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| Density at 80 km is |
|| Altitude || Density || Scale Height || Average || || km || kg/m3 || km || km to 80 || || 80 km || 1.85e-5 || 6.33 || 6.33 || || 70 km || 8.28e-5 || 7.14 || 6.7 || || 60 km || 3.10e-4 || 8.02 || 7.1 || || 50 km || 1.03e-3 || 8.14 || 7.5 || |
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|| $ 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 ~~=~ a ~ t ~~~~~~~~~~~~~~~ t_e ~=~ v_e / a $ $ v^2 ~=~ 2 ~ a ~ x ~~~~~~~~~~~~ {v_e}^2 ~=~ 2 ~ a ~ x_e $ Chosen heating rate over the launch run, ramping up proportional to time and velocity: $ H ~~=~ H_e ~ t / t_e ~=~ H_e ~ v / v_e $ thus $ H ~~=~ { 1 \over 2 } \rho(z) ~ v^3 ~=~ { 1 \over 2 } \rho(z) ~ ( 2 ~ a ~ x) ~ ( a ~ t ) ~=~ \rho(z) ~ a^2 ~ x ~ t $ $ H = { 1 \over 2 } \rho_e ~ {v_e}^3 ~ t / t_e ~=~ \rho(z) ~ a^2 ~ x ~ t $ $ H_e ~=~ { 1 \over 2 } \rho_e ~ {v_e}^3 \rho(z) ~ a^2 ~ x_e ~ t_e $ FIXME ABOVE, TO GET TO THE BELOW EQUATION: rearranging: $ \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 blows up as $ x $ approaches zero, to infinite 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. Over a small range of $z$ around $z_0$, the function $ \rho(z) $ will approximate an 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. |
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 |
Scale Height |
Average |
km |
kg/m3 |
km |
km to 80 |
80 km |
1.85e-5 |
6.33 |
6.33 |
70 km |
8.28e-5 |
7.14 |
6.7 |
60 km |
3.10e-4 |
8.02 |
7.1 |
50 km |
1.03e-3 |
8.14 |
7.5 |
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 ~~=~ a ~ t ~~~~~~~~~~~~~~~ t_e ~=~ v_e / a
v^2 ~=~ 2 ~ a ~ x ~~~~~~~~~~~~ {v_e}^2 ~=~ 2 ~ a ~ x_e
Chosen heating rate over the launch run, ramping up proportional to time and velocity:
H ~~=~ H_e ~ t / t_e ~=~ H_e ~ v / v_e
thus
H ~~=~ { 1 \over 2 } \rho(z) ~ v^3 ~=~ { 1 \over 2 } \rho(z) ~ ( 2 ~ a ~ x) ~ ( a ~ t ) ~=~ \rho(z) ~ a^2 ~ x ~ t
H = { 1 \over 2 } \rho_e ~ {v_e}^3 ~ t / t_e ~=~ \rho(z) ~ a^2 ~ x ~ t
H_e ~=~ { 1 \over 2 } \rho_e ~ {v_e}^3 \rho(z) ~ a^2 ~ x_e ~ t_e
FIXME ABOVE, TO GET TO THE BELOW EQUATION:
rearranging:
\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 blows up as x approaches zero, to infinite 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.
Over a small range of z around z_0, the function \rho(z) will approximate an 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.