Elastic Launch Loop (Space Elevator)

Terraformer · 2013-12-03 03:52:50

You mean, you can build a ring of silicon carbide 20,000km up?

Tom Kalbfus · 2013-12-03 06:39:23

My point is you don't need the whole ring to be nonorbiting, only the outside of the ring, the part one might hang things from, put a track on the outside and you can accelerate things to orbital velocity. On the other hand why not simply have an orbiting ring? The whole ring is in a circular orbit, on the inside (the side facing the Earth) is a maglev track, speeding along the maglev track is a car running counter to the orbital velocity of the ring and matching the velocity of the Earth's surface, and from that, you simply lower a cable 200 km to the Earth's surface. The track is so designed to support the weight of the car and cable under the Earth's gravity. Once the payload is at the top of the cable, it flips over to the top of the maglev track (the side facing away from the Earth) where another maglev car awaits, the maglev car then decelerates relative to the track and accelerates to orbital velocity or higher and the payload is released. No super materials are required in this design, you perhaps need superconductors, but nothing superstrong.

Last edited by Tom Kalbfus (2013-12-03 06:40:19)

JoshNH4H · 2013-12-03 10:40:37

Terraformer, well... yes

Tom: The issues with that kind of structure are cost-related, insofar as you still have to build a ring around the earth. It's physically possible, though, which is nice.

Tom Kalbfus · 2013-12-03 11:35:40

It could be build with extraterrestrial materials and robots, which could be by the way teleoperated from Earth at this close distance. Lets make a conservative estimate of the mass of this thing, lets use iron, which has a density of 7.96 tons per cubic meter, lets assume the track is 1 meter thick and 4 meters wide. The Earth is about 6,400 km in radius, add another 200 km and we have a ring that is 6,600 km in radius. The formular for circumference is 2*Pi*radius which is 41,469.023 km. convert this to 41,469,023 and multiply this by 4 times 1 times 7.96 tons and we get 5,281,494,769.28 tons, approximately 5.3 billion metric tons! Phobos by contrast has 11 trillion metric tons with a diameter of 22 km. The Asteroid Eros is orbiting between Earth and Mars, it has a mass of 7 trillion metric tons, if we could move Eros into Earth orbit, we'd have over 1000 times as much material as we would need to build this track!

JoshNH4H · 2013-12-03 12:12:41

Yes, but with a structure that large you have all sorts of nasty effects with resonance and such. Those are annoying and very difficult to deal with. What you'll probably need is a truss structure with a very large effective diameter, even if it's mostly empty space. This will still cost mass, though.

Tom Kalbfus · 2013-12-03 12:23:24

If we change its dimensions to 10 meters thick and 40 meters wide as a truss structure while keeping the mass the same, and some sort of folding effect to counteract resonance over distance Average density would then be about 79.6 kg per meter cubed. with multiple tensoned cables threaded throughout, It would be a sideways space elevator, it would spin at slightly faster than orbital velocity in order to lift up the weight of the maglev car and the cable hanging down from it.

Tom Kalbfus · 2013-12-03 13:04:23

By the way a similar device can be built out of Phobos in Mars orbit, a cable extending down about 6000 km would reach that planet's surface. Venus could be given a moon similar to phobos by capturing an asteroid, there are a few that get close to Venus orbit for this. A cable could be lowered down towards Venus' surface, it at least its upper atmosphere, and a separate vehicle from that could reach its surface, the planet's lack or rotation is not a factor as it would be with a space elevator.

Tom Kalbfus · 2013-12-03 16:31:41

Another use for this would be interplanetary transfer. Lets say on the outer track we accelerated the payload to the point where it experiences 1 g of outward force at 200 km altitude (6,600 km radius), and then released it so that it is on a tangential intercept course to a similar ring around Mars. How long would this interplanetary trip be, that is when the two planets are closest and when they are farthest.

The first step would be to figure out what the tangential velocity would be when released from the ring.

Second step is to figure out the departure velocity from Earth after subtracting the local escape velocity from this altitude

third step is to figure out the approach velocity as the object approaches Mars.

Assume there is a similar ring at the altitude of Phobos orbit

forth step is to figure out the objects velocity as it falls toward Mars and tangentially approaches the Mars ring

fifth step: A maglev car on the Mars ring then accelerates to match the tangential velocity of the infalling object and grabs it

The sixth step: The maglev car then slows down to match the velocity of the Martian surface, and the payload is then lowered down a cable to the Martian surface.

Terraformer · 2013-12-03 16:48:35

Well, 8km/s *is* quite fast, but you still have to decelerate, and high velocity rendezvous is... tricky. Of course, with that sort of capability, you can throw a fully tanked spacecraft, which gives you plenty of radiation shielding as well.

Tom Kalbfus · 2013-12-09 12:03:15

All velocity is relative, its the difference in velocity that is important. You have a spaceship moving at 8 km per second approaching a maglev car moving down the track in space at 8 km per second, relative to each other, the difference in velocity is not very much at all. All the maglev car has to do is stay on the track while moving at 8 km as the incoming spaceship docks with it. The maglev car, with its new payload then magnetically breaks, and then slows down to match the velocity of the MArtian surface, then the payload/spaceship transfers to an elevator which then lowers it to the surface of Mars. The Solar System could eventually have maglev tracks ringing each of the 8 planets of the Solar System for quick and easy interplanetary travel.

Terraformer · 2013-12-09 12:21:32

You're still trying to hit a small spot on a structure that is, on the whole, moving at 8km/s relative to you.

Tom Kalbfus · 2013-12-10 01:29:17

No different than the Space Shuttle docking with the Space Station. Both are moving very fast relative to the Earth, but relative to each other the approach is slow. In the case of approaching Mars, the spaceship docks with a car that is already moving on the track. The speed relative to the track may be 8 km/sec, but not relative to the car that's moving in the same direction as the space ship on that track. The spaceship can also make course corrections, it would be much the same as an airplane landing on an aircraft carrier in principle.

Spaniard · 2013-12-11 14:33:23

For interplanetary, we could use "space jumpers". Electromagnetic cannons, enough large to avoid extreme accelerations.
The cannons would be moved backward, but because it would be more massive, the new orbit could be enough high to avoid reenter on the atmosphere. Then, using a slow but efficient propulsion (solar - ion, magnetic sails...) could return to the original orbit for reuse.
The spaceship could use aerobraking on destination, and ion propulsion for correction after the aerobraking and make a stable low orbit.

Spaniard · 2013-12-11 15:54:02

Tom Kalbfus wrote:

The whole ring is in a circular orbit, on the inside (the side facing the Earth) is a maglev track, speeding along the maglev track is a car running counter to the orbital velocity of the ring and matching the velocity of the Earth's surface, and from that, you simply lower a cable 200 km to the Earth's surface...

This structure has an advantage. Because the speed of the orbit could be different from the relative speed of the ground, it is possible to use different planes and reduce the distance to the minimum.
For example, it is viable to create a orbital ring on the moon, and a small, faster cable to move between Moon and the ring compared to a normal space elevator.
We could use a ring in polar orbit instead a perpendicular elevator to the planet/moon axis too.

The ring could be used as a electromagnetic cannon for cargo once in orbit.

Of course, in terms of material, any ring around a big body needs a lot of mass. A future tech not for today.

But interesting ideas, as materials nowadays are strong enough to make a cable. Only plenty ones (iron, oxygen, etc.) are needed for the ring. With self-replicating tech and space mining, projects like this could be viable in this century.

And we don't need even reach the surface (to avoid high winds and excesive tensions). A platform reachable with hydrgen balooms in the sky could be enough.
Perhaps not enough useful on Earth but Venus.

Last edited by Spaniard (2013-12-11 15:54:35)

Tom Kalbfus · 2013-12-11 17:43:29

Suppose we built a ring around the Sun (at 1 AU) and used it as a circular mass driver? If we aligned this ring such that a tangential velocity towards Alpha Centauri were possible. How fast could we accelerate a spaceship along this ring before centrifugal force became too great for the astronauts to withstand? The ring is in orbit of course, but what's spinning around in the ring could be moving faster. It could of course be solar powered. If not confined to accelerating cargo with passengers, it could accelerate pellets to much higher velocity.

1211 km/s gives 1 g
2423 km/s gives 4 g
5000 km/s gives 17 g
10,000 km/s gives 68 g
50,000 km/s gives 1,704 g
100,000 km/s gives 6,816 g
200,000 km/s gives 27,265 g

A stream of pellets could be fired at Alpha Centauri, and a spaceship could intercept those pellets, each pellet would tranfer it momentum to the spaceship accelerating it faster.

If a pellet hits the starship at an initial relative velocity of 1000 m/s, causing the starship to go from 0 m/s to 1 m/sec, traveling 0.5 m after 1 second.
The next pellet hits the starship at 1001 m/sec and so on.

JoshNH4H · 2013-12-11 20:58:15

Have you calculated the mass and tensile strength requirements to build this massive loop?

Terraformer · 2013-12-12 05:43:31

Hold on there. Iron is just about good enough to build a Halo, if you use up the asteroid belt, and that spins at 7km/s to provide 1g on the surface. You're talking about building a full on ringworld, except subjecting it to even more force...

Of course, if you locate it on a planet, you could use the planets gravity to hold it together I guess.

JoshNH4H · 2013-12-12 10:05:05

Here's an idea... build the damn thing a couple hundred thousand km above the surface of the Sun. It'll work out just fine if you've perfectly reflective materials.

Tom Kalbfus · 2013-12-12 10:12:48

JoshNH4H wrote:

Have you calculated the mass and tensile strength requirements to build this massive loop?

Depends on the mass of it compared to the mass of the object being accelerated. We all know that something like a ringworld is impossible using materials we know about, but that's only because an entire ringworld spinning around a sunlike star would be spinning at 1211 km/s to produce 1g at 1 AU's distance. (The actual gravity of the particular star matters very little at this distance.) But in my example the ring is not spinning at 1211 km/s, it is simply a maglev track in orbit around the Sun, you could accelerate a car along this track, and so long as the outward force is less than the combined tensile strength of the orbiting track, it would hold itself together as the mass of the track itself under centrifugal force doesn't come into play, only the mass of the car levitated off the track. If the payload is not made out of flesh and blood, one can accelerate it up to much higher velocities. Lets say we could accelerate a pellet at 27,265 g, that is 272,650 meters per second squared, we can represent this as 272.7 km/sec^2, that means after 1 second of such acceleration the pellet would be traveling at 272.7 km/sec and would have traveled 136.35 km. After 10 seconds that same pellet would have reached 2,727 km per second, and would have traveled a distance of 13,635 km. After 100 seconds, the pellet would have reached 27,270 km per second and would have traveled a distance of 1,363,500 km. After 733.4 seconds under such acceleration the pellet would be traveling at 200,000 km per second and would have traveled a distance of 73,340,667.4 km. Seems to me that it would be easier to make such an accelerator in the shape of a ring around the Sun, which at 150,000,000 km would be 942,477,796 km in circumference, and it could collect 1450 watts per square meter from the Sun along its entire length Of course if we built in it Earth's Solar System, we should consider the Earth in its orbit, so we could build it inside Earth's orbit for greater solar concentration. The Earth at its closest to the Sun is 147 million km, 0.95 au is 142,500,000 km from the Sun, the circumference is 895,353,906.3 km, this distance gets 1606 watts per square meter, you could fit the acceleration distance within this circumference 12.2 times., and I think it would be easier to build it along the length of its orbit around the Sun, than at a radial orentation towards the Sun. If it was pointing at Alpha Centauri, once it moves in its orbit, it would no longer be, a circular accelerator track would always have part of it oriented towards Alpha Centauri with access to 1606 watts of sunlight per square meter, make a solar collecting array large enough and it should be enough to accelerate the stream of pellets,

JoshNH4H · 2013-12-12 12:05:37

You will at some point (above 250,000 km/s or so, I would guess) want to take into account relativistic effects in your equation for spin gravity. I have no idea how to do this, though.

Good point about the maglev track, though. You're basically thinking of using the same physics as involved in a Space Fountain, right?

It's very important to the design that you're suggesting that gravity (or at least some kind of counter-force) should be present, so that the "centrifugal" force of the rotating ring is cancelled out by the inward force of gravity.

Because the centripetal acceleration falls off proportionally to the radius, the amount of material required increases proportionally to the radius, and gravity falls off in proportion to the square of the radius, the amount of material required as a counterweight increases with the square of the radius, meaning that you need less material if you build closer to the Sun. Another benefit of building closer to the Sun is that you can actually use light pressure to counter the rotational inertia added to the loop by actually launching something.

The equatorial radius of the Sun is 700,000 km. If we build at twice that, 1,400,000 km, the solar insolation will be about 15.7 MW/m^2, and the gravity will be 68.5 m/s^2 (7 g). Avoiding solar flares also becomes an issue, but with good magnetic shielding it should be possible.

Tom Kalbfus · 2013-12-12 12:10:38

From information here
http://bado-shanai.net/astrogation/astrogalpcen.htm
we get the eclipic coordinates of Alpha Centauri of Ecliptic Latitude; -42.587 deg -0.1862t deg. and Ecliptic Longitude; 239.488 deg +0.8803t deg.
This means the circular track must tilt -42.587 degrees with respect to Earth's orbit, An asteroid stationed along the circular track will provide the material for the pellets, and each pellet would be released once it is pointing at these ecliptic coordinates.

Lets say the starship is poised in front of this stream of pellets, the first pellet is released at a velocity of 1 km per second, it travels 1 km in one second and hits a pusher plate on the starship accelerating the starship by 1 meter per second in one second, thus traveling 0.5 meters in that second, the next pellet travels at 1.001 km per second it travels 1.0005 km and then hits the starship 0.9995 seconds later, the starship accelerates by another meter per second and is now traveling at 2 meters per second, and has traveled a distance in 2 seconds of 2 meters.

Another pellet is released the next second, it is traveling at 1.002 km per second, it travels 1.002 km to hit the starship 1 second later and the starship accelerates by another 1 meter per second to 3 meters per second on the third second it has traveled 4.5 meters.

Another pellet is released the very next second, it is traveling at 1.003 meters per second, it travels 1.0045 km to hit the starship 1.0015 seconds later, the starship accelerates by 1 meter per second to 4 meters per second, which in 4 seconds time means it has traveled 8 meters, the next pellet will be traveling at 1 meter per second faster than the previous one,it will have to travel 1.008 meters to reach the starship, which means it will take 1.0030 seconds to cross that distance and hit the starship accelerating it by another 1 meter per second which by 5 seconds it will have traveled 12.5 meters from its original start position.

Another pellet is released the next second at 1.006 km per second, it crosses the 1.0125 km in 1.0065 seconds the starship accelerates another 1 meter per second to 6 meters per second

Pellet velocity: distance traveled: pellet travel time: new starship velocity: new starship position: pellet arrival time.
1.006 km/s: 1.018 km: 1.0112 seconds: 7 meters per second: 24.5 meters
1.007 km/s: 1.0245 km: 1.0174 seconds: 8 meters per second: 32 meters
1.008 km/s: 1.032 km: 1.0238 seconds: 9 meters per second: 40.5 meters
1.009 km/s: 1.0405 km: 1.0312 seconds: 10 meters per second: 50 meters
... 100 seconds into trip. (1 min, 40 sec)
1.100 km/s: 6 km: 5.45 seconds: 101 meters per second: 5.1005 km
... 1000 seconds into trip. (16 min. 40 sec)
2 km/s: 501 km: 250.5 seconds: 1001 meters per second: 501.0005 km
2.001 km/s: 502.0005 km: 250.87 seconds: 1002 meters per second: 502.002 km
(pellet arrives 1.37 seconds after previous one)
... 10,000 seconds into trip (2 hr. 46 min. 40 sec)
11 km/s: 50,001 km: 4,545.55 seconds: (75 minutes. 46 seconds) 10,001 meters per second: 50,010.0005 km
11.001 km/s: 50,011.0005 km: 4,546.04 seconds: (75 minutes. 46 seconds) 10,002 meters per second: 50,020.002 km
(pellet arrives 1.49 seconds after the previous one)
... 100,000 seconds into trip (1 day. 3 hrs. 46 min. 40 seconds)
101 km/s: 5,000,001 km: 49,504.96 seconds: (13 hours. 45 minutes. 4 seconds) 100.001 km per second: 5,000,100.0005 km
101.001 km/s: 5,000,101.0005 km: 49,505.46 seconds (13 hours. 45 minutes. 5 seconds) 100.002 km per second: 5,000,200.002 km
(pellet arrives 1.5 seconds after previous one)
... 10,000,000 seconds into trip (115 days. 17 hours. 46 minutes. 40 seconds)

That's all I have time for, but as you can see the lag time of each pellet after the previous one gets longer the more time elapses, probably the pellets will have to get successively more massive to transfer more kinetic energy as arrival times between pellets becomes increasingly spaced out. This example provides an acceleration of 1 meter per second squared to the starship or about 0.1 g acceleration, each time the pellet hits the pusher plate of the starship, it hits at a relative velocity of 1 km/sec.

Tom Kalbfus · 2013-12-12 12:24:19

JoshNH4H wrote:

You will at some point (above 250,000 km/s or so, I would guess) want to take into account relativistic effects in your equation for spin gravity. I have no idea how to do this, though.
Good point about the maglev track, though. You're basically thinking of using the same physics as involved in a Space Fountain, right?
It's very important to the design that you're suggesting that gravity (or at least some kind of counter-force) should be present, so that the "centrifugal" force of the rotating ring is cancelled out by the inward force of gravity.
Because the centripetal acceleration falls off proportionally to the radius, the amount of material required increases proportionally to the radius, and gravity falls off in proportion to the square of the radius, the amount of material required as a counterweight increases with the square of the radius, meaning that you need less material if you build closer to the Sun. Another benefit of building closer to the Sun is that you can actually use light pressure to counter the rotational inertia added to the loop by actually launching something.
The equatorial radius of the Sun is 700,000 km. If we build at twice that, 1,400,000 km, the solar insolation will be about 15.7 MW/m^2, and the gravity will be 68.5 m/s^2 (7 g). Avoiding solar flares also becomes an issue, but with good magnetic shielding it should be possible.

Yep I realize that, that's why I stopped at 200,000 km per second, I think this is fast enough for a stream of pellets, 2/3rds light speed is fast enough to get some reasonable travel times to Alpha Centauri. Slowing down might depend on using a magnetic sail to brake against the stellar winds of the Alpha Centauri stars, this is only good for slowing down though, as the stellar winds will be traveling over 200,000 km/sec relative to the approaching starship, but once the ship brakes the outgoing stellar winds will be much slower than that, so it won't be much good for returning to Earth.

JoshNH4H · 2013-12-12 13:31:11

I've always wondered about aerobraking techniques at such high speeds. There's nothing fundamentally impossible about them, but of course in the real world they end up being extremely difficult.

I also wonder about things like plasma-based lightsails. If you can get your sail to be opaque to just 1% of the light that hits it you're doing pretty well in terms of force.

For example, if a 10 tonne spacecraft can produce a 1% opaque lightsail that has a diameter of 1000 km at 1 AU, it will generate a force of 36 kN. Because plasmasails generate a constant force at all distances from the Sun, this will be a constant force. If you keep your plasma sail on for fifteen billion kilometers (From the Sun to Eris), you'll have a total energy in your spacecraft of 5.4e17 J. This corresponds to a velocity of 5,000,000 m/s, nothing to sneeze at. There will be a corresponding breaking effect at the target star.

Tom Kalbfus · 2013-12-12 13:54:40

You can never accelerate to faster than the speed of the stellar winds though. If you were to start at 1 au in orbit around the Sun, the Stellar winds are traveling less that 200,000 km/s. An artificially generated stream or pellets can bring you up to almost the velocity of the fastest pellet. Then your starship is moving at 200,000 km/s. If the outward velocity of the centari stellar winds is 1000 km/s for example (just throwing out numbers here) then the incoming starship will meet stellar winds that are moving 201,000 km/s relative to it, when it slows down by 5000 km/sec, those incoming stellar winds would then be moving at 196,000 km/s relative to the starship, once it enters the Alpha Centauri system and is moving at planetary speeds, the outgoing stellar winds will only be 1000 km/sec, if it uses those winds to head back to Earth, it can only travel up to 1000 km/s. 4.4 light years times 300 is 1360 years! the trip to Alpha Centauri if we ignore acceleration and deceleration times would be 6.6 years for 4.4 light years.

JoshNH4H · 2013-12-12 14:05:18

Actually, because it's a plasma lightsail, your speed limit isn't 400 km/s (speed of the stellar wind), but 299,792 km/s (the speed of light). It uses a similar technology as the M2P2 plasma sail, but instead of using Helium to make your plasma bubble, you instead use elements like Lithium, Sodium, Potassium, Mercury, etc., all of which have absorption bands within the peak spectrum of the Sun's emission spectrum. The addition of some methane, water, and CO2 to the mix for infrared absorption would be a plus too.

It's a plasma lightsail, so it combines the low mass of a plasmasail with (I can't believe I'm about to say this) the high thrust density of a lightsail.

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