01-07-2004 04:36									#61
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Now, du is the extra mass density required to support the "weight" of the r to r+dr section.

Therefore du * s = u * dr * (GM/(r^2) - r*w^2)

or du/dr = (u/s) * (GM/(r^2) - r*w^2)

Now to solve this differential equation

u(r) = A * exp(-GM/(r*s) - (0.5s^-1) * (r^2) * (w^2))

with A the undetermined constant.

Set u(R0) = u0

Get A = m/s*(GM/R0^2 - R0*w^2) * exp(GM/(R0*s) + (0.5s^-1) * (R0^2) * (w^2))

EDIT: dropped a factor of 1/s

Last edited by KrazyHorse on 01-07-2004 at 04:58

	01-07-2004 04:37									#62
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quote: Originally posted by DanS KH is a show-off.

Not at all. Trying to get some form of solution for total length of cable with anchoring mass m'.

	01-07-2004 05:08									#63
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Now to get the total force on the the cable we integrate from R0 to Rf (the radius of the anchoring mass) the expression:

u(r) * (GM/(r^2) - r*w^2)

or A * exp(v(r)) * dv/dr * s

So

F1 = A * s * (exp(v(Rf)) - exp(v(R0)))

Now add to this the force on the anchoring mass m'

F2 = m' * (GM/(r^2) - r*w^2)


F1 + F2 = 0 for minimal stable height.

	01-07-2004 05:14									#64
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If we have no anchoring mass F2 = 0, therefore F1 = 0

Therefore v(Rf) = v(R0)

Therefore -GM/R0 - 0.5(R0^2)*(w^2) = -GM/Rf - 0.5(Rf^2)*(w^2)

	01-07-2004 05:19									#65
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I'll take your word for it. I've been out of school too long.

	01-07-2004 05:39									#66
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or numerically

-6.260 * 10^7 = -3.986 * 10^14/Rf - (2.644 * 10^-9)Rf^2

Which (solving the cubic equation) gives a distance (from centre of the earth to end of cable) of

1.57 * 10^8 metres or 157 000 km. The length is this distance minus 6400 km (radius of the earth) or 150 600 km or so.

For comparative purposes the distance from the centre of the Earth to geosynchronous orbit is when

GM/(r^2) - r*w^2 = 0

or r = 42 200 km

So the height of geosynchronous orbit is about 35 800 km above the surface of the earth.

In other words the cable would be about 4 times longer than the height of geosynchronous orbit.

	01-07-2004 05:45									#67
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Freak.

	01-07-2004 05:46									#68
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the figure listed in the article in Discover is 62,000 miles long. Roughly 8 times the Earth's diameter.

That alone sounds far fetched. How would you coil that much ribbon. Even if it were thinner than paper, it would still take up a lot of space. How would you get it all up in space.

And what would happen if it got tangled up . Who would untangle all that?

	01-07-2004 05:49									#69
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62 000 miles (assuming I haven't dropped any factors here) would require an anchoring mass. Otherwise it would fall down...

	01-07-2004 08:17									#70
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If this thing is ever built, all that came before it will be ancient history.

	01-07-2004 08:19									#71
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KrazyHorse get the hell to work and make it happen.

	01-07-2004 09:28									#72
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quote: Originally posted by KrazyHorse To what? You definitely need to put your "anchor" further away than geosynchronous orbit (~40 000 km IIRC) And the closer your anchor is to geosynchronous the more mass you need to put there. It's a simple question of requiring that the net force on the upper part balances out the net force on the lower end.

I don't think it's efficient to use a Space Elevator to put things into geosynchronous orbit. Wastes too much energy. Much better just move a rocket into low orbit and fire it from there.

	01-07-2004 09:55									#73
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We don't need a space elevator, we have Jerry Springer! Who needs a space elevator when we have the mundane to keep us eternally occupied?

Just imagine if we found alien artifacts that showed an ancient civilisation preoccupied with the worst of their species?

	01-07-2004 09:57									#74
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quote: Originally posted by KrazyHorse 62 000 miles (assuming I haven't dropped any factors here) would require an anchoring mass. Otherwise it would fall down...

The whole point of the space elevator is that it won't. The centrifugal force would negate it's weight.

***Edit: Lazy Lung didn't read the rest of the thread! I hereby banish thyself to silence for the next 5 minutes!

	01-07-2004 11:06									#75
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KH - I dont belive you took into account the Thickness (mass per unit of length) of the Cable above Geosyncronus Orbit. Balancing the elevator is like a See-saw. If your using another person on the other end of the see-saw then the closer they are to the Fulcrum (Geosyncronus orbit) the heavier they need to be. Conversly the further away they are the lighter they can be. With a long enough plank you dont even need someone sitting on it at all. Also you must take into account that every part of the Cable is a differnt distance from the Earth and thus its weight is differnt.

A construction senario I read about involved bringing a 20 metric Ton spool of material up to GSO with conventional Rockets and then lowering the thread down to Earth. At 20,000 Kilos and aproimatly 20,000 Kilometers thats 1 Kilo per Kilometer or 1 gram per Meter. Thats obviusly to thin but if a large number of such filiments were spun together like silk from a spider they might be able to do the job.

	01-07-2004 11:06									#76
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quote: KH - I dont belive you took into account the Thickness (mass per unit of length) of the Cable above Geosyncronus Orbit.

Yes I did.

	01-07-2004 11:08									#77
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quote: Also you must take into account that every part of the Cable is a differnt distance from the Earth and thus its weight is differnt.

Duh.

How much training do you have in physics, beautiful?

	01-07-2004 11:13									#78
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I'll give you a couple of hints:

GM/(r^2) is the acceleration due to gravity at any distance from a spherical mass

- r*w^2 is the fictitious acceleration provided by the fact that the rotating frame is noninertial

	01-07-2004 11:15									#79
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And as for not taking into account effects of mass above geosynchronous:

quote: Now to get the total force on the the cable we integrate from R0 to Rf

This integral is not limited by Rg the radius of geosynchronous orbit. It's limited by the unknown Rf which I solved for by assuming a balancing of net force.

	01-07-2004 11:22									#80
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quote: Originally posted by Urban Ranger I don't think it's efficient to use a Space Elevator to put things into geosynchronous orbit. Wastes too much energy. Much better just move a rocket into low orbit and fire it from there.

?

This makes no sense.

I have the feeling you don't understand how this thing works.

	01-07-2004 20:24									#81
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Yeah, the hard part is getting into orbit at all.

Although the power source I'm still concerned about. To get up in orbit they want to beam a laser onto a solar panel on the underside. This technology is proven. But clouds and other debris could cause problems. And imagine the poor bird that flies through the laser . And I believe they have solar panels on the top as well. This would be used for the higher stages.

	01-07-2004 20:30									#82
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quote: Originally posted by KrazyHorse I'll give you a couple of hints: GM/(r^2) is the acceleration due to gravity at any distance from a spherical mass - r*w^2 is the fictitious acceleration provided by the fact that the rotating frame is noninertial

AH! So these cables are attached to a rotating frame? Presumably circular with a radius of 60,000 miles or so?

Yep, definitely pie in the sky.

	01-07-2004 21:01									#83
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The earth is the rotating frame

	01-07-2004 22:56									#84
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quote: Originally posted by Cruddy AH! So these cables are attached to a rotating frame? Presumably circular with a radius of 60,000 miles or so? Yep, definitely pie in the sky.

frame = frame of reference

i.e. not a physical construct

	01-07-2004 23:11									#85
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Out behind the Philippine house, past the banana farm, pretty close to the equator, we own a small lot that isn't being used. I suggest we start the space elevator there in September.

There is a Sari Sari store with cold beer which would be useful for the extensive planning needed.

Are you with me?

	01-07-2004 23:42									#86
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This is a cool exercise, but this scientist is genuinely off his rocker if he thinks it will be done in 15 years at a cost of $10 billion.

For one, buckytube rope has just recently been fabricated and it doesn't yet come anywhere near the required strength. Further, this stuff is insanely expensive right now. $500 per gram in really short tubes.

It seems to me that very high towers and skyscrapers will be built with the stuff before people are able to wrap their heads around space elevators. Buckytubes are much stiffer than steel, so people wouldn't get sick on 459th floor from swaying in the wind.

Last edited by DanS on 01-07-2004 at 23:48

	02-07-2004 00:52									#87
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can it handle a plane crash into it?

	02-07-2004 03:27									#88
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Depends.

	02-07-2004 04:08									#89
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I dont think this material will be very usefull for conventional Buildings. Its got great TENSILE strength but as far as I know little or no COMPRESSIVE strength. Steel and Concreete will still be the best materials for compressive strength. The first aplications will be in replacing steel cables in Bridges or cranes, the elimination of Rust will be icing on the cake.


As for a plance Crash (asuming you ment lateraly into a piece of taut cable) would be a test of shear strength which I dont realy know about. Potentialy the plane might be sliced in half like cutting cheese with a wire or the cable might stretch for hundreds of miles. Or the plane would explode on impact and the fire ball would likly cosume and burn the carbon in the cable.

Last edited by Impaler[WrG] on 02-07-2004 at 04:21

	02-07-2004 04:23									#90
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