Can we now build the space elevator?

snipped-for-privacy@yahoo.com (Robert Clark) wrote in message


So the "screen measured diagonally" effect is at work here too!?

I think it far more likely that a single factor of correction will be introduced, based on the inflated ant-carries-1000-times-its-body-weight figures for a single tube. Plus another factor for packing density. Plus another factor for inevitable defect concentrations. Plus another factor or stress concentration at cracks. And for engineering margin of safety.

Just how bad is the overestimation of practical tensile strength by using the wall area rather than the base area?
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snipped-for-privacy@yahoo.com (Robert Clark) wrote in message

Thats kinda funny :-)
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snipped-for-privacy@netzero.com (Edward Green) wrote in message strength.

Here's one article that discusses how the tensile strength is calculated:
Mechanical properties of carbon nanotubes: theoretical predictions and experimental measurements. C. R. Physique 4 (2003) 9931008 http://www.academie-sciences.fr/publications/comptes_rendus/pdf/CRPhys_article9.pdf
It calculates the tensile strength based on a nanotube wall thickness of .34 nm, which they take from the thickness of the individual layers in graphene. I've seen discussed nanotubes with diameters of 1.4 nm. The cross-sectional area of a solid cylinder of this dimeter would be PI*.7^2 = 1.54. To find the area of the cross-section of the hollow tube of wall thickness .34 nm, subtract .34 from .7 giving .36 and calculate the area of a disk with this radius: PI*.36^2 = .41. Then subtract 1.54-.41 = 1.13, this is the area of the annulus with outer diameter 1.4 and inner diameter 2*.36=.72. The ratio of 1.13 to 1.54 is .73, so this is not a very significant difference. However, for a nanotube of 10 nm diameter, the calculations are: area of disk = Pi*5^2 = 78.5 radius of inner disk = 5 -.34 = 4.66 area of inner disk = Pi*4.66^2 = 68.19 area of annulus = 78.5 -68.19 = 10.31
In this case the hollow tube has only .13 times the cross-sectional area of the filled-in tube. The proportion gets worse as the tube diameter increases. Larger diameter tubes may be easier to work with, but for greater strength the thinner ones look like the best way to maintain the calculated tensile strength (unless arbitrarily wide layered nanotubes can be produced.)
Bob Clark
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