Sunday, March 02, 2008

GETTING THERE: PRACTICALITY

LIST OF TABLES
Our thought experiment notionalizes a spaceship which constantly accelerates at g, acceleration due to near Earth gravity. This "g-force" acceleration brings spacecraft to great velocities to greatly shorten trip duration; it also simulates Earth gravity for ship crew and passengers. Crew and passengers could travel in Earth like conditions to nearby planets in just days and to neighboring stars in years. Thus, interstellar travel can become practical.
To get the energy for g-force propulsion,
our notional spacecraft uses a particle accelerator 
to speed fuel particles to near light speed.
Thus, a relatively small mass of exhaust particles
could accelerate a larger spaceship at g-force
to simulate gravity and gain enormous speeds.
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:
ffExh×VExh = Mship×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 (ffExh) be one kilogram (kg)
  • Let exhaust particle velocity (VExh) be 30 million meters per second, or about 10% c, light speed.
  • Let spacecraft mass (MShip) be about 3 million kgs (3,000 metric Tonnes, mTs). Using more precise values for c and g; then, MShip = 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).
    1 kg×.1 c= 3,057 mT  ×g

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 g-force requires a lot of fuel.

For one day (86,400 sec) of g-force propulsion, 1 kg/sec fuel flow (ffsec) will become a daily fuel flow (ffdy) 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:
ffDay = 86.4 mT/3,057 mT = 2.826%TOGW.

Thus, this thought experiment constructs several helpful tables. First table lists typical g-force 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.  
TABLE-1. G-force to Nearby Planets:
Total Fuel Requirements
g = 0.5 AU/day2ffDay=86,400×ffsec 
dist.timeFuel
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
ObservedTypicald

g
t × ffday

 MShip
For TABLE-1, let ffsec = 1.0 kg and MShip = 3,057 mT.
Let ffsec 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 ffDay is fuel flow per day; then,
ffDay= ffsec × 86,400 secs/day

Required fuel is trip time times daily fuel flow:
 F = t × ffDay 

Therefore,
F = 2 × (d/g) × ffsec × 86,400 secs/day

To express fuel as percentage weight of spaceship:
%TOGW =  F / MShip = (t × ffDay) / MShip
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.

ffExh =ffSec

(1 - v2Exh/c2)
To apply LT to our space borne particle accelerator, make the following term replacements. Replace m0 with the term, ffSec, original sized mass of fuel flow per second. This is fuel consumed per second at the start of the particle accelerator.  Replace mr with the term, Exhaust fuel flow (ffExh), which is ffSec 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 (vExh, which replaces vr). Thereafter, exhaust particles (ffExh) exit propulsion system and "push" spacecraft in opposite direction.

 ffExh  =  n × ffsec


n= 1

(1 - v2Exh/c2)
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 - dc2c2/c2)
vExh can be expressed as "dc × c" or "dc c" where decimal component (dc) is a decimal between zero and one, and c is light speed. Note that dc = vExh/c. Thus, dc2 = v2Exh/c2, and substitution is made as shown.
TABLE-2a. Relativity and Fuel Particle Mass
VExhdcnffExh
(decimal c)Dec. Comp. Mass Mult. Exhaust Flow
.10 c.11.0051.005 ffsec
.20 c.21.0211.021 ffsec
.30 c .31.0481.048 ffsec
.40 c.41.0911.091 ffsec
.50 c.51.155 1.155 ffsec
.60 c.61.250 1.250 ffsec
.70 c.71.400 1.400 ffsec
.80 c.81.667 1.667 ffsec
.866 c.8662.0002.000 ffsec
GivenVExh

c
1

√(1 - dc2)
(n × ffsec)
TABLE-2a 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

(vExh, shown as decimal light speed).

ffsec is original fuel mass (per second) at rest.
Fuel is consumed at rest.


ffExh 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.



mr =mo

(1 - v2r/c2)
Lorentz Transform (LT)
quantifies relativistic mass increase of high speed particles.EXAMPLE: Let one gram of original mass, mo, achieve a relativistic velocity, vr = .866c = 86.6% light speed; then, relativistic mass, mr , will be 2.000 grams.


n= 1

(1 - dc2)
After few axiomatic substitutions,
terms eventually reduce
to equation at left.
TABLE-2b restates momentum equation to determine ship's mass.
Since relativistic speed increases the mass of exhaust fuel particles, ffExh can be expressed as a multiple of original fuel quantity per second n × ffSec This slightly changes TE momentum equation:
n × ffSec × vExh = Mship × g

Rearrange TABLE-2a; for independent variable (IV), 2b uses fuel particle multiple (n) as a range of integers. Fuel particle exhaust velocity (VExh) could be any of an infinite number of values between zero and light speed, c. However, TABLE-2b uses subset of VExh , selected velocity values such that exhaust particle (ffExh) grows to integer multiple, n, of consumed fuel flow per second (ffSec).
EXAMPLE: Let VExh = .943c, then, relativity triples the size of consumed fuel particle:
 n × ffSec = ffExh = 3 × ffSec

Transform from solving for n to solving for dc.
n= 1

(1 - dc2)
n2= 1

1 - dc2
Square both sides.
1 - dc2 = 1

n2
-1 + dc2 = -1

n2
Rearrange as shown.
dc2 = 1 + -1

n2
dc2 =  n2 - 1

n2
dc =  (n2 - 1)

n
Solve for decimal component of light speed, dc.

Given sent of integers {n}, map to corresponding set of decimal values {dc} which readily show required particle speeds required to grow mass a certain multiple, n.

TABLE-2b. Relativity & MShip Capacity.
 ffExh × VExh / g =  Mship = n × ffsec × dc × c/g
ndcMShip
Mass Mult.(decimal c)(mega-ffsec)
1 .0 n/a
2 .866 52.95 mega-ffsec
3 .943 86.46 mega-ffsec
4 .968 118.40 mega-ffsec
5 .980 149.76 mega-ffsec
6 .986 180.85 mega-ffsec
7 .990 211.80 mega-ffsec
Given(n2 - 1)

n
c

g
√(n2-1)×ffSec
ffSec× n × √(n2-1)

n
c

g
=  Mship =c

g
√(n2-1)×ffSec
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:
 ffExh = 2 × ffSec .
Thus, ship's mass: 
c/g×√(n2-1)  ffSec = Mship = 30.57×106 × √(22-1)ffSec
Thus,
Mship = 52.95×106 ffSec = 52.95 million ffSec = 52.95 mega-ffSec
EXAMPLE: Let ffSec= 1.0 kg; then,
MShip = 52.95 million kg = 52,950 mT
3. APPROXIMATE RANGE: Division

TABLE-3a. Daily Difference (∇) Decreases.
nMShip∇ 
(integer)(mega-ffsec)(%TOGW)
2
52.95 mega-ffsec
0.163%
3
86.46 mega-ffsec
0.100%
4
118.40 mega-ffsec
0.073%
Given
c

g
(n2-1)  ffSec
.002826

√(n2-1)
Simplify computations: Let n = ffExh/ffsec
∇ = Percent of ship mass required for daily fuel consumption ("daily diff").
=ffDay

MShip
86,400sec×ffsec

c/g√(n2-1)×ffSec
=86,400 sec

30.57×106sec×√(n2-1)
=0.002826

(n2-1)
Sec per Day = 86,400 secs/day
c= 299,792,458 m/sec
g = 9.80665 m/sec2
c/g = 30.57×106 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.

TABLE-3b. Theoretical Approx. Increases
nMShip∇ R100% 
(integer)(mega-ffsec)(%TOGW)(days)
2
52.95 mega-ffsec
0.163%613.5 dy
3
86.46 mega-ffsec
0.100% 1,000 dy
4
118.40 mega-ffsec
0.073%1,370 dy
Given
c

g
(n2-1)  ffSec
.002826

√(n2-1)
100%

R100% = 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.
R100% =MShip /ffDay = 100% /
NOT FEASIBLE
TABLE-3c. Approximate Feasible Increases
nMShip∇ R50% 
(integer)(mega-ffsec)(%TOGW)(days)
386.5 mega-ffsec0.100%500 dy
4118.4 mega-ffsec0.073%685 dy
5149.7 mega-ffsec0.057%866 dy
Given
c

g
(n2-1)×ffSec
.002826

√(n2-1)
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 g-force propulsion for many days (R50%).






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 g-force vessel's interplanetary range. To demonstrate, arbitrarily select a daily fuel consumption rate (∇) of 1% ship's gross weight (GW). Recall ∇= .002826/(n2-1); thus, rearrange to solve for n:
n = (1 +.0028262

2
)= (1 +.0028262

.012
)
Let ∇ = 1% GW; next, determine mass multiple n (ffExh = n × ffsec):     n= 1.07986276 = 1.0392

Let velocity of exhaust particle be expressed as dc × c,
determine dc as follows:
dc =  (n2 - 1)

n
=  (1.03922 - 1)

1.0392
dc = .272; VExh = .272 c
Artificially assume 100% efficiency
(practical efficiency to be discussed in TABLE-5.)

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.
GWt = (1-∇)t × GW0 
Continuing current example where exhaust particle speed is .272c, then (1-∇) = .99. Let ship's original Gross Weight (GW0) 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 TABLE-4a.
GW68.9 = (.99)68.9 × 100,000mT = 50,034 mT
TABLE-4a. Interplanetary Feasibility (1% solution)
ndcVExh
% GWMass Mult. Dec. Comp.Exhaust Vel
1%
1.0392
.272
.272c
Given
(1 +.0028262

2
)
(n2 - 1)

n
(dc ×c)
GWt = TOGW×(1-∇)t
GW50= TOGW×(.99)50 =.605TOGW
GW60= TOGW×(.99)60 =.547TOGW
GW67= TOGW×(.99)67 =.5100TOGW
GW68= TOGW×(.99)68=.5049TOGW
GW68.7= TOGW×(.99)68.7 =.50135TOGW
GW68.8= TOGW×(.99)68.8=.50984TOGW
GW68.9= TOGW×(.99)68.9=.50034TOGW
GW69= TOGW×(.99)69=.4998TOGW
GW70= 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 g-force 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)
Re-express 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-∇)
TABLE-4b. Interstellar Feasibility
t = log(1-%TOGW) / log(1-∇)
nt
(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
Given0.2826%

 (n2 - 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 g-force ship's propulsion system consumes one million water molecules (H2O) 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 g-force 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 (1-E).
For this example, inefficiency:
(1-E) = 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?
TABLE-5a. Inefficiency
nVExhE'E
2.866c50%50% 
3.934c45%55%
4.968c40%60%
5.980c36%64%
6.986c33%67%
7.990c30%70%
8.992c27%73%
Given(n2-1)×c

n
.5×.9n-21-E'
Like all planes, trains and ocean vessels, TE's g-force ship must divert part of it's output energy for non-propulsion 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 g-force 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 × .9n-2.
.... efficiency factor
TABLE-5b. E-factor
nE'Eε
345%55%1.82
440%60%1.68
536%64%1.57
633%67%1.49
730%70%1.42
827%73%1.36
924%76%1.31
Given.5×.9n-21-E'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 g-force propulsion.



It might be useful to reconsider example.
Q. If g-force 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
ffDay = ∇ × ε
1) ffsec , fuel flow per second, required for g-force propulsion. Can be determined from momentum conservation.
2) ffDay , fuel flow per day, required for g-force propulsion. Compute with factor 86,400 sec/day.
3)∇, daily percentage of ship's mass, which must exit ship's exhaust to produce g-force acceleration. In a perfectly efficient g-force vessel, ∇ = ffDay ; however, ffDay must exceed to provide auxiliary and peripheral power needs in addition to propulsion.
4) ε,  efficiency factor, coefficient required to determine how much ffDay must exceed ∇.
ε = 1 ÷ .75 = 1.333
ffsec = 1,000,000 par/sec × 1.333 = 1,333,333 particles/sec





TABLE-5c. Interstellar Practicality
t = log(1-%TOGW) / log(1-ε∇)
nεt
(Mass Mult.)(Daily Diff. )(E-factor)(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
Given0.2826%

√(n2 - 1)
1

1-(.5×.9n-2)
log (1-.5)

log(1-ε∇)
Find time (t in days) such that GWt = 50% TOGW.
t = log .5 /log(1-εΔ)
n, mass multiple, maps directly to particle exhaust speed, VExh. 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 g-force.
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 (ffsec) 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 (ffDay) can be approximated by 86,400 × ffsec (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 g-force propulsion.
  • Exhaust particle velocity  (VExh) To achieve g-force propulsion, accelerator must take particles to near light speed, c.
  • Exhaust Particle Flow (ffExh). Lorentz Transform determines relativistic increase in mass as fuel particle speeds up from rest to propulsion exhaust.
    ffExh =ffsec

    (1 - V2Exh/c2)
  • Light speed, decimal component (dc) is the ratio of vExh/c.  This is a handy way of expressing VExh in terms of decimal light speed or percent c.
  • Mass Multiple (n) quantifies mass increase of exhaust fuel particles. ffExh= n × ffsec
    n = 1

    (1 - (dc2c2)/c2)
    = 1

    (1 - dc2)
  • Practical range acknowledges far less then 100% efficiency of mass conversion into propulsion energy.  
t = log (1-%TOGW)

log(1 - ε∇)
1: Momentum Exchange
TABLE-1. G-force to Nearby PlanetsLet exhaust particle velocity (VExh) be 30 million meters per second, or about 10% c, light speed
2: Relativistic Impact
TABLE-2a. Relativity
Fuel Particle Increases
Observe some mass increase as speeds increase to 86.6%c where mass grows.
TABLE-2b. Relativity & MShip Capacity.Observe significant mass increase as speeds increase well beyond 86.6% c.
3: APPROXIMATE RANGES: Division
TABLE-3a. Daily Diff (∇) Decreases.As exhaust particle speed increases, less fuel is needed for g-force propulsion.
TABLE-3b. Theoretical Range IncreasesNotional range if a spaceship could convert 100% mass to energy.
TABLE-3c. Feasible Range IncreasesNotional range if a spaceship could convert 50% mass to energy.
4: ACCURATE RANGES: Logarithms
TABLE-4a. Interplanetary Feasibility More accurate feasibility from exponentials.  Fast particle velocities can take us to the planets.
TABLE-4b. Interstellar
Feasibility
More convenient feasibility from logarithms.  Very fast particle velocities can take us to the stars.
5: PRACTICAL RANGES: Efficiency
TABLE-5a.
Inefficiency
Inevitable inefficiency acknowledges necessary diversion of particles for non-propulsion purposes.
TABLE-5b.
E-factor
To achieve g-force propulsion, must consume enough fuel to cover inevitable inefficiencies.
TABLE-5c. Interstellar PracticalityNeed to cover four years of g-force propulsion plus considerable margin.
CONCLUSION
To make theoretical maximum ranges more feasible, consider following factors:
  1. Limit fuel availability to only half the spaceship's mass.
  2. Use logarithms to model ship's ever decreasing gross weight due to fuel consumption.
  3. For practicality, consider inevitable inefficiencies.



VOLUME I: ASTEROIDAL
VOLUME II: INTERPLANETARY
VOLUME III: INTERSTELLAR



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