Space Elevators Do Too Stay Put
Apr. 23rd, 2007 11:23 pmIn honor of Pixel-Stained Techno-Peasant Wretch Day, I'm showing some engineering work I did for free. This isn't anything I'd actually get paid for, at least not in this decade (or possibly century) but it does show how an engineer can analyze a practical problem.
Back in July,
pauldrye wrote a Pyramid article about an alternate history setting for GURPS Infinite Worlds. One colorful detail he included was a space elevator anchored at the equatorial island of Batam.
Fanboys being what they are, someone immediately took offense at this skyhook not being precisely on the equator and claimed it would wind up oscillating north and south. I pointed out the flaws in his logic and it got heated. This is when I wound up flaming someone for dissing centrifugal force. Eventually it was clear I'd have to run the numbers to end the back and forth.
The first step was how to model the elevator. For this discussion the only question was whether the elevator would hold a constant position relative to the equator or oscillate north-south. That lets us use a simple model: a single point mass attached by a weight-less string to Batam Island. If I had a finite-element-model simulation handy, I might use the more complex model of a series of point masses connected to each other with strings, but that would be a lot more work to do by equations, and require a bunch of assumptions about how the mass is distributed over the length of the skyhook.
The model only requires two dimensions, with the z-axis along the North Pole and x-axis extending from the center of the Earth through the equator at Batam's longitude. The elevator's counterweight weight isn't detailed in the article, so we'll assume it's far enough out to put the elevator's center of gravity at 70,000 km from the center of the Earth (r). That's where we put the point mass. The weightless string attaches it to Batam Island at 1.15 degrees North latitude (Rz, Rx).

The mass has three forces acting on it: tension in the string (T), Earth's gravity (G), and centrifugal force (C). Since all three have a constant value they'll come to an equilibrium:
T + G + C = 0
It's easier to solve the math if we break out the equations for each axis, each force having an x and z component:
C = Tx + Gx
Tz = Gz
Note that centrifugal force is entirely normal to the Earth's axis and has no z component. Its magnitude is C = w r2, or 37 cm/s2. I'm neglecting the mass of the elevator here since it doesn't matter in the force balance. G is a function of the radius from the center of the Earth. At 70,000 km it's a modest 8.1 cm/s2.
Breaking G into its x and z components requires the value of a:
Gx = G cos a
Gz = G sin a
Since we don't know what a is, we'll have to iterate on it until we find a value that lets all the equations agree. We're still short a few more equations:
Tx = T cos e
Tz = T sin e
b = r tan a
e = atan( (Rz - b) / (r - Rx) ) (point for anyone who spots the simplifying assumption there)
That's enough to solve the problem. But to make the make a little easier we'll combine some of them:
T = (C - Gx) / cos e
Now we can plug in a value of a and see if Gz = Tz. If not, we try a new value. This can be tedious, but Excel's Goal Seek function doesn't mind. The results:
a = 0.083 degrees
b = 101.9 km
That puts the elevator's CG 26 km closer to the equator than the anchor point (Batam Island) but effectively over the equator, since that's a very small amount of latitude at that altitude. As C, G, and T are all constant forces the elevator will hold that position.
So, no, a space elevator anchored off the equator will not oscillate north-south. A satellite in geosynchronous orbit which is inclined to the equator will BECAUSE IT'S NOT FRIGGIN' ANCHORED! Sorry. Did I mention this was originally done for a flamewar?
Back in July,
Fanboys being what they are, someone immediately took offense at this skyhook not being precisely on the equator and claimed it would wind up oscillating north and south. I pointed out the flaws in his logic and it got heated. This is when I wound up flaming someone for dissing centrifugal force. Eventually it was clear I'd have to run the numbers to end the back and forth.
The first step was how to model the elevator. For this discussion the only question was whether the elevator would hold a constant position relative to the equator or oscillate north-south. That lets us use a simple model: a single point mass attached by a weight-less string to Batam Island. If I had a finite-element-model simulation handy, I might use the more complex model of a series of point masses connected to each other with strings, but that would be a lot more work to do by equations, and require a bunch of assumptions about how the mass is distributed over the length of the skyhook.
The model only requires two dimensions, with the z-axis along the North Pole and x-axis extending from the center of the Earth through the equator at Batam's longitude. The elevator's counterweight weight isn't detailed in the article, so we'll assume it's far enough out to put the elevator's center of gravity at 70,000 km from the center of the Earth (r). That's where we put the point mass. The weightless string attaches it to Batam Island at 1.15 degrees North latitude (Rz, Rx).
The mass has three forces acting on it: tension in the string (T), Earth's gravity (G), and centrifugal force (C). Since all three have a constant value they'll come to an equilibrium:
T + G + C = 0
It's easier to solve the math if we break out the equations for each axis, each force having an x and z component:
C = Tx + Gx
Tz = Gz
Note that centrifugal force is entirely normal to the Earth's axis and has no z component. Its magnitude is C = w r2, or 37 cm/s2. I'm neglecting the mass of the elevator here since it doesn't matter in the force balance. G is a function of the radius from the center of the Earth. At 70,000 km it's a modest 8.1 cm/s2.
Breaking G into its x and z components requires the value of a:
Gx = G cos a
Gz = G sin a
Since we don't know what a is, we'll have to iterate on it until we find a value that lets all the equations agree. We're still short a few more equations:
Tx = T cos e
Tz = T sin e
b = r tan a
e = atan( (Rz - b) / (r - Rx) ) (point for anyone who spots the simplifying assumption there)
That's enough to solve the problem. But to make the make a little easier we'll combine some of them:
T = (C - Gx) / cos e
Now we can plug in a value of a and see if Gz = Tz. If not, we try a new value. This can be tedious, but Excel's Goal Seek function doesn't mind. The results:
a = 0.083 degrees
b = 101.9 km
That puts the elevator's CG 26 km closer to the equator than the anchor point (Batam Island) but effectively over the equator, since that's a very small amount of latitude at that altitude. As C, G, and T are all constant forces the elevator will hold that position.
So, no, a space elevator anchored off the equator will not oscillate north-south. A satellite in geosynchronous orbit which is inclined to the equator will BECAUSE IT'S NOT FRIGGIN' ANCHORED! Sorry. Did I mention this was originally done for a flamewar?
no subject
Date: 2007-04-24 04:59 am (UTC)Since the tether material is so strong and light, however, it affords us another opportunity — the chance to build not only an Earth-to-GEO transportation system, but an anywhere-to-anywhere transportation system. Using the tether material and the sky station, we can suspend bridhes from continent to continent. After all, a material that can support 148,000 km length of its own weight isn't going to have any trouble holding up a 12,000 mile length of bridge material — say, a railway from LA to Tokyo. Once the cities of the world are connected by these suspension bridges, and the bridges in turn to the sky station, it will become possible to travel from anywhere on Earth to anywhere in space by rail.
That's right — rail. In my opinion, it's a mistake to think of the orbital tether system as a "space elevator". People hate elevators — they're cramped, claustrophobic, boring, and smelly. But trains — well, who doesn't like trains? Instead of a space elevator, what we want to build is a space railway, with all the romance that implies — and all the comforts of luxury rail travel.
A space railway system would do for the New Frontier what the Transcontinental Railroad did for the old: it will enable the settlement of ordinary people in a new land. Traveling from Lima or Lisbon or Little Rock, settlers will catch their trains — first to SkyStation, then to Port Terra at GSO. Once there, they will board ships that will take them out into space, and to the new worlds that await.
And someday, when we put the wormhole gate at Port Terra, they can take the train all the way!
no subject
Date: 2007-04-24 05:49 pm (UTC)It's a mistake - maybe - to think of it as a space elevator but we're pretty stuck with the name. SE it is and SE it will always be.
Mind you - 'rails to the stars' is a keen metaphor.
no subject
Date: 2007-04-25 12:35 am (UTC)The distance from LA to Tokyo is about 4,298 miles. The fastest train in current regular service has a top speed of around 356 mph. Allowing time for speed-up, slow-down, and switching, let's say our hypothetical intercontinental train has an average speed of 300 mph. That puts the total trip time from Union Station in LA to Tokyo Station in Marunouchi at fourteen hours, twenty minutes. And your luggage stays with you all the way.
Nonstop on American Airlines, LAX to Tokyo Narita: eleven hours, thirty minutes.
Plus three hours for check-in at LAX.
Plus the trip to and from each airport (Narita is 60km from central Tokyo — an hour by express train.)
And who knows if your luggage will arrive with you — or in what condition?
And so we see that our imaginary LA-Tokyo train is actually a better choice than an airplane. It takes you from downtown to downtown with no airline-style security humiliation and no chance of losing your bags. And a train can offer amenities no airline can match: on a train, one can get up, walk around, eat a gourmet meal off a real china plate accompanied by a fine wine, enjoy drinks in the club car, watch a movie, play cards, sleep in a real bed with sheets, make love, and even enjoy a smoke afterwards. On a plane, all one can do is sit. And sit. And sit.
(And all this is assuming we use conventional rail technolgy on our intercontinental train. In reality, we'd probably use a maglev system, which would allow our train to travel at 600 mph or better — roughly as fast as a jetliner — further increasing its speed advantage.)
It's settled. Next time, I'm taking the train!
no subject
Date: 2007-04-24 11:22 pm (UTC)http://www.sjgames.com/gurps/books/allstarjam2004/
no subject
Date: 2007-04-24 04:20 pm (UTC)(finds a piece of string)
(ties it to a pencil)
(swings it gently around his waist, 45-degree angle to the vertical)
...well, gravity's interfering big time with my experiment, but I can still see how it would work.
no subject
Date: 2007-04-24 04:32 pm (UTC)But I think we shouldn't make them until we're damn sure the thing won't break and whip around the planet, shattering the crust. It's one of those advanced toys that we shouldn't play with until we're proven we can be responsible with lesser toys (http://news.google.com/news?hl=en&ned=&q=crash&btnG=Search+News).
no subject
Date: 2007-04-24 05:31 pm (UTC)no subject
Date: 2007-04-24 05:45 pm (UTC)If they could LiftPort would not build one and y'all would be right to not let us.
Just curious
Date: 2007-04-24 10:30 pm (UTC)2) What about precession?
Re: Just curious
Date: 2007-04-24 11:20 pm (UTC)Needed for space elevator tether: ~65–120 GPa (http://en.wikipedia.org/wiki/Space_elevator)
What about precession? 25,800 years is a lot of time to adjust to changes.
Re: Just curious
Date: 2007-04-27 12:09 am (UTC)