PUSH TOWARD INTERSTELLAR
TE assumes an interstellar exhaust velocity (V_{Exh}) of at least 86.6% light speed, .866 c. Thus, as exhaust velocity approaches every closer to c; spaceship range increases. When we achieve this; interstellar flights will become feasible. However, for interstellar flights to become practical, we must overcome significant problems.
Reliability becomes ever more important for interstellar voyages. Precise quantities of near light, exhaust particles must exit vessel for every second of every day.
Flight duration expands by several orders of magnitude. Exhaust particles must maintain near light speeds for years while interplanetary need only days of much slower exhaust speeds.
Flight duration expands by several orders of magnitude. Exhaust particles must maintain near light speeds for years while interplanetary need only days of much slower exhaust speeds.
EXPAND ENVELOPE  

from INTERPLANETARY particle accelerator  ... will transform several grams/sec of water to ions and expel them at speeds from 10% to 50%c. This will produce equivalence and quick flights to nearby planets (days to weeks). 
to INTERSTELLAR particle accelerator  ... will transform several kilograms/sec of water to ions and expel them at growth factors from 2 to 11 (i.e. corresponding particle exhaust velocities of 86.6%c to 99.6%c). This will produce gforce acceleration for lengthy durations from 100 days to over a year.. 
...can consume reasonable quantities of fuel particles, original fuel flow per second (ff_{sec}), and accelerate them into a constant flow of exhaust particles (ff_{Exh}) .
For this initial example, further assume particles' exhaust speed (V_{Exh}) is 86.6% c. One can express exhaust velocity as a product of decimal component (d_{c}) and light speed, c.
EXAMPLE: V_{Exh} = d_{c }× c = .866 c
Thus, d_{c} = .866 = V_{Exh }/ c
m_{r} =  m_{o} √(1d_{c}^{2}) 

Particle Size
Lorentz Transform quantifies the relativistic growth due to particle speed. Thus, fuel flow per second (ff_{sec}) is the original mass at rest, and exhaust fuel flow (ff_{Exh}) is the relativistic mass at an accelerated velocity.
Notional fuel quantity increases per following table.
ff_{Exh} = n × ff_{sec}
Original Fuel Flow  Exhaust Velocity  Growth Factor  Exhaust Fuel Flow 

ff_{sec}  V_{Exh} =d_{c}×c  n  ff_{Exh} 
1 kg  .8667c  2.0  2.0 kg 
Given  1 √(1d_{c}^{2})  n×ff_{sec} 
Original Fuel Flow  Exhaust Speed  Growth Factor  Ship Mass 

ff_{sec}  V_{Exh} = d_{c}×c  n  M_{Ship} 
1 kg  .943 c  3  83,623 mT 
2 kg  .943 c  3  173,246 mT 
Given  Given  1 √(1d_{c}^{2})  √(n^{2}1)×30.57×10^{6}ff_{sec} 
d_{c}  =  V_{Exh}/c 

n  =  1/√(1d_{c}^{2}) 
n^{2}  =  1/(1d_{c}^{2}) 
1d_{c}^{2}  =  1 / n^{2} 
d_{c}^{2}  =  1  (1 / n^{2}) = (n^{2 } 1) / n^{2} 
d_{c}  =  √(n^{2 } 1) / n 
...comes from momentum exchange. One second of fuel flow's small mass times enormous speed equals spaceship's huge mass times 9.8065 m/sec velocity increase for one second (acceleration due to near Earth gravity, g).
M_{Ship} × g = ff_{Exh} × V_{Exh} 

Both right side terms can be reexpressed. Exhaust fuel mass (ff_{Exh}) can be rewritten as growth factor, n, times fuel flow per second (n×ff_{sec}). Particle exhaust velocity (V_{Exh}) can be rewritten as decimal component times light speed (d_{c}×c).
M_{Ship} × g = (n×ff_{sec}) × (d_{c}×c) 

Rewrite equation as shown below. Note that left panel shows decimal component, d_{c}, defined in terms of n, growth factor. or d_{c} = √(n^{2 } 1) / n.
M_{Ship} = (n×ff_{sec}) × (√(n^{2 } 1) / n×(c/g) 

c = 299,792,458 m/sec g= 9.80665 m/sec^{2}
c/g = 30,570,323 sec
c/g = 30,570,323 sec
The two constants, light speed (c) and acceleration due to gravity (g) can combine for a third constant (c/g), 30.57 megasec. The two "n"s cancel out and the implicit "/sec" of ff_{sec }cancels out the sec from c/g.
M_{Ship} = √(n^{2 } 1) × 30.57 megaff_{sec} 

At the same exhaust speed, the greater the fuel mass, the greater the ship's initial mass which increases 9.8065 m/sec for every second of powered flight (aka "gforce").
EXAMPLE: If ship's gforce propulsion system consumes 1 kg/sec anytime during a particular day; then, one could further assume that day's consumption as about 86,400 kg. We don't know exact amount of consumption per second for any given ship; however, we can designate that value as an variable, ff_{sec}.
(One day = 24 hours × 3,600 sec /hour = 86,400 seconds.)
ff_{day} =day × ff_{sec} =86,400 × ff_{sec} 



c =

299,792,458 m/sec


g =

9.80665 m/sec^{2}

Day =

86,400 sec

Day × g /c =  .002826 = .2826% 
Conclusion: Inspection shows daily decrease depends on particle size, n (multiple of original ff_{o}), which depends on particle speed, d (decimal portion of c, light speed).
THUS, vessel's range depends
not on actual quantity of ff_{sec}
nor actual ship's gross weight, GW.
THUS, vessel's range depends
not on actual quantity of ff_{sec}
nor actual ship's gross weight, GW.
Particle Exhaust Speed  Exhaust Fuel Flow  Ship's Gross Weight  Daily Decrease 

V_{Exh}= d_{c}×c  ff_{Exh}=n×ff_{sec}  GW  ∇ 
.866 c  2.00 kg  52,949 mT  0.16 % 
.943 c  3.00 kg  86,465 mT  0.10 % 
.968 c  4.00 kg  118,397 mT  0.07 % 
Given  ff_{sec} √(1d_{c}^{2})  ff_{sec}×√(n^{2}1)×c g  Day × ff_{sec} GW 
Relativistic Growth Factor  Decimal Component Light Speed  Daily Decrease 

n  d_{c}  ∇ 
2  .866  0.16 % 
3  .943  0.10 % 
4  .968  0.07 % 
Given  √(n^{2}1) n  .2826% √(n^{2}1) 
Efficiency, (E), will likely improve as humanity learns to design better gforce propulsion systems. Assume inefficiency (E') decreases with growth factor, n; perhaps, E'=.5×.9^{n2}. Subsequently, Efficiency (E) will increase (1E'); efficiency factor (ε = 1/E) will decrease.
Daily exhaust flow,∇, is the amount of charged particles needed to achieve gforce momentum.
Daily consumption rate. ε∇, is the amount of charged particles needed to ensure required exhaust flow. It accounts for inevitable inefficiencies. AXIOMATIC: ε∇ always exceeds ∇.
Percent Take Off Gross Weight (%TOGW) is the portion of ship's initial mass allocated for fuel.
Relativistic Growth Factor  Decimal Component Light Speed  Vessel's Propulsion Exhaust Rate  Forecast Efficiency Factor  RANGE: Propulsion Time 

ff_{Exh}=n×ff_{sec}  _{VExh= dc ×c}  ∇=ff_{Day}/GW  ε  t_{p} 
2  .866  0.163%  2.000  212 days 
3  .943  0.100%  1.818  381 days 
4  .968  0.073%  1.681  565 days 
5  .980  0.058%  1.574  763 days 
6  .986  0.048%  1.488  975 days 
7  .990  0.041%  1.419  1,197 days 
Given  √(n^{2}1) n  .2826% √(n^{2}1)  1 1 (.5×.9^{n2})  log(1%TOGW) log(1ε∇) 
CONCLUSION. As particle exhaust speed increases, vessel's range increases.
EXAMPLE: Independently vary ship's %TOGW (portion of ship's mass dedicated to fuel) from 40% to 60%. Assume particle's exhaust speed as constant 99.0% c; then, previous work leads us to determine following values. Decimal component (d_{c}) is .990; growth factor (n) is 7; and daily exhaust flow, ∇, is 0.041% Ship's GW per day.
Efficiency factor (ε) is inverse of efficiency (1/E). For interstellar performance, Thought Experiment arbitrarily assumes an efficiency model: ε = 1/E = 1/(1.5×.9^{n2}). (NOTE: There's no way of knowing the actual efficiency of future gforce propulsion systems, but we're sure their efficiency will improve.)
Daily exhaust flow,∇. Previous work approximates this value by dividing .2826% by the term, √(n²1).
Daily consumption rate. ε∇, product of efficiency factor and daily exhaust flow. ε∇ ensures sufficient quantity for daily exhaust by consuming more sure than needed. Design flaws and peripheral needs will compel a consumption rate greater than exhaust flow.
Percent Take Off Gross Weight (%TOGW). For any given particle exhaust speed, range (R) increases with %TOGW. (See following table.)
Ship's Fuel  Exhaust Speed  Growth Factor  Exhaust Rate  Consume Rate  RANGE: Prop. Time 

%TOGW  d_{c}  n  ∇  ε∇  t_{p} 
40.00%  0.866  2  0.163%/day  .326%/day  156.3 days 
50.00%  0.866  2  0.163%/day  .326%/day  212.1 days 
60.00%  0.866  2  0.163%/day  .326%/day  280.3 days 
Given  Given  1 √(1d_{c}^{2})  .2826% √(n^{2}1) 
∇
1(.5×.9^{n2}) 
log(1%TOGW)
log(1ε∇) 
CONCLUSION. As fuel load (%TOGW) increases, range increases.
V. Assume interstellar flight profile ....
REASON: To maintain Earthlike gravity (gforce) for the pax and crew throughout the multiyear voyage, the vessel must use two methods: 1) propulsion via a high speed ion stream and 2) centrifugal force by spin. Initially, ship uses ion stream to accelerate at g (g = 9.80665 m/sec²^{ }= 0.489 AU/day²) for a feasible duration (at least 100 days). Even after a year of such gforce, ship will be far short of midway between any two stars. However, TE assumes fuel limits require first few star ships to stop propulsion after only 100 days; thus, it must then cruise at constant velocity (approx .25 c) until it reaches the deceleration point 100 days prior to destination. During cruise, it must change into a habitat (roughly cylindrical shape) and spin about its longitudinal axis at an exact angular velocity to produce centrifugal gforce at inside of outer hull. At deceleration point, ship will again emit ion stream to slow vessel down to orbiting velocity at destination.
Growth Factor  At relativistic speeds, particle grows by multiple, n. 
Particle Velocity  Each multiple, n, maps to a specific, relativistic speed (v_{Exh}). 
Exhaust Flow  To achieve gforce acceleration, ship must consistently expel numerous high speed particles which collectively decrements ship's GW, expressed as a daily percentage. 
Consume Rate  Due to inevitable inefficiencies, consumption rate (ε∇) must exceed exhaust flow as determined by an Efficiency Factor (ε). 
Prop. Time  Total of 200 days. TE assumes initial interstellar vessels will gforce accelerate for 100 days, cruise for years; then, gforce decelerate for 100 days; thus, total propulsion time would be 200 days. 
Accel. Time  Divide propulsion time (t_{p}) by 2; thus, 200 days/2 = 100 days. 
MaxVelocity  TE assumes 100 days of gforce produces cruise velocity of about .25c derived from exponential (t) as shown. V_{t} = c(1(1Δ)^{t}) For more.... 
Accel Distance  TE assumes 100 days of gforce travels about .0353 LY, derived from natural logarithm. d = c×t + V/ [ln(1Δ)]. For more.... 
Gro. Fact.  Part. Vel.  Exh. Flow  Effic. Factor  Cons. Rate  200 Day Cons. 

n  v_{Exh}  ∇  ε  ε∇  %TOGW 
2  .866 c  .163% day 
2.00  .326% day  47.99% 
3  .943 c  ,100%/Day  1.818.  .182%/Day  30.49% 
4  .968 c  .073%/Day  1.681.  .123%/Day  21.76% 
5  .980 c  .058%/Day  1.574.  .091%/Day  16.61% 
6  .986 c  .048%/Day  1.488.  .071%/Day  13.26% 
7  .990 c  .041%/Day  1.419  .058%/Day  10.93% 
8  992 c  .036%/Day  1.362  .048%/Day  9.24% 
9  994 c  .032%/Day  1.314  ..042%/Day  7.97% 
10  995 c  .028%/Day  1.274  .036%/Day  6.98% 
11  996 c  .026%/Day  1.240  .032%/Day  6.2% 
Given  √(n^{2}1) n  .2826% √(n^{2}1)  1 1 (.5×.9^{n2})  ε × ∇  100%(1ε∇)^{200} 
0 Comments:
Post a Comment
<< Home