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.

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