1. Momentum Exchange 
Our notional spacecraft uses particle accelerators to output a continuous stream of exhaust particles at near light speed. The resultant constant force propels spacecraft at ever increasing velocities.
Recall Momentum Equation from previous work. After some manipulation, we get:
ff_{Exh}  ×  V_{Exh}  =  M_{ship}  ×  g 
EXAMPLE:
1 kg
sec  ×  30,000,000m
sec  =  3,000,000kg  ×  10m
sec² 
For one second of powered flight,
 Let exhaust fuel particle mass (ff_{Exh}) be one kilogram (kg)
 Let exhaust particle velocity (V_{Exh}) be 30 million meters per second, or about 10% c, light speed.
 Let spacecraft mass (M_{Ship}) be about 3 million kgs (3,000 metric Tonnes, mTs). Using more precise values for c and g; then, M_{Ship} = 1 kg × .1 c/g = 3,057 mT.
 Given above three terms, spaceship can increase its velocity about 10 m/sec (happens to be same value as g, acceleration due to near Earth gravity, about 10 m/sec²).
Such a propulsion system can enable spacecraft to reach nearby planets in days and nearby stars in years; further, the spacecraft can simulate Earth gravity. However, constant gforce requires a lot of fuel.
For one day (86,400 sec) of gforce propulsion, 1 kg/sec fuel flow (ff_{sec}) will become a daily fuel flow (ff_{dy}) of 86.4 mTs. Expressed as a percent of ship's mass, daily fuel consumption is about 2.8% Take Off Gross Weight (TOGW). Exact value:
ff_{Day} = 86.4 mT/3,057 mT = 2.826%TOGW.
Thus, this thought experiment constructs several helpful tables. First table lists typical gforce times to nearby planets. Short durations indicate minimal fuel requirements. If our notional propulsion system accelerates fuel particles to 10% light speed, fuel requirements for a few days of flight time could be a reasonable percentage of the ship's mass.

TABLE1. Gforce to Nearby Planets: Total Fuel Requirements
 g = 0.5 AU/day^{2}  ff_{Day}=86,400×ff_{sec } 
dist.  time  Fuel 
d

t_{ }

F

Mars

2 AU

4.0dy

11.31% TOGW

Jupiter

6 AU

6.92dy

19.58% TOGW

Saturn

10 AU

9.94dy

28.13% TOGW

Uranus

20 AU

12.64dy

35.77% TOGW

Neptune

30 AU

15.46dy

43.75% TOGW

Kuiper Belt

40 AU

17.89dy

50.62% TOGW

Observed  Typical  2×√d
√g  t_{ }× ff_{day}
_{} M_{Ship} 
For TABLE1, let ff_{sec }= 1.0 kg and M_{Ship }= 3,057 mT.
Let ff_{sec} be an unspecified quantity of fuel needed to propel our spaceship 10m/s faster for each second of space flight.
Since 1 day = 86,400 secs and ff_{Day} is fuel flow per day; then,
ff_{Day}= ff_{sec} × 86,400 secs/day
Required fuel is trip time times daily fuel flow:
F = t_{ }× ff_{Day}
Therefore,
F = 2 × √(d/g) × ff_{sec} × 86,400 secs/day
To express fuel as percentage weight of spaceship:
%TOGW = F / M_{Ship} = (t_{ }× ff_{Day}) / M_{Ship}

2. Relativistic Impact 
Can an interstellar vessel carry enough fuel?
Consider following points:
Technology Growth Continues. We expect initial space borne particle accelerators to have their problems; however, we also expect human engineers to continually improve them. After many decades of interplanetary travel, space borne particle accelerators will become very reliable and efficient with much greater performance. Since current particle accelerators routinely achieve burst speeds much greater than .99c, our thought experiment reasonably assumes reliable, efficient particle exhaust streams with speed exceeding .9c. For practical interstellar travel; such exhaust speeds will be essential. Einstein's special theory of relativity describes changes at near light speed velocities.
ff_{Exh} =  ff_{Sec}
√(1  v^{2}_{Exh}/c^{2}) 
To apply LT to our space borne particle accelerator, make the following term replacements. Replace m _{0} with the term, ff _{Sec}, original sized mass of fuel flow per second. This is fuel consumed per second at the start of the particle accelerator. Replace m _{r} with the term, Exhaust fuel flow (ff _{Exh}), which is ff _{Sec} after relativistic growth. Thus, one second of original fuel flow per second becomes relativistic exhaust fuel per second at the end of the particle accelerator cycle when particles achieve exhaust velocity (v _{Exh}, which replaces v _{r}). Thereafter, exhaust particles (ff _{Exh}) exit propulsion system and "push" spacecraft in opposite direction.
n  =  1
√(1  v^{2}_{Exh}/c^{2}) 
LT can be restated with exhaust velocity equals fuel flow (per sec) at rest times a multiple (n). Multiple can be expressed as shown.
n  =  1
√(1  d_{c}^{2}c^{2}/c^{2}) 
v _{Exh} can be expressed as "d _{c} × c" or "d _{c} c" where decimal component (d _{c}) is a decimal between zero and one, and c is light speed. Note that d _{c }= v _{Exh}/c. Thus, d _{c}^{2 }= v ^{2}_{Exh}/c ^{2}, and substitution is made as shown.

TABLE2a. Relativity and Fuel Particle Mass
V_{Exh}  d_{c}  n  ff_{Exh} 
(decimal c)  Dec. Comp.  Mass Mult.  Exhaust Flow 
.10 c  .1  1.005  1.005 ff_{sec} 
.20 c  .2  1.021  1.021 ff_{sec} 
.30 c  .3  1.048  1.048 ff_{sec} 
.40 c  .4  1.091  1.091 ff_{sec} 
.50 c  .5  1.155  1.155 ff_{sec} 
.60 c  .6  1.250  1.250 ff_{sec} 
.70 c  .7  1.400  1.400 ff_{sec} 
.80 c  .8  1.667  1.667 ff_{sec} 
.866 c  .866  2.000  2.000 ff_{sec} 
Given  V_{Exh}
c  1
√(1  d_{c}^{2})  (n × ff_{sec}) 
TABLE2a shows range of speeds from .1c with negligible mass increase to .866c when fuel particle mass doubles
(mass multiple (n) =2).
Apply Lorentz Transform
to determine mass growth due to particle exhaust speed
(v_{Exh}, shown as decimal light speed).
ff_{sec }is original fuel mass (per second) at rest. Fuel is consumed at rest.
ff_{Exh} is same particle mass after ingestion into particle accelerator, transformation into plasma (ions) and accelerated to particle exhaust speed. Fuel particles impart momentum at exhaust speed.
m_{r} =  m_{o}
√(1  v^{2}_{r}/c^{2}) 
Lorentz Transform (LT) quantifies relativistic mass increase of high speed particles.EXAMPLE: Let one gram of original mass, m_{o}, achieve a relativistic velocity, v_{r} = .866c = 86.6% light speed; then, relativistic mass, m_{r }, will be 2.000 grams.
After few axiomatic substitutions, terms eventually reduce to equation at left.

TABLE2b restates momentum equation to determine ship's mass.
Since relativistic speed increases the mass of exhaust fuel particles, ff_{Exh} can be expressed as a multiple of original fuel quantity per second n × ff_{Sec} This slightly changes TE momentum equation:
n × ff_{Sec} × v_{Exh} = M_{ship} × g 
Rearrange TABLE2a; for independent variable (IV), 2b uses fuel particle multiple (n) as a range of integers. Fuel particle exhaust velocity (V _{Exh}) could be any of an infinite number of values between zero and light speed, c. However, TABLE2b uses subset of V _{Exh }, selected velocity values such that exhaust particle (ff _{Exh}) grows to integer multiple, n, of consumed fuel flow per second (ff _{Sec}).
EXAMPLE: Let V _{Exh }= .943c, then, relativity triples the size of consumed fuel particle:
n × ff_{Sec} = ff_{Exh} = 3 × ff_{Sec}
Transform from solving for n to solving for d _{c}.
^{}Square both sides.

1 + d_{c}^{2}  =  1
n^{2} 

Rearrange as shown.

d_{c}^{2}  =  n^{2 } 1
n^{2} 


Solve for decimal component of light speed, d_{c}.
Given sent of integers {n}, map to corresponding set of decimal values {d_{c}} which readily show required particle speeds required to grow mass a certain multiple, n. 
TABLE2b. Relativity & M_{Ship }Capacity.
ff_{Exh} × V_{Exh} / g = M_{ship} = n × ff_{sec} × d_{c} × c/g 
_{n}  d_{c}  M_{Ship} 
Mass Mult.  (decimal c)  (megaff_{sec}) 
1  .0  n/a 
2  .866  52.95 megaff_{sec} 
3  .943  86.46 megaff_{sec} 
4  .968  118.40 megaff_{sec} 
5  .980  149.76 megaff_{sec} 
6  .986  180.85 megaff_{sec} 
7  .990  211.80 megaff_{sec} 
Given  √(n^{2}  1)
n 

ff_{Sec}× n ×  √(n^{2}1)
n  c
g  = M_{ship} =  c
g  √(n^{2}1)×ff_{Sec} 
Restate momentum equation
to determine ship's mass
EXAMPLE: Let mass multiple, n, be 2 which maps to an exhaust particle velocity of 86.67% light speed c.
Then, fuel particle mass doubles:
ff_{Exh }= 2 × ff_{Sec }.
Thus, ship's mass:
c/g×√(n^{2}1) ff_{Sec }= M_{ship} = 30.57×10^{6} × √(2^{2}1)ff_{Sec}
Thus,
M_{ship} = 52.95×10^{6} ff_{Sec }= 52.95 million ff_{Sec }= 52.95 megaff_{Sec}
EXAMPLE: Let ff_{Sec}= 1.0 kg; then,
M_{Ship} = 52.95 million kg = 52,950 mT

3. APPROXIMATE RANGE: Division 
TABLE3a. Daily Difference (∇) Decreases.
_{n}  M_{Ship}  _{}∇ 
(integer)  (megaff_{sec})  (%TOGW) 
2

52.95 megaff_{sec}

_{0.163%}

3

86.46 megaff_{sec}

_{0.100%}

4

118.40 megaff_{sec}

_{0.073%}

Given 
 .002826
√(n^{2}1) 
 Simplify computations: Let n = ff_{Exh}/ff_{sec}
∇ = Percent of ship mass required for daily fuel consumption ("daily diff").
∇  =  ff_{Day}
M_{Ship}  =  86,400sec×ff_{sec}
c/g√(n^{2}1)×ff_{Sec} 
∇  =  86,400 sec
30.57×10^{6}sec×√(n^{2}1)  =  0.002826
√(n^{2}1) 
 Sec per Day = 86,400 secs/day 
 c= 299,792,458 m/sec 
 g = 9.80665 m/sec^{2} 
 c/g = 30.57×10^{6 }sec 
Then quantity,  (Sec per day) / (c/g) = .002826/day 

100% SOLUTION: Theoretical
If entire ship's mass was available; then, max range can be theoretically be determined by dividing daily difference (∇) into 100%.
Unfortunately, this doesn't leave much room for crew, passengers, payload or even infrastructure (the ship itself).
Thus, this method proves infeasible.

TABLE3b. Theoretical Approx. Increases
_{n}  M_{Ship}  ∇  R_{100%} 
(integer)  (megaff_{sec})  (%TOGW)  (days) 
2

52.95 megaff_{sec}
 _{0.163%}  _{613.5 dy} 
3

86.46 megaff_{sec}
 _{0.100%}
 _{1,000 dy}

4

118.40 megaff_{sec}
 _{0.073%}  _{1,370 dy} 
Given 
 .002826
√(n^{2}1)  100%
∇ 
R_{100%} = Range in Days (onboard fuel enables powered flight for many days). This simple method (100%/∇) approximates maximum range if 100% of ship's mass was fuel; of course, not possible.
R_{100%}  =  M_{Ship} /ff_{Day} = 100% / ∇ 
NOT FEASIBLE

TABLE3c. Approximate Feasible Increases
_{n}  M_{Ship}  ∇  R_{50%} 
(integer)  (megaff_{sec})  (%TOGW)  (days) 
3  86.5 megaff_{sec}  _{0.100%}  _{500 dy} 
4  118.4 megaff_{sec}  _{0.073%}  _{685 dy} 
5  149.7 megaff_{sec}  _{0.057%}  _{866 dy} 
Given 
 .002826
√(n^{2}1)  50%
∇ 
On a daily basis, mass conversion to kinetic energy is an extremely small percent of ship's mass (∇).
EXAMPLE: At exhaust particle speed of 98% c, particle mass grows 5 times, and daily fuel consumption is a tiny .057% of ship's mass.
Thus, 50% ship's mass, also known as 50 Percent Take Off Gross Weight (%TOGW), can produce gforce propulsion for many days (R_{50%}).

50% Solution: Feasible
Like current day cruise vessels, interstellar spaceship would carry passengers, living quarters on decks, lots of equipment and a large propulsion system; but it would have a lot of empty space. Thus, fuel comprising 50% ship's TOGW could easily fit into a relatively thin shell (perhaps 2 meters thick) next to outer hull 100 m in diameter.
While it might be reasonable to plan initial fuel load as any amount between 10% and 90%; Thought Experiment (TE) arbitrarily picks 50% as the initial fuel load for this notional spaceship example.

4. FEASIBLE PRECISE RANGE: Logarithms 
Exponentials can more accurately determine a gforce vessel's interplanetary range. To demonstrate, arbitrarily select a daily fuel consumption rate (∇) of 1% ship's gross weight (GW). Recall ∇= .002826/ √(n ^{2}1); thus, rearrange to solve for n:
n =  √(1 +  .002826^{2}
∇^{2}  )  =  √(1 +  .002826^{2}
.01^{2}  ) 
Let ∇ = 1% GW; next, determine mass multiple n (ff_{Exh} = n × ff_{sec}): n= √1.07986276 = 1.0392
Let velocity of exhaust particle be expressed as d_{c} × c,
determine d_{c} as follows:
d_{c}  =  √(n^{2 } 1)
n   =  √(1.0392^{2 } 1)
1.0392 
d_{c} = .272; V_{Exh} = .272 c
Artificially assume 100% efficiency
(practical efficiency to be discussed in TABLE5.)
GW – Gross Weight decreases throughout powered flight due to continuing expenditure of fuel particles. Daily Gross Weights (GWs) can be expressed by following formula where t is the number of days traveled.
GW_{t} = (1∇)^{t} × GW_{0} 
Continuing current example where exhaust particle speed is .272c, then (1∇) = .99. Let ship's original Gross Weight (GW _{0}) be 100,000 mT. After a given duration (t, in days), ships gross weight will decrease to exactly 50,000 metric Tonnes (mTs). To determine value of t, one could go through several iterations of guessing as shown by TABLE4a.
GW_{68.9} = (.99)^{68.9} × 100,000mT = 50,034 mT 

TABLE4a. Interplanetary Feasibility (1% solution)
∇  _{n}  d_{c}  V_{Exh} 
% GW  Mass Mult.  Dec. Comp.  Exhaust Vel 
1%

1.0392

.272

.272c

Given 
 √(n^{2 } 1)
n  (d_{c} ×c) 
GW_{t} = TOGW×(1∇)^{t}
GW_{50}  =  TOGW×(.99)^{50 }  =  .605TOGW 
GW_{60}  =  TOGW×(.99)^{60 }  =  .547TOGW 
GW_{67}  =  TOGW×(.99)^{67 }  =  .5100TOGW 
GW_{68}  =  TOGW×(.99)^{68}  =  .5049TOGW 
GW_{68.7}  =  TOGW×(.99)^{68.7 }  =  .50135TOGW 
GW_{68.8}  =  TOGW×(.99)^{68.8}  =  .50984TOGW 
GW_{68.9}  =  TOGW×(.99)^{68.9}  =  .50034TOGW 

GW_{69}  =  TOGW×(.99)^{69}  =  .4998TOGW 

GW_{70}  =  TOGW×(.99)^{70 }  =  .495TOGW 
Daily fuel consumption is 1% ship's mass (∇) with a 50%TOGW fuel load, iteratively improve range precision: ≈70 days, 69 days, 68.9 days.
Exponentials
determine quicker and more accurate ranges
than simple division.
After several interpolations,
more closely approximate range
from certain daily decrements
and initial fuel load.

Logarithms. While ∇ of 1%GW provides ample interplanetary range, we need much greater performance for interstellar travel. Logarithms can quickly provide more precise ranges as ship's performance increases so that much lesser fuel consumption can propel gforce vessels.
Derive applicable method as follows:
Variable (t) range duration (days) is exponent of left side expression.

(1∇)^{t}  =  (1  %TOGW) 
Take log of both sides:

log (1∇)^{t}  =  log (1  %TOGW) 
Reexpress left side as log times exponent (t).

t × log (1∇)  =  log (1  %TOGW) 
Solve for t, range of %TOGW fuel load with fuel consumption as percentage of ship's mass (∇).

t  =  log (1  %TOGW)
log (1∇) 
 TABLE4b. Interstellar Feasibility
t = log(1%TOGW) / log(1∇) 
n  ∇  t 
(Mass Mult.)  (Daily Diff. )  (Range) 
3

0.100% GW

693 Days

4

0.074% GW
 950 Days 
5

0.059% GW
 1,201 Days 
6

0.049% GW
 1,451 Days 
7

0.042% GW
 1,699 Days 
8

0.036% GW
 1,946 Days 
9

0.032% GW
 2,193 Days 
Given  0.2826%
√(n^{2}  1)  log (1  .5)
log(1∇) 

5. PRACTICAL RANGE: Consider Efficiency 
What is efficiency?
Let Efficiency (E) be ratio of output to input:
E = out : in,
For convenience, assume that gforce ship's propulsion system consumes one million water molecules (H_{2}O) during a given second of powered flight. If system is perfectly efficient, it would output all one million molecules, and the ratio of out/in would be one million : one million for E = 1.0 (or 100%).
However, no process is perfectly efficient, and prudent practicality predicts propulsion system will be much less than 100% efficient.
EXAMPLE: Of the one million molecules input to the propulsion, let 750,000 exit as exhaust particles which contribute to ship's gforce propulsion. Thus, E = 750,000 particles / 1,000,000 particles = .75 = 75%.
The remaining 250,000 particles go elsewhere and makeup the inefficiency (E').
Inefficiency (E') is complement of Efficiency (1E).
For this example, inefficiency:
(1E) = 25% = E'
Since output can never exceed input, both efficiency and inefficiency must have values between 0 and 100%.
No human system will ever be perfectly efficient. For example, a typical automobile must divert part of it's energy output to peripheral equipment such as radio, air conditioning as well as essential components such the car's generator which provides essential electrical sparks to ignite the gasoline; thus, some output must divert as part of the auto's energy cycle.

Will inefficiency decrease?
TABLE5a. Inefficiency
_{n}  V_{Exh}  E'  E 
2  .866c  50%  50% 
3  .934c  45%  55% 
4  .968c  40%  60% 
5  .980c  36%  64% 
6  .986c  33%  67% 
7  .990c  30%  70% 
8  .992c  27%  73% 
Given  √(n^{2}1)×c
n  .5×.9^{n2}  1E' 
Like all planes, trains and ocean vessels, TE's gforce ship must divert part of it's output energy for nonpropulsion requirements. For example, it has peripheral needs such as life support, communications, navigation. It will also have a primary component of the propulsion cycle which requires diversion of some of the accelerated high speed particles to superheat water particles into plasma ions for subsequent input into ship's particle accelerator, propulsion system. Without this feed, particle acceleration cycle cannot be sustained.
The auto's combustion engine has continuously improved for over a hundred years, and the gforce spaceship's accelerator will similarly improve with continuously reduced inefficiency.
Arbitrarily assume baseline inefficiency (E') of 50% for n=2; from that base, let E' reduce by 10% for every integer increase of mass multiple (n); thus, E' = .5 × .9^{n2}.

.... efficiency factor
TABLE5b. Efactor
_{n}  E'  E  ε 
3  45%  55%  1.82 
4  40%  60%  1.68 
5  36%  64%  1.57 
6  33%  67%  1.49 
7  30%  70%  1.42 
8  27%  73%  1.36 
9  24%  76%  1.31 
Given  .5×.9^{n2}  1E'  1
E 
Above example shows us if a ship consumes one million water particles with 75% efficiency; that only 750,000 particles actually exit the ship and contribute to gforce propulsion.
It might be useful to reconsider example.
Q. If gforce propulsion requires 1 million particles to exit the ship each second of powered flight, how many particles must be consumed each second to accommodate both the propulsion and the requisite inefficiency???
A. To determine required consumption rate, TE uses an efficiency factor (ε), the reciprocal of E, efficiency.
ε = 1/ E
ff_{Day} = ∇ × ε
1) ff_{sec }, fuel flow per second, required for gforce propulsion. Can be determined from momentum conservation.
2) ff_{Day }, fuel flow per day, required for gforce propulsion. Compute with factor 86,400 sec/day.
3)∇, daily percentage of ship's mass, which must exit ship's exhaust to produce gforce acceleration. In a perfectly efficient gforce vessel, ∇ = ff_{Day} ; however, ff_{Day }must exceed ∇ to provide auxiliary and peripheral power needs in addition to propulsion.
4) ε, efficiency factor, coefficient required to determine how much ff_{Day }must exceed ∇.
ε = 1 ÷ .75 = 1.333
ff_{sec} = 1,000,000 par/sec × 1.333 = 1,333,333 particles/sec
 TABLE5c. Interstellar Practicality
t = log(1%TOGW) / log(1ε∇) 
n  ∇  ε  t 
(Mass Mult.)  (Daily Diff. )  (Efactor)  (Range) 
4

0.073%GW

1.68

565 Days

5

0.058%GW
 1.57 
763 Days

6

0.048%GW
 1.49 
975 Days

7

0.041%GW
 1.42 
1,197 Days

8

0.036%GW
 1.36 
1,429 Days

9

0.032%GW
 1.31 
1,669 Days

10

0.028%GW
 1.27 
1,915 Days

Given  0.2826%
√(n^{2}  1)  1
1(.5×.9^{n2})  log (1.5)
log(1ε∇) 
Find time (t in days) such that GW_{t} = 50% TOGW.
t = log .5 /log(1εΔ)
n, mass multiple, maps directly to particle exhaust speed, V_{Exh}. For example, for particle to grow in mass a multiple of 10 times, it must achieve a relativistic velocity of 99.5%c. TE assumes that technology will mass mult. (n) will increase as relevant technology improves.
∇, daily difference, is amount of ship's mass converted to ions and ejected from the ship's exhaust to propel ship forward at gforce.
ε, efficiency factor, times ∇ indicates how many particles must be consumed to ensure sufficient particles for propulsion plus other requirements. (Compute ε with assumed efficiency baseline and assumed inefficiency decrease model.)
t, range (time in days), is more practical because new range model considers real limitations such as limited room on vessel as well as inevitable inefficiencies. (Assuming technology will achieve mass multiple of 10, reasonable assumptions predict a range of 4 years with margin.)

SUMMARY 
Key Concepts
 Fuel flow per second (ff_{sec}) is amount of fuel required to increase speed of spaceship an additional 10 m/s for next second of flight. This rate must accommodate inevitable inefficiencies.
 Daily fuel (ff_{Day})_{ }can be approximated by 86,400 × ff_{sec} (recall that one has 86,400 is number of seconds).
 Daily Difference (∇) is the per cent decrease of the space vehicle's gross weight due to one day's gforce propulsion.
 Exhaust particle velocity (V_{Exh}) To achieve gforce propulsion, accelerator must take particles to near light speed, c.
 Exhaust Particle Flow (ff_{Exh}). Lorentz Transform determines relativistic increase in mass as fuel particle speeds up from rest to propulsion exhaust.
ff_{Exh} =  ff_{sec }
√(1  V^{2}_{Exh}/c^{2}) 
 Light speed, decimal component (d_{c}) is the ratio of v_{Exh}/c. This is a handy way of expressing V_{Exh} in terms of decimal light speed or percent c.
 Mass Multiple (n) quantifies mass increase of exhaust fuel particles. ff_{Exh}= n × ff_{sec}
n =  1
√(1  (d_{c}^{2}c^{2})/c^{2})  =  1
√(1  d_{c}^{2}) 
 Practical range acknowledges far less then 100% efficiency of mass conversion into propulsion energy.
t =  log (1%TOGW)
log(1  ε∇) 

1: Momentum Exchange 
TABLE1. Gforce to Nearby Planets  Let exhaust particle velocity (VExh) be 30 million meters per second, or about 10% c, light speed 
2: Relativistic Impact 
TABLE2a. Relativity
Fuel Particle Increases  Observe some mass increase as speeds increase to 86.6%c where mass grows. 
TABLE2b. Relativity & M_{Ship} Capacity.  Observe significant mass increase as speeds increase well beyond 86.6% c. 
3: APPROXIMATE RANGES: Division 
TABLE3a. Daily Diff (∇) Decreases.  As exhaust particle speed increases, less fuel is needed for gforce propulsion. 
TABLE3b. Theoretical Range Increases  Notional range if a spaceship could convert 100% mass to energy. 
TABLE3c. Feasible Range Increases  Notional range if a spaceship could convert 50% mass to energy. 
4: ACCURATE RANGES: Logarithms 
TABLE4a. Interplanetary Feasibility  More accurate feasibility from exponentials. Fast particle velocities can take us to the planets. 
TABLE4b. Interstellar
Feasibility  More convenient feasibility from logarithms. Very fast particle velocities can take us to the stars. 
5: PRACTICAL RANGES: Efficiency 
TABLE5a.
Inefficiency  Inevitable inefficiency acknowledges necessary diversion of particles for nonpropulsion purposes. 
TABLE5b.
Efactor  To achieve gforce propulsion, must consume enough fuel to cover inevitable inefficiencies. 
TABLE5c. Interstellar Practicality  Need to cover four years of gforce propulsion plus considerable margin. 

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