It's a trivial exercise in rotational kinematics to prove Parsons wrong (again).
I'll scan the pages of working and upload them (if I'm feeling generous, or if anybody else wants to see them).
I couldn't remember numbers, so I assumed we were talking about a 1 gigaton (metric) asteroid attached to the elevator platform, and sitting at a distance of 50,000 km's from the platform.
It turns out that if you do the maths, the asteroid has a velocity of 6.7 km/s. Now, compare this to the fact that a satelite in a stable orbit at this altitude has an orbital velocity of 0.0469 m/s. Take a moment to compare those to numbers. 6,700 m/s versus 0.0469 m/s. The asteroids velocity is 10,000 times greater then it would be if the asteroid were in a stable orbit.
It also turns out that, when you do the maths, the earth exerts a gravitational pull on the asteroid of 4.7 meganewtons... Sounds big right? But the centripetal force acting on the asteroid is 2 orders of magnitude greater still, at 490 meganewtons. This means that the asteroid exerts a net upwards pull on the cable and the platform of 485 meganewtons.
The other thing that you find out, when you do the math, is that the effect that this has on the earth is to accelerate the earth towards the elevator in just the way that parson suggested...
At a rate of 8.21x10^-17 m/s/s.
To put this in perspective, this is equivalent to 0.08 fermi/s/s.
To put that in perspective, most atomic nuclii have a diameter of about 10 fermi.
This means that it would take about 30 seconds for the earth to traverse the width of an atomic nucleus.
Of course, then there's the fact that over the course of a day (and over the course of a year) the acceleration cancels itself out.
The other thing that I feel vaguely compelled to point out is the fact that if we hang the 1 gigaton asteroid a distance of 110,000 kms from the elevator platform, then the velocity of asteroid is actually greater then the escape velocity of the earth at sea level (I know that escape velocity has a dependence on height, but I can't be bothered with the math, and escape velocity is highest at sea level anyway), so even should the cables snap, the asteroid, and everything it is attached to would simply be flung out of the earth-moon system.
So, there we go, Parsons is horribly, trivially, demonstrably wrong (again) in each and every one of his assertions.
So basically - although the counterweight may have a linear velocity of 6,700 m/s, it still has the same angular velocity as the surface of the earth, or a geostationary orbit. Essentially it's the same as getting a meter ruler, holding it at arms length, and spinning on the spot. The tip of the ruler has a higher linear velocity then say your nose, or even the base of the ruler, but the ruler doesn't twist itself out of shape does it?
PIATLAS
11th March 2008 - 01:33 AM
Trippy, analyze my calculations.
My Math:
The Earth has a radius of 6,378 Km at the Equator. The conversion factor km to miles is 1.609. The Earths radius would be 3963.95 miles.
Geo-stationary orbit is at 35,786 Km above the Equator. That translates to 22,241.1 miles above the equator.
The Earth spins 25,000 miles in 24 hours. That’s 1041.67 miles per hour.
The Geo-stationary orbit is around the center of the Earth point at the Equator at `3963.95 + 22241.1’ = 26,205.1 miles
To have an Asteroid at that radius from the center of the Earth Geo-Stationary then as the Earth travels 25,000 miles in one day the Asteroid would with it travel 2 times pi times 26,205.1 miles in 24 hours. That’s 164,651/24 equals 6,860.48 miles per hour
Now with the space elevator the only way to reach `centrifugal force escape velocity’ at altitude, would be to take the elevator beyond 100 mile up and all the way up to 22,241.1 miles above the surface of the Equator.
Even though the payload at rest on the ground is spinning with the Earth at 1041.67 miles per hour, as the payload is winched up to 22,241.1 miles up (That’s a long haul) It would experience carousel forces acting on it causing tremendous stress on the tether as it has it’s velocity around the center of the Earth increases from 1,041.67 to 6860.48 miles per hour.
Enthalpy
11th March 2008 - 02:16 AM
Piatlas:
"10,000km = 1,000,000 cm"
this is the kind of math that would make the rope almost possible...
Just in case you want to check:
TiAl6V (common titanium alloy) has a strength of some 800MPa (or N/mm2).
So just correct the conversion above and see it's completely impossible. The form won't change this conclusion.
By the way, titanium is outperformed by good steel (at equal mass), and both are brutally outperformed by fibers like aramide, polyethylene and LCP. 30 years ago, titanium was a wonderful material, but fact is that the steel industry makes fast progress and the titanium industry makes none, pity.
I put an upper bound to the rope's feasibility using the strength of chemical bonds because real materials, for instance alloys, are an awful lot weaker than their chemical bonds.
Completely and hopelessly impossible. Sorry.
PIATLAS
11th March 2008 - 04:51 AM
Sunlight temperature without atmospheric shielding in orbit acting on the tether could be high and would affect the tensile strength of the tether
Trippy
11th March 2008 - 05:43 AM
QUOTE (PIATLAS+Mar 11 2008, 02:33 PM)
Trippy, analyze my calculations.
My Math:
The Earth has a radius of 6,378 Km at the Equator. The conversion factor km to miles is 1.609. The Earths radius would be 3963.95 miles.
Geo-stationary orbit is at 35,786 Km above the Equator. That translates to 22,241.1 miles above the equator.
The Earth spins 25,000 miles in 24 hours. That’s 1041.67 miles per hour.
The Geo-stationary orbit is around the center of the Earth point at the Equator at `3963.95 + 22241.1’ = 26,205.1 miles
To have an Asteroid at that radius from the center of the Earth Geo-Stationary then as the Earth travels 25,000 miles in one day the Asteroid would with it travel 2 times pi times 26,205.1 miles in 24 hours. That’s 164,651/24 equals 6,860.48 miles per hour
Now with the space elevator the only way to reach `centrifugal force escape velocity’ at altitude, would be to take the elevator beyond 100 mile up and all the way up to 22,241.1 miles above the surface of the Equator.
Even though the payload at rest on the ground is spinning with the Earth at 1041.67 miles per hour, as the payload is winched up to 22,241.1 miles up (That’s a long haul) It would experience carousel forces acting on it causing tremendous stress on the tether as it has it’s velocity around the center of the Earth increases from 1,041.67 to 6860.48 miles per hour.
None of which is actually relevant - the point I was making is this: All points of the space elevator complete one revolution around the earth in 23hrs, 56 min, 4.1 sec.
Therefore, no lateral displacement, no sideways forces.
Grumpy
11th March 2008 - 11:03 AM
Enthalpy
QUOTE
Bacon was meticulous: Though he had produced the first carbon fibers (as graphite whiskers) in 1958, he didn't publish a paper in the Journal of Applied Physics until two years later, after thoroughly studying his find. In the process, some scientists suggest that Bacon may even have produced the first carbon nanotubes without realizing it.
The fibers that Bacon discovered possessed an unusually high tensile strength (a measure of the amount of force with which a fiber can be pulled before it breaks) and Young's modulus (a measure of a material's stiffness--its ability to resist elongation under load). Steel has a tensile strength of 1–2 gigapascals (GPa) and a Young's modulus of 200 GPa; Bacon's fibers were measured at 20 GPa and 700 GPa, respectively. But they were also expensive to make, and even commercial versions of carbon fibers are haunted to this day by high prices.
The first commercial high-performance carbon fibers were available by 1963, based on a process discovered at Parma for heat-treating rayon. High-modulus fibers became available within a few more years, when Bacon and Wesley Schalamon used a "hot-stretching" process to stretch carbon yarn during heat-up, not afterward.
In 1970, Singer made progress on another kind of carbon fiber at Parma while studying pitch, a tarlike mixture of hundreds of branched compounds with differing molecular weights formed by heating petroleum or coal. He used a "taffy-pulling" apparatus to align molecules in pitch's mesophase (its liquid-crystal state) and then heated the resulting mixture to produce a highly oriented carbon fiber.
These graphitized mesophase-pitch fibers had an even higher Young's modulus than Bacon's fibers (approaching 1,000 GPa) and showed high thermal conductivity, making them ideal for applications such as brakes and electronic circuits.
http://pubs.acs.org/cen/NCW/8142acsn.htmlThis was in 2003, much has happened since then.
QUOTE (->
| QUOTE |
Bacon was meticulous: Though he had produced the first carbon fibers (as graphite whiskers) in 1958, he didn't publish a paper in the Journal of Applied Physics until two years later, after thoroughly studying his find. In the process, some scientists suggest that Bacon may even have produced the first carbon nanotubes without realizing it.
The fibers that Bacon discovered possessed an unusually high tensile strength (a measure of the amount of force with which a fiber can be pulled before it breaks) and Young's modulus (a measure of a material's stiffness--its ability to resist elongation under load). Steel has a tensile strength of 1–2 gigapascals (GPa) and a Young's modulus of 200 GPa; Bacon's fibers were measured at 20 GPa and 700 GPa, respectively. But they were also expensive to make, and even commercial versions of carbon fibers are haunted to this day by high prices.
The first commercial high-performance carbon fibers were available by 1963, based on a process discovered at Parma for heat-treating rayon. High-modulus fibers became available within a few more years, when Bacon and Wesley Schalamon used a "hot-stretching" process to stretch carbon yarn during heat-up, not afterward.
In 1970, Singer made progress on another kind of carbon fiber at Parma while studying pitch, a tarlike mixture of hundreds of branched compounds with differing molecular weights formed by heating petroleum or coal. He used a "taffy-pulling" apparatus to align molecules in pitch's mesophase (its liquid-crystal state) and then heated the resulting mixture to produce a highly oriented carbon fiber.
These graphitized mesophase-pitch fibers had an even higher Young's modulus than Bacon's fibers (approaching 1,000 GPa) and showed high thermal conductivity, making them ideal for applications such as brakes and electronic circuits. |
http://pubs.acs.org/cen/NCW/8142acsn.htmlThis was in 2003, much has happened since then.
Completely and hopelessly impossible. Sorry
Grumpy
11th March 2008 - 12:08 PM
Enthalpy
MWNT 0.8-0.9Young's 150GPa tensile.
Stainless Steel ~0.2 ~0.65-1
Grumpy
Enthalpy
12th March 2008 - 02:29 AM
Dear Grumpy, you haven't read what I wrote.
A single molecule isn't a rope. You can't compare their strengths.
Your last post is difficult to understand. Oh yes, the figures are copied from the article in Wikipedia for which you give the link. You should have copied the names and the units as well. Without forgetting the T for theoretical prediction (=false) and E for experimental.
Again, the Wiki article compares a single molecule to a usable bulk material.
Meaningless.
And as we observe this tremendous loss of performance (magnitudes) for any fiber when making a rope of it, I repeat: Completely impossible.
Grumpy
12th March 2008 - 11:58 AM
Enthalpy
QUOTE
MWNT 0.8-0.9Young's 150GPa tensile.
Stainless Steel ~0.2 ~0.65-1
Multi-wall nano tube=0.8-0.9 Young's modulus(a measure of stiffness) 150 GIGA Pascals(a measure of tensile strength)
Stainless Steel=~0.2 Young's and ~0.65-1GPa
200GPa is the theoretical limit for carbon.
Yes, that is single molecular strands forming a rope, each strand going from one end to the other, just like the cables of the Golden Gate bridge are steel wires that stretch from one side of the bridge to the other. But, for a good space elevator you would need strands a few hundred thousand MILES long and a few gazzilion of them. The fibers would not necessarily need such strength in non-critical sections of the cable.
Grumpy
Capracus
13th March 2008 - 04:19 PM
I've been thinking of an idea along the lines of the space tether concept. Instead of a tether suspended from an orbital platform, the tether or tethers would be anchored to the Moon itself.
The tethers would descend into the Earth's atmosphere to an optimum altitude and suspend an aerodynamic platform. The suspended platform would have have a relative airspeed at sea level of over 1000 mph, with speed increasing with altitude. Due to the platform's aerodynamic characteristics and airspeed, it would be able to generate lift and be capable of unloading the stress on the tether, and climbing to the upper atmosphere.
The platform would drop to a prescribed altitude to meet load carrying aircraft that could transfer their loads in flight to the platform. Once loaded, the platform would climb to it's maximum altitudes and the loads would then be transported Moon-ward via a climber. As the climber moved up the tether, it could park loads in an L1 orbit, or continue to the surface of the Moon.
Grumpy
13th March 2008 - 04:48 PM
Capracus
Talk about ambitious!!! Let's get to Low Earth Orbit and Geostationary first.
Grumpy
Grumpy
14th March 2008 - 10:47 AM
Capracus
Your post yesterday got me thinking. At roughly 8,000 miles diameter, thats roughly 25,000 miles circumference/28 days is a little less than 1,000 miles/day "drift" as we could call it.
25,000 miles cir./24 hours/day=a little over 1000 mph. Managable at say 125,000+ feet altitude or a ballistic trajectory just outside the atmosphere.
We could do this with materials we have now, the highest stresses are avoided, a tether on the moon could be made of Titanium or even steel.
The same sort of thing could be developed into a space craft that could reach down from orbit and snatch loads from anywhere by using a tether/counterweight system, or catch them from orbit and deposit them on the ground, especially from the moon or other airless bodies.
adoucette
14th March 2008 - 02:31 PM
QUOTE (Capracus+Mar 13 2008, 11:19 AM)
Due to the platform's aerodynamic characteristics and airspeed, it would be able to generate lift and be capable of unloading the stress on the tether,
Sound's like a free lunch.
Arthur
Capracus
14th March 2008 - 04:22 PM
QUOTE (Grumpy+Mar 14 2008, 10:47 AM)
The same sort of thing could be developed into a space craft that could reach down from orbit and snatch loads from anywhere by using a tether/counterweight system, or catch them from orbit and deposit them on the ground, especially from the moon or other airless bodies.
Grumpy, here's an entire website devoted to this very subject.
http://spacetethers.com/
Capracus
14th March 2008 - 05:00 PM
QUOTE (adoucette+Mar 14 2008, 02:31 PM)
Sound's like a free lunch.
Arthur
I wouldn't call it a free lunch. I know the gravity load would be negated through aerodynamic lift, for the platform at least, but you would still be left with the drag of the platform and tether. I haven't worked out the lift over drag advantage.
My whole point of the concept was to somehow take advantage of the relative motion of the Earth and Moon.
Another added advantage of the relative motion of the the tether, is that if the tether is a an electrical conductor, it will generate current as it passes through the Earth's magnetic field. This current could be used to electromagnetically accelerate loads up a properly configured tether system, or into various orbits.
adoucette
14th March 2008 - 06:26 PM
Of course its a free lunch.
You expect this aerodynamic platform to be both PULLED aginst the airflow by the tether to generate lift and then for this lift to UNLOAD the tether.
That's the same as picking yourself up by pulling on your bootstraps.
Arthur
Capracus
15th March 2008 - 11:19 AM
QUOTE (adoucette+Mar 14 2008, 06:26 PM)
Of course its a free lunch.
You expect this aerodynamic platform to be both PULLED aginst the airflow by the tether to generate lift and then for this lift to UNLOAD the tether.
That's the same as picking yourself up by pulling on your bootstraps.
Arthur
Well actually if configured as you state, it would unload the tether, but in doing so the platform would have to carry the accumulating weight of the tether as it climbed.
My original thinking was to have the platform move up the tether, with the end of the tether hanging down. I realize that in order to impart relative motion to the platform, a counterweight or airfoil system would be necessary to counter the drag of the platform on the tether.
adoucette
15th March 2008 - 02:10 PM
QUOTE (Capracus+Mar 15 2008, 06:19 AM)
Well actually if configured as you state, it would unload the tether,
No, it wouldn't.
In your version you have an airfoil CREATING energy.
But you know that is not the way they work.
They bend the local airflow it to create an opposite force that is perpendicular to their path.
But, as in any mechanical device there are LOSSES in its operation.
The force on the tether created by the airfoil HAS to be greater than the lift, else you are getting more energy (lift) out of the airfoil than you put in (drag).
The VECTOR of the forces on the tether might change, but putting an airfoil on the end of the tether would not reduce the forces on the tether UNLESS the airfoil had its OWN power supply.
Arthur
Grumpy
15th March 2008 - 02:38 PM
adoucette
I would think the pull of the cable would only have to overcome drag, which is usually much less than the full weight of the aircraft(glider, in this case). The end would need to be above 125,000-150,000 feet and would be being dragged through thin air at about Mach 1 for that altitude(1000 mph). Which would be a perfect target for advanced air breathing single stage to tether aircraft.
Grumpy
adoucette
15th March 2008 - 02:59 PM
QUOTE (Grumpy+Mar 15 2008, 09:38 AM)
adoucette
I would think the pull of the cable would only have to overcome drag, which is usually much less than the full weight of the aircraft(glider, in this case). The end would need to be above 125,000-150,000 feet and would be being dragged through thin air at about Mach 1 for that altitude(1000 mph). Which would be a perfect target for advanced air breathing single stage to tether aircraft.
Grumpy
Well, like I explained to Capracus, and as you are well aware of, in the energy game, there is no free lunch.
The energy used to turn the airsteam to create LIFT has to come from somewhere as the airfoil is simply an unpowered means to that end.
That energy, in this set up, has to come from the tether as that is the only source of energy the airfoil has to work with to create this LIFT.
The drag is different and it represents the LOSS in the system, which represents the excess energy ABOVE the energy required to produce LIFT that the tether also has to supply.
Arthur
Capracus
16th March 2008 - 12:45 PM
QUOTE (adoucette+Mar 15 2008, 02:59 PM)
Well, like I explained to Capracus, and as you are well aware of, in the energy game, there is no free lunch.
None claimed.
QUOTE
The energy used to turn the airsteam to create LIFT has to come from somewhere as the airfoil is simply an unpowered means to that end.
Agreed.
QUOTE (->
| QUOTE |
| The energy used to turn the airsteam to create LIFT has to come from somewhere as the airfoil is simply an unpowered means to that end. |
Agreed.
That energy, in this set up, has to come from the tether as that is the only source of energy the airfoil has to work with to create this LIFT.
Agreed.
QUOTE
The drag is different and it represents the LOSS in the system, which represents the excess energy ABOVE the energy required to produce LIFT that the tether also has to supply.
In this case they are the same.
As you know, with an efficient airfoil, drag is only a fraction of lift.
So to pull the platform up the tether using lift, a portion of the force generated by the lift of the airfoil has to be vectored against the drag. The method for doing so depends on the properties of the tether.
I have no idea what the properties of a 240,000 mile tether system would be. I doubt that it would be analogous to the tow line of a glider, but for simplification we can use that example.
Say you had a 1200 lb glider being tow behind a powered aircraft at a steady rate. The glider has an L/D of 40:1, this means that you need a force of greater than 30 lbs to overcome the drag on the towline. If the towline was fed through a pulley attached to the CG of the plane, and attached to a 40 lb weight that was allowed to play out from the pulley, the sailplane would be drawn to the tow aircraft. So by lowering less than 1/10th of the useful load, the plane can be pulled forward up the towline.
As far as the space tether is concerned, if a counterweight is necessary to allow it to gain vertical access into the atmosphere, then this property of the tether system would serve the purpose of the counterweight in the above example. If the tether system turned out to be more rigid, then the counterweight would not be necessary.
The combined gravity of the Earth and the Moon, coupled with the centrifugal force of the Moon's rotation, would effect the stiffness of the the system. The rotational speed of the tether at the Earth end, relative to the Moon's center would be about 2300 mph.
adoucette
16th March 2008 - 03:20 PM
The weight of air that is accelerated and turned by an airfoil in steady flight is EQUAL to the weight of the object supported by that airfoil.
The energy to overcome the force of gravity is the SAME (if you ignore losses due to drag) if the object is using a wing to support itself or if it is just hanging from the end of the tether.
There is NO FREE LUNCH and the only source of energy to counteract the force of gravity is the TETHER.
You can't, like a glider, use the conversion of potential energy into kinetic energy because you don't want your platform to go DOWN.
Arthur
Grumpy
16th March 2008 - 05:05 PM
adoucette
Any extra energy needed would cause the counter-weight to lose orbital velocity. If that counterweight was the Moon, the loss would not be noticeable. Even if the counterweight was only a small asteroid, the orbital loss would be minimal, and solar/nuclear ion/mass driver/magnetic drive could then boost the orbit after each launch(or several launches, with a orbital energy loss for each one). That is another reason a rotating counterweight/tether system, where energy could be stored as rotational energy. Artificial gravity for the crew would also be a plus. Ice could be used as both counterweight and shielding, plus it would provide hydrogen/oxygen/carbon as well as reaction mass.
Grumpy
adoucette
16th March 2008 - 05:54 PM
QUOTE (Grumpy+Mar 16 2008, 12:05 PM)
adoucette
Any extra energy needed would cause the counter-weight to lose orbital velocity. If that counterweight was the Moon, the loss would not be noticeable. Even if the counterweight was only a small asteroid, the orbital loss would be minimal, and solar/nuclear ion/mass driver/magnetic drive could then boost the orbit after each launch(or several launches, with a orbital energy loss for each one). That is another reason a rotating counterweight/tether system, where energy could be stored as rotational energy. Artificial gravity for the crew would also be a plus. Ice could be used as both counterweight and shielding, plus it would provide hydrogen/oxygen/carbon as well as reaction mass.
Grumpy
Totally agree.
I'm just pointing out that an airfoil does not create energy, so putting an airfoil on the platform at the end of the tether can't reduce the energy demands of the platform on the tether.
Arthur
N O M
17th March 2008 - 12:45 AM
It's a fairly safe bet that there will be lunar mining prior to an elevator being built. It would make sense to use the elevator to lower some of the lunar-mined resources (He3, precious metals, cheese) as a counter-balance to whatever is coming up the elevator.
Another answer would e to construct the counterweight out of some incredibly dense material, such as Mr.Robin.Parsons
TheDoc
17th March 2008 - 12:48 AM
QUOTE (N O M+)
Another answer would e to construct the counterweight out of some incredibly dense material, such as Mr.Robin.Parsons
Now there's an idea, NOM!
Capracus
17th March 2008 - 10:34 AM
QUOTE (adoucette+Mar 16 2008, 03:20 PM)
The weight of air that is accelerated and turned by an airfoil in steady flight is EQUAL to the weight of the object supported by that airfoil.
The energy to overcome the force of gravity is the SAME (if you ignore losses due to drag) if the object is using a wing to support itself or if it is just hanging from the end of the tether.
There is NO FREE LUNCH and the only source of energy to counteract the force of gravity is the TETHER.
You can't, like a glider, use the conversion of potential energy into kinetic energy because you don't want your platform to go DOWN.
Arthur
Arthur, I think there must be some confusion as the the configuration I attempted to describe. This diagram may clear things up.
http://www.box.net/shared/ggtdu4fc4s
adoucette
17th March 2008 - 01:30 PM
QUOTE (Capracus+Mar 17 2008, 05:34 AM)
Arthur, I think there must be some confusion as the the configuration I attempted to describe. This diagram may clear things up.
http://www.box.net/shared/ggtdu4fc4s
No confusion.
Do you agree that to reduce the gravitational load of the platform on the tether you must provide a SOURCE of ENERGY to the platform to do so?
If not, then you are claiming a "free lunch"
If so,
What is this source of energy?
How does it get to the platform?
Arthur
Capracus
17th March 2008 - 04:49 PM
QUOTE (adoucette+Mar 17 2008, 01:30 PM)
No confusion.
Do you agree that to reduce the gravitational load of the platform on the tether you must provide a SOURCE of ENERGY to the platform to do so?
The platform is an aerodynamic body capable of generating lift. It is not meant to be fixed to the tether in the lower atmosphere. The platform climbs to a maximum altitude dependent on air density, at which time it becomes fixed to the tether. From there an alternate transport system, e.g. a traction device or electromagnetic acceleration system, moves up the tether into space.
QUOTE
What is this source of energy?
How does it get to the platform?
The rotation of the Earth, which in turn moves the atmosphere relative to Moon/tether position, creates an airstream that provides lift to the platform. The platform would encounter a 1000 mph airflow. After overcoming the drag induced frictional forces between the tether and platform, the platform would climb.
Trippy
17th March 2008 - 05:39 PM
QUOTE (Capracus+Mar 18 2008, 05:49 AM)
The platform is an aerodynamic body capable of generating lift. It is not meant to be fixed to the tether in the lower atmosphere. The platform climbs to a maximum altitude dependent on air density, at which time it becomes fixed to the tether. From there an alternate transport system, e.g. a traction device or electromagnetic acceleration system, moves up the tether into space.
The rotation of the Earth, which in turn moves the atmosphere relative to Moon/tether position, creates an airstream that provides lift to the platform. The platform would encounter a 1000 mph airflow. After overcoming the drag induced frictional forces between the tether and platform, the platform would climb.
But the plat form is not moving relative to the surface of the earth (in the horizontal direction).
Capracus
17th March 2008 - 06:21 PM
QUOTE (Trippy+Mar 17 2008, 05:39 PM)
But the plat form
is not moving relative to the surface of the earth (in the horizontal direction).
Trippy, imagine an airplane flying in line with the equator. Extend a line from the equator up through the airplane and out to the equator of the Moon. This is the orientation of the tether system. Check the diagram.
http://www.box.net/shared/ggtdu4fc4s
adoucette
17th March 2008 - 06:26 PM
QUOTE (Capracus+Mar 17 2008, 11:49 AM)
The platform is an aerodynamic body capable of generating lift. It is not meant to be fixed to the tether in the lower atmosphere. The platform climbs to a maximum altitude dependent on air density, at which time it becomes fixed to the tether.
What is the source of energy that allows the platform to climb to this maximum altitude, you know while it is not fixed to the tether?
oh, and what do you recon the speed over the earth's surface will be if your tether is attached to the moon?
I'm not sure myself, but if the speed of movement of the umbra from a solar eclipse is any indication of the relative motion, you're gona need a FAST platform.
Arthur
Trippy
17th March 2008 - 06:46 PM
QUOTE (Capracus+Mar 18 2008, 07:21 AM)
Trippy, imagine an airplane flying in line with the equator. Extend a line from the equator up through the airplane and out to the equator of the Moon. This is the orientation of the tether system. Check the diagram.
http://www.box.net/shared/ggtdu4fc4s
I already checked out the diagram.
It's called a Geostationary orbit for a reason.
It's because an object in this orbit has the same angular velocity as an object on the surface of the earth.
An object in a Geostationary orbit appears stationary in the sky from the ground hence the name
Tell me. If something appears stationary from the ground, then how can it be moving relative to THE GROUND?
Trippy
17th March 2008 - 07:17 PM
Oh and for the record?
The idea of attaching a space elevator to the moon goes beyond madness, beyond Lunacy even.
The difference between apogee and perigee is 42,592 km.
That means the platform end is going to move up and down by 42,592 km, so when the moon is at perigee (and moving it's fastest) that's fine, bang, over and done.
But what about at Apogee? When the platform is at an altitude that's almost high enough for Geostationary orbit?
adoucette
17th March 2008 - 08:14 PM
Good point.
And its actually a bit worse than that (since the moon varies in its orbit)
On Dec 12th of this year the moon will be 356,567 km away but on Dec 26 it will be a whopping 406,600 km away.
This means it would have to spool out over 50,000 km of tether (enough to go around the world TWICE, just to keep the platform stable (and spool it out at a rate of ~ 150 km/hr)
Of course, by Jan 10th the moon is back to 357,500 km away, so now you would have to reel 49,000 km of tether back in or else the platform crashes into the earth.
Arthur
Trippy
17th March 2008 - 11:50 PM
QUOTE (adoucette+Mar 18 2008, 09:14 AM)
Good point.
And its actually a bit worse than that (since the moon varies in its orbit)
On Dec 12th of this year the moon will be 356,567 km away but on Dec 26 it will be a whopping 406,600 km away.
This means it would have to spool out over 50,000 km of tether (enough to go around the world TWICE, just to keep the platform stable (and spool it out at a rate of ~ 150 km/hr)
Of course, by Jan 10th the moon is back to 357,500 km away, so now you would have to reel 49,000 km of tether back in or else the platform crashes into the earth.
Arthur
Capracus
18th March 2008 - 01:08 PM
QUOTE (Trippy+Mar 17 2008, 06:46 PM)
I already checked out the diagram.
It's called a Geostationary orbit for a reason.
It's because an object in this orbit has the same angular velocity as an object on the surface of the earth.
An object in a Geostationary orbit appears stationary in the sky from the ground
hence the nameTell me. If something appears stationary from the ground, then how can it be moving relative to
THE GROUND?
No Trippy, the Moon is not in a geostationary orbit around the Earth. Otherwise you would not see the Moon move from horizon to horizon on a given night. An object in geostationary orbit would remain in a fixed position relative to a location on the Earth.
Capracus
18th March 2008 - 01:23 PM
QUOTE (Trippy+Mar 17 2008, 07:17 PM)
Oh and for the record?
The idea of attaching a space elevator to the moon goes beyond madness, beyond Lunacy even.
Almost as loony as believing that the Moon is in a geostationary orbit around the Earth.
QUOTE
The difference between apogee and perigee is 42,592 km.
That means the platform end is going to move up and down by 42,592 km, so when the moon is at perigee (and moving it's fastest) that's fine, bang, over and done.
Believe it or not, I think there may be a conservation of momentum solution to this problem. If not, the tether could always be replaced by a giant bungee chord.
Capracus
18th March 2008 - 01:36 PM
QUOTE (adoucette+Mar 17 2008, 08:14 PM)
Good point.
And its actually a bit worse than that (since the moon varies in its orbit)
On Dec 12th of this year the moon will be 356,567 km away but on Dec 26 it will be a whopping 406,600 km away.
This means it would have to spool out over 50,000 km of tether (enough to go around the world TWICE, just to keep the platform stable (and spool it out at a rate of ~ 150 km/hr)
Of course, by Jan 10th the moon is back to 357,500 km away, so now you would have to reel 49,000 km of tether back in or else the platform crashes into the earth.
Arthur
Can you please keep quiet about this minor detail, I'm still having problems getting funding for the project.
adoucette
18th March 2008 - 01:45 PM
QUOTE (Capracus+Mar 18 2008, 08:08 AM)
No Trippy, the Moon is not in a geostationary orbit around the Earth. Otherwise you would not see the Moon move from horizon to horizon on a given night. An object in geostationary orbit would remain in a fixed position relative to a location on the Earth.
I'll answer for Trippy because I'm 100% positive he doesn't believe that the moon is in Geostationary orbit.
He is obviously referring to the oft discussed proposal of using a large mass that IS at Geo distance as the space anchor for the platform, and thus the tether does stay over one spot on the earth (in fact its tied to a fixed spot on the earth).
Which brings me to another problem with your moon based solution. A little checking into the moons path over the surface of the earth shows that the tether, over time, would encounter a significant number of the satellites below it.
Maybe this is another thing you should keep from your backers?
Arthur
Capracus
18th March 2008 - 01:59 PM
QUOTE (adoucette+Mar 18 2008, 01:45 PM)
I'll answer for Trippy because I'm 100% positive he doesn't believe that the moon is in Geostationary orbit.
He is obviously referring to the oft discussed proposal of using a large mass that IS at Geo distance as the space anchor for the platform, and thus the tether does stay over one spot on the earth (in fact its tied to a fixed spot on the earth).
Which brings me to another problem with your moon based solution. A little checking into the moons path over the surface of the earth shows that the tether, over time, would encounter a significant number of the satellites below it.
Maybe this is another thing you should keep from your backers?
Arthur
I think I'll just take the $5 billion I already have and leave the country.
adoucette
18th March 2008 - 02:59 PM
Good plan.
Capracus
18th March 2008 - 04:11 PM
QUOTE (Trippy+Mar 17 2008, 07:17 PM)
Oh and for the record?
The idea of attaching a space elevator to the moon goes beyond madness, beyond Lunacy even.
The difference between apogee and perigee is 42,592 km.
That means the platform end is going to move up and down by 42,592 km, so when the moon is at perigee (and moving it's fastest) that's fine, bang, over and done.
But what about at Apogee? When the platform is at an altitude that's almost high enough for Geostationary orbit?
I may not have to flee to North Korea after all.
Trippy has demonstrated a way to move payloads from the lower atmosphere to a distance 30,000 miles into space.
Once a month you could load the fixed tether platform and send it on it's way.
During it's transition within the atmosphere it could pick up and drop off cargo daily across the globe. During it's entire transition it could be used for Earth observation.
All of the original features of the system would still apply as far as moving loads to and from space and the Moon, but gladly, the cumbersome aerodynamic platform goes away.
adoucette
18th March 2008 - 04:24 PM
QUOTE (Capracus+Mar 18 2008, 11:11 AM)
I may not have to flee to North Korea after all.
That's NEVER a good plan.
Oh, and if there is any STRETCH to your tether, I think your peak altitude is likely to be quite a bit higher.
Arthur
Trippy
18th March 2008 - 05:37 PM
QUOTE (Capracus+Mar 19 2008, 02:08 AM)
No Trippy, the Moon is not in a geostationary orbit around the Earth. Otherwise you would not see the Moon move from horizon to horizon on a given night. An object in geostationary orbit would remain in a fixed position relative to a location on the Earth.
I thought even an idiot would have been able to work out that I was clearly adressing a slightly different idea (perhaps I misread the diagram the first time I looked at it).
You're the only person I've heard propose this crazy-stupid idea of attaching a sopace elevator to the moon.
Trippy
18th March 2008 - 05:42 PM
QUOTE (Capracus+Mar 19 2008, 02:23 AM)
Almost as loony as believing that the Moon is in a geostationary orbit around the Earth.
I'm sorry, but where,
precisely did I state that I believed the moon was in a geostationary orbit around the earth?
Oh wait, that's right.
I DIDN'T.QUOTE (Capracus+Mar 19 2008, 02:23 AM)
Believe it or not, I think there may be a conservation of momentum solution to this problem. If not, the tether could always be replaced by a giant bungee chord.
I don't believe it.
Trippy
18th March 2008 - 05:44 PM
QUOTE (adoucette+Mar 19 2008, 02:45 AM)
I'll answer for Trippy because I'm 100% positive he doesn't believe that the moon is in Geostationary orbit.
'S right.
I'd even go as far as saying that I know for sure that it isn't.
Capracus
18th March 2008 - 06:09 PM
QUOTE (Trippy+Mar 18 2008, 05:37 PM)
I thought even an idiot would have been able to work out that I was clearly adressing a slightly different idea (perhaps I misread the diagram the first time I looked at it).
Don't be so hard on yourself.
QUOTE
You're the only person I've heard propose this crazy-stupid idea of attaching a sopace elevator to the moon.
Well since the subject of this thread is about taking advantage of orbital energies, the orbiting Moon and it's relatively close proximity to the Earth sticks out like a sore thumb.
I really haven't been doing much more than addressing the narrow aspects of the idea, because like you, I don't really take the idea that seriously.
It is interesting though to play with the various observations you guys come up with. Thanks for the input.
Capracus
18th March 2008 - 06:41 PM
QUOTE (Trippy+Mar 18 2008, 05:42 PM)
I'm sorry, but where,
precisely did I state that I believed the moon was in a geostationary orbit around the earth?
Oh wait, that's right.
I DIDN'T.Your own words say otherwise.
QUOTE (Trippy+)
I already checked out the diagram.
It's called a Geostationary orbit for a reason.
It's because an object in this orbit has the same angular velocity as an object on the surface of the earth.
An object in a Geostationary orbit appears stationary in the sky from the ground hence the name
Well since the object in the diagram maintains the same relative position as the Moon, by inference you assume the Moon to be geostationary.
I figured something got lost in the translation. That's the reason I sketched the diagram, I was having similar problems with Arthur. I don't seriously take you for the dummy you think I am.
Trippy
18th March 2008 - 07:27 PM
QUOTE (Capracus+Mar 19 2008, 07:41 AM)
Your own words say otherwise.
QUOTE (Trippy+)
I already checked out the diagram.
It's called a Geostationary orbit for a reason.
It's because an object in this orbit has the same angular velocity as an object on the surface of the earth.
An object in a Geostationary orbit appears stationary in the sky from the ground hence the name
Well since the object in the diagram maintains the same relative position as the Moon, by inference you assume the Moon to be geostationary.
I figured something got lost in the translation. That's the reason I sketched the diagram, I was having similar problems with Arthur. I don't seriously take you for the dummy you think I am.
Don't be an idiot.
Arthur figured out what I meant.
I already stated that I might have misread the diagram.
Get over yourself.
You're being blatantly dishonest, trying to twist my words into something that I never meant, that I have explicitly stated I never meant, and in the context of my very next post makes it implicit (that I never meant it).
Oh, and for the record (And this is one of my pet peeves in this forum) describing an idea as ridiculous does not neccessarily imply anything about the poster.
Smart people have dumb ideas to.
yor_on
18th March 2008 - 08:42 PM
Although I agree with Enthalpy I can't help but ask :)
How about synthetic spider silk.
http://tc.engr.wisc.edu/Steuber/papers/2002/spider_p_1st.doc
Capracus
18th March 2008 - 09:06 PM
QUOTE (Trippy+Mar 18 2008, 07:27 PM)
Don't be an idiot.
Arthur figured out what I meant.
I already stated that I might have misread the diagram.
Get over yourself.
You're being blatantly dishonest, trying to twist my words into something that I never meant, that I have explicitly stated I never meant, and in the context of my very next post makes it implicit (that I never meant it).
Oh, and for the record (And this is one of my pet peeves in this forum) describing an idea as ridiculous does not neccessarily imply anything about the poster.
Smart people have dumb ideas to.
Down boy! Sit! Go munch a doggy downer.
Take your own advice and don't get offended when someone points out a personal error.
I don't think less of you for the criticism of the subject matter, but this wounded puppy routine is getting tired.
Trippy
18th March 2008 - 11:18 PM
QUOTE (Capracus+Mar 19 2008, 10:06 AM)
Down boy! Sit! Go munch a doggy downer.
Take your own advice and don't get offended when someone points out a personal error.
I don't think less of you for the criticism of the subject matter, but this wounded puppy routine is getting tired.
And yet you were the one making ostensibly derogatory comments about it.
Oh well, go figure.
Capracus
19th March 2008 - 12:02 AM
QUOTE (Trippy+Mar 18 2008, 11:18 PM)
And yet you were the one making ostensibly derogatory comments about it.
Oh well, go figure.
Woof!
Trippy
19th March 2008 - 04:09 AM
QUOTE (Capracus+Mar 19 2008, 01:02 PM)
Woof!
This is just ridiculous.
Real mature by the way.
So what, you're resorting to name calling because I pointed out the glaringly obvious flaw in your plan?
Or you're resorting to name calling because I pointed out that your statement was down to your poor interpretation of my posts?
Or do you just enjoy acting like a 5 year old?
Capracus
19th March 2008 - 10:52 AM
QUOTE (Trippy+Mar 19 2008, 04:09 AM)
This is just ridiculous.
Real mature by the way.
So what, you're resorting to name calling because I pointed out the glaringly obvious flaw in your plan?
Or you're resorting to name calling because I pointed out that your statement was down to your poor interpretation of my posts?
Or do you just enjoy acting like a 5 year old?
I welcome the criticism, but not this unreasonable defense of your ego that you continue to display as evidenced by your last post.
If that's who you are then so be it, I'll know what to expect in the future.
Just don't be surprised if I react with sarcasm.
Latrosicarius
19th March 2008 - 05:37 PM
I think a space elevator could happen given the right materials. It would not have to support its own weight. It would be counter-weighted in high orbit.
It could probably not have just one anchor point, but would probably have to have millions of nanotube "ropes" so to speak, fanning out from the bottom. That's because I'd imagine there will be an immense upwards pull on the tether due to centripetal force in order for it to remain taught, therefore it would have to anchor into as much earth as possible to keep it attached to the planet.
I see two other problems with it (besides the obvious):
1.) It would be like an antenna, traveling through the Earth's magnetic field, at a very high rate of speed. Last I heard, this creates all kinds of heat and electricity.
2.) I hope it would not be heavy enough to slow the earth's rotation over time, due to extending the planet's radius/weight. Remember angular momentum must be conserved.
Trippy
19th March 2008 - 05:58 PM
QUOTE (Capracus+Mar 19 2008, 11:52 PM)
I welcome the criticism, but not this unreasonable defense of your ego that you continue to display as evidenced by your last post.
If that's who you are then so be it, I'll know what to expect in the future.
Just don't be surprised if I react with sarcasm.
I think you're reading more into my posts then I'm actually putting into them.
Trippy
19th March 2008 - 06:08 PM
For the record, I wasn't (and am not) trying to defend my ego.
However, we're still waiting on this 'conservation of momentum solution' you've alluded to.
And it occurs to me that if you were to carry this idea out, you'd have to start constructing it from the Lunar L1 Lagrange point.
Capracus
19th March 2008 - 07:27 PM
QUOTE (Trippy+Mar 19 2008, 06:08 PM)
For the record, I wasn't (and am not) trying to defend my ego.
For the record, I was trying to be a smart a$$.
QUOTE
However, we're still waiting on this 'conservation of momentum solution' you've alluded to.
After you guys brought up the non circular orbit issue, the stationary tether configuration gets more complicated.
I think the monthly perigee pick up idea makes a lot more sense.
I'll give the conservation of momentum solution some thought just for curiosity sake, but I should concentrate my efforts on the real project, I've got these restless investors to deal with.
QUOTE (->
| QUOTE |
| However, we're still waiting on this 'conservation of momentum solution' you've alluded to. |
After you guys brought up the non circular orbit issue, the stationary tether configuration gets more complicated.
I think the monthly perigee pick up idea makes a lot more sense.
I'll give the conservation of momentum solution some thought just for curiosity sake, but I should concentrate my efforts on the real project, I've got these restless investors to deal with.
And it occurs to me that if you were to carry this idea out, you'd have to start constructing it from the Lunar L1 Lagrange point.
I agree that an L1 start point would make sense in that it being neutral gravitationally, the system could be built in both directions and remain gravitationally balanced.
Ideally materials would Lunar mined and processed, so in this case the project would depend on a lunar industrial capability. All this assumes you could find all of the necessary materials on the Moon. Otherwise it may have to wait for Grumpy's Beanstalk to bring materials from Earth.
Latrosicarius
19th March 2008 - 07:53 PM
Does not the L1 Lagrange point fall between the Earth and the Moon, which is NOT a geosynchronous point above the equator???
Capracus
19th March 2008 - 08:06 PM
QUOTE (Latrosicarius+Mar 19 2008, 07:53 PM)
Does not the L1 Lagrange point fall between the Earth and the Moon, which is NOT a geosynchronous point above the equator???
This Earth/Moon L1 Lagrange point is geosynchronous with the Lunar equator. This particular configuration we're discussing is anchored to the Moon.
Latrosicarius
19th March 2008 - 08:23 PM
QUOTE (Capracus+Mar 19 2008, 03:06 PM)
This Earth/Moon L1 Lagrange point is geosynchronous with the Lunar equator. This particular configuration we're discussing is anchored to the Moon.
Um, okay... but that's even worse.
The altitude from the surface of the moon required for a geosynchronous orbit would lie beyond the moon's Hill Sphere because it rotates so slow (remember, it's tidal-locked with the earth, and rotates one time for every 27 earth rotations).
There would not be enough centripetal force to keep a moon tether taught.
Capracus
19th March 2008 - 08:40 PM
QUOTE (Latrosicarius+Mar 19 2008, 08:23 PM)
Um, okay... but that's even worse.
The altitude from the surface of the moon required for a geosynchronous orbit would lie beyond the moon's Hill Sphere because it rotates so slow (remember, it's tidal-locked with the earth, and rotates one time for every 27 earth rotations).
There would not be enough centripetal force to keep a moon tether taught.
It gets even worse. The tether is anchored on the Moon and terminates in low Earth atmosphere.
See diagram.
http://www.box.net/shared/ggtdu4fc4s
Latrosicarius
19th March 2008 - 09:34 PM
QUOTE (Capracus+Mar 19 2008, 03:40 PM)
It gets even worse. The tether is anchored on the Moon and terminates in low Earth atmosphere.
See diagram.
http://www.box.net/shared/ggtdu4fc4s
Okay, well in that case, it will be taught, not because of centripetal action, but because it passes far beyond the L1 point, and the Earth's gravity will capture the counterweight.
(I can't see that page because it is blocked by my company.)
But I'd imagine you would then run into another problem. The moon's orbit is not circular around the earth.
It's pericynthion is 363,104 km and it's apocynthion is 405,696, meaning there is about a 43,000 km variation in it's orbital eccentricity.
Do you think the anchor on the moon's surface will be able to withstand such differences in gravity from the Earth?
Capracus
19th March 2008 - 09:54 PM
QUOTE (Latrosicarius+Mar 19 2008, 09:34 PM)
Okay, well in that case, it will be taught, not because of centripetal action, but because it passes far beyond the L1 point, and the Earth's gravity will capture the counterweight.
(I can't see that page because it is blocked by my company.)
But I'd imagine you would then run into another problem. The moon's orbit is not circular around the earth.
It's pericynthion is 363,104 km and it's apocynthion is 405,696, meaning there is about a 43,000 km variation in it's orbital eccentricity.
Do you think the anchor on the moon's surface will be able to withstand such differences in gravity from the Earth?
Trippy already noticed that minor detail. Plan #2 is a monthly perigee pickup.
Question, would the centrifugal force and gravity be additive?
Latrosicarius
19th March 2008 - 10:40 PM
QUOTE (Capracus+Mar 19 2008, 04:54 PM)
Trippy already noticed that minor detail. Plan #2 is a monthly perigee pickup.
Sorry, what do you mean by "pickup"?
QUOTE (Capracus+Mar 19 2008, 04:54 PM)
Question, would the centrifugal force and gravity be additive?
Yes, but it would also be detractive on the opposite side (although, for the Moon, this would not be an issue, since it's always facing the Earth).
Capracus
20th March 2008 - 12:14 AM
QUOTE (Latrosicarius+Mar 19 2008, 10:40 PM)
Sorry, what do you mean by "pickup"?
Yes, but it would also be detractive on the opposite side (although, for the Moon, this would not be an issue, since it's always facing the Earth).
The tether at perigee would dip into the earth's atmosphere to an altitude of say 40,000 ft to receive in flight payloads.
Once the Moon's orbit raised the tether out of the atmosphere, an electromagnetic acceleration system ganged with tether would accelerate the payloads to orbital velocities.
Nuclear and or solar generation from the Moon would power the system. Electromagnetic Induction from continuous transition through Earth's magnetic field could also be a source of power generation.
Trippy
20th March 2008 - 05:10 AM
QUOTE (Capracus+Mar 20 2008, 08:27 AM)
For the record, I was trying to be a smart a$$.
Sorry - guess I'm too used to dealing with sub-par humanoids who think that that sort of "attack" is a valid defense of their idea.
Consider this an apology.
Capracus
20th March 2008 - 10:41 AM
QUOTE (Trippy+Mar 20 2008, 05:10 AM)
Sorry - guess I'm too used to dealing with sub-par humanoids who think that that sort of "attack" is a valid defense of their idea.
Consider this an apology.
I appreciate your understanding and insight.
Trippy
20th March 2008 - 12:12 PM
QUOTE (Capracus+Mar 20 2008, 11:41 PM)
I appreciate your understanding and insight.
Of course, it also occurs to me that the whole structure is going to have to be insanely thick, probably at the point where it attaches to the Lunar surface.
Latrosicarius
20th March 2008 - 05:26 PM
QUOTE (Capracus+Mar 19 2008, 07:14 PM)
The tether at perigee would dip into the earth's atmosphere to an altitude of say 40,000 ft to receive in flight payloads.
Once the Moon's orbit raised the tether out of the atmosphere, an electromagnetic acceleration system ganged with tether would accelerate the payloads to orbital velocities.
Hmm... I do not mean to be pessimistic, but ...
The length of the tether you specify is approximately 110 times longer than the diameter of the moon. Extending mass that far may slow the rotation of the moon due to angular momentum.
Also factoring in that the end of the tether will be inside Earth's atmosphere and magnetic field, considerable drag (not to mention heat) will be imparted on the tether.
QUOTE (Trippy+Mar 20 2008, 07:12 AM)
Of course, it also occurs to me that the whole structure is going to have to be insanely thick, probably at the point where it attaches to the Lunar surface.
It could probably not have just one anchor point, but would probably have to have millions of nanotube "ropes" so to speak, fanning out from the bottom. That's because I'd imagine there will be an immense upwards pull on the tether (in most cases due to centripetal force, but in the case of the Moon, because it is passing the L1 Lagrange point and the Earth is pulling on the other end) in order for it to remain taught, therefore it would have to anchor into as much ground as possible to keep it attached to the surface.
Trippy
25th March 2008 - 05:20 AM
Something occured to me today, in the other thread Parsons raised the point that the tether might build up a signicant voltage as it moves through the earths magnetosphere.
But what occured to me was this: Will it?
The shuttle experiment was not conducted in a geostationary orbit, it was conducted in a low orbit, however, the tether, by definition, must co-rotate with the earths surface, and therefore, the earths magnetic field, meaning that the tether isn't moving through the earths geomagnetic field.
So the only way a current can be generated in the tether is as the earths magetic field moves up and down (or in and out, if you prefer).
Capracus
25th March 2008 - 09:55 AM
QUOTE (Trippy+Mar 25 2008, 05:20 AM)
Something occured to me today, in the other thread Parsons raised the point that the tether might build up a signicant voltage as it moves through the earths magnetosphere.
But what occured to me was this: Will it?
The shuttle experiment was not conducted in a geostationary orbit, it was conducted in a low orbit, however, the tether, by definition, must co-rotate with the earths surface, and therefore, the earths magnetic field, meaning that the tether isn't moving through the earths geomagnetic field.
So the only way a current can be generated in the tether is as the earths magetic field moves up and down (or in and out, if you prefer).
Trippy
25th March 2008 - 05:53 PM
QUOTE (Capracus+Mar 25 2008, 10:55 PM)
Electrodynamic tether
http://en.wikipedia.org/wiki/Electrodynamic_tether
I'm aware of the concept, but this doesn't answer my question.
Latrosicarius
3rd April 2008 - 01:57 PM
QUOTE (Trippy+Mar 25 2008, 12:20 AM)
Something occured to me today, in the other thread Parsons raised the point that the tether might build up a signicant voltage as it moves through the earths magnetosphere.
But what occured to me was this: Will it?
The shuttle experiment was not conducted in a geostationary orbit, it was conducted in a low orbit, however, the tether, by definition, must co-rotate with the earths surface, and therefore, the earths magnetic field, meaning that the tether isn't moving through the earths geomagnetic field.
So the only way a current can be generated in the tether is as the earths magetic field moves up and down (or in and out, if you prefer).
if you are talking about a lunar tether that dips into the earth's atmosphere, then yes.
If you're talking about a geo tether, then maybe.
The earth has 2 layers of magnetic field. The inner layer is fairly circular and oriented in a fairly fixed mode around the poles.
The outer layer is regulated by the sun. it acts like the tail of a comet and is cast out far in the opposite direction of the sun by solar interaction. So the outer magnetic field changes its orientation as the earth rotates around the sun.
At least that is the impression I get from this:
http://www.physforum.com/index.php?showtopic=20630&hl=byw, I mentioned this back on
page 5QUOTE
It would be like an antenna, traveling through the Earth's magnetic field, at a very high rate of speed. Last I heard, this creates all kinds of heat and electricity.
Matt Holck
8th April 2008 - 04:47 AM
Near Orbit Platforms?
voltic com/0/i/Harmonics/Orbit_Ring_Rail.gif
I'm not allowed to post a link to the picture as a new member troll
add a .
Latrosicarius
8th April 2008 - 08:19 PM
QUOTE (Matt Holck+Apr 7 2008, 11:47 PM)
http://voltic.com/0/i/Harmonics/Orbit_Ring_Rail.gif
I don't get it. What does the track do, and what do you mean it "moves" faster than it's orbit?
Capracus
9th April 2008 - 08:04 AM
QUOTE (Latrosicarius+Apr 8 2008, 08:19 PM)
I don't get it. What does the track do, and what do you mean it "moves" faster than it's orbit?
I believe the track configuration is meant to ensure equal opposition to the pull of gravity, allowing the system to maintain it's orbital position and to oppose the inward pull of the platforms.
The purpose of the tether seems to be to position the platforms and for transport between Earth and platforms.
For the track to remain stable, it would need some degree of rotation in order to overcome gravitational variation and precession. This would require a system to provide acceleration to the track and additional energy to power it.
Matt Holck
9th April 2008 - 02:55 PM
Thank You for your explanation
QUOTE
For the track to remain stable, it would need some degree of rotation in order to overcome gravitational variation and precession. This would require a system to provide acceleration to the track and additional energy to power it.
I hadn't thought of that
The "tidal" pull of of the moon would stretch the track distorting its shape
I think some track elasticity would provide a good deal of corrective acceleration
The elasticity could store energy when the ring stretches
oh dear feedback score -1
Latrosicarius
9th April 2008 - 07:31 PM
QUOTE (Capracus+Apr 9 2008, 03:04 AM)
... explanation ...
Okay, well why is a track necessary? Pardon my ignorance, but is this "track" a structure that has to be built above the entire circumference of the equator?
What's wrong with using counterweight and centripetal force?
QUOTE (Matt Holck+Apr 9 2008, 09:55 AM)
oh dear feedback score -1It was probably your use of large font. Obnoxious formatting pisses some people off. (Just FYI)
Capracus
10th April 2008 - 10:37 AM
QUOTE (Latrosicarius+Apr 9 2008, 07:31 PM)
Okay, well why is a track necessary?
It opposes the gravitational pull on the platforms.
QUOTE
Pardon my ignorance, but is this "track" a structure that has to be built above the entire circumference of the equator?
That would seem to be the case.
QUOTE (->
| QUOTE |
| Pardon my ignorance, but is this "track" a structure that has to be built above the entire circumference of the equator? |
That would seem to be the case.
What's wrong with using counterweight and centripetal force?
This configuration allows for geostationary position in LEO.
Matt Holck
10th April 2008 - 02:15 PM
The primary problem with the classical beanstalk is whether a materiel exists that can withstand the pull of such a long cord with a counter weight out past geo-synchronous orbit. My solution is not requiring that the orbit need not be out past geo synchronous orbit.
In low orbit, objects circle the earth at a much faster rate than the earth’s spin. This means that the platforms traveling in sync with the surface of the planet would fall to the surface at low orbit so they would have to be supported by a track. The track itself can circle the earth in low orbit at a higher speed so that it would not fall to earth. Since their would be platforms pushing down against the track, the track ring would have to be traveling faster than its stable low orbit velocity to counter the push of the platforms moving in sync with the earth’s spin at that low orbit.
cream-pye'czar-hour
10th April 2008 - 02:49 PM
You could draw a ring around the Earth encircling the globe at whatever altitude even 2 feet from the ground or 100 miles altitude. It wouldn't stay up. Gravity would lop-side the loop track to touch the Earth unless it was big enough to be in geo-sych orbit allowing centrifugal force to balance it equally at every point in its perimeter in the Earths gravitational field. Otherwise you'll be as frustrated a a kid wit a magnet and a small steel ring that cant find a way to spin it around the magnet without it at some point touching the magnet by the forces of attraction.
56 miles up is considered orbit, plane wise at a certain velocity. Now balance the ring if you can. Please Continue...........
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