# "Diverse Configurations of the Space Cable"

A coil gun is not how I would do things. But this is John's paper. The following is intended only to analyze some things that were not mentioned in the paper, perhaps leading to a better paper.

This is a private page, not publication quality, so some of the math markup is not pretty. I hope it allows a two way conversation that email or published papers do not. A spreadsheet (open office ODF format) is attached.

Thinking about these things have been fruitful for the launch loop. In particular, it should be possible to lower the west end of the launch loop from 80km to 50km, adding a gradual (logarithmic, not parabolic) curve to the launch track from west station to the middle of the track. That will permit a shorter west incline, a shorter elevator ride to west station, more mass at west station, and extra lift capacity for a vehicle test track. Beyond a few meters of clearance improvement, launch loop east station will stay at 80km height.

The ferrite track magnet poles. Improvements in stability analysis. I'm grateful for the opportunity to learn your thoughts, and the parallax view that our non-identical viewpoints make possible.

## Vertical Gee forces on the Vehicle, Near Release

It is unclear whether the section III coil-gun version of the space cable has a parabolic or half-circle track. I will assume parabolic shape, described by:

y = 50000 - { x^2 \over 50000 } ~ ~ ~ (1)

• y measured from the ground, in meters
x measured from the center, in meters
"flat earth" approximation, no extra curvature

I will assume a very light vehicle and payload; otherwise, the vertical distortion of the track feeds back iteratively into the analysis. The resulting complexity is beyond my pay grade!

I assume the payload release, and the maximum velocity point, is at the top of the parabola. Other release points result in earth-intersecting orbits. The following analysis is simplified, for clarity:

The horizontal position of the payload near the top of the parabola is:

x = V t ~ ~ ~ (2)

• t is time relative to the release time, negative before release
V is the release velocity, 10000 m/s at release

So near the top,

y = 50000 - { ( V t )^2 \over 50000 } ~ ~ ~ (3)

\dot { y } = { - 2 V^2 t \over 50000 } ~ ~ ~ (4)

\ddot { y } = { - 2 V^2 \over 50000 } ~ ~ ~ (5)

\ddot { y } = -4000 m/s^2 ~ ~ ~ (6)

... Which is quite a bit larger than the horizontal acceleration. The track must have a fairly deep vertical kink, pulling the payload down, to compensate for the vertical force on the vertically accelerating payload. When the payload is released, the kink remains. Cleverly modulated vertical velocity at the ramps, 40 seconds before release, might be able to restore the track to its proper shape. Even so, a few milliseconds of error in the release time leaves quite a large uncompensated perturbation. The vehicle acceleration begins 10 seconds before release, so we must be compensating for release 30 seconds before acceleration even starts.

• As an alternative, Perhaps a system of tensioning cables can form a truss structure under the arch of the track. Diagonal stringers can be released into a low tension mode as the payload passes overhead, and retightened as compensating forces arrive from the ramps. This adds a lot of complexity, but reduces the effects of bolt propagation delay.

## Track distance, height, and velocity

As before, I will assume a very light vehicle, and use the release point as zero time and zero x position. For ease of computation. I will also assume a constant "tangent to track" acceleration of 1000 m/s2, even though the perpendicular acceleration and total acceleration may be larger.

The tangential distance L from position x to the top of a parabola can be computed from:

\partial L ^ 2 = \partial y ^ 2 + \partial x ^ 2 ~ ~ ~ ~ (7)

d L = - \sqrt { ( { { d y } \over { d x } } ) ^ 2 + 1 } ~ ~ d x ~ ~ (8) (note, length L increases as x goes negative)

For our parabola, define b == -x / 25000 . Integrating (8), we get:

L = 12500 ( b \sqrt { b^2 + 1 } + \ln( b + \sqrt { b^2 + 1 } ) ) ~ ~ ~ ~ (9)

y = 50000 ( 1 - 0.25*b^2 ) (10)

At the start of acceleration, L = 50000 meters. There is no closed form solution for b, but a bit of spreadsheet voodoo approximates it as 1.5278533266 , resulting in starting x = -38196 meters and starting y = 20820 meters. With a tangential acceleration of 1000 m/s2, the velocity is

V = 10000 \sqrt { 1 - L / 50000 } ~ ~ ~ ~ (11)

## Aerodynamic Pressure on the Payload

D is the atmospheric density in arbitrary units, scaled so that D = 1 at the top of the parabola.

Q is the dynamic payload pressure in arbitrary units, scaled so that Q = 1 at the top of the parabola.

Assume an atmospheric density scaling height of 7500 meters. Thus,

D = \exp { ( { { 50000 - y } \over 7500 } ) } (12)

Q = D ( V / 10000 ) ^ 2 (13)

D is 1 and V is 10000 m/s at the top of the parabola

As |x| and L increase, V decreases and P increases. Near the top of the parabola, V decreases with |x|, but y and P are barely changing, so the dynamic pressure diminishes with |x| for a short range. Then, as the parabola descends more rapidly with |x|, the drop in y brings an exponential increase in P and a somewhat smaller increase in Q (V is decreasing). At x = -32764m, Q is maximized, 3.303 arbitrary units. At this point, L = 40567 meters, y = 28530 meters, D = 17.507 units, and V = 4343.57 meters per second.

Comparison to rockets: Rocket fairings and structure must be designed to withstand "Max Q", maximum dynamic pressure. A rocket, accelerating vertically at a uniform 60 m/s2, will have a velocity of 2449 m/s at an altitude of 50000 meters. Assuming the same drag coefficient, Q = 0.06 at that altitude. Max Q for that rocket is at one scale height, 7500 meters, where D = 289.07 and V=948.7 m/s, so Max Q is 2.602 units. In fact, rocket acceleration profiles start out slow. They climb more slowly out of the dense atmosphere, and throttle up after max Q. A more realistic acceleration profile might start at 20 meters per second (3 gees perceived), with throttle up commencing above 8 km altitude. Max Q = 0.9, lower than the lowest dynamic pressure anywhere on a 50 km high coil-gun space cable.

Of course, a step rocket will be bigger and heavier, with a larger frontal area. On the other hand, there will not be a region of highly sheared gas as there is between the track and the vehicle magnets. This comparison is not to point out deficiencies in either, but to point out a similarity, and an opportunity to explore a rich existing body of knowledge.

## Suggested alternative to reduce Q and acceleration

These space cable dynamic pressures change a lot if the height and length are increased by 20%. Redefine b = x / 30000 . At 10km higher altitude, D at the top is reduced by a factor of 0.263 . L = 50km occurs at x = -40112 meters, y = 33813 meters. Max Q = 0.6887 occurs at x = -33793 meters, y = 40967 meters, L = 39971 meters, V = 4479 meters/second, D = 3.335 . The peak vertical acceleration at the top is reduced by 20% to 3333 m/s2. However, because the drag is lower, the V necessary to punch through the upper atmosphere is reduced - that will reduce vertical forces somewhat more. The down side is that the transit time for bolts, and the necessary correction anticipation times, increase by 20%.

DiverseSpaceCable (last edited 2011-02-19 03:22:01 by KeithLofstrom)