3891
Comment:

4554

Deletions are marked like this.  Additions are marked like this. 
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Assume a cable with a density of $ \rho $ and a static tension $ F_T $, with a spool at $ l = 0 $ and an infinitely strong attachment at $ l = L $. For a step increase in force of $ \delta F $ away from the attachment, the strain wave propagates down the cable towards the attachment at the speed of sound $ V_C $, reflecting off it and propagating back, returning to the spool in time $ T_{RT} = L / 2 V_C $. The cable in front of the strain wave moves away from the attachment at velocity $ V_S $, and the force accelerates a new segment of length $ d l $ in time $ d t $. The segment length $ d l = V_C d t $, the segment mass $ d m = \rho d l = \rho V_C d t $ and the force is $ \delta F = \rho V_C V_S $. Assuming $ \delta F \ll F_T $, the power expended spooling the cable is $ P = F_T V_S = \delta F F_T / \rho V_C $. The total energy is $ E = 2 \delta F F_T / \rho {V_C}^2 $.  Assume a cable with a linear mass density of $ \rho_L $ and a static tension $ F_T $, with a spool at $ l = 0 $ and an infinitely strong attachment at $ l = L $. For a step increase in force of $ \Delta F $ away from the attachment, the strain wave propagates down the cable towards the attachment at the speed of sound $ V_C $, reflecting off it and propagating back, returning to the spool in time $ T_{RT} = L / 2 V_C $. The cable in front of the strain wave moves away from the attachment at velocity $ V_S $, and the force accelerates a new segment of length $ d l $ in time $ d t $. The segment length $ d l = V_C d t $, the segment mass $ d m = \rho_L d l = \rho_L V_C d t $ and the force is $ \Delta F = \rho_L V_C V_S $. Assuming $ \Delta F \ll F_T $, the power expended spooling the cable is $ P = F_T V_S = \Delta F F_T / \rho_L V_C $. The total energy is $ E = 2 \Delta F F_T L/ \rho_L {V_C}^2 $. 
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Assume the cable is stressed at $ K S $, where $ S $ is the breaking strength and $ K $ is the safety factor. If the cross section area of the cable is $ A $, then $ F_T = A K S $ . If the volume mass density of the cable material is $ \rho $, then $ \rho_L = A \rho $. The propagation velocity is derived from $ V_C = \sqrt { Y / \rho } $ where $ Y $ is Young's modulus (often called E, but we don't want to confuse it with the energy) . The total energy becomes $ E = 2 \Delta F A K S L/ A \rho ( Y / \rho ) = 2 \Delta F K S L / Y $. Note that a material with a higher breaking strain ( higher $ S / Y $ ) will store more energy. 
Linear Cables
Stabilization and elevator cables on the launch loop are very long, and propagation delay is a big issue. In most systems people are familiar with, cables are short enough and forces change slowly enough that propagation delay is not a major issue. With a launch loop, forces can change rapidly (milliseconds) while the propagation delays are 10s of seconds.
For example, when an instantaneous force change is applied to one end of a very long cable, the end does not stretch a little, it moves, and keeps moving until the force has had time to propagate to a stationary attachment and back. For a 100km Kevlar 49 stabilization cable, that can be 22 seconds, in which time many meters of cable moves.
So to increase force in a cable, the cable must be spooled, and energy applied to spool it. This energy is stored in the elastic strain of the cable material. A thinner cable stretches more per unit of force, and moves faster, so more energy must be expended to spool it.
Assume a cable with a linear mass density of \rho_L and a static tension F_T , with a spool at l = 0 and an infinitely strong attachment at l = L . For a step increase in force of \Delta F away from the attachment, the strain wave propagates down the cable towards the attachment at the speed of sound V_C , reflecting off it and propagating back, returning to the spool in time T_{RT} = L / 2 V_C . The cable in front of the strain wave moves away from the attachment at velocity V_S , and the force accelerates a new segment of length d l in time d t . The segment length d l = V_C d t , the segment mass d m = \rho_L d l = \rho_L V_C d t and the force is \Delta F = \rho_L V_C V_S . Assuming \Delta F \ll F_T , the power expended spooling the cable is P = F_T V_S = \Delta F F_T / \rho_L V_C . The total energy is E = 2 \Delta F F_T L/ \rho_L {V_C}^2 .
Assume the cable is stressed at K S , where S is the breaking strength and K is the safety factor. If the cross section area of the cable is A , then F_T = A K S . If the volume mass density of the cable material is \rho , then \rho_L = A \rho . The propagation velocity is derived from V_C = \sqrt { Y / \rho } where Y is Young's modulus (often called E, but we don't want to confuse it with the energy) . The total energy becomes E = 2 \Delta F A K S L/ A \rho ( Y / \rho ) = 2 \Delta F K S L / Y . Note that a material with a higher breaking strain ( higher S / Y ) will store more energy.
Material 
density 
elastic 
strength 
CTE 
Vsound 
Support 
100km 
Therm exp 
notes 


modulus 



Length 
Round Trip 
100Km*100K 


gm/cm^{3} 
GPa 
GPa 
um/mK 
km/s 
km 
seconds 
m 











Steel SAE980x 
7.9 
200 
0.65 
12 
5.0 
8.4 
40 
120 

Pure Kevlar 
1.44 
124 
3.62 
2.7 
9.3 
250 
22 
27 

Pure Spectra 
0.97 
168 
2.58 
12 
13.2 
270 
15 
120 
continuous creep 
Pure Diamond 
3.52 
1140 
>60 
1.2 
18.0 
1700 
11 
12 

Pure nanotube 
~1.4 
~1000 
~60 
9? 
26.0 
4400 
8 
90 

composites: 









80% Kevlar 









80% Spectra 









80% nanotube 









note: Nanotube properties are controversial. The CTE simulation by Prakash is used here, but other simulations differ wildly.