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## comment, see HelpOnProcessingInstructions

Lowering the West End of the Track

Most of the acceleration track of the launch loop should be at 80 kilometers altitude, to avoid most atmospheric drag. However, near the start of acceleration, the vehicle is moving relatively slowly, and can tolerate higher air density. This is an opportunity to lower west station, permitting a shorter west incline, shorter stabilization cables, and a heavier incline. This detail is not essential to the launch loop, but it will save money and improve operations considerably.

The velocity along the track is V^2 ~ = ~ 2 a L . The drag is proportional to density and velocity squared, so for constant drag, the density can increase inversely proportional to track distance L+0 . The scale height H_S varies between 50km and 80km geometric altitude, but averages 7200 meters. So the height is H ~ = ~ H_0 + H_S \ln { L / L0 } , until we get so close to the west station and the west incline that the angle becomes limited.

The slope of the track is d H / d L ~ = ~ H_S / L . The curvature of the track is d^2 H / d L^2 ~ = ~ - H_S / L^2

Given a total track length of 2000 km, height versus track length might vary as:

First approximation

<-5:>

L

H

slope

angle

curvature

density

V^2

drag

km

km

degrees

km-1

normalized

2000.0

80.00

0.0036

0.21

-1.8E-6

1.00

1.000

1.000

1000.0

75.00

0.0072

0.41

-7.2E-6

2.00

0.500

1.000

500.0

70.02

0.0144

0.83

-2.9E-5

4.00

0.250

1.000

250.0

65.03

0.0288

1.65

-1.2E-4

8.00

0.125

1.000

125.0

60.04

0.0576

3.30

-4.6E-4

16.00

0.063

1.000

62.5

55.05

0.1152

6.57

-1.8E-3

32.00

0.031

1.000

31.3

50.06

0.2304

12.97

-7.4E-3

64.00

0.016

1.000

15.6

45.07

0.4608

24.74

-2.9E-2

128.00

0.008

1.000

That last entry is problematic - the incline starts at 20 degrees, and curves downwards, so the angle at west station is quite a bit less than 20 degrees. So the curve of the launch track must stop increasing somewhere above 50 km altitude. This is a "flat earth" calculation. The earth's curvature is -1 / R_E , or 1.57E-4 km-1, so significant extra weight and cable tension is needed to curve the track and rotor as it approaches west station. Lets stop dropping the track quite so steeply at 150 kilometers out, limiting the curvature to 3.2E-4 km-1. This puts the east end of West station is about 50.6 km high. This also means the first 150 km of track can mass 3 * ( ( 3.2/1.57 + 1 ) * ( 14/7.67 )^2 - 1 ) = 30 kg/m rather than 7 kg/m .

At 3 gees, the payload will spend about 100 seconds accelerating over this 150 km section. Up to 70 seconds after launch, this is a good place to abort and return a payload to west station if necessary. Since the track is sloping downwards, plain old gravity can return an aborted payload in about 15 minutes, while a 0.5 gee acceleration can return it in about 6 minutes.

The extra weight of the track can support an abort braking system, perhaps even some kind of semi-passive "velocity transformer" coil system to accelerate the payload with less drag on the rotor, and more pull on the stress-relieved track. Wild hand-waving here!

Second approximation

L

H

slope

angle

curvature

density

V^2

drag

km

km

degrees

km-1

normalized

2000.0

80.00

0.0036

0.21

-1.8E-6

1.00

1.000

1.000

1000.0

75.00

0.0072

0.41

-7.2E-6

2.00

0.500

1.000

500.0

70.02

0.0144

0.83

-2.9E-5

4.00

0.250

1.000

250.0

65.03

0.0288

1.65

-1.2E-4

8.00

0.125

1.000

Curvature becomes constant closer than 150km

150.0

61.35

0.0480

2.75

-3.2E-4

13.33

0.075

1.000

100.0

58.55

0.0640

3.66

-3.2E-4

19.67

0.050

0.984

50.0

54.95

0.0800

4.57

-3.2E-4

32.43

0.025

0.811

0.0

50.55

0.0960

5.48

-3.2E-4

44.80

0.000

0.000

WORK IN PROGRESS, MORE LATER

LowerWestIncline (last edited 2017-01-26 08:12:54 by KeithLofstrom)