If gforce is great, is super gforce even greater?
Expectations continually adjust to ever higher levels. For example, it's very possible that someone will win one million dollars in a lottery; you'd think she'd be happy forever. However, relevant experience indicates such happiness lasts for a much shorter time.
Our neophyte millionairess will soon learn that a million dollars is not infinite, and materialistic desires will likely grow to accommodate that finite amount. For instance, she can now fly first class, but she can't yet buy a private jet.
In similar fashion, consistent gforce propulsion will prove a great boon to humankind due to very short travel times to nearby planets as well as inflight comfort from simulated Earthlike gravity. However, humans will soon want more stuff shipped even faster. Humans probably won't travel at greater than one gforce, but shipments of essential materials could be greatly expedited if they could fly at superg forces.
Near Earth's surface, a free falling object will accelerate at rate g, approx. 10 m/sec^{2}.
Thus, an acceleration of 2 g would be 20 m/sec^{2}.
Using Newtonian motion equations, we've previously determined that first second of 1g free fall will move object 5 meters. Extend to one day of constant gforce acceleration to move vessel to .25 Astronomical Units (AUs).

Expand our thought experiment's envelope to exceed gforce acceleration for selected vessels. Assume no humans, but Artificial Intelligence (AI) devices will guide a vessel with essential cargo.
Express 2g as different units of distance/time.
20 m
sec^{2}  ≈  2g ≈  20 m
sec^{2}     ×  (86,400 sec)^{2}
day^{2}     ×  1 AU
1.5×10^{11}m    ≈  2g ≈  1.0 AU
day^{2}  

If double gforce does not damage ship or contents; then, much shorter travel time would be an enormous benefit.
d  =  a×t^{2}
2  ≈  1.0 AU
day^{2}  ×  (1 day)^{2}
2  =  0.5 AU 
let a = 2g = 1.0AU/day²
Note that one second of 2g free fall moves an object for 10 meters; thus, extend duration to one day (86,400 seconds) for distance of 0.5 AU.


Describe greater accelerations as integer (i) times g, free fall acceleration experienced near Earth's surface.
DISTANCE EXAMPLE: After one day of 2gforce acceleration, object travels a great distance:
d_{Accel }= .5 × a × t^{2} = .5 × 2g × t^{2 }= .5 × 2(.5AU/day^{2}) × day^{2} = .5 AU


ACCELERATE TO GREAT SPEEDS QUICKER!
After one day (=24 hrs × 3,600 sec/hr =86,400 sec) of 2×gforce acceleration,
V_{Mid}  ≈  2g × t ≈  20 m
sec^{2} 


 ×  86,400 sec 


 =  1,728,000 m
sec 

vessel increases velocity to 1,728 km/sec, an enormous speed!
For current example, let this happen at midpoint.
V_{Mid}  ≈  1,728 ,000 m
sec 


 ×  1 km
1,000 m  =  1,728 km
sec


Thus, vessel must SLOWDOWN during 2nd half of trip,
which takes a day of 2g deceleration.
During 2nd day, vessel's speed decreases at 2g to 0 km/sec.
V_{Dest}  ≈  V_{Mid } 2g × t ≈  1,728,000 m
sec    20 m
sec^{2}  ×  86,400 sec 

RESTATE: Accelerate at negative rate (2×g) for 2nd day
V_{Dest}  ≈  1,728 km
sec    1,728 km/sec
day 


 ×  1 day  =  0 km
sec 

to return to initial velocity.

QUICKLY DECELERATE TO ORBITAL SPEED!
AXIOMATIC: 2nd day deceleration distance from midway will equal first day’s acceleration distance to midway:
d_{Decel}  ≈  .5×a×t^{2}= .5×2g×t^{2} ≈  .5  ×  2  .5AU
day^{2}  ×  1day^{2}  =  .5AU  

ABSOLUTE DISTANCE:
[d_{Decel}] = .5 AU = d_{Accel} 

Thus, the sum of acceleration distance (for first half of voyage) and deceleration distance (for final half) is the total distance; or total could be restated as twice the acceleration distance.
d_{Accel} + [d_{Decel }]= d_{Ttl }= 2 (.5×a×t^{2}) 

Determine total distance for a 2g vessel to accelerate for one day and decelerate for 2nd day.
d_{Ttl}  ≈  2(.5×2g×t^{2}) ≈  2  [.5×2×  .5AU
day^{2} 
 ×1day^{2}]  =  1.0AU  

Gforce vessels as well as super gforce vessels both require: 1) acceleration to midway for first half of trip. 2) equal propulsive force in opposite direction for 2nd half. Thus, vessel attains useful, orbital speed at destination.

Distances: Different Gforces
One Day Accelerate; 2nd Day Decelerate
Acceleration (a)  Velocity (V= a×t)  Distance (a×t^{2}/2) 
Same value, different units.  After one day's acceleration.  d_{Accel}  d_{Decel}  d_{Ttl} 
i × g  m/sec^{2}  AU/day^{2}  km/sec  AU/day  % c  AU  AU  AU 
1 g  10  0.5  864  0.5  0.288%c  0.25  0.25  0.5 
2 g  20  1.0  1,728  1.0  0.576%c  0.50  0.50  1.0 
3 g  30  1.5  2,592  1.5  0.864%c  0.75  0.75  1.5 
Given  i × 10 m
sec^{2}  i × .5 AU
day^{2}  i×(864km/s)×t
day  i×(.5AU)×t
day^{2}  i×(.288%c)×t
day  i×(.5AU)t^{2}
2×day^{2}  = d_{Accel}  = d_{Accel}
+ d_{Decel} 
 gravity force (g): experienced by all who reside on Earth's surface.
 10 m/sec²: acceleration experienced by freely falling objects near Earth's surface.
 .5 AU/day²: 10 m/sec² is transformed into units more convenient for space travel.

 km/sec: 1 g speed increases 10 m/sec for every sec of entire day (86,400×10=864 km/sec); an enormous speed.
 AU/day: for a more convenient measure, transform 864 km/sec to 0.5 AU/Day.
 %c: 1 day of gforce achieves small portion of light speed (c). 864 km/sec ÷ c = .288%,

d_{Accel}: Acceleration distance
d_{Decel}: Decel. dist. = Accel dist.
d_{Ttl}: d_{Accel }+ d_{Decel } = 2 × d_{Accel}

Determine distance after one day's acceleration followed by one day's deceleration.

For a given set of SuperG's
determine flight time for a given distance, d.
t_{Accel} = √(d/a), estimates acceleration time to halfway.
EXAMPLE: Typical distance from Earth to Jupiter: d_{ }= 5AU.
d _{Accel} = d _{Decel} = d/2 = 2.5 AU
t_{Accel}  =  √  2 × d_{Accel}
a  =  √  d
3g  = 
 √  5 AU
1.5AU/day^{2}  =  1.826 days 
let a = 3g = 1.5AU/day²
t_{Accel} + t_{Decel} = t = 2 × t_{Accel} = 2√(d/(3g))= 3.652 days

Accelerate to midway for half of distance. Decelerate to destination for remaining half.
Travel Times for Different Gforces 
 Acceleration (i ×g)  1 g  2 g  3 g 
Dest  Typical Distance  Flight Time 
NEO  1 AU  2.8 day  2.0 day  1.6 day 
Mars  2AU  4.0 day  2.8 day  2.3 day 
Jupiter  5AU  6.3 day  4.5 day  3.7 day 
Given  Observed 


Total time considers acceleration to halfway: √(d/a)
plus equal time for deceleration. 
Fuel Consumptions: Different Gforces
Both acceleration and deceleration require powered flight, which requires fuel.

Assume for interplanetary flights, particle exhaust speed,
V_{Exh} = .866 c
Thus, decimal component of light speed:
d_{c} = V_{Exh} /c = .866
From Lorentz Transform, growth factor:
n = 1/√(1d_{c}^{2}) = 1/√(1.866^{2}) = 2
At particle velocity of 86.6%c,
exhaust flow is twice fuel consumption at rest:
ff_{Exh} = 2 × ff_{sec}
Previous work leads to following equation:
∇ =  i × g × day
c × √(n^{2}1)  =  i × .288%
√3  =  i × .1663% 
"i" can be a single digit integer (i.e., 1,2, 3, ...) as shown below.
Fuel Consumptions for Different Gforces 
 Acceleration (n×g)  1 g  2 g  3 g  ∇=.17%TOGW  ∇=.33%TOGW  ∇=.50%TOGW 
Dest  Typical Distance  Flight Time  Corresponding Fuel Consumed 
NEO  1 AU  2.8 day  2.0 day  1.6 day  0.94%TOGW  1.32%TOGW  1.69%TOGW 
Mars  2AU  4.0 day  2.8 day  2.3 day  1.32%TOGW  1.88%TOGW  2.30%TOGW 
Jupiter  5AU  6.3 day  4.5 day  3.7 day  2.10%TOGW  2.98%TOGW  3.70%TOGW 
Given  Observed 
 Fuel: f = t × ε ×∇ 
Efficiency Factor (ε) arbitrarily chosen as 2.0 for interplanetary travel.
 Define variables:
1) ∇, daily difference = daily percentage of ship's mass needed to convert to exhaust particles to propel ship in opposite direction.
2) g, from expression "gforce", g is acceleration due to Earth surface gravity. For convenience, round g's value to 10 m/sec^{2}.
3) day = quantity of seconds per day = 86,400 sec.
4) c, light speed. Round to 3×10^{8}m/sec.
5) Given above approximations, g×day/c = .00288 = .288%.
6) n, growth factor, is the amount of relativistic mass increase due to high speeds achieved by accelerated particles. Thus, exhaust fuel flow, ff_{Exh} is greater then ff_{sec }, original fuel flow per second:
ff_{Exh}= n × ff_{sec}
ff_{Exh} =  ff_{sec}
√(1d_{c}^{2})     
7) d_{c} is decimal component of V_{Exh}, exhaust particle speed expressed as decimal light speed. Thus,
d_{c} = V_{Exh}/c
While particle exhaust velocity (V_{Exh})_{ }can range from zero to just short of c, assume for interplanetary voyages one reasonable value of V_{Exh} = 86.6% c = .866 c.

Operational Flight Plan Different SuperG's
Determine time of flight after halfway acceleration followed by halfway deceleration.
 Let V_{Exh} = .866c  ∇= i × .1663% 
a = 1 × g
∇=0.17%
 a = 2 × g
∇=0.34%
 a = 3 × g
∇=0.50%
 a = 4 × g
∇=0.67%
 a = 5.5 × g
∇=0.91%

Dest  Dist. (d)  Time(t)  Fuel (f)  Time  Fuel  Time  Fuel  Time  Fuel  Time  Fuel 
Typical  days  %TOGW  days  %TOGW  days  %TOGW  days  %TOGW  days  %TOGW 
NEO  1 AU  2.83  0.94%  2.00  1.33%  1.63  1.62%  1.41  1.88%  1.31  2.20% 
Mars  2 AU  4.00  1.32%  2.83  1.87%  2.31  2.29%  2.00  2.64%  1.71  3.10% 
Jupiter  5 AU  6.32  2.09%  4.47  2.94%  3.65  3.60%  3.16  4.15%  2.70  4.86% 
Saturn  10 AU  8.94  2.94%  6.32  4.13%  5.16  5.05%  4.47  5.81%  3.81  6.80% 
Uranus  20 AU  12.65  4.13%  8.94  5.79%  7.30  7.06%  6.32  8.12%  5.39  9.49% 
Neptune  30 AU  15.49  5.03%  10.95  7.05%  8.94  8.58%  7.75  9.85%  6.61  11.48% 
Kuiper Belt  40 AU  17.89  5.79%  12.65  8.10%  10.33  9.84%  8.94  11.29%  7.63  13.14% 
Given  Observed 


For even faster interplanetary transport, consider super Gforces even greater than 3g. With acceleration as i × g, "i" values could include both integers and decimals.

100 AU from Sol
This is much further than the Kuiper Belt (KB), but only a miniscule portion of distance from Sol to the edge of the Solar System.
NOTE: Many people consider the KB to be the "edge" of the Solar System, as we believe there are no planets beyond this belt of asteroids and comets. However, Sol's influence extends well beyond the KB to the Oort Cloud (OC) which contains billions of comets. OC is about 1 LY in diameter (about 65,000 AU) and it's about 1/4 of distance to nears stellar neighbor, Alpha Centauri stellar system.

SCENARIO: Normal passenger travel typically takes about 28 days to travel from Earth to this site at 1 g. Flight requires fuel of about 9% of vessels TOGW. However, nonpassenger transport could travel at higher g's to expedite transport of essential cargo.
WHY ORBIT AT 100AU? Why deploy a habitat to where there are no planets, no moons, perhaps a rare glimpse of an asteroid/comet?
ANSWER: At 100 AU, this habitat could become an extremely valuable resource; it can store and forward large volumes of communications from the inner solar system to ships heading farther out, eventually to the stars.
 Quick look at following table tells us that a 4 g vessel can make same trip in half the time of an one g flight while consuming twice the fuel
 a = 1 × g
∇=0.17%
 a = 2 × g
∇=0.33%
 a = 3 × g
∇=0.50%
 a = 4 × g
∇=0.67%

Dist  time  fuel  time  fuel  time  fuel  time  fuel 
1  2.83  0.94%  2.00  1.33%  1.63  1.62%  1.41  1.88% 
100  28.28  8.99%  20.00  12.50%  16.33  15.11%  14.14  17.26% 

CONCLUSION: Interplanetary flight profiles for both gforce and super gforce propulsions
can be closely approximated with Newtonian motion equations. 
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