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In the countless calculations and discussions concerning the "space rope" I've never found any addressing its capability to resist winds.

Consider, as in most current works, it's a 1m wide ribbon, attached near equator, going straight up past geostationary orbit. It's being projected for 20 tons of load on top of its own weight. But never in the calculations did I see the drag put by trade winds typical to that area on what is essentially good several thousand meters of a sail.

This will be a considerable force. A ship of displacement of 400 tons, like this one is capable of travel of 30km/hour against the drag of water with sail surface of 1200m^2. Of course it is projected to gather as much force with the sails as possible, but the tower will have at least an order of magnitude more of the "sail surface".

How would one go about calculating what lateral strength would be put on that tower by the wind (trade winds are up to 8m/s but I'm not sure if it's applicable for higher altitudes), on its part that is still immersed in atmosphere? 1m wide ribbon, going straight up, assuming pessimistic scenario of being oriented perpendicular to wind direction.

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  • $\begingroup$ Remember jet stream winds can be over 50m/s, and they are unpredictable in their route. Admittedly the strongest ones are closer to the pole, but be aware of loops of high altitude (>30000m), high speed winds in any calculation. $\endgroup$
    – Rory Alsop
    Commented Jan 8, 2013 at 11:59
  • $\begingroup$ Good question. I can imagine there could be technical solutions along the lines of how oscillations are addressed in airfoils and tall buildings. $\endgroup$
    – Mike Dunlavey
    Commented Jan 8, 2013 at 12:58
  • $\begingroup$ @RoryAlsop: Due to lower air density drag caused at these altitudes will be lower than of proportional low altitude winds. OTOH, the speeds may be significantly higher. $\endgroup$
    – SF.
    Commented Jan 8, 2013 at 13:13
  • $\begingroup$ I'd be more concerned with the oscillations set up in the ribbon. All a constant drag force will do is displace the space elevator from the true vertical (and I think it's not exactly vertical to begin with). The oscillations induced by the wind, though, will travel in waves up and down the whole cable, if not damped somehow. $\endgroup$
    – Christoph
    Commented Jan 20, 2013 at 12:06
  • $\begingroup$ @Christoph: Displace enough of it from true vertical - pull it down far enough - and you pull it off balance and it falls down. $\endgroup$
    – SF.
    Commented Jan 20, 2013 at 16:54

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Somewhat unsurprisingly, someone already has written a paper about this: Approximating Aerodynamic Response of the Space Elevator to Lower Atmospheric Wind I have only skim-read it, but I think you should find a lot of information in there.

The most interesting tidbits: Max. horizontal deflection was about 200km for typhoon-strength winds (but only O(10km) for more normal conditions), some configurations lead to near-horizontal angles near the ground, and the presence of a climber has significant influence on the shape attained.

There's also an acompanying website, apparently: http://spaceelevatorwiki.com/wiki/index.php/Aerodynamic_Response

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You can integrate the drag equation vertically: dF = 1/2*dA*Cd*rho(z)*v(z)^2.

dA = 1m^2/m = 1m in your example; dA is the width of the ribbon. cD ~1.2 for a flat square plate going into the wind, for a long thin plate I am not sure exactly but it should be around 1. Rho = density. It drops off by a factor of 2 every ~5000m and starts at 1.2kg/m^3.

Main assumptions: 1. High Re and low mean-free path: valid for all but the (unimportant) tenuous upper atmosphere. 2. The change in v and rho is gradual on the scale of the width, which is valid except for possibly a few points of extreme wind-shear (these exceptions won't add up to much, however).

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