Differences between revisions 9 and 10
 ⇤ ← Revision 9 as of 2009-07-11 22:53:10 → Size: 4554 Editor: KeithLofstrom Comment: ← Revision 10 as of 2009-07-11 23:44:02 → ⇥ Size: 6227 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 12: Line 12: 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. 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.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. A realizable spooling system will have inertia, and will take time and energy to start and stop, so the onset of forces will not beimmediate. Quicker-acting forces, for brief periods of time, can be achieved by moving counterweights.

# 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.

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 mm2, 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.

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. Quicker-acting 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/cm3 GPa GPa um/m-K 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.

LinearCables (last edited 2009-08-14 20:56:24 by KeithLofstrom)