Read "Moon is a harsh Mistress" by R.A. Heinlein for a Sci-fi view of this. You folks who are talking about bringing the ore back to earth seem to miss the point - it's better to keep this stuff up in orbit to you can use it instead of bringing it back to Earth. I forget the price per Kilogram of sending something up in the space shuttle, but it's something like $20,000/kg +.
Of course, if you bang an asteroid with a nice meaty chunk of Platinum or Paladium, there's a lot of healthy uses for that back home.
This article also talks about one of the biggest advantages of mining asteroids - getting mass (ore, water, aggregate) off an asteroid is trivial in terms of energy and thrust required. If you're looking to build spaceships or space stations, there's a big advantage in using the materials already up there instead of bringing them out of the Earth's gravity well. The Moon and Mars have the same advantage to a lesser degree.
The Buoyant force, as discovered by Archimedes, is the force due to the pressure from the gas (or liquid) displaced by an object. It is:
Fb = DVg
Where D=density of the gas (or liquid), V is the volume of the object and g is 9.8m/s^2 on Earth.
So, an object like a helium balooon doesn't get it's lift from the density of the helium inside the balloon, it gets it from the displaced air.
What actually happens as a helium balloon rises is that the pressure outside the balloon decreases and thus the balloon expands in volume. This sounds great, since more volume implies more lift, but it doesn't work that way because the density of the diplaced air is going in the opposite direction. The balloon needs to expand in order to continue climbing. Ultimately, as the density of the air outside the balloon drops to zero (in the vacuum of space), it is impossible for the balloon climb any higher. These guys are trying to approach that theoretical limit.
If you've ever seen a video of the launch of a weather balloon, you might have noticed that the balloon looks funny and under-inflated. This is because the engineers have left a lot more balloon skin remaining so that this expansion can take place.
Robert A Heinlein once wrote that once you're in earth orbit, you're half way to anywhere in the solar system. Unfortunately, sitting on a balloon at a high altitude is hardly in earth's orbit. The main problem is that you're not traveling very fast, and it takes a lot of kinetic energy to be in orbit, and that's one thing our ballooning friends will not have (until they jump;-) ).
Ok, here's the math:
132,000 feet = 40 x 10^3 m
This is VERY small compared with the earth's radius:
6.4x10^6m.
So you're actually only pulling away slightly from the earth's gravitational influence. Normally, one calculates the force of gravity for these problems with:
FG=G*m1*m2/r^2
but since we're so close to the earth, we can approximate the whole problem with our old terrestrial gravity without introducing too much error:
Fg=mg where g=9.8m/s^2
So, the potential energy gain by ballooning up that high for a 1kg block is
PE = Fg * h = mgh = 1kg*9.8m/s^2 * 4*10^4m =
= 3.9 * 10^5 Joules
That sounds like a lot, but consider how much KINETIC energy is required to be in a circular orbit at that low altitude:
Fcentripetal = Fgravity
mv^2/r = mg (remember, we're still modeling terretrial gravity - r = radius of earth v=speed)
mv^2=mgr
1/2mv^2=1/2*mgr = KE
KE = 1/2* 1kg * 9.8m/s^2 * (6.4*10^6m)
KE = 3.2*10^6 Joules
or approximately 10 times the PE gain by ballooning up that high.
So the bottom line is, nope you don't gain that much because you still need to be traveling really fast to be in orbit and even faster yet to pull away from the earth's gravity. You do save a little more by escaping the dragging atmosphere, but I doubt it adds up to that much.
Nice idea though. I'm sure NASA guys ran the math a long while back and junked the idea. What they haven't junked it the idea of traveling really fast with a jet within the atmosphere (to Mach 8+) and then using rockets to launch into orbit.
2)Correct the legal structures that currently allow industries to externalze costs. Just to give a timely example, a gallon of gas would cost alot more than $1.50 if the oil companies had to foot, say, 25% of the nation's defense budget every year to preserve access to the oil (the ethical considerations notwithstanding, of course.) As it is, the taxpayers pick up the tab instead. A whole lot of "fringe" and "green" technologies would be much more in demand if the users of current technology had to pay the true costs of that technology.
Say it loud, Brother. I'd like to add to that:
3) The nuclear industry has managed to externalize the cost of both the disposal of their horribly long half-life byproducts and also the insurance costs of their plants (which the government (i.e. you, the taxpayer) picks up since no private insurance company would touch their policies with a ten-foot pole).
The real reason why solar power isn't here now is because almost all of the costs are upfront and burdened by the buyer of the panels, not a third-party victim.
If you like Nuke Energy, tap into the biggest fusion source within 4 light-years: go solar. Consider:
All of the US's current electricity could be generated by a 10mi x 10 mi solar grid (100 square miles).*
The Yuk-nuclear waste reserve is approximate 1800 square miles.
A field of glass or a 100,000 year + nuke hot spot - it's your choice.
* 3702 US billion kWh/ year : ( source )
164 average Watts/m^2 * 24 hour / day @ 40 deg Lat.
Assume ~ 10% efficiency and run the math. (+/- 20% error)
Re:Ternary has been known to be efficient...
on
Ternary Computing
·
· Score: 1
I wasn't too impressed by the claims of efficiency. Mathematically, optimizing the rw product is interesting, but pushing binary's value of 40 (for number for 0-999,999) to ternary's 39 is a dubious advantage.
Before we turn the computing industry upside down, perhaps we could convince people to use Metric units.
My 60GB Hungary IBM drive died 1 week into my new job last month. This was the first hard drive crash I've ever had. I got a 75GB replacement IBM drive from the company that sold it to me that was made in Taiwan. I guess I should think about swapping that out as well.
Although I have never seen it or heard of it previously (it sounds awesome to watch), I believe it. Here are the two dominant reasons according to Physics:
1) The buoyant force of air. Air has a density of ~1kg/m^3. Displacing 1m^3 of air produces 9.8 Newtons of upward lifting force (about 2.2 lbs of force). If your 1m^3 balloon has a mass of less than 1kg, it floats. This is reason why hydrogen gas balloons/blimps float.
2) When falling, an object reaches it's terminal (maximum) velocity when the gravitation force pulling it down is equal to the viscous air resistance (and buoyant force) opposing it balance out.
where D is the density of the object. So, assuming the same object shape (which results in the same "b"), the closer an object's Density gets to the atmospheric density (1kg/m^2 on earth), the slower it falls. If it drops below the atmospheric density, it floats.
Sorry for the lecture folks, I used to be a physics teacher.
The game consists of a 3d battlefield of, say, 1600 pixels cubed. There is a missile which is launched somewhere in that space which is guranteed to hit above a certian height before heading back down. Your job is to code a script or program (which the game will interpret and run) that has access to:
The truth is, Adam, I had exactly the same thought: Code this up in a simulation. In one version of the simulation, you have access to x(t), y(t), z(t) for the incoming missile, in the other version, you have noisy sensors and possibly stealth warheads - there's a big difference in the two.
In the GPS-assist case, since ballistic missiles pretty much "coast" all the way down, the future trajectory of the missile is pretty much known. Sure, it's not quite as simple as modeling the earth as a uniform sphere with a small projectile in orbit, but the Military has already solved the additional complications (e.g. non-uniform earth, atmospheric drag during final approach, etc.) - that's how they can land a missile on top of an enemy silo with 10 meters to spare.
The point is, hitting that known trajectory with another missile is absolutely trivial in a simulation. Hitting the missile in the test with the known trajectory, while impressive, is more a matter of your rockets going where you want them to go.
The real challenge is, what do you do with your noisy sensor data when that missile releases 20 metallic coated balloons - perhaps one enshrouding the actual warhead? How about if the warhead is outfitted with a stealth hull. You can't differentiate them with radar, and since they're traveling in a relative vacuum, they're all moving with the same speed.
Net result: It would take a massively redundant system and a lot of wasted missiles to take care of all the decoys, never mind the backups for if you miss the real thing.
Surely there's a cheaper way to eliminate the threat from North Korea and Iraq. If we want to spend a ton of money on self-preservation, why don't we just buy these countries?
A few people have commented on the meteor fear, with the moon's battered surface being the example of fear. Most of the moon's major craters are from a long-long time ago (I forget the estimates 3-4 Billion years I think, during the formation of the solar system). The reason why they're there is there is no weathering on the moon.
And, of course, our pretty atmosphere doesn't do to much to a big comet coming in at 42 km/s from the outer solar system.
Re:why can't this be built in full gravity?
on
Hotel on the Moon
·
· Score: 1
While it is true that g (local gravitational constant) is factored out of the torque equilibrium equations, the sheer on the lever arm structure IS dependent on g. Imagine the hotel as a light wooden baton jimmied under a really heavy rock with a fulcrum. As you push down on the baton, you are simulating an increase in g (not quite, as you would have to push down on every chunk of mass of the baton, but if you push from the center of gravity of the exposed piece, it's a decent approximation). Obviously, a sufficiently strong (simulated) force of gravity will break the baton/building.
Anyway, at the rate Space Exploration is going, The Diamond Age will have arrived by the time we're building hotels on the moon, so we'll already have this thing on Earth.
Another less well known Zelazny
on
Lord of Light
·
· Score: 1
Nice to get a review of a less well known Zelazny. Unlike most fantasy authors, Zelazny actually knows how to write. If you liked Lord of Light, you might want to hunt down a copy of Zelazny's "Isle of the Dead" which may or may not still be in print. It has a similar framework and has one of the most interesting alien races from a psychological standpoint. The book is not perfect, but parts of it are pure poetry (as is Lord of Light - especially Sam's confrontation with his ex-girlfriend). It also discusses futuristic shaping of solar systems (and terraforming) which is no dissimilar to what has recently been discussed to save the Earth from frying in the heat of a more energetic sun.
Slashdot Mirror
$600 for 512 MB (Keychain)
compared to
$0.25 cents for 640 MB (CD-Rom)
Enjoy, gadget freaks. I think I'll keep slipping my backups into my coat pocket instead of my jeans.
Read "Moon is a harsh Mistress" by R.A. Heinlein for a Sci-fi view of this. You folks who are talking about bringing the ore back to earth seem to miss the point - it's better to keep this stuff up in orbit to you can use it instead of bringing it back to Earth. I forget the price per Kilogram of sending something up in the space shuttle, but it's something like $20,000/kg +.
Of course, if you bang an asteroid with a nice meaty chunk of Platinum or Paladium, there's a lot of healthy uses for that back home.
This article also talks about one of the biggest advantages of mining asteroids - getting mass (ore, water, aggregate) off an asteroid is trivial in terms of energy and thrust required. If you're looking to build spaceships or space stations, there's a big advantage in using the materials already up there instead of bringing them out of the Earth's gravity well. The Moon and Mars have the same advantage to a lesser degree.
Alright, quick lesson on Buoyancy.
The Buoyant force, as discovered by Archimedes, is the force due to the pressure from the gas (or liquid) displaced by an object. It is:
Fb = DVg
Where D=density of the gas (or liquid), V is the volume of the object and g is 9.8m/s^2 on Earth.
So, an object like a helium balooon doesn't get it's lift from the density of the helium inside the balloon, it gets it from the displaced air.
What actually happens as a helium balloon rises is that the pressure outside the balloon decreases and thus the balloon expands in volume. This sounds great, since more volume implies more lift, but it doesn't work that way because the density of the diplaced air is going in the opposite direction. The balloon needs to expand in order to continue climbing. Ultimately, as the density of the air outside the balloon drops to zero (in the vacuum of space), it is impossible for the balloon climb any higher. These guys are trying to approach that theoretical limit.
If you've ever seen a video of the launch of a weather balloon, you might have noticed that the balloon looks funny and under-inflated. This is because the engineers have left a lot more balloon skin remaining so that this expansion can take place.
Robert A Heinlein once wrote that once you're in earth orbit, you're half way to anywhere in the solar system. Unfortunately, sitting on a balloon at a high altitude is hardly in earth's orbit. The main problem is that you're not traveling very fast, and it takes a lot of kinetic energy to be in orbit, and that's one thing our ballooning friends will not have (until they jump ;-) ).
Ok, here's the math:
132,000 feet = 40 x 10^3 m
This is VERY small compared with the earth's radius:
6.4x10^6m.
So you're actually only pulling away slightly from the earth's gravitational influence. Normally, one calculates the force of gravity for these problems with:
FG=G*m1*m2/r^2
but since we're so close to the earth, we can approximate the whole problem with our old terrestrial gravity without introducing too much error:
Fg=mg where g=9.8m/s^2
So, the potential energy gain by ballooning up that high for a 1kg block is
PE = Fg * h = mgh = 1kg*9.8m/s^2 * 4*10^4m =
= 3.9 * 10^5 Joules
That sounds like a lot, but consider how much KINETIC energy is required to be in a circular orbit at that low altitude:
Fcentripetal = Fgravity
mv^2/r = mg (remember, we're still modeling terretrial gravity - r = radius of earth v=speed)
mv^2=mgr
1/2mv^2=1/2*mgr = KE
KE = 1/2* 1kg * 9.8m/s^2 * (6.4*10^6m)
KE = 3.2*10^6 Joules
or approximately 10 times the PE gain by ballooning up that high.
So the bottom line is, nope you don't gain that much because you still need to be traveling really fast to be in orbit and even faster yet to pull away from the earth's gravity. You do save a little more by escaping the dragging atmosphere, but I doubt it adds up to that much.
Nice idea though. I'm sure NASA guys ran the math a long while back and junked the idea. What they haven't junked it the idea of traveling really fast with a jet within the atmosphere (to Mach 8+) and then using rockets to launch into orbit.
Former Physics Teacher
Say it loud, Brother. I'd like to add to that:
3) The nuclear industry has managed to externalize the cost of both the disposal of their horribly long half-life byproducts and also the insurance costs of their plants (which the government (i.e. you, the taxpayer) picks up since no private insurance company would touch their policies with a ten-foot pole).
The real reason why solar power isn't here now is because almost all of the costs are upfront and burdened by the buyer of the panels, not a third-party victim.
If you like Nuke Energy, tap into the biggest fusion source within 4 light-years: go solar. Consider:
All of the US's current electricity could be generated by a 10mi x 10 mi solar grid (100 square miles).*
The Yuk-nuclear waste reserve is approximate 1800 square miles.
A field of glass or a 100,000 year + nuke hot spot - it's your choice.
* 3702 US billion kWh/ year : ( source )
164 average Watts/m^2 * 24 hour / day @ 40 deg Lat.
Assume ~ 10% efficiency and run the math. (+/- 20% error)
I wasn't too impressed by the claims of efficiency. Mathematically, optimizing the rw product is interesting, but pushing binary's value of 40 (for number for 0-999,999) to ternary's 39 is a dubious advantage.
Before we turn the computing industry upside down, perhaps we could convince people to use Metric units.
>The Number of The Beast, wherein the prevalence >of 3 phase wiring led to trinary computing.
;-)
The article overlooked another advantage of base-3 - the sinister 666 is reduced to the pathetic 22020.
My 60GB Hungary IBM drive died 1 week into my new job last month. This was the first hard drive crash I've ever had. I got a 75GB replacement IBM drive from the company that sold it to me that was made in Taiwan. I guess I should think about swapping that out as well.
Backup-backup-backup...
Although I have never seen it or heard of it previously (it sounds awesome to watch), I believe it. Here are the two dominant reasons according to Physics:
1) The buoyant force of air. Air has a density of ~1kg/m^3. Displacing 1m^3 of air produces 9.8 Newtons of upward lifting force (about 2.2 lbs of force). If your 1m^3 balloon has a mass of less than 1kg, it floats. This is reason why hydrogen gas balloons/blimps float.
2) When falling, an object reaches it's terminal (maximum) velocity when the gravitation force pulling it down is equal to the viscous air resistance (and buoyant force) opposing it balance out.
Here's the math with "up" being positive:
@ terminal velocity:
Fnet = ma = -mg + Fbuoyant +Fviscous = 0
Fbuoyant = (1kg/m^2)Vg (on earth - V=volume of object)
Fviscous ~ bv (b= constant , v = speed of object (simple model)
bv = mg - (1kg/m^2)Vg = DVg - (1kg/m^2)Vg
= (D - (1kg/m^2))Vg
velocity = (D - (1kg/m^2))Vg/b
where D is the density of the object. So, assuming the same object shape (which results in the same "b"), the closer an object's Density gets to the atmospheric density (1kg/m^2 on earth), the slower it falls. If it drops below the atmospheric density, it floats.
Sorry for the lecture folks, I used to be a physics teacher.
And if you opt-out, you can ride in the cargo hold.
Seriously, if being anonymous is that important to you, drive.
The game consists of a 3d battlefield of, say, 1600 pixels cubed. There is a missile which is launched somewhere in that space which is guranteed to hit above a certian height before heading back down. Your job is to code a script or program (which the game will interpret and run) that has access to:
The truth is, Adam, I had exactly the same thought: Code this up in a simulation. In one version of the simulation, you have access to x(t), y(t), z(t) for the incoming missile, in the other version, you have noisy sensors and possibly stealth warheads - there's a big difference in the two.
In the GPS-assist case, since ballistic missiles pretty much "coast" all the way down, the future trajectory of the missile is pretty much known. Sure, it's not quite as simple as modeling the earth as a uniform sphere with a small projectile in orbit, but the Military has already solved the additional complications (e.g. non-uniform earth, atmospheric drag during final approach, etc.) - that's how they can land a missile on top of an enemy silo with 10 meters to spare.
The point is, hitting that known trajectory with another missile is absolutely trivial in a simulation. Hitting the missile in the test with the known trajectory, while impressive, is more a matter of your rockets going where you want them to go.
The real challenge is, what do you do with your noisy sensor data when that missile releases 20 metallic coated balloons - perhaps one enshrouding the actual warhead? How about if the warhead is outfitted with a stealth hull. You can't differentiate them with radar, and since they're traveling in a relative vacuum, they're all moving with the same speed.
Net result: It would take a massively redundant system and a lot of wasted missiles to take care of all the decoys, never mind the backups for if you miss the real thing.
Surely there's a cheaper way to eliminate the threat from North Korea and Iraq. If we want to spend a ton of money on self-preservation, why don't we just buy these countries?
A few people have commented on the meteor fear, with the moon's battered surface being the example of fear. Most of the moon's major craters are from a long-long time ago (I forget the estimates 3-4 Billion years I think, during the formation of the solar system). The reason why they're there is there is no weathering on the moon. And, of course, our pretty atmosphere doesn't do to much to a big comet coming in at 42 km/s from the outer solar system.
While it is true that g (local gravitational constant) is factored out of the torque equilibrium equations, the sheer on the lever arm structure IS dependent on g. Imagine the hotel as a light wooden baton jimmied under a really heavy rock with a fulcrum. As you push down on the baton, you are simulating an increase in g (not quite, as you would have to push down on every chunk of mass of the baton, but if you push from the center of gravity of the exposed piece, it's a decent approximation). Obviously, a sufficiently strong (simulated) force of gravity will break the baton/building. Anyway, at the rate Space Exploration is going, The Diamond Age will have arrived by the time we're building hotels on the moon, so we'll already have this thing on Earth.
Nice to get a review of a less well known Zelazny. Unlike most fantasy authors, Zelazny actually knows how to write. If you liked Lord of Light, you might want to hunt down a copy of Zelazny's "Isle of the Dead" which may or may not still be in print. It has a similar framework and has one of the most interesting alien races from a psychological standpoint. The book is not perfect, but parts of it are pure poetry (as is Lord of Light - especially Sam's confrontation with his ex-girlfriend). It also discusses futuristic shaping of solar systems (and terraforming) which is no dissimilar to what has recently been discussed to save the Earth from frying in the heat of a more energetic sun.