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The launch loop uses round cables for the elevator. The electric field $ E_0 $ will be highest at the edge of the cable: '''The launch loop''' uses round cables for the elevator. The electric field $ E_0 $ will be highest at the edge of the cable:
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$ E_0 = 2 \pi \epsilon_0 \lambda / r_{cable} $ $ E_0 ~ = ~ 2 \pi \epsilon_0 ~ \lambda / r_{cable} $
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$ \lambda = r_{cable} * E_{max} / 2 \pi \epsilon_0 $ $ \lambda ~ = ~ r_{cable} ~ E_{max} / 2 \pi \epsilon_0 $
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$ f ~ = ~ \lambda ~ \dot ~ 2 \pi \epsilon_0 ~ \lambda / r_{space} $

Or, defined in terms of the maximum electric field:

$ f ~ = ~ 2 \pi \epsilon_0 ~ ( r_{cable}^2 / r_{space} ) ~ E_{max}^2 $

For the numbers above, with $ r_{cable} $ = 8.6E-3 meters, $ r_{space} $ = 40 meters, and E_max = 100KV/meter , the force per meter is 1.1 microNewton / meter . On the other hand, if perturbation forces push the cables within 2 centimeters of each other (center to center, nearly touching), the force per meter increases to 2200 micronewtons/meter .

If the tension was uniform along the length of the cable, then a uniform force-per-length f will deflect each cable in a parabola. In actual fact, the tension is higher at the top than at the bottom, so the deflection is less, and the deflection bulges towards the bottom. That is a difficult calculation to make (hint! hint! - somebody please make a differential equation and solve it!), so the following analysis averages the top and bottom tension and assumes a parabola.

The total force along the entire length of the cable is $ F ~ = ~ f L $ Half the force at each end deflects the cable at an angle of $ \alpha ~ = ~ 0.5 F / T $, where T is the cable tension. That angle times one quarter of the length is the total deflection:

$ deflection ~ = ~ 0.125 F L / T ~ = ~ 0.125 f L^2 / T ~ = ~ 0.25 pi \epsilon_0 ~ L^2 ~ r_{cable}^2 ~ E_{max}^2 / T ~ r_{space} $

For a 40 meter spacing and an average tension of 310 kN, the deflection is 2 millimeters. If the spacing is 2 cm, the deflection is 4 meters (increasing the spacing and reducing the force, obviously). So, with small perturbations over a long distance, the cables will resist being pushed together. The electrostatic forces are NOT strong compared to wind gusts affecting the cables differentially, however.

More Later
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The space elevator uses wide and very thin strips. The limiting fields will we quite high at the edges
===The space elevator--- uses wide and very thin strips of CNT for the tether. The limiting fields will we quite high at the edges

Pulley Elevators

for the Launch Loop and the Space Elevator

A pulley elevator is a simple-appearing way to provide both support and lift in a space elevator or a launch loop. The vehicle can be passive, containing no more than some kind of automatically actuated clamp so that it can release from one moving elevator cable and clamp onto another.

Real life is never that simple. The main drawback for an elevator cable is that if it is tall enough and the cables are closely spaced, then they can easily come close enough to abrade each other. The design must prevent this. The other problem is the same as all long cables, where force changes are rapid compared to the speed-of-sound propagation time; the analysis is complicated, and involves not just the static stretch of the cable, but the inertia and mass flow rate.

Megastructure cables and elevators face yet another complication; their interaction with a nonlinear gravitational field in a rotating frame. Very long cables may move in ways not anticipated by a designer focused on elasticity and strength of materials, particularly movements radially from the expected straight-line path. Electrostatic repulsion may also play a role.

Parameters for two pulley elevators

Launch Loop

SE1000

unit

Height

68

1000

km

pulley diam

40

40 m

m

speed

400

m/s

altitude0

4

0

km

altitude1

72

1000

km

g0

9.79

9.81

m/s2

g1

9.59

7.33

m/s2

Tension 0

200000

N

Tension 1

420000

N

Material

Kevlar 49

CNT

Density

1440

1300

kg/m3

Modulus

127

630

GPa

strength

3.6

150

GPa

safety

2.0

area

2.33

cm2

diameter

1.72

---

cm

width

---

thickness

---

mass

46000

kg

Limiting Case - a moving cable without external tension

More Later

Limiting Case - a non-moving cable around pulleys

More Later

Electrostatic Repulsion

The launch loop uses round cables for the elevator. The electric field E_0 will be highest at the edge of the cable:

E_0 ~ = ~ 2 \pi \epsilon_0 ~ \lambda / r_{cable}

Where \lambda is the charge density in Colombs/meter and \epsilon_0 = 8.85 E-12 farad/meter is the permittivity of vacuum (and air, to a good approximation). Alternately, for a given maximum electric field (say, 100KV/m, 3% of the breakdown field of air) then \lambda is given by:

\lambda ~ = ~ r_{cable} ~ E_{max} / 2 \pi \epsilon_0

The actual voltage on the cables might be a few hundred kilovolts, depending on the geometry of the bottom pulley in relation to ground. If both cables have the same voltage, then they will have the same charge, and repel. The repulsion force per length f is given by the charge density times the electric field at the spacing distance r_{space} :

f ~ = ~ \lambda ~ \dot ~ 2 \pi \epsilon_0 ~ \lambda / r_{space}

Or, defined in terms of the maximum electric field:

f ~ = ~ 2 \pi \epsilon_0 ~ ( r_{cable}^2 / r_{space} ) ~ E_{max}^2

For the numbers above, with r_{cable} = 8.6E-3 meters, r_{space} = 40 meters, and E_max = 100KV/meter , the force per meter is 1.1 microNewton / meter . On the other hand, if perturbation forces push the cables within 2 centimeters of each other (center to center, nearly touching), the force per meter increases to 2200 micronewtons/meter .

If the tension was uniform along the length of the cable, then a uniform force-per-length f will deflect each cable in a parabola. In actual fact, the tension is higher at the top than at the bottom, so the deflection is less, and the deflection bulges towards the bottom. That is a difficult calculation to make (hint! hint! - somebody please make a differential equation and solve it!), so the following analysis averages the top and bottom tension and assumes a parabola.

The total force along the entire length of the cable is F ~ = ~ f L Half the force at each end deflects the cable at an angle of \alpha ~ = ~ 0.5 F / T , where T is the cable tension. That angle times one quarter of the length is the total deflection:

deflection ~ = ~ 0.125 F L / T ~ = ~ 0.125 f L^2 / T ~ = ~ 0.25 pi \epsilon_0 ~ L^2 ~ r_{cable}^2 ~ E_{max}^2 / T ~ r_{space}

For a 40 meter spacing and an average tension of 310 kN, the deflection is 2 millimeters. If the spacing is 2 cm, the deflection is 4 meters (increasing the spacing and reducing the force, obviously). So, with small perturbations over a long distance, the cables will resist being pushed together. The electrostatic forces are NOT strong compared to wind gusts affecting the cables differentially, however.

More Later

===The space elevator--- uses wide and very thin strips of CNT for the tether. The limiting fields will we quite high at the edges

More Later

The Steady State General Case - moving cables around pulleys, without velocity change

More Later

The Dynamic General Case, with velocity change

More Later

PulleyElevator (last edited 2009-08-29 21:06:27 by HSI-KBW-078-043-047-065)