Track Slope

Assume a maximum exit velocity a little larger than Earth escape (11.2 km/s) at 6458 km radius, 80 km altitude, near the equator. At that radius, the Earth's rotation velocity is 0.47 km/s ( 2π × 6458 km / 86414s ), so the maximum atmosphere-relative velocity (ignoring wind and adding drag loss) is 10.8 km/s. We will compute track slope backwards from that.

Loop inclination - the angle to the equator - will probably be between 10 and 30 degrees, To Be Determined.

Atmosphere Density Model

Linear Heating Profile

• Assume constant 3 gee acceleration ( a = 29.4 m/s²) to earth-relative escape velocity at exit altitude z_e = 80 km ( v_x = 10.8 km/s ), and a heating rate proportional to time on the track.

• Assume that the track altitude is lower near the beginning of the acceleration run, and is designed to increase with distance (and thus velocity) to produce that heating profile.

• Assume classical drag heating H = { 1 \over 2 } \rho v^3 proportional to density and velocity cubed.

An escape velocity run to v_e = 10800 m/s will last t_x = 10800/29.4 = 367 seconds. Most launches will be to high earth orbits, a slightly shorter launch run.

v^2 ~=~ 2 a x

Heating rates, during acceleration and at exit:

H ~~=~ { 1 \over 2 } \rho(z) ~ v^3

H_e ~=~ { 1 \over 2 } \rho(z_e) ~ {v_e}^3

Arbitrarily chosen heating rate over the launch run, ramping up proportional velocity:

H ~~=~ H_e ~ v / v_e

H ~~=~ { 1 \over 2 } \rho(z) ~ v^3

H_e ~=~ { 1 \over 2 } \rho(z_e) ~ {v_e}^3

Thus

H ~=~ { 1 \over 2 } \rho(z) ~ v^3 ~=~ { 1 \over 2 } \rho(z_e) ~ {v_e}^3 ~ ( v / v_e ) ~=~ { 1 \over 2 } \rho(z_e) ~ {v_e}^2 ~ v

Striking common factors from both sides:

\rho(z) ~=~ \rho(z_e) ~ {v_e}^2 / v^2

Substituting v^2 ~=~ 2 a x and v_e^2 ~=~ 2 a x_e :

\rho(z) ~=~ \rho(z_e) ~ 2 ~ a x_e / 2 ~ a ~ x ~=~ \rho(z_e) ~ ( x_e / x )

So, given the exit velocity v_e , the exit density \rho(z_e) , and the distance x along the track from the starting point, we can compute the (higher) density at distance x , and use that to determine the lower altitude at that point.

This oversimplified model blows up as x approaches zero, to infinite \rho(z) density. At some point (relatively close to the start) the track angle dz / dx starts increasing rapidly, where it makes sense to limit the track angle and run at a linear path down to west station, at a lower altitude. Also, the actual distance travelled is larger, (dx^2 + dy^2 )^{1 \over 2} , or dx / cos( i ) where i is the inclination of the track from horizontal.

Over a small range of z around z_0, the function \rho(z) will approximate a simple exponential function:

\rho(z) \approx \rho_0 e^{-(z-z_0)/l_0} ~~~~~ where \rho_0 = \rho(z_0) and l_0 = { { \Large \rho(z_0) } \over { { { \LARGE \partial\rho(z) } \over { \LARGE z } } } { \large ~at~ z~=~ z_0 } } , a log-linear fit, also called the '''density scale height''' around a z_0 .

Density scale height varies between 5.89 km and 8.25 km between 30 and 80 kilometers altitude; logarithmically average that to 7.24 km, spanning the range.

It's best not to "waste too much deflection angle" on the ramped portion of the loop between west station and the main launch run; as a wild guess, i = 5 degrees above west station seems sufficient. 1/cos( 5° ) = 1.0038 = dl / dx , good enough for rough estimates. The lower ramped section will have a fairly thick aluminum outer sheath in order to conduct extreme lightning strikes without too much heating or voltage drop.

Altitude vs. Loop Distance

Using the simplified density model, a pure exponential with a H = 7.24 km density scale height, we can compute the change in altitude from the velocity relative to the exit velocity:

\Delta z ~=~ - H ~ ln( \rho(z) / \rho(z_e) ) ~=~ - H ~ ln( {v_e}^2 / 2 a x ) ~=~ - H ~ ln( x_e / x )

More later

Slope: { \Large { 1 \over \cos( \theta ) } } ~=~ { \Large { { \partial x } \over { \partial z } } } ~=~ { { \Large \left( { - x_e } \over H \right) } ~ exp { \Large { \left( { { z_e - z } \over H } \right) } } }