g requires the current
> total energy consumption of humanity
Let's call it 4g in the spacecraft's frame, or a force of 4g*m (rest mass). Doing a quick special-relativistic approximation to the process, I get:
v = c * tanh(4gt/c)
as the spacecraft's velocity in the earth's frame as a function of time. Then its position is:
s = c^2/(4g) * ln(cosh(4gt/c))
or in other words:
cosh(4gt/c) = exp(s*4g/c^2)
Plugging in s = 2.19 ly (half the distance to Alpha Centauri) we get:
t = 2.36 years
or so. At that point you turn over and start decelerating.
The peak power draw in this setup will be at this t=2.36 year mark, at which point we have v = 0.99999999c if my calculations are right. dE/dt = mva*gamma or so. Gamma is 8500 or so in this case. So we're looking at about 1e14 watts per kilogram. The space shuttle's mass is about 2e6 kilograms, so you're talking 2e20 watts.
World average power consumption in 2008 was 1.5e13 watts according to . So in fact peak power draw for your proposed constant-4g trip would be about 10,000,000 times more than the power consumption of all of humanity today.
Never underestimate the energy of macroscopic objects traveling at near-lightspeed.
> and why 100Mbps, it seems so excessive, I find my 5Mbps adequate.
I have 3MBps right now, and it's definitely a bottleneck day-to-day for me (e.g. I'm bandwidth-gated on simple things like binary searches on Gecko nightlies to find when a bug appeared, necessitating that I keep a large local cache of nightlies).
It's "up to" because they also offer slower plans (25/25, 15/5, etc) at lower prices than their 50/50 plan.