#format jsmath
= 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 } } $, a log-linear fit, also called the [[ Atmosphere | '''density scale height''' ]] around a$ 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.