Question about space elevators:

Originally posted by DeepThought
Your space elevator would have to cope with the fairly crazy orbit the moon has. Supermoons are caused when the moon is closer to the earth than normal. It's average orbital distance increases by a centimetre per year. The orbit is perturbed by the sun, and all the other planets, although obviously Jupiter is the most important. A space elevator to t ...[text shortened]... t you meant. As long as the rotation is non-zero you just need the counterweight far enough up.

That would be a trick, since the moon rotates at a speed of about 4 meters per second. I suppose it would not be impossible, although I doubt the need for it since the escape velocity of the moon is about 1/6th that of Earth, you don't need much propulsion in the first place.

Also, since there is no atmosphere to speak of on the moon, that lends itself to other technologies, like magnetic sleds on the slope of a mountain, where you would have a magnetically supported drive and a few kilometers of run up the hill, you only need a couple of km/sec to get off the moon, and in that case solar energy could supply all the energy, no chemical reactions needed at all there. It looks like a 60 km long run at 3 g's would get you up to lunar escape in about one minute! A magnetic sled would have almost zero friction so it would be very efficient in terms of energy required. Assuming 100% efficiency converting some form of energy to kinetic energy, say a magnetic sled, it takes 32 Hp to accelerate 550 pounds at one g, so 96 hp to accelerate 550 pounds at 3 g's.

So less than 1,000 hp to get a payload of 5500 pounds at 3 g's, less than 3/4 megawatt for one minute. If you had a nice capacitor bank at one hundred to one, charge the cap for 100 minutes, you only need 7.5 kilowatts to do that for one run every two hours.

Pretty cheap, eh. If we go with those figures, you could sling off 30,000 kilograms a day with that one device for the cost of less than 10 kw of power.

You could get that much power with a solar array of about 6 by 6 meters in direct sunlight on the airless surface of the moon.

Originally posted by sonhouse
That would be a trick, since the moon rotates at a speed of about 4 meters per second. I suppose it would not be impossible, although I doubt the need for it since the escape velocity of the moon is about 1/6th that of Earth, you don't need much propulsion in the first place.

Also, since there is no atmosphere to speak of on the moon, that lends itself t ...[text shortened]... with a solar array of about 6 by 6 meters in direct sunlight on the airless surface of the moon.

As a hack's argument. A month is about 2.5 million seconds (1E6 seconds is ~11.5 days) so 2*pi/T ~6.5 / 2.5 E6 ~ 2.5 E-6 1/s;
a = w^2r
=> r = a/w^2 = 1.6m/s² / (2.5 E -6 1/s)^2 ~ 2.5E12m = 2.5E9 km. So it's not going to work, as that distance is much further than the distance between the moon and the earth.

But you could have a weight closer to earth than the Lagrange point and use earth's gravity to keep the line taut. Of course, we'd be trapped between two disasters. If the line snapped the weight would plummet to earth and the line to the moon.

As a practical engineering solution (or even remotely feasible engineering solution) your rail gun idea is clearly better though. Space elevators are just not going to work. Almost any other solution to getting into orbit is better.

Originally posted by DeepThought
Edit: A second look at your post indicates this (elevator to lunar orbit) is probably what you meant. As long as the rotation is non-zero you just need the counterweight far enough up.

Sorry, my post was not clear. I was asking whether a lunar space elevator was even possible - and sonhouse says no, its not.
Then the second part of my post was a separate question as to whether a equatorial train on the earth would make a earth orbit space elevator more viable given that the tether could be shorter.

Originally posted by DeepThought
As a hack's argument. A month is about 2.5 million seconds (1E6 seconds is ~11.5 days) so 2*pi/T ~6.5 / 2.5 E6 ~ 2.5 E-6 1/s;
a = w^2r
=> r = a/w^2 = 1.6m/s² / (2.5 E -6 1/s)^2 ~ 2.5E12m = 2.5E9 km. So it's not going to work, as that distance is much further than the distance between the moon and the earth.

But you could have a weight closer to ...[text shortened]... levators are just not going to work. Almost any other solution to getting into orbit is better.

Well, one thing in the lunar elevator's favor: The tension would not be near as great as one for Earth, since the gravity is only 1/6 of Earth. That is a big gain right there. It means maybe you don't need superscience and nanotech ribbons 200 X stronger than steel, maybe get away with normal materials.

Just a guess, didn't do any calc's on that though.

There was an idea put forward in the science section of Analog Sci Fi magazine a few years ago where you use a sling like I mentioned and have a city sized satellite in orbit around Earth and the sling sends stuff from the moon to Earth and the gravitation of Earth gives you basically free energy, capturing the incoming in another kind of sled that turns the kinetic energy of the incoming to electrical energy and used to power the city and send some to Earth.

The object leaving the moon goes out at less than 2 K/sec and then picks up all that free gravitational energy going down Earth's gravity well and has to be decelerated from about 40Km/second to zero in a very short distance, a whole lot of minus g's to do that but you get 6 times the energy of the same mass of gasoline and no chemicals are burned doing it.

Seems like free energy to me.

It is not against any laws of physics, just a hairy engineering job.

They would use raw materials shipped from the moon to add to the city.

Originally posted by DeepThought
As a hack's argument. A month is about 2.5 million seconds (1E6 seconds is ~11.5 days) so 2*pi/T ~6.5 / 2.5 E6 ~ 2.5 E-6 1/s;
a = w^2r
=> r = a/w^2 = 1.6m/s² / (2.5 E -6 1/s)^2 ~ 2.5E12m = 2.5E9 km. So it's not going to work, as that distance is much further than the distance between the moon and the earth.

But you could have a weight closer to ...[text shortened]... levators are just not going to work. Almost any other solution to getting into orbit is better.

If the line snapped the weight would plummet to earth and the line to the moon.

Would it?

Conservation of angular momentum, as it falls it's angular speed increases.

The moon is ~ 384,000km away.

L1 is ~ 85% that distance. ~ 326,000km, but the counter weight would be closer,
so lets say that is 75% that distance ~ 288,000km. [lets call it 280,000km]


It is going around the earth once every ~28 days [2,419,200s]

and travelling ~1.76E9m [assuming circular orbit]

Giving an angular velocity of ~727m/s

Specific angular momentum = r*v

r1 = 280,000,000m
v1 = 727m/s

r2 = 6,400,000m
v2 = ?

r1*v1 = r2*v2

v2 = [r1*v1]/r2

v2 ~ 31,800m/s

This is greater than the orbital velocity at that altitude. Thus the counterweight would
not hit the Earth, it would go into orbit around it.


Given that the counter weight is also at most likely to be of the order of a million tons,
and likely to be deliberately constructed out of things that wont survive re-entry and could
also contain explosives to fragment it on atmospheric entry... It wouldn't pose much of a threat
anyway.

The cable similarly, it's very long but unbelievably light, and not likely to do much damage.

Originally posted by twhitehead
Sorry, my post was not clear. I was asking whether a lunar space elevator was even possible - and sonhouse says no, its not.
Then the second part of my post was a separate question as to whether a equatorial train on the earth would make a earth orbit space elevator more viable given that the tether could be shorter.

It is, but you can only build [at most] two, one reaching to Earth-Moon-L1 [EM-L1]
And the other reaching to EM-L2

If you are using a maglev train set-up you might as well just build a rail gun anyway.

EDIT: I didn't spot the bit about the train being on the Earth... I don't see that helping
unless you build the track above atmospheric height... at which point you have already
built an epic mega-structure before you even started.

If you were going to go that route you might as well just build a ~100km tall tower/ramp
and fire spacecraft out of a maglev vacuum tube at the top.

Originally posted by twhitehead
Sorry, my post was not clear. I was asking whether a lunar space elevator was even possible - and sonhouse says no, its not.
Then the second part of my post was a separate question as to whether a equatorial train on the earth would make a earth orbit space elevator more viable given that the tether could be shorter.

Don't worry, it's not easy to write clear posts and this is unlikely to be a controversial thread.

I'm not sure, I'm quite taken with the idea of using the earth's gravity to pull on a tether from the moon. The elevator can be stabilized providing the top of the lunar elevator was closer to earth than the L1 point. The thing is there are a lot simpler ways of getting between the moon and the earth. Sonhouse's plan of using a rail gun looks plausible.

I was wondering what you wanted the train for. The problem with space elevators from earth is air-resistance. If it's geostationary then wind will cause the elevator to vibrate and if it hits a resonant frequency that could be a real problem. If it's not geostationary then the wind speed is huge, to illustrate suppose we have the train going at twice the speed the earth rotates at, so relative to the earth's surface the train has to go at the speed the earth rotates at - this reduces the tether length by a factor of 4:

1 day is 3600*24 = 10,800 * 8 = 86400 seconds (note this is a solar day not a full rotation, but I can't be bothered to make the correction).
ω = 2π/T = 7.27 E -5 hertz (inverse seconds)
The radius of the earth at the equator is 6378100 metres, the speed of sound is 330 m/s
v = ωr = 6378100*7.27E-5 = 463 m/s = 1,668 km/h ~ 1,043 mph ~ Mach 1.4

So your supersonic train with a cable going from it is going to face an awful lot of drag. To make the tether smaller by any worthwhile fraction is going to introduce a massive wind resistance problems.

Originally posted by DeepThought
Don't worry, it's not easy to write clear posts and this is unlikely to be a controversial thread.

I'm not sure, I'm quite taken with the idea of using the earth's gravity to pull on a tether from the moon. The elevator can be stabilized providing the top of the lunar elevator was closer to earth than the L1 point. The thing is there are a lot simpl ...[text shortened]... her smaller by any worthwhile fraction is going to introduce a massive wind resistance problems.

The problem with space elevators from earth is air-resistance. If it's geostationary
then wind will cause the elevator to vibrate and if it hits a resonant frequency that could
be a real problem.

You don't bring the cable down to sea level anyway, you tether to a 20~25km tall tower
to get you above ~80% of the atmosphere.

And you actually induce vibrations into the cable [carefully] to bend it around passing
debris/satellites.

Originally posted by googlefudge

If the line snapped the weight would plummet to earth and the line to the moon.

Would it?

Conservation of angular momentum, as it falls it's angular speed increases.

The moon is ~ 384,000km away.

L1 is ~ 85% that distance. ~ 326,000km, but the counter weight would be closer,
so lets say that is 75% that distance ~ 288,000km. ...[text shortened]...

The cable similarly, it's very long but unbelievably light, and not likely to do much damage.

I didn't really think about that sentence when I typed it in, but you are right. Calculating the distance of the other end of the ellipse confirms it:

E = p^2/2m - GMm/r

If the radial component of the momentum is zero, which it is at the ends of the ellipse, then we can replace p with (L / r). Without loss of generality, let the mass of the weight be 1.

E = E(r) = (L^2/2 - GMr)/r^2

E(r1) = E(r2), where r1 and r2 are the distances of the endpoints from the earth.

After a little algebra:

1/r_1 + 1/r_2 = 2GM/L^2 ( = 2GMm^2/L^2 if we leave m in).

Using your figure for the specific angular momentum and some numbers from Wikipedia I got to: 63,800 km or about 10 earth radii. So there's no chance of a collision. It wouldn't even bother satellites. It'd even return to it's starting point when they're ready to reattach it (up to orbital decay, precession, etc.). So I don't think a self-destruct mechanisms really necessary.

Originally posted by googlefudge

The problem with space elevators from earth is air-resistance. If it's geostationary
then wind will cause the elevator to vibrate and if it hits a resonant frequency that could
be a real problem.

You don't bring the cable down to sea level anyway, you tether to a 20~25km tall tower
to get you above ~80% of the atmosphere.

And ...[text shortened]... ally induce vibrations into the cable [carefully] to bend it around passing
debris/satellites.

This 25km tall tower. Isn't it going to melt the rocks underneath it? There's basic limitations on how tall mountains can be and the tower can't be mountain shaped. I think there's a lot of technological assumptions in this.

Originally posted by DeepThought
This 25km tall tower. Isn't it going to melt the rocks underneath it? There's basic limitations on how tall mountains can be and the tower can't be mountain shaped. I think there's a lot of technological assumptions in this.

It's not a 25km tall pyramid, think more like the Eiffel Tower, it's mostly hollow.

The people working on this have already done the calculations for a steel and
concrete structure of that height and it's structurally fine.

The limiting factor on steel and concrete skyscrapers isn't the structural strength,
it's fitting in al the elevators.

Given that we probably wouldn't build this from steel and concrete [at least not the
upper levels] this isn't a problem from a physics/engineering standpoint.

How you actually go about building something this tall, with workers in space suites.
That's another matter.

In terms of support, given the best place to put one of these is in the oceans,
much of the weight can be supported with buoyancy chambers, with a minimum
translated to the ocean floor.

Originally posted by googlefudge
How you actually go about building something this tall, with workers in space suites.
That's another matter.

You build it from the bottom up and jack it up (or as build it in the sea and use those buoyancy chambers to lift it.)

Also not mentioned is guy wires are typically used on very tall structures other than human habitable sky scrapers.

The Warsaw radio mast had remarkably thin supports:
http://en.wikipedia.org/wiki/Warsaw_radio_mast

The vertical steel tubes forming the vertices of the mast had a diameter of 245 millimetres (10 in); the thickness of the walls of these tubes varied between 8 and 34 mm (0.31 to 1.33 in) depending on height.

also:

Its weight was debated: Polish sources claimed 420 tonnes (930,000 lb).

So, no melting rocks.

Originally posted by twhitehead
You build it from the bottom up and jack it up (or as build it in the sea and use those buoyancy chambers to lift it.)

Also not mentioned is guy wires are typically used on very tall structures other than human habitable sky scrapers.

This structure has to withstand vastly greater loadings in all directions, and would
thus need to be more substantial.
We are talking about something that is probably several miles across at the base.

Originally posted by googlefudge
This structure has to withstand vastly greater loadings in all directions,

Why?