>With/. being one of the largest content delivery systems on the net, I'd be curious to find out how much revenue they generate based upon subscribers alone.
Well, right now, there are 183 comments, of which 9 have a small star behind the author's name. Stats for some other articles:
SCO Targets US Government, TiVo: 2 of 1392
Real Announce Helix Grant Program, Player: 4 of 156
Former Intel Engineer Pleads Guilty To Taliban Aid: 4 of 1023
Novell To Cease NetWare Development?: 9 of 167
Sinclair's Answer To The Segway: 0 of 208
Given these numbers, it seems that about 1% of the active/. readers is a subscriber. If they have a maximum of 30 ad suppressions per day, they each contribute US$ 0.15 per day. The big unknown is of course the number of active/. readers. If it's 50000, then that's 500 subscribers, or US$ 75 per day.
[about holograms]
> I don't know why the eyeglasses companies haven't latched onto this. Cheap, light eyeglasses that don't need to be ground or anything. Just cut out the shape to fit the eyepiece, patch it in, and go.
It's possible to make a holographic lens, but the focal length is inversely proportional to the wavelength. The wavelengths of visible light are in the range 400-700 nm, so that would be quite useless unless you only live/work under the monochromatic illumination (e.g. low-pressure sodium lamps).
There is another issue which people probably refer to as "memory effect", though it technically is not.
With NiCd and to a lesser extent NiMH as well, the voltage tends to drop a bit if the cell is not fully discharged for a long time. E.g., at 50% charge level, a fresh NiXX cell should give 1.15 V (or whatever number). If the cell is never discharged to below 50%, then the voltage at 40% will get a bit lower, e.g. 1.10 V. This can upset the battery indicator of electronic devices. One almost-full discharge to 0.9 V per cell will cure the problem. See this page for more info.
You can build a simple discharger with a 1 amp silicon diode in series with a 0.5 ohm resistor (min. 500 mW). The diode will make sure that the discharge is not beyond 0.7 volt.
> Ummm, has someone told those SETI guys this? Maybe
that's why we haven't found anything yet...
I think they're hoping to detect a transmission that is
meant to be detected, in the range 1.4--1.7 GHz. In that range, the thermal background of the sun is
about 1e10 watt, so only a very directional narrow-band
transmission has a chance to be noticed.
I remember that people have tried to send a message to a
few nearby stars a few years ago with a powerful directional
transmitter. The message was a series of pictures,
explaining how we look like, how we count, what our solar
system looks like, etc. I can't remember what it was called,
but that's the kind of transmission that we might receive.
>However, wouldn't the amount of noise in these bands
be much higher than you would expect to occur naturally?
Let's try to estimate that. Consider the frequency range
around 1 MHz, the MW AM broadcast band which has been in use
for quite some time. The range is something like 0.7--1.5
MHz, or 200--400 m wavelength. I'm not so sure about the
total power installed in the world; say 500 kW for every
200x200 km area of populated land. That is maybe 200 MW of
total installed power (that feels like a high estimate).
How much does the sun radiate in the 200-400 m wavelength
region? The Planck
black-body equation says that the sun (temperature 5700
K) radiates 1e-17 W/m2 in this wavelength range. Multiplied
by the surface (5e18 m2) of the sun: 50 W.
So, yes, our Earthian broadcasts will easily overpower
the noise caused by the sun in the MW radio range. The alien
astronomists will be very confused about the radiation
emanating from that star with an intensity that is periodic
in 24 hours and construct theories about ionized particles
that emit radiation in mentioned band while interacting with
the star's magnetic field.:)
Disclaimer: I don't know enough about astronomy to know
what sources other than blackbody radiation exist in the MW
frequency range.
>We've already been sending out artificial radio waves
from Earth for around 100 years.
Most of those radio waves are drowned in noise, alone by
the fact that there are too many transmitters that share the
same frequencies. All those powerful broadcasts for radio
and TV are radiated into outer space, but from a long
distance one would see a superposition of all those signals
for different TV and radio stations, i.e. noise.
The exception is the short-wave radio band. Because one
can receive a powerful SW station over the whole world, the
frequencies in that band do not overlap. Unfortunately, the
ionosphere that reflects the SW broadcasts will also prevent
aliens to receive them.
Re:Semi-Log; Diameter; Thickness; Mass
on
Making Change
·
· Score: 1
> (...) the stockmarket which recently abolished fractions (down to what, 1/64, 1/128?) in favor of decimal stock prices. (...) Ahh, if nerds were running the world, things would be so damn efficient...
If humans had evolved with 12 fingers instead of 10, then metric units and imperial units would be the same. 10 would be divisible by 2, 3, 4, and 6 and a foot would be 10 inches, or a meter would be 100 (i.e. 90hex) cm.
> The thing that isn't recoverable isn't time
smearing,
I don't see how time smearing can be solved due to sound
sources at different locations. 1-D example: two walls at
x=0 and x=1. Assume that the sound velocity is v=1. A
microphone is placed at x=0.2 and one sound source (#1) is
at x=0.4 and another (#2) at x=0.6. Suppose that the
microphone is only sensitive for sound travelling towards
negative x.
The IIR for source #1 is (pairs of time, amplitude)
something like u1(t) = (0.2;1) (1.4;1/2) (2.2;1/2) (3.4;1/4) (4.2;1/4)...
And for source #2: u2(t) = (0.4;1) (1.2;1/2) (2.4;1/2) (3.2;1/4) (4.4;1/4)...
If you know that the sound source is at x=0.4, then the
measured signal f(t) is the convolution (* is the
convolution operator) f(t) = f1(t) * u1(t)
of the source signal f1(t), which can be inverted to obtain
f1(t). However, if there are two sources, then the measured
signal is
f(t) = f1(t) * u1(t) + f2(t) * u2(t).
What you want to reconstruct is f1(t), or a weighted sum
f1(t) + a f2(t). I don't see how you can do something like
that with only one microphone. In principle, f2(t)
could be a strange function that creates a sound field
that exactly mimics the impulse response of sound source #1,
e.g. the solution of the equation
f2(t) * u2(t) = u1(t).
If you measure f(t)=u1(t), it's impossible to say
whether that's caused by f1(t)=delta(t) (delta is the
dirac-delta function) or by f2(t) as above.
>Just
record a loud click-type noise (eg let a balloon burst). This is the
"finite impulse response" (FIR) of your room.
Unfortunately, this impulse response will be dependent on where exactly the sound source is located in the room, because the time intervals between the click and its reflections are dependent on the distance to the wall. If one microphone picks up sounds from several sound sources at the same time, it is impossible to reconstruct the original signal.
Even if the sound source is localized, there is the problem that the sound source used for calibration should deliver a true delta pulse, i.e. an infinitely short (or at least less than 25 microseconds) pressure pulse which has a flat frequency spectrum from 0 to 20 kHz. Otherwise, certain frequencies that were absent or weaker in the calibration pulse will be emphasized in the reconstruction process. A bursting balloon will certainly not do the job.
The only thing you might do is measure the general frequency characteristic of the room and compensate for that. A digital algorithm might do that a bit better than a graphical equalizer, but you can't get rid of the time smearing caused by reverberation.
With dvorak one gets a bit lazy regarding long excursions of the fingers. It is quite strenuous to reach with the little finger the CTL keys that are all the way in the corners.
> alternating between hands (with a spaced out layout like QWERTY) could actually help speed;
Dvorak is designed for alternation. That's why the vowels are on the left hand and the consonants on the right. Your example:
....... alternating between hands with a spaced out layout
dvorak: lrrlrrlrlrr rlrrllr rlrrr rlrr l rllrlr lll rlrllr
qwerty: lrlllrllrrl llllllr rlrll lrlr l lrllll rrl rlrrrl
Words such as between brings back memories of my qwerty past where I often encountered letter clusters that had to be done not by one hand but even by one finger (that was in the Dutch language).
>the little research there is that's solidly pro-Dvorak was done by advocates.
According to these Dvorak pages, there isn't either that much solid research which is contra-Dvorak.
>But how difficult have your life become after the switch due to QWERTY everywhere else? Do you have to use other PCs often? How easy/difficult is it to constantly switch from Dvorak to qwerty?
Switching back and forth between US-qwerty and Dvorak is reasonably easy, although I don't do that very often nowadays. All computers that I use regularly have Dvorak installed. It takes a few phrases to readjust. I'm still not used to Swedish qwerty, which has all the punctuation marks elsewhere as well as an extra physical key, but that's partly because I never use it.
Strangely, I have much bigger problems with small differences, such as the position of CapsLock and Ctrl, or the location of the "\". On my Dvorak version, ctrl is left of the "A" and caps is moved to one of the Windows keys that I never use. I'd rather type with standard qwerty than a standard Dvorak with a Caps where I expect a Ctrl. Who uses a caps lock anyway?
So in your situation I'd choose one layout (Dvorak with extensions for accented letters) for latin alphabet and one for cyrillic and carry a floppy with me with drivers/keytables to convert every computer that I work with.
> I wouldn't like to have to put that kind of effort in retraining myself with a new keyboard layout, and I'm not sure I could ever become as good with it as I am with this one.
Well, I actually did make the switch. My experience and that of other Dvorak users (as reported on the net) is that
it's less effort to switch than to learn typing from scratch. You don't have to retrain the finger coordination, i.e. to move your fingers independently and hit the keys (remember how hard it is for a non-touch-typist to type "asdfghjkl;" a few times in a row). You just have to rememorize which key is where.
it does take a few weeks to regain typing speed. Yes, that requires some motivation (in my case RSI).
the final Dvorak typing speed is slightly faster than qwerty.
it is much less strainful to type Dvorak compared to qwerty. That's the biggest advantage.
> [...] awkward motions required by a QWERTY key layout than is strictly necessary to get the job done -- hence the Dvorak key layout.
According to and old poll (posted in which year actually?), 25% of Slashdot thinks Dvorak is better than querty. But who actually uses it? I've not met any fellow Dvorak enthousiasts in my work environment (but neither that many/. readers).
>Laptop LCDs are made on purpose to have a narrow viewing angle so the guy next to you on the airplan can't see your screen.
Actually, I believe that the narrow angle is to reduce power consumption and doesn't have that much to do with the production costs. If most of the light is emitted in forward direction, a lower power for the light source will be sufficient. Or: no light is wasted in illuminating the cheeks of the guy next to you.
I'd say that it isn't expensive anyway. It's a sheet of transparent plastic with an embossed pattern that acts as an array of small lenses.
>Say a proton hits the propellor. It will bounce, move in a semicircle in toward the center,
You are basically saying that no momentum is transferred in a collision: if the propellor is in thermal equilibrium with the particles, then after each collision, the particles have a velocity and hence an angular momentum that is on the average the same as before.
A cold propellor would work, but then it's not a PM anymore.:-)
>Now inject a hot plasma of some sort into the device.
Electrons in the plasma move in tight little
counterclockwise circles because of the field. Protons move
in much wider clockwise circles (they're heavier), so they
hit the propellor blades preferentially in one direction and
make it rotate.
Hmm, some kind of Maxwell's demon. My intuition says that
the transfered momentum per unit of time is the same for all
particles, but I can't seem to prove it mathematically. The
particles make circular trajectories with a radius r =
mv/Bq, where m is the particle mass, v the
velocity in the plane of the rotor, B the magnetic
field, and q the charge. A particle in such a trajectory has an angular momentum J = mvr = m^2v^2/Bq. The torque
that colliding particles exert on the rotor is
T = nvAJ = nAm^2v^3/Bq, (eq. 1)
where n is the number of particles per unit volume
and A is the surface of the rotor. (I assume that
every collision transfers the full angular momentum from the
particle to the rotor) Now, thermodynamics say that
<mv^2> = 2kT, where k is Boltzmann's
constant and T is the temperature. If we substitute
in eq. (1), we obtain
T = nA m^(1/2) (2kT)^(3/2) / Bq (eq. 2)
which unfortunately still contains the mass. If the
numbers of positive and negative particles are the same,
then still the protons would cause more momentum because of
the proportionality to m^(1/2). So this attempt to
solve the problem has failed.:-( I need m^2 v^4 instead of m^2 v^3 in eq. (1).
Either I did something wrong in the reasoning, or I overlooked a more fundamental issue, but I tried.:-) Now please tell us the
real solution!
solution.
[article] Everyone
in the audience who thinks they're going to be using the same word processor
in ten years, raise your hand
>this is a point that (La)TeX users have been arguing for years.
Indeed, I've exclusively been using LaTeX for document prepration over the past 14 years. But even though XML documents are not binary, it's not evident to me that it will be easy to render/interpret one of them in ten years from now; doing something useful with it comprises more than just being able to validate the syntax.
>Still, I don't think it's too hard to believe that the cable will be capable of withstanding the tensile forces of its own weight pulling on it. Individually, a nanofiber might break under its own weight, but collectively, as a woven mesh / rope, it will have a much higher tensile strength.
I would say that the enforcing effect of a cable instead of a single beam is that
1. a local rupture cannot propagate; just one fiber breaks, but not the whole cable.
2. the system gets some elasticity such that shocks are absorbed.
3. the tensile force is evenly distributed over the cross-section of the cable.
These effects result in a cable that is more robust against typical causes of failure, but I don't see why the total tensile strength would increase.
Another issue is how to cope with shockwaves in the cable. When you change the force on the cable because you're trying to lift something heavy, this increase in tension will propagate with a finite velocity along the cable. A few hours later, the wave arrives at the satellite, causes damage, and reflects such that the wave can inflict damage onto the elevator cabin. I recall a documentary on Discovery about the lifting of a wreck on the ocean floor where that was a serious problem.
>The reason the numbers don't seem to add up is that the
gravitational pull will drop off with altitude. [...] gravity drops
off rapidly (1/r^2)
Yes, gravity drops off with 1/r^2 where r is the distance to the
center of the earth. The first 300 km are not really
significant compared to the radius of the earth which is 6400 km.
Anyhow, let's do it exactly. A segment of cable with length dr and
a density (per length unit) rho has a mass dm=rho*dr. On this segment,
two forces act:
1. Gravitional force (downwards)
d Fg = dm g0 (r0/r)^2,
where g0 is the gravitation constant at sea level (9.8 N/kg), r0 is
the radius of the earth (6400 km), and r is the distance to the center
of the earth.
2. Centrifugal force (upwards)
dFc = dm w r,
where w is the angular velocity (2pi/24 hours = 7.3e-5 rad/s). We
ignore the fact that g0 at sea level incorporates a neglegible
centrifugal effect.
The total force F of this cable extending from sea level (r=r0) to
some altitude R is
F = INTEGRAL(r0..R) (dFg - dFc) =
INTEGRAL(r0..R) rho dr (g0 r0^2 r^-2 - w^2 r), or
rho [g0 r0^2 (1/r0 - 1/R) + w(r0^2 - R^2)/2]. Assume that rho=1e-3
kg/m; R=3.5e7 m, r0=6.4e6 m. Then F = 4.8e4 N, the equivalent of 4810
kg of mass hanging at sea level, all to be supported by this
1-mm-thick supercable. If we would have assumed that the gravity
didn't decrease with altitude and that the centrifugal effect didn't
play a role, then 30,000 km cable would weigh 2.8e5 N, so it does
help.
The tensile
strength of steel is around 800 MPa, i.e. 1 sq. mm will break at
800 N (that corresponds to around 10 km hanging at sea level). If the
hypothetical carboncable has a tensile strength that is 30x higher,
then it would be able to support 800*30=2.4e4 N, which is at least in
the same order of magnitude as what we need for the space cable.
>There will be no breakage "under its own weight" with the space elevator, because there won't be any weight pulling the elevator down to earth.
Incorrect reasoning. Suppose that you have your rock locked in a geostationary orbit. Then you unwind the cable: the cable hangs down from the GS orbit. (the roundtrip time for an orbit decreases with decreasing altitude; since everything is forced into a 24-h roundtrip time, the cable does HANG or otherwise it will fall down). In other words, the cable needs at least to support its own weight, which is quite challenging with the 30,000 km that are below the GS orbit, even if the gravity is not that strong at the higher altitudes.
>The lift ribbon will not need to support the weight of the whole system; on the contrary, centrifugal force will hold it aloft (i.e., the whole thing is effectively in orbit). [...]
So, the material needs tensile strength, not weight-bearing capacity. Think carbon nanotubes, not "diamond beams".
A hanging steel rod or fibre will break under its own weight already at a length of about 10 km. The article mentions that the carbon nanofibers are 30x stronger than steel, which means that you get 300 km. Maybe they meant 30x stronger by volume, not by mass, in which case it adds another factor 8 for the difference in density between carbon and steel, resulting in 2400 km before this hypothetical material collapses under its own weight. The geostationary orbit is around 30,000 km above the earth, so I don't see how this is supposed to work, even in theory.
Another problem is that 100,000 km of cable represents quite a large surface. Suppose that the cable is 1 mm thick; that is a total surface of 100,000 square meters. Just one microscopically small dust particle that hits the cable at 10 km/s can break the cable. With such a big surface, the chance of that happening is quite large. Oops, there goes the space elevator!
Slashdot Mirror
Well, right now, there are 183 comments, of which 9 have a small star behind the author's name. Stats for some other articles:
SCO Targets US Government, TiVo: 2 of 1392
Real Announce Helix Grant Program, Player: 4 of 156
Former Intel Engineer Pleads Guilty To Taliban Aid: 4 of 1023
Novell To Cease NetWare Development?: 9 of 167
Sinclair's Answer To The Segway: 0 of 208
Given these numbers, it seems that about 1% of the active /. readers is a subscriber. If they have a maximum of 30 ad suppressions per day, they each contribute US$ 0.15 per day. The big unknown is of course the number of active /. readers. If it's 50000, then that's 500 subscribers, or US$ 75 per day.
> I don't know why the eyeglasses companies haven't latched onto this. Cheap, light eyeglasses that don't need to be ground or anything. Just cut out the shape to fit the eyepiece, patch it in, and go.
It's possible to make a holographic lens, but the focal length is inversely proportional to the wavelength. The wavelengths of visible light are in the range 400-700 nm, so that would be quite useless unless you only live/work under the monochromatic illumination (e.g. low-pressure sodium lamps).
With NiCd and to a lesser extent NiMH as well, the voltage tends to drop a bit if the cell is not fully discharged for a long time. E.g., at 50% charge level, a fresh NiXX cell should give 1.15 V (or whatever number). If the cell is never discharged to below 50%, then the voltage at 40% will get a bit lower, e.g. 1.10 V. This can upset the battery indicator of electronic devices. One almost-full discharge to 0.9 V per cell will cure the problem. See this page for more info.
You can build a simple discharger with a 1 amp silicon diode in series with a 0.5 ohm resistor (min. 500 mW). The diode will make sure that the discharge is not beyond 0.7 volt.
I found it: it is the Cosmic Call project, it has been discussed here before.
I think they're hoping to detect a transmission that is meant to be detected, in the range 1.4--1.7 GHz. In that range, the thermal background of the sun is about 1e10 watt, so only a very directional narrow-band transmission has a chance to be noticed.
I remember that people have tried to send a message to a few nearby stars a few years ago with a powerful directional transmitter. The message was a series of pictures, explaining how we look like, how we count, what our solar system looks like, etc. I can't remember what it was called, but that's the kind of transmission that we might receive.
Let's try to estimate that. Consider the frequency range around 1 MHz, the MW AM broadcast band which has been in use for quite some time. The range is something like 0.7--1.5 MHz, or 200--400 m wavelength. I'm not so sure about the total power installed in the world; say 500 kW for every 200x200 km area of populated land. That is maybe 200 MW of total installed power (that feels like a high estimate).
How much does the sun radiate in the 200-400 m wavelength region? The Planck black-body equation says that the sun (temperature 5700 K) radiates 1e-17 W/m2 in this wavelength range. Multiplied by the surface (5e18 m2) of the sun: 50 W.
So, yes, our Earthian broadcasts will easily overpower the noise caused by the sun in the MW radio range. The alien astronomists will be very confused about the radiation emanating from that star with an intensity that is periodic in 24 hours and construct theories about ionized particles that emit radiation in mentioned band while interacting with the star's magnetic field. :)
Disclaimer: I don't know enough about astronomy to know what sources other than blackbody radiation exist in the MW frequency range.
Most of those radio waves are drowned in noise, alone by the fact that there are too many transmitters that share the same frequencies. All those powerful broadcasts for radio and TV are radiated into outer space, but from a long distance one would see a superposition of all those signals for different TV and radio stations, i.e. noise.
The exception is the short-wave radio band. Because one can receive a powerful SW station over the whole world, the frequencies in that band do not overlap. Unfortunately, the ionosphere that reflects the SW broadcasts will also prevent aliens to receive them.
If humans had evolved with 12 fingers instead of 10, then metric units and imperial units would be the same. 10 would be divisible by 2, 3, 4, and 6 and a foot would be 10 inches, or a meter would be 100 (i.e. 90hex) cm.
I don't see how time smearing can be solved due to sound sources at different locations. 1-D example: two walls at x=0 and x=1. Assume that the sound velocity is v=1. A microphone is placed at x=0.2 and one sound source (#1) is at x=0.4 and another (#2) at x=0.6. Suppose that the microphone is only sensitive for sound travelling towards negative x.
The IIR for source #1 is (pairs of time, amplitude) something like ...
...
u1(t) = (0.2;1) (1.4;1/2) (2.2;1/2) (3.4;1/4) (4.2;1/4)
And for source #2:
u2(t) = (0.4;1) (1.2;1/2) (2.4;1/2) (3.2;1/4) (4.4;1/4)
If you know that the sound source is at x=0.4, then the measured signal f(t) is the convolution (* is the convolution operator)
f(t) = f1(t) * u1(t)
of the source signal f1(t), which can be inverted to obtain f1(t). However, if there are two sources, then the measured signal is
f(t) = f1(t) * u1(t) + f2(t) * u2(t).
What you want to reconstruct is f1(t), or a weighted sum f1(t) + a f2(t). I don't see how you can do something like that with only one microphone. In principle, f2(t) could be a strange function that creates a sound field that exactly mimics the impulse response of sound source #1, e.g. the solution of the equation
f2(t) * u2(t) = u1(t).
If you measure f(t)=u1(t), it's impossible to say whether that's caused by f1(t)=delta(t) (delta is the dirac-delta function) or by f2(t) as above.
Unfortunately, this impulse response will be dependent on where exactly the sound source is located in the room, because the time intervals between the click and its reflections are dependent on the distance to the wall. If one microphone picks up sounds from several sound sources at the same time, it is impossible to reconstruct the original signal.
Even if the sound source is localized, there is the problem that the sound source used for calibration should deliver a true delta pulse, i.e. an infinitely short (or at least less than 25 microseconds) pressure pulse which has a flat frequency spectrum from 0 to 20 kHz. Otherwise, certain frequencies that were absent or weaker in the calibration pulse will be emphasized in the reconstruction process. A bursting balloon will certainly not do the job.
The only thing you might do is measure the general frequency characteristic of the room and compensate for that. A digital algorithm might do that a bit better than a graphical equalizer, but you can't get rid of the time smearing caused by reverberation.
And to sample 240 lines, that is 240 lines of black with 240 lines of white inbetween, you still need 480 pixels horizontally.
With dvorak one gets a bit lazy regarding long excursions of the fingers. It is quite strenuous to reach with the little finger the CTL keys that are all the way in the corners.
Dvorak is designed for alternation. That's why the vowels are on the left hand and the consonants on the right. Your example:
....... alternating between hands with a spaced out layout
dvorak: lrrlrrlrlrr rlrrllr rlrrr rlrr l rllrlr lll rlrllr
qwerty: lrlllrllrrl llllllr rlrll lrlr l lrllll rrl rlrrrl
Words such as between brings back memories of my qwerty past where I often encountered letter clusters that had to be done not by one hand but even by one finger (that was in the Dutch language).
>the little research there is that's solidly pro-Dvorak was done by advocates.
According to these Dvorak pages, there isn't either that much solid research which is contra-Dvorak.
Switching back and forth between US-qwerty and Dvorak is reasonably easy, although I don't do that very often nowadays. All computers that I use regularly have Dvorak installed. It takes a few phrases to readjust. I'm still not used to Swedish qwerty, which has all the punctuation marks elsewhere as well as an extra physical key, but that's partly because I never use it.
Strangely, I have much bigger problems with small differences, such as the position of CapsLock and Ctrl, or the location of the "\". On my Dvorak version, ctrl is left of the "A" and caps is moved to one of the Windows keys that I never use. I'd rather type with standard qwerty than a standard Dvorak with a Caps where I expect a Ctrl. Who uses a caps lock anyway?
So in your situation I'd choose one layout (Dvorak with extensions for accented letters) for latin alphabet and one for cyrillic and carry a floppy with me with drivers/keytables to convert every computer that I work with.
Well, I actually did make the switch. My experience and that of other Dvorak users (as reported on the net) is that
According to and old poll (posted in which year actually?), 25% of Slashdot thinks Dvorak is better than querty. But who actually uses it? I've not met any fellow Dvorak enthousiasts in my work environment (but neither that many /. readers).
Actually, I believe that the narrow angle is to reduce power consumption and doesn't have that much to do with the production costs. If most of the light is emitted in forward direction, a lower power for the light source will be sufficient. Or: no light is wasted in illuminating the cheeks of the guy next to you.
I'd say that it isn't expensive anyway. It's a sheet of transparent plastic with an embossed pattern that acts as an array of small lenses.
You are basically saying that no momentum is transferred in a collision: if the propellor is in thermal equilibrium with the particles, then after each collision, the particles have a velocity and hence an angular momentum that is on the average the same as before.
A cold propellor would work, but then it's not a PM anymore. :-)
Hmm, some kind of Maxwell's demon. My intuition says that the transfered momentum per unit of time is the same for all particles, but I can't seem to prove it mathematically. The particles make circular trajectories with a radius r = mv/Bq, where m is the particle mass, v the velocity in the plane of the rotor, B the magnetic field, and q the charge. A particle in such a trajectory has an angular momentum J = mvr = m^2v^2/Bq. The torque that colliding particles exert on the rotor is
T = nvAJ = nAm^2v^3/Bq, (eq. 1)
where n is the number of particles per unit volume and A is the surface of the rotor. (I assume that every collision transfers the full angular momentum from the particle to the rotor) Now, thermodynamics say that <mv^2> = 2kT, where k is Boltzmann's constant and T is the temperature. If we substitute in eq. (1), we obtain
T = nA m^(1/2) (2kT)^(3/2) / Bq (eq. 2)
which unfortunately still contains the mass. If the numbers of positive and negative particles are the same, then still the protons would cause more momentum because of the proportionality to m^(1/2). So this attempt to solve the problem has failed.Either I did something wrong in the reasoning, or I overlooked a more fundamental issue, but I tried. :-) Now please tell us the
real solution!
solution.
>this is a point that (La)TeX users have been arguing for years.
Indeed, I've exclusively been using LaTeX for document prepration over the past 14 years. But even though XML documents are not binary, it's not evident to me that it will be easy to render/interpret one of them in ten years from now; doing something useful with it comprises more than just being able to validate the syntax.
I forgot to copy the square of omega (w^2) from my scratch paper to the comment. The correct equations:
dFc = dm w^2 r,
F = INTEGRAL(r0..R) (dFg - dFc)
= INTEGRAL(r0..R) rho dr (g0 r0^2 r^-2 - w^2 r)
= rho [g0 r0^2 (1/r0 - 1/R) + w^2(r0^2 - R^2)/2].
With rho=1.3e-3 kg/m (updated value); R=3.5e7 m, r0=6.4e6 m, g0=9.8 N/kg, w = 7.3e-5 rad/s, we get F = 6.3e4 N for the tension on the cable.
I would say that the enforcing effect of a cable instead of a single beam is that
1. a local rupture cannot propagate; just one fiber breaks, but not the whole cable.
2. the system gets some elasticity such that shocks are absorbed.
3. the tensile force is evenly distributed over the cross-section of the cable.
These effects result in a cable that is more robust against typical causes of failure, but I don't see why the total tensile strength would increase.
Another issue is how to cope with shockwaves in the cable. When you change the force on the cable because you're trying to lift something heavy, this increase in tension will propagate with a finite velocity along the cable. A few hours later, the wave arrives at the satellite, causes damage, and reflects such that the wave can inflict damage onto the elevator cabin. I recall a documentary on Discovery about the lifting of a wreck on the ocean floor where that was a serious problem.
Yes, gravity drops off with 1/r^2 where r is the distance to the center of the earth. The first 300 km are not really significant compared to the radius of the earth which is 6400 km.
Anyhow, let's do it exactly. A segment of cable with length dr and a density (per length unit) rho has a mass dm=rho*dr. On this segment, two forces act:
1. Gravitional force (downwards) ,
d Fg = dm g0 (r0/r)^2
where g0 is the gravitation constant at sea level (9.8 N/kg), r0 is the radius of the earth (6400 km), and r is the distance to the center of the earth.
2. Centrifugal force (upwards)
dFc = dm w r,
where w is the angular velocity (2pi/24 hours = 7.3e-5 rad/s). We ignore the fact that g0 at sea level incorporates a neglegible centrifugal effect.
The total force F of this cable extending from sea level (r=r0) to some altitude R is
F = INTEGRAL(r0..R) (dFg - dFc) =
INTEGRAL(r0..R) rho dr (g0 r0^2 r^-2 - w^2 r), or
rho [g0 r0^2 (1/r0 - 1/R) + w(r0^2 - R^2)/2]. Assume that rho=1e-3 kg/m; R=3.5e7 m, r0=6.4e6 m. Then F = 4.8e4 N, the equivalent of 4810 kg of mass hanging at sea level, all to be supported by this 1-mm-thick supercable. If we would have assumed that the gravity didn't decrease with altitude and that the centrifugal effect didn't play a role, then 30,000 km cable would weigh 2.8e5 N, so it does help.
The tensile strength of steel is around 800 MPa, i.e. 1 sq. mm will break at 800 N (that corresponds to around 10 km hanging at sea level). If the hypothetical carboncable has a tensile strength that is 30x higher, then it would be able to support 800*30=2.4e4 N, which is at least in the same order of magnitude as what we need for the space cable.
Incorrect reasoning. Suppose that you have your rock locked in a geostationary orbit. Then you unwind the cable: the cable hangs down from the GS orbit. (the roundtrip time for an orbit decreases with decreasing altitude; since everything is forced into a 24-h roundtrip time, the cable does HANG or otherwise it will fall down). In other words, the cable needs at least to support its own weight, which is quite challenging with the 30,000 km that are below the GS orbit, even if the gravity is not that strong at the higher altitudes.
A hanging steel rod or fibre will break under its own weight already at a length of about 10 km. The article mentions that the carbon nanofibers are 30x stronger than steel, which means that you get 300 km. Maybe they meant 30x stronger by volume, not by mass, in which case it adds another factor 8 for the difference in density between carbon and steel, resulting in 2400 km before this hypothetical material collapses under its own weight. The geostationary orbit is around 30,000 km above the earth, so I don't see how this is supposed to work, even in theory.
Another problem is that 100,000 km of cable represents quite a large surface. Suppose that the cable is 1 mm thick; that is a total surface of 100,000 square meters. Just one microscopically small dust particle that hits the cable at 10 km/s can break the cable. With such a big surface, the chance of that happening is quite large. Oops, there goes the space elevator!