Differences between revisions 41 and 86 (spanning 45 versions)
Revision 41 as of 2020-08-01 01:12:17
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Revision 86 as of 2020-08-07 16:14:30
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Deletions are marked like this. Additions are marked like this.
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Loop inclination - the angle to the equator - will probably be between 10 and 30 degrees, To Be Determined. Loop inclination - the angle to the equator - will probably be between 10 and 30 degrees, To Be Determined.  
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|| 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 ||
|| 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 ||
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$ v ~~=~ a ~ t ~~~~~~~~~~~~~~~ t_e ~=~ v_e / a $ $ v^2 ~=~ 2 a x $
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$ v^2 ~=~ 2 ~ a ~ x ~~~~~~~~~~~~ {v_e}^2 ~=~ 2 ~ a ~ x_e $ Heating rates, during acceleration and at exit:
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Chosen heating rate over the launch run, ramping up proportional to time and velocity: $ H ~~=~ { 1 \over 2 } \rho(z) ~ v^3 $
 
$ H_e ~=~ { 1 \over 2 } \rho(z_e) ~ {v_e}^3 $
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$ H ~~=~ H_e ~ t / t_e ~=~ H_e ~ v / v_e $ Arbitrarily chosen heating rate over the launch run, ramping up proportional velocity:
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thus $ H ~~=~ H_e ~ v / v_e $
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$ 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(z) ~ v^3 $
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$ H =  { 1 \over 2 } \rho_e ~ {v_e}^3 ~ t / t_e ~=~ \rho(z) ~ a^2 ~ x ~ t $ $ H_e ~=~ { 1 \over 2 } \rho(z_e) ~ {v_e}^3 $
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$ H_e ~=~ { 1 \over 2 } \rho_e ~ {v_e}^3 \rho(z) ~ a^2 ~ x_e ~ t_e $ Thus
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FIXME ABOVE, TO GET TO THE BELOW EQUATION: $ 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 $
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rearranging: 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 ) $
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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. 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.
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Over a small range of $z$ around $z_0$, the function $ \rho(z) $ will approximate an exponential function: Over a small range of $z$ around $z_0$, the function $ \rho(z) $ will approximate a simple exponential function:
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$ \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 { { { \Huge \partial\rho(z) } \over { \Huge z } } } { \Large ~at~ z~=~ z_0 } } $ $ \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.

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.

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