The sticker on the back doesn't tell you how much it uses in practice, it tells you the maximum it will ever use. It's useful for sizing circuits and picking fuses, but not for estimating running costs. The label on the back doesn't tell you if it uses 80W or 1W in standby. It doesn't tell you if the maximum rating applies during normal viewing, or only for two seconds at startup.
To find out how much power something is using you really need a "Kill-A-Watt" meter. Google it. They are quite handy!
-Rob
Sorry Sabbath, but that AC is right. 'Solar Variability' refers to changes in the actual output from the sun and is not related to seasonal and diurnal cycles. consider it a term of art.
-Rob
you are off by a factor of 2. you are correct that a 5km drop gets you ~30 seconds of freefall (0.5at^2). but you could start at the bottom of the shaft and launch up (with a brief high-g acceleration) and then be in freefall all the way up and all the way down, for ~60 seconds of uninterupted freefall. The high-g at lauch would be the same as the breaking g at the end of the trip. you could get 30 seconds of freefall with only 1.125 km worth of shaft. just a nitpick, but... -Rob
From your description, it sounds as if WoW glider effectively runs WoW in a virtual machine (or something similar).
Running a program in a virtual machine is not copyright infringement. Or at very least, VMware et. al. should take an interest in the case lest a judge creates a law saying that VM's are illegal!
-Rob
What about the force of the explosion? With no air resistance isn't just as likely that some pieces (of both the satellite and the missile) will end up in higher orbit thus the concern for collision with other satellites.
No. The satellite is low, really in the upper atmosphere. No matter what delta-V you add to the debris, the orbit for the debris will go through the same point. You have to change the velocity TWICE to put the debris into a higher orbit! with a single impulse, essentially all the debris will go lower (and into the atmosphere) at some point on it's orbit.
Debris really wont be a problem.
-Rob
Not quite.
Watts is a unit of power (which is energy per unit time).
Watt hours (power * time) is a unit of energy.
Watts per hour (power / time) measures energy/time^2. That unit would not come up very often!
-Rob
I was in the Gifted & Talented program when i was in junior high and high school. All the GT's in the whole city (Baton Rouge, La.) went to one magnet school. There were non-GT's at the same school, but they didn't share any classes except P.E. and electives such as shop and music. The social structure was similar but with important differences. You still had cliques and such; the geekier ones and the cooler guys. The geekiest of the geeks would get picked on a little, but nothing too malicious. The most important difference was that being smart (not the same as geeky) got respect. If you set the curve on the calculus mid-term, the other kids were envious.
In all the years since, i have never run with a brighter crowd. I did my undergrad in physics, and the typical physics student was pretty bright; i got my PhD, and the average physics grad student was even brighter. But in the GT program, the dumb kids had an IQ over 130. There aren't too many places like that!
-Rob
Sigh. The Carnot engine is the idealized 'perfect' heat engine and gives the highest possible efficiency allowed by the laws of thermodynamics. Any real heat engine will have a lower efficiency.
The ideal Carnot engine can never have 100% efficiency for any finite temperature difference (meaning it will have to dump waste heat into the 'environment' heatsink).
The Kalina cycle might actually achieve about half the efficiency of the Carnot cycle. So if the geothermal heatsource is at 200 degC (that is pretty hot) and the sink is at 10 degC, then the Carnot cycle would give 190/(200+273) = ~40% efficiency, so the Kalina might get about 20% efficiency.
There is ALWAYS excess heat. I didn't mean to sound patronizing, but i still don't see how someone who understands Carnot's theorum could claim that there is no waste heat.
-Rob
I call BS.
Go look up the Carnot engine. it defines the maximum theoretical effeciecy for converting thermal energy to non-thermal energy. there is ALWAYS excess.
Now, a geothermal plant located near the ocean (or even on a platform 30 miles offshore) could use a much larger (and colder) heatsink.
-Rob
naturally, it depends on the operations per I/O byte of the problem.
Example: your CPU can do 3gflops and you have a network doing 10gbps (or ~300M single precision floats per second).
if on average you need to do >> 10 operations per sp float input, then it will likely pay to share the problem in a cluster.
-Rob
goodbye to flight sims in an online multiplayer world?
how silly.
The flight sim WarBirds was a massively multiplayer online game before it was an acronym!
I was playing that a lot ~10 years ago. It really changed what a combat flight sim was about. It's one thing to shoot down a cheesy AI flown plane... gets old pretty quick. But in WarBirds you knew that when you shot down a plane, somewhere on the planet there was a real live person cursing and throwing his joystick across the room (smile).
-Rob
actually, that is the most accurate fraction with a denomenator 10,000.
When you go to 5 digit denomenators the best is 312689 / 99532
with 6 digits : 3126535 / 995207
with 7 digits : 5419351 / 1725033
with 8 digits you hit the magic 245850922 / 78256779. the error is -7.8179366199075e-17 so the difference is zero in double precision arithmatic.
-Rob
OK, this is cool. I fired up Google Earth and went looking. sure enough, it seems like there are craters everywhere. An Example: looks like a perfectly good crater at Lat = 22.16, Lon = 19.45 What do you think. it falls approximately on the same line, is it a crater or isn't it? seems to be a big one at Lat = 22.04 Lon = 19.22 right next door. It looks like a whole shotgun blast of small craters north of there on the same line
Lat = 27.317 Lon = 19.391
Lat = 27.51 Lon = 19.81
Lat = 27.28 Lon = 19.66 Those look different than the others though? salt domes? I got dozens of "craters" placemarked. this is fun, but do i have the crater fever and i'm seeing them where they don't exist? -Rob
Slashdot Mirror
To find out how much power something is using you really need a "Kill-A-Watt" meter. Google it. They are quite handy! -Rob
Sorry Sabbath, but that AC is right. 'Solar Variability' refers to changes in the actual output from the sun and is not related to seasonal and diurnal cycles. consider it a term of art. -Rob
+1 insightful
you are off by a factor of 2.
you are correct that a 5km drop gets you ~30 seconds of freefall (0.5at^2). but you could start at the bottom of the shaft and launch up (with a brief high-g acceleration) and then be in freefall all the way up and all the way down, for ~60 seconds of uninterupted freefall. The high-g at lauch would be the same as the breaking g at the end of the trip.
you could get 30 seconds of freefall with only 1.125 km worth of shaft.
just a nitpick, but...
-Rob
From your description, it sounds as if WoW glider effectively runs WoW in a virtual machine (or something similar). Running a program in a virtual machine is not copyright infringement. Or at very least, VMware et. al. should take an interest in the case lest a judge creates a law saying that VM's are illegal! -Rob
No. The satellite is low, really in the upper atmosphere. No matter what delta-V you add to the debris, the orbit for the debris will go through the same point. You have to change the velocity TWICE to put the debris into a higher orbit! with a single impulse, essentially all the debris will go lower (and into the atmosphere) at some point on it's orbit. Debris really wont be a problem. -Rob
+1 informative
mod up -Rob
+1 funny
Just to be clear: It shouldn't be Watt hours or Watts per hour. It should just be Watts. -Rob
Not quite. Watts is a unit of power (which is energy per unit time). Watt hours (power * time) is a unit of energy. Watts per hour (power / time) measures energy/time^2. That unit would not come up very often! -Rob
I was in the Gifted & Talented program when i was in junior high and high school. All the GT's in the whole city (Baton Rouge, La.) went to one magnet school. There were non-GT's at the same school, but they didn't share any classes except P.E. and electives such as shop and music. The social structure was similar but with important differences. You still had cliques and such; the geekier ones and the cooler guys. The geekiest of the geeks would get picked on a little, but nothing too malicious. The most important difference was that being smart (not the same as geeky) got respect. If you set the curve on the calculus mid-term, the other kids were envious.
In all the years since, i have never run with a brighter crowd. I did my undergrad in physics, and the typical physics student was pretty bright; i got my PhD, and the average physics grad student was even brighter. But in the GT program, the dumb kids had an IQ over 130. There aren't too many places like that!
-Rob
Sigh. The Carnot engine is the idealized 'perfect' heat engine and gives the highest possible efficiency allowed by the laws of thermodynamics. Any real heat engine will have a lower efficiency. The ideal Carnot engine can never have 100% efficiency for any finite temperature difference (meaning it will have to dump waste heat into the 'environment' heatsink). The Kalina cycle might actually achieve about half the efficiency of the Carnot cycle. So if the geothermal heatsource is at 200 degC (that is pretty hot) and the sink is at 10 degC, then the Carnot cycle would give 190/(200+273) = ~40% efficiency, so the Kalina might get about 20% efficiency. There is ALWAYS excess heat. I didn't mean to sound patronizing, but i still don't see how someone who understands Carnot's theorum could claim that there is no waste heat. -Rob
I call BS. Go look up the Carnot engine. it defines the maximum theoretical effeciecy for converting thermal energy to non-thermal energy. there is ALWAYS excess. Now, a geothermal plant located near the ocean (or even on a platform 30 miles offshore) could use a much larger (and colder) heatsink. -Rob
naturally, it depends on the operations per I/O byte of the problem. Example: your CPU can do 3gflops and you have a network doing 10gbps (or ~300M single precision floats per second). if on average you need to do >> 10 operations per sp float input, then it will likely pay to share the problem in a cluster. -Rob
goodbye to flight sims in an online multiplayer world? how silly. The flight sim WarBirds was a massively multiplayer online game before it was an acronym! I was playing that a lot ~10 years ago. It really changed what a combat flight sim was about. It's one thing to shoot down a cheesy AI flown plane... gets old pretty quick. But in WarBirds you knew that when you shot down a plane, somewhere on the planet there was a real live person cursing and throwing his joystick across the room (smile). -Rob
actually, that is the most accurate fraction with a denomenator 10,000. When you go to 5 digit denomenators the best is 312689 / 99532 with 6 digits : 3126535 / 995207 with 7 digits : 5419351 / 1725033 with 8 digits you hit the magic 245850922 / 78256779. the error is -7.8179366199075e-17 so the difference is zero in double precision arithmatic. -Rob
OK, this is cool. I fired up Google Earth and went looking. sure enough, it seems like there are craters everywhere.
An Example: looks like a perfectly good crater at Lat = 22.16, Lon = 19.45
What do you think. it falls approximately on the same line, is it a crater or isn't it?
seems to be a big one at Lat = 22.04 Lon = 19.22 right next door.
It looks like a whole shotgun blast of small craters north of there on the same line
Lat = 27.317 Lon = 19.391
Lat = 27.51 Lon = 19.81
Lat = 27.28 Lon = 19.66
Those look different than the others though? salt domes?
I got dozens of "craters" placemarked. this is fun, but do i have the crater fever and i'm seeing them where they don't exist?
-Rob