# 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

 Altitude Density Density Scale Average km kg/m3 Height km km to 80 80 km 1.85e-5 5.89 6.33 70 km 8.28e-5 7.43 6.7 60 km 3.10e-4 8.25 7.1 50 km 1.03e-3 8.03 7.5 40 km 4.00e-3 6.86 7.4 30 km 1.84e-2 6.45 7.24

## Linear Heating Profile

 a vehicle acceleration x vehicle distance along track x_e vehicle exit distance v vehicle velocity along track v_e vehicle exit velocity H vehicle heating (relative) H_e vehicle exit heating t time from start of launch run t_e exit time \rho atmospheric density \rho_e exit atmospheric density z vehicle/track altitude z_e exit altitude
• 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 )

\rho(z) = \rho(z_e) {v_e}^2 / 2 a 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.

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

{ \Large \theta } ~=~ arcsin { \Large { \left( { H \over x } \right) } }

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.

## Constant Heating Profile

By similar reasoning, if the heating is constant over the entire run, then the slope increases more rapidly.

TrackSlope (last edited 2020-08-30 23:40:35 by KeithLofstrom)