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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 the attachment and back. For a 100km stabilization cable, that can be 20 seconds, in which time many meters of cable moves. 
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 $. The total spooling length is $ \Delta L = \Delta F L / A Y $ . Note that a thinner material with a higher breaking strain ( higher $ S / Y $ ) will store more energy. Let's put some numbers on that. Assume a 100km vertical Kevlar 49 cable with numbers from the table below, no vertical taper, with a spool at the top and an attachment at the bottom. For a 2 cm diameter round cable, and a $ K $ of 0.5 (2x safety factor), the static tension $ F_T $ is 570 kN, or about 58 tonnes. The mass of the cable is 45 tonnes. The vertical round trip propagation time is 22 seconds. If $ \delta F $ is 1 KN = 1 kiloNewton, then the spooling length is 2.6 meters, the spooling power is 133 kiloWatts and the spooling energy is 2.9 !MegaJoules. 1 kN is the gravitational weight of 102 kilograms or 224 pounds at sea level. 
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A realizable spooling system will have inertia, and will take time and energy to start and stop, so the onset of forces will not be immediate. Quickeracting forces, for brief periods of time, can be achieved by moving counterweights. 

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 Material  density  modulus  strength  CTE  Vsound  Length  Round Trip  100Km*100K    gm/cm^3^  GPa  MPa  um/mK  m/s  m  seconds  m   Steel SAE980x  7.9  200  650  12  5000  8400  40  120   80% Kevlar   80% Spectra   80% nanotube  
 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   Zylon HM  1.56  270  5.8  6  13.2  380  15  60    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 [[ http://etd.lib.fsu.edu/theses/available/etd08262005003434/.../Thesis.pdf  Prakash ]] is used here, but other simulations differ wildly. ==== Space elevator considerations ==== Consider a much longer linear space elevator cable made of diamond (without taper, yeah yeah, not possible, but it makes the math easier). The cable is 40 thousand kilometers long, 4e7 meters. The cable has a cross section of 10 mm^2^, and a $ K $ of 1 at the top. The mass of the cable is 1400 tonnes. The static tension $ F_T $ is 600 kN, or about 61 tonnes. The vertical round trip propagation time is 4400 seconds. If $ \Delta F $ is 1 KN = 1 kiloNewton, then the spooling length is 3500 meters, the spooling power is 950 kiloWatts and the spooling energy is 4.2 !GigaJoules. Note that the power and energy are higher than before because the cable is thinner and the spooling length is higher. Actually, a space elevator cable must be tapered, a lot thicker in the middle than at the ends. Only the bottom end is subject to heavy gravity, so the weight of the thinner lower cable is responsible for most of the tension in the system. However, the thinner lower cable is also less strong, so it will tend to stretch much faster than the middle. The fat stiff cable in the middle, and the weak cable at the bottom, becomes a classic spring resonator, with a resonance period measured in days. This will propagate "high frequency" (hoursscale) forces to the ground poorly, and will be subject to all sorts of destructive resonances; for example, it may get pumped by the moon. The dynamic analysis of tapered space elevator cables is "interesting" ( see the papers by [[ http://staff.polito.it/nicola.pugno/  Nicola Pugno ]] ) but we need not go into that here. 
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 . The total spooling length is \Delta L = \Delta F L / A Y . Note that a thinner material with a higher breaking strain ( higher S / Y ) will store more energy.
Let's put some numbers on that. Assume a 100km vertical Kevlar 49 cable with numbers from the table below, no vertical taper, with a spool at the top and an attachment at the bottom. For a 2 cm diameter round cable, and a K of 0.5 (2x safety factor), the static tension F_T is 570 kN, or about 58 tonnes. The mass of the cable is 45 tonnes. The vertical round trip propagation time is 22 seconds. If \delta F is 1 KN = 1 kiloNewton, then the spooling length is 2.6 meters, the spooling power is 133 kiloWatts and the spooling energy is 2.9 MegaJoules. 1 kN is the gravitational weight of 102 kilograms or 224 pounds at sea level.
A realizable spooling system will have inertia, and will take time and energy to start and stop, so the onset of forces will not be immediate. Quickeracting forces, for brief periods of time, can be achieved by moving counterweights.
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 
Zylon HM 
1.56 
270 
5.8 
6 
13.2 
380 
15 
60 

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.
Space elevator considerations
Consider a much longer linear space elevator cable made of diamond (without taper, yeah yeah, not possible, but it makes the math easier). The cable is 40 thousand kilometers long, 4e7 meters. The cable has a cross section of 10 mm^{2}, and a K of 1 at the top. The mass of the cable is 1400 tonnes. The static tension F_T is 600 kN, or about 61 tonnes. The vertical round trip propagation time is 4400 seconds. If \Delta F is 1 KN = 1 kiloNewton, then the spooling length is 3500 meters, the spooling power is 950 kiloWatts and the spooling energy is 4.2 GigaJoules. Note that the power and energy are higher than before because the cable is thinner and the spooling length is higher.
Actually, a space elevator cable must be tapered, a lot thicker in the middle than at the ends. Only the bottom end is subject to heavy gravity, so the weight of the thinner lower cable is responsible for most of the tension in the system. However, the thinner lower cable is also less strong, so it will tend to stretch much faster than the middle. The fat stiff cable in the middle, and the weak cable at the bottom, becomes a classic spring resonator, with a resonance period measured in days. This will propagate "high frequency" (hoursscale) forces to the ground poorly, and will be subject to all sorts of destructive resonances; for example, it may get pumped by the moon. The dynamic analysis of tapered space elevator cables is "interesting" ( see the papers by Nicola Pugno ) but we need not go into that here.