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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 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 $. |
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 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 .
Material |
density |
elastic |
strength |
CTE |
Vsound |
Support |
100km |
Therm exp |
notes |
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modulus |
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Length |
Round Trip |
100Km*100K |
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|
gm/cm3 |
GPa |
GPa |
um/m-K |
km/s |
km |
seconds |
m |
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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 |
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composites: |
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80% Kevlar |
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80% Spectra |
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80% nanotube |
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note: Nanotube properties are controversial. The CTE simulation by Prakash is used here, but other simulations differ wildly.