Posted on 06/25/2004 2:21:35 PM PDT by Junior
Back from vacation UBER ping. Love this stuff.
Wow, that's pretty good! Last time I checked they were still stuck at a few microns. Excellent progress.
Show me the differential equation that says this thing works.
I use the Bonder-Farrel Equations, personally.
The Maxwell Equations are my personal choice.
Indeed they do, however, the more persistant problems for a geostationary orbit is the non sphericity of the earth:
Geosynchronous Orbit
A geosynchronous orbit is an orbit that has the same period (single revolution) that is equal to the time it takes the Earth to complete one revolution about its axis (one sidereal day). A sidereal day is measured with respect to the stars as apposed to the sun (one solar day). This is approximately 23 hours and 56 minutes. The semi-major axis for a circular orbit that has this period is approximately 42,164 kilometers and a mean altitude of approximately 35,790 kilometers above mean sea level. One of the unique features of this orbit is that as the inclination approaches zero (stays on the equator) and the orbit is circular, the object orbiting will stay over the same location on the Earth due to the fact it is moving at the save speed as the Earth is turning under it. This special type of geosynchronous orbit is called a Geostationary Orbit (stationary with respect to the surface of the Earth). As the inclination increases for a geosynchronous orbit, the ground trace of the orbit on the Earth plots a figure eight (8) pattern.
A more in depth discussion of geostationary orbits
First, from the above paragraph, you may have deduced that a geosynchronous orbit is not necessarily a geostationary orbit. However a geostationary orbit must be a geosynchronous orbit. These terms are often used interchangeably since most geosynchronous orbits are also geostationary. However, that is not always the case. It is the zero (0) degree inclination that makes it that special orbit called the geostationary orbit.
I used the term sidereal day for describing geosynchronous orbits. How do we measure a day? Usually we measure it in reference to the sun being in the same position from one day to the next (i.e. noon to noon). However, that is not the same time it takes the Earth to rotate once on its axis. Remember the Earth is also in orbit around the sun requiring it to travel just a tiny bit further in its rotation for the same spot on the Earth to be pointing towards the sun each day. This is the difference between the Mean Solar Day (our normal 24 hour day) and the Sidereal Day. The difference is approximately 4 minutes per day.
For a geosynchronous orbit, this orbit must be synched to the actual rotation period of the Earth (sidereal day). Even though a satellite is place in a near geostationary orbit upon launch there are forces that act upon the satellite that increase the orbital inclination. Remember an inclination of zero (0) for a geosynchronous orbit is also a geostationary orbit. The primary cause of this is that the equatorial plane is coincident with the ecliptic. So both the sun and the moon slowly over time increase the satellite’s orbital inclination. Also since the Earth is not a true sphere, the geosynchronous satellites drift (in-track) towards two stable equilibrium points over the Earth’s equator. This is why “station keeping” is required for geostationary satellites. Satellites are typically maintained within a band that is approximately 0.10 degrees. When station keeping is no longer possible (all the fuel is used) or there is a satellite malfunction, most geostationary satellites are boosted into a higher orbit (end of life orbit boost) so they will not drift into an area where another geostationary satellite is operating.
yngvi is a louse!
Well, I'll be damned. It never occured to me that a day wasn't a day.
At that altitude and speed, the object in orbit will remain over a single point on Earth forever.
But, the author wants to move the orbiting object out to 62,000 miles. To remain stationary over a single point on the Earth, the speed of the object would have to be increased to around 17,000 mph. At 62,000 miles, this speed is triple the speed necessary to keep it in orbit. As a result, the object wants to pull away from the Earth.
This would create a tremendous force on the "overcooked pasta", depending on the mass at the end. Certainly, when that 13 ton elevator arrives at the "top", its mass is now adding to the stress on the pasta.
I want fishing line made of this stuff, you would need binoculars to see where your bait landed.
Too Cool
Well, not entirely free. I believe the difference would come out of the rotational energy of the Earth, wouldn't it? Like when you use a planet to slingshot a space-craft. Not that we would have to worry about moving to a 25-hour day anytime real soon...
The part of the elevator out beyond geostationary orbit acts as a counterweight. (62,000 miles sounds too far, however.) The pull that it generates is needed to counterbalance the tendency of the part of the elevator below geostationary orbit to fall back upon the Earth under its own weight. The key fact of the elevator is that its center of mass is at the geostationary point.
Start with a small object in free-fall in a geostationary orbit. Now let it grow a bit longer. Before it gets very long, the tidal force grabs it and it points towards the Earth. It's still in free-fall, but it's under tension from the tidal force, and its rotation has become phase-locked with its orbital period. The object as a whole is still in free-fall, though.
Now let it continue to grow until it one end of it reaches almost to the ground. It's under tension, but it doesn't fall to the ground and it doesn't fly off into space, because it's still in free-fall, as a whole.
This would create a tremendous force on the "overcooked pasta", depending on the mass at the end.
But that's mostly a tidal force. There is some additional tension caused by the rotational angular momentum of the object, which also has a period of one cycle per sidereal day. That's probably enough to forget pasta as a building material, but all the rest of the angular momentum has dropped out of the equation: we've already accounted for the orbital angular momentum when we stipulated that the object as a whole is in freefall.
Certainly, when that 13 ton elevator arrives at the "top", its mass is now adding to the stress on the pasta.
Yes, the stresses change in response to the position of the carriage. I imagine, though, that the mass of the "stalk" is huge compared to the mass of the carriage plus payload, so the position of the center of mass doesn't change very much. Moreover, you could operate two carriages in opposite directions to preserve the balance.
lowly, earth-bound placemarker
The "perturbations" that can be solved rather easily reminds me of the "perturbations" that we realized all along that would be involved with fusion energy machines, particularly tokomaks. Those "perturbations" rather quickly come to dominate all other considerations when you get to technologies that are this far from our walk-around experiences, and the fusion world has not yet managed to succeed. We have been able here in the particle accelerator world to overcome some similar problems, and optical astronomers have been similarly successful with new active mirror designs, but it has required HUGE amounts of both human, intellectual, and monetary capital to overcome those perturbations.
Your way certainly makes more sense, and two carriages would cancel out the "vertical" effects.
But it's the horizontal effect that presents the problem. Yes, angular speed is constant at 15 degrees/hr. (360 degrees/24 hrs). But horizontal velocity has to increase from 1000 mph at Earth's surface to 6,880 mph at geostationary orbit (22,236 miles). The rising carriage would place a big "horizontal" force on the ribbon which cannot be canceled out by another carriage. As the other carriage is descending (and slowing horizontally), the ribbon will take on the shape of a huge "S".
Not true. See my post #247
Well, that depends what you're using it for. If you want to get into orbit, the geostationary point is your destination. The outer part of the stalk could be used, though, for slinging something out into deep space. Once you get past the geostationary point, it doesn't cost any energy to slide outwards. When you get to the desired point on the stalk, you just let go, and off you fly to Jupiter. Certainly the farther out you go, the smaller the weight limit for the payload, because of the stress on the stalk.
The rising carriage would place a big "horizontal" force on the ribbon which cannot be canceled out by another carriage.
True, but the horizontal force depends on how fast you move the carriage. And it could be compensated, in part, by having several carriages going up and several going down at all times. The angular momentum shed by a descending carriage would, on average, be picked up by an ascending carriage.
As the other carriage is descending (and slowing horizontally), the ribbon will take on the shape of a huge "S".
I've seen pictures where space elevators are depicted as having a shape like a car antenna in a stiff breeze, bowing in the spinward direction and then arching backwards. I don't know whether that's right, though. It would be interesting to see the correct calculation.
True, only if the ribbon were stiff. A flexible ribbon will form the letter "S" -- descending carriage pushing the ribbon one way, and the ascending carriage pushing the the ribbon the other.
As to how fast? Even going upwards at 1000 mph will take 24 hours to arrive at orbit. Sub-sonic? Two days. 100 mph? 10 days.
I just noticed that even forces that locally cancel out can still have a drift that grows with the square root of the time. Lots of small impulses impart Brownian motion to things. What I was wondering about is how beanstalk handles drifts in the Earth's axis of rotation. The equator isn't exactly constant otherwise UT0 and UT1 would be the same. Perhaps these are small enough to be corrected by gyroscopes.
What about wind? Few places are really quiet. This would seem to be a problem for the root of the beanstalk.
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