EXAMPLE: Let particle exhaust speed = 56.5% light speed; then, d_{c} = .565, and daily difference (∇) = 0.5% GW to produce gforce acceleration. For practicality, TE arbitrarily assumes inevitable inefficiencies increase actual consumption rate to twice daily diff (2×∇=1% GW) to cover design flaws and peripheral power needs.
REQUIRED FUEL
f = 1  (12∇)^{t}

RANGE
R =  log(1%TOGW)
log(12∇) 

1) NEWTONIAN. Gforce time for given distance to destination (d): t = √(2×d/g)
2) ACCELERATE to halfway (d/2). Determine gforce time: t = √(2×(d/2)/g) = √(d/g)
3) PLUS DECELERATE to destination, Double time from 2) : t = 2√(d/g)
4) PLUS RETURN. Double time from 3) . Total 2 way Travel time: t = 4√(d/g)

1% Solution to the Planets
Interplanetary Gforce solution: 2∇ = 1% GW/day
Daily GW remainder is: (12∇) = 1  .01 = 99%.
Thus, every day's GW is 99% of previous day's GW.
Determine fuel consumed:
initial GW (100%) minus expected GW(.99^{t}).
Destination

Typical
distance (d)

Travel
time (t)

Required
fuel (f)

Jupiter

6 AU

13.9 day

13.0% TOGW

Saturn

10 AU

17.9 day

16.5% TOGW

Uranus

20 AU

25.3 day

22.5% TOGW

Neptune

30 AU

31.0 day

26.8% TOGW

Kuiper Belt

40 AU

35.8 day

30.2% TOGW

Observed  4×√d
√.5 AU/day²  1  .99^{t} 
Range: R = log(1%TOGW)/log(12∇)
R = log(1.5)/log(1.01) = 68.97 days
1% consumption rate leaves plenty of margin for two way
interplanetary destinations; however, ...

...interstellar destinations need much better performance. 
As particles go faster, mass increases.
Relativistic particles grow greatly.
Thus, exhaust fuel flow (f_{Exh}) exceeds consumed fuel flow (ff_{sec}).
Restate ff_{sec}with n, mass multiple: ff_{Exh} = n × ff_{sec}
ff_{Exh} =
 ff_{Sec}
√(1 –^{ }d_{c}^{2}) 
Lorentz Transform (LT) quantifies relativistic mass increase of high speed particles. EXAMPLE: Let one second of consumed fuel flow, ff_{sec}, = 1 gm; let it achieve a relativistic velocity, d_{c}×c =.700c = 70.0% light speed prior to exhaust;
n 
=
 1
√(1  d_{c}^{2}) 
=
 1
√(1  .700^{2})  =  1.4 
then, 1.0 gm of ff_{sec} will become 1.4 gm of exhaust fuel flow, ff_{Exh},just before it exits vessel and contributes to forward momentum.
Thus, slightly change ship's mass equation:
M_{Ship} =  n × ff_{Sec} _{ }×  d_{c}× c
g  =  d_{c}
√(1  d_{c}^{2})  × 30.57×10^{6} ff_{Sec} 
However, TE still assumes daily consumption is
ff_{Day}  =  86,400sec×ff_{Sec} 
because fuel is consumed at rest.
Thus, reaccomplish daily diff. equation:
∇=  ff_{Day}
M_{Ship}  =  86,400sec×ff_{Sec}
ff_{Sec}×d_{c}×c/(g×√(1d_{c}^{2}))  =  86,400 sec × g/c
d_{c}/√(1d_{c}^{2})  =  .2826% GW
n × d_{c} 
EXAMPLE: Let particle's exhaust speed (V_{Exh}) be .7c; then, light speed's decimal component, d_{c }, is .7; mass multiple, n, is 1.4; and ship's gross weight daily difference, 2∇, computes as 0.58% GW. Subsequently, compute the daily gross weight remainder (100% 2∇)= (1.0058) = .9942.
During powered flight, the gforce vessel's GW is ever decreasing at the daily rate of 2∇; thus, the quantity, 12∇, is a recurring remainder, and the exponential, t, an compute ship's gross weight (GW_{t}) after t days of powered flight. EXAMPLE; Let accel time (t) be 30 days.
GW_{t }= GW_{0}(12∇)^{t}= (1.0058)^{30 }GW_{0} = 84.1% GW_{0}
If t, acceleration time, is known (Example: t = 60 days) ; then, determine fuel required:
f _{t}= GW_{0}(1(12∇)^{t}) = (1(.9942)^{30})GW_{0} = 15.9% TOGW
Under above conditions, vessel must expend 15.9% of initial GW to maintain gforce for 30 days (NOTE: Artificially estimates efficiency as 50%.)

Relativity Increases Gforce Effect 
ff_{Exh} × V_{Exh} / g = M_{Ship} = n × ff_{sec}× d_{c} × c / g 
_{d}_{c} 
n 
∇ 
t_{Prop} 
t_{Acc} 
_{V}_{Cru} 
d_{Acc} 
dec. comp. 
Mass Mult. 
Daily Diff. 
Prop.
time 
Accel time 
Cruise
Veloc. 
Accel
Dist. 
.300  1.05  0.90% GW  38.2 days  9.6 days  4.6 AU/dy  28 AU 
.400  1.09  0.65% GW  53.2 days  13.3 days  6.4 AU/dy  43 AU 
.500  1.15  0.49% GW  70.5 days  17.6 days  8.4 AU/dy  75 AU 
.600  1.25  0.38% GW  91.6 days  22.9 days  10.9 AU/dy  126 AU 
.700  1.40  0.29% GW  119.9 days  30.0 days  14.1 AU/dy  214 AU 
.800  1.67  0.21% GW  163.2 days  40.8 days  18.9 AU/dy  392 AU 
.866  2.00  0.16% GW  212.7 days  53.2 days  24.2 AU/dy  659 AU 
Given  1
√(1  d_{c}^{2})  .2826%GW
n× d_{c}  log(1%TOGW)
log(1ε∇)  _{t}_{Prop}
4  c[1R_{c}^{tAcc}] 

Propulsion Time (t_{Prop}): Propulsion system converts fuel to energy; thus, propulsion time is the ship's range, available travel time (days) due to completely converting ship's available fuel to forward momentum.
%TOGW: Relative amount of vessel's initial fuel is often called: "Percent Take Off Gross Weight (%TOGW)".
EXAMPLE1: If vessel's initial GW is 100 mTs and initial fuel load is 60 mTs; then %TOGW is 60%, and ship's minimum weight is 40%TOGW or 40mT.
EXAMPLE2: If %TOGW is 50%; then, ship's minimum weight (fuel depleted) is 50%. For convenience, this table assumes %TOGW = 50%.
Daily Difference (∇): To maintain gforce propulsion, ship's exhaust requires a certain daily amount (∇) to exit ship's exhaust at certain high velocity to properly contribute to ship's forward momentum.
Efficiency Coefficient (ε): To overcome inevitable inefficiencies, ship's daily fuel consumption must exceed ∇ to ensure gforce momentum. ε is TE's best guess as to how much to increase. For this table, TE conveniently assumes a single value of 2 for ε. For following tables, TE artificially assumes a dynamic efficiency model as described below.
Acceleration Time: Typical interstellar flight profile can divide into four phases; thus, P =4. Phases are described below; for each phase,: t = t_{Prop}/4.
Phase1) Accelerate to midpoint. Fuel = GW_{0}(1(1∇)^{t}) = 15.9% TOGW
Phase2) Decelerate to dest. Fuel = GW_{0}(1(1∇)^{2t}) = 29.3% TOGW
Phase3) Accelerate back to midpoint. Fuel = GW_{0}(1(1∇)^{3t}) = 40.5% TOGW
Phase4) Decelerate and return to dept. Fuel = GW_{0}(1(1∇)^{4t}) = 50.0% TOGW
AXIOMATIC: All four phases require equal time/distance.
Cruise Velocity: After t days of gforce propulsion, vessel accelerates to velocity which it will maintain throughout the cruise portion of flight to nearby star. Daily Light Speed Increment due to gforce (Δ): AXIOMATIC: After one day of gforce, the vessel accelerates to a velocity of 86,400 sec × 9.8065 m/s² = 847.2 km/sec = 0.2826% c = Δ . Due to relativity, TE further assumes that Δ = 0.2826% c for any gforce day.
Daily Remainder till light speed (R_{c}): Thus, onboard observer measures c(1Δ) = 0.997174 c, remaining velocity until light speed, c for any gforce day. TE assumes that earthbound observer can model ship's velocity as difference between c ( 299,792,458 m/s = 173.145 AU/day) and recurring remainder (1Δ)^{t }which can be determined with exponential, t, as number of gforce days.
Acceleration Distance: TE uses a calculus textbook to determine a straightforward method to calculate distance from above velocity equation.

E 
Efficiency (E) is output divided by input.
It's always less then one.
Assume propulsion system superheats some water into 1,000 plasma ions for the propulsion system. The goal is to output all 1,000 particles from the exhaust manifold at very high speeds and contribute to the ship's forward momentum. However, inevitable inefficiencies will prevent some of the 1,000 from fulfilling their destiny. If only 750 ions make it out the exhaust, efficiency is 75%:
E = out/in = 750/1000 = 75%.

E'  Inefficiency (E') is loss divided by input. In our example of the 1,000 water ions, 250 ions don't make to the exhaust.
Note that input equals output + loss; thus, E + E' = 100%.
Since E and E' are complementary,
we can determine a system's efficiency by measuring its inefficiency.
E = 1 E'


E' 
TE assumes a model of decreasing inefficiency to enable a corresponding model of increasing efficiency. Assume that ever improving technology will increase propulsion performance while simultaneously decreasing inefficiency. TE arbitrarily assumes a baseline of 50% inefficiency when propulsion performance enables a mass multiple (n) of 2. For every integer increase above n = 2, TE arbitrarily assumes an inefficiency decrease of 10%: E' = (.5×.9^{n2}) Thus,
E = 1  E' = 1  (.5×.9^{n2})


ε 
Efficiency Factor is Inverse of Efficiency: ε = 1/E.
If you need 1,000 high speed ions to move vessel at gforce for one second, but systemic inefficiency only allows 750 ions through the propulsion system; then, the system needs to offset the input with enough ions to bring the output to the full 1,000. Thus, planners can use efficiency factor to to do this. Until experience gives us actual efficiency factor; arbitrarily assume:


Let one second of consumed fuel flow, ff_{sec},
be 1,000 gms (one kilogram);
let it achieve a relativistic velocity, d_{c}×c =.992c = 99.2% light speed;
then, 1.0 kg of ff_{sec} will become 8.0 kgs of exhaust fuel flow, ff_{Exh}, just before it exits vessel and contributes to forward momentum.
For convenience, note following identities:
 Dynamic Efficiency Factor (ε): As time passes, propulsion systems will likely improve with both better performance and better efficiency. If exhaust particle speed is a good performance measure; then, relativistic mass growth, n, is also a good measure since it maps directly to speed. Furthermore, Thought Experiment (TE) artificially assumes a dynamic efficiency model using same n value to reflect ever improving efficiency.
Increases Gforce Efficiency 
∇=  ff_{Day}
M_{Ship}  =  86,400sec×ff_{Sec}
n×ff_{Sec}×d_{c}×c/g  =  .86,400×g/c
n × d_{c}  =  .2826% GW
√(n^{2}1) 

n 
∇ 
_{ε} 
_{t}_{Acc} 
_{V}_{Cru} 
_{d}_{Acc} 
Mass Mult. 
Daily Diff. 
Effic. Fact. 
Accel time 
Cruise
Vel. 
Accel dist. 
2  0.163%  2.00  53 days  24.1 AU/dy  655 AU 
3  0.100%  1.82  95 days  40.9 AU/dy  2,038 AU 
4  0.073%  1.68  141 days  57.0 AU/dy  4,295 AU 
5  0.058%  1.57  191 days  72.3 AU/dy  7,510 AU 
6  0.048%  1.49  244 days  86.3 AU/dy  11,709 AU 
7  0.041%  1.42  299 days  98.9 AU/dy  16,870 AU 
8  0.036%  1.36  357 days  110.1 AU/dy  22,939 AU 
Given  .2826%
√(n^{2}1)  1
1(.5×.9^{n2})  log(1%TOGW)
4×log(1ε∇) 
c[1R_{c}^{tAcc}]

c×t_{Acc}  +  V_{Cru}
ln(R_{c}) 

Percent Take Off Gross Weight (%TOGW) explained in previous table. For above table, TE assumes %TOGW= 50%. Daily Remaining velocity (R_{c}) explained in previous table. TE assumes onboard observer measures R_{c} = c(1Δ) = 0.997174 c,


Acceleration: After one year (365.25 days) of gforce propulsion, vessel will achieve velocity of 64.43% light speed (c).
Time: Acceleration duration is not entirely arbitrary. Vessel's range must accommodate acceleration to cruise, deceleration from cruise to destination and repeat for return leg of journey. Thus, one quarter of vessel's range can be considered available for acceleration duration.
Velocity: TE assumes gforce velocity determined by: V_{Cru}=c[1  R_{c}^{t}]
Distance: TE assumes gforce distance determined by: d = c[t + V_{Cru}/ln(R_{c})]
Deceleration: To maintain gforce while decelerating from 64.43%c to essentially zero, deceleration requires same duration/distance as acceleration. 

Interstellar Profile: Consistent Acceleration Time
Destination  Acceleration Time  Accel.
Velocity  Accel.
distance  Cruise distance  Cruise time  Total Time 
Star  D  t  t_{Day}  V_{Cru}  d  d_{Cru}  t_{Cru}  t_{r}  T 
Name  LY  Yr  day  % c  LY  LY  Yrs  Yrs  Yrs 
α Cent  4.365  1.0  365.3  64.43%  0.38  3.61  5.60  3.34  7.60 
Bar...Sta  5.970  1.0  365.3  64.43%  0.38  5.21  8.08  4.83  10.08 
Luh...16  6.600  1.0  365.3  64.43%  0.38  5.85  9.07  5.41  11.07 
OBSERVED  Given  365.25t  See previous tables to see how to compute these values.  D
 2d  d_{Cru}
V_{Cru}  t_{Cru}
n  t_{Cru}
+ 2t 
Cruise: Due to vessel's finite range, interstellar vessel must stop acceleration well prior to the interstellar midpoint and maintain constant velocity until deceleration.
Cruise Velocity achieved during acceleration remains vessel's velocity during cruise duration.
Distance: determined by subtracting acceleration distance (d) and deceleration distance (d) from total trip distance (D). Since acceleration distance equals deceleration distance; cruise distance can be expressed: d_{Cru} = D  2 d
Cruise Time can be expressed as normal time and relativistic.
Normal Time 
At vessel velocity of 64.43%c,
Earth bound observer measures 1.55 years
for each LY of cruise distance.
 t_{Cru}=d_{Cru}/V_{Cru} 
Relativistic Time 
Due to relativistic effects,
ship bound chronometer measures:1.19 years
for each cruise LY.
 t_{r}=t_{Cru}×√(1V_{Cru}^{2}/c^{2}) 

Destination  Acceleration Time  Accel.
Velocity  Accel.
distance  Cruise distance  Cruise time  Total Time 
Star  D  t  t_{Day}  V_{Cru}  d  d_{Cru}  t_{Cru}  t_{r}  T 
Name  LY  Yr  day  % c  LY  LY  Yrs  Yrs  Yrs 
α Cent  4.365  0.8  292.2  52.26%  0.26  3.85  6.85  5.66  8.45 
Bar...Sta  5.970  0.9  328.7  60.56%  0.31  5.34  8.82  7.02  10.62 
Luh...16  6.600  1.0  365.3  64.43%  0.38  5.85  9.07  6.94  11.07 
Wolf..359  7.786  1.1  401.8  67.92%  0.4  6.90  10.16  7.46  12.36 
Lal...21185  8.291  1.2  438.3  71.07%  0.51  7.27  10.22  7.19  12.62 
Sirius  8.58  1.3  474.8  73.91%  0.58  7.41  10.03  6.76  12.63 
OBSERVED  Given  365.25t  See previous tables to see how to compute these values.  D
 2d  d_{Cru}
V_{Cru}  t_{Cru}
n  t_{Cru}
+ 2t 

Interstellar Profile: Tailor Acceleration Times
As destination distances increase, consider increasing acceleration times to increase cruise velocities. Thus, overall trip time would decrease.

TRADEOFFS. For interstellar flights, planners will consider the tradeoffs between available fuel and available range.
1) Longer propulsion durations means accelerating to a faster cruise velocity and shorter overall trip time.  2) Shorter propulsion times accelerate the vessel to a slower cruise velocity; thus, cruise duration increases. However, it requires less fuel and decreases the risk of inflight fuel depletion. If ship runs out of fuel at a high velocity; then, vessel and occupants are stranded at that velocity with no way to slow down. 
Tailor Propulsion Performance
to accommodate destination distances. Increase mass multiple (n) to increase acceleration time (thus, increasing cruise velocity) and decrease cruise time.
n  d_{c}  ∇  ε  Accel. Time (t)  Cruise Vel. (V_{Cru})  Accel. Dist. (d) 
8  0.9922  0.036%GW  1.36  357 dy  0.98Yr  110.2AU/dy  .636c  22,939AU  0.36LY 
9  0.9938  0.032%GW  1.31  417 dy  1.14Yr  120,0AU/dy  .693c  29,842AU  0.47LY 
10  0.9950  0.028%GW  1.27  479 dy  1.31Yr  128.5AU/dy  .742c  37,490AU  0.59LY 
11  0.9959  0.026%GW  1.24  542 dy  1.48Yr  135.8AU/dy  .784c  45,793AU  0.72LY 
12  0.9965  0.024%GW  1.21  605 dy  1.66Yr  141.9AU/dy  .820c  54,661AU  0.86LY 
13  0.9970  0.022%GW  1.19  670 dy  1.83Yr  147.2AU/dy  .850c  64,010AU  1.01LY 
14  0.9974  0.020%GW  1.16  735 dy  2.01Yr  151.5AU/dy  .875c  73,763AU  1.17LY 
15  0.9978  0.019%GW  1.15  801 dy  2.19Yr  155.2AU/dy  .896c  83,850AU  1.33LY 
Given  √(n^{2}1)
n  0.2826%
√(n^{2}1)  1
1(.5×.9^{n2})  log(1%TOGW)
4×log(1ε∇)  c[1  R_{c}^{t}] 


Destination  Mass Mult.  Accel time  Cruise Vel.  Accel. dist.  Cruise dist.  Cruise Time  Total Time 
Star  D  n  t  V_{Cru}  d  d_{Cru}  t_{Cru}  t_{r}  T 
Name  LY  Yr  dec. c  LY  LY  Yr  Yr  Yr 
α Cent  4.365  8  0.98  0.64 c  0.36  2.42  3.80  2.93  5.65 
Bar...Sta  5.970  9  1.14  0.69 c  0.47  3.69  5.33  3.84  7.61 
Luh...16  6.600  10  1.31  0.74 c  0.59  3.99  5.37  3.60  7.99 
Wolf..359  7.786  11  1.48  0.78 c  0.72  4.83  6.16  3.82  9.13 
Lal...21185  8.291  12  1.66  0.82 c  0.86  4.99  6.08  3.48  9.40 
Sirius  8.583  13  1.83  0.85 c  1.01  4.92  5.79  3.05  9.46 
Luy.7268  8.728  14  2.01  0.88 c  1.17  4.71  5.38  2.60  9.41 
Ross 154  9.681  15  2.19  0.90 c  1.33  5.30  5.92  2.62  10.30 
OBSERVED  See green table in left panel.  D
2d  d_{Cru}
V_{Cru}  t_{Cru}
n_{Cru}  t_{Cru} +2t 

CONCLUSION: Many ways to get there!!! 


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