FINITE FUEL
One tank of gas
can only take your car so far.
can only take your car so far.
Like your car, thought experiment's
spaceship has a finite range.
spaceship has a finite range.
Note Following Assumptions
1) Inevitable inefficiency enables us to introduce some practicality into this model. Previous work introduced the term, ε (Greek "epsilon") as an efficiency factor. Examples: To artificially represent the unachievable 100% efficiency for fuel flow conversions, ε would equal 1.0. For the Thought Experiment (TE) model, we choose a much more practical 10% fuel flow efficiency; thus,
ε = 10% = 0.1
Someday, actual empirical data will determine actual efficiency factors.
2) Marginal Fuel Requirements  it makes sense to plan for "spare fuel"; it's always a good policy to plan the mission to end with some fuel remaining. Since this fuel is not planned for consumption, it might be better to consider marginal fuel as part of ship's infrastructure. Thus, thought experiment disregards spare fuel for computing range of spacecraft.3) Assume continuous gforce acceleration due to constant particle exhaust; TE calls this "powered flight" in contrast to orbiting, spinning habitats which are traveling in "nonpowered flights".
Rearrange Equation 1 (Eq1):

Recall previous work:
Particle exhaust speed, v_{Exh}, can be described as decimal light speed: d_{c}c
Example: Let v_{Exh} = .27c ; thus, d_{c} = .27
Exhaust particle mass, ff_{Exh}, will be larger then originally consumed fuel, ff_{Sec}, due to relativistic growth. This growth factor can be determined by Lorentz Transform (LT) which determines relativistic mass increase, m_{r}.
Exhaust particles with velocity of .27c will grow 4% due to relativity.
ff_{Exh} = m_{r }× ff_{sec}= 1.04 × ff_{sec} Thus, Equation 2 (Eq2) can be rewritten:
 
Recall fuel flow per second (ff_{sec}) is the quantity of at rest fuel consumed per second; however, typical gforce flight durations can be reckoned in days. Accordingly, TE determines daily fuel consumption. Thus, let Day_{Sec} = 86,400 seconds per day; then, typical day's fuel consumption can be described as:
F_{Day} = Day_{Sec} × ff_{sec}
TE now introduces a new term, Percent TakeOff Gross Weight for first day of powered flight, %TOGW_{Day1}. Simple substitutions and rearrangements give us Equation 4 (Eq4).
COMBINE CONSTANTS:
c (light speed), g (acceleration due to Earth surface gravity), and Day_{Sec} (seconds per day) are shown below:
 
 
 
Rewrite expression, m_{r}×d_{c}:
 
Eq6 Determines Daily Difference in Gross Weight
%GW_{Day} is daily percent decrease of vessel's GW. NOTE: Vessel's GW ever decreases due to fuel consumption; however, daily percentage of GW used for fuel remains constant.
Work gives us following points:
Practical Daily Difference (Δ)
INTRODUCE SOME PRACTICALITY: Use Efficiency Factor (ε) to account for inevitable inefficiencies. For this example, assume 10% fuel efficiency; thus, let ε = 1/E = 1/10% = 10
 
Apply Rate
If ship's initial gross weight (GW_{0}) is 100 metric Tonnes (mT); then, ship will consume 10 mTs of fuel on first day of powered flight.
F_{Day} = Δ × GW_{0} = 10% × 100mT = 10mT
GW_{1} = GW_{0}  F_{Day} = 100mT  10mT = 90mT
Assume we devote half of ship's mass to fuel; (%TOGW = 50%). Then, ship starts with 50 tons of fuel, and a rate of 10%/day seems to give us 5 days of powered flight. Thus, 5 days can be an initial approximation of ship's range.
Fortunately, ship's actual range is slightly better.
At beginning of trip, no fuel is consumed.
Thus, the initial value of F_{Day} ( = Δ × GW_{0}) is really only good for first day of flight. After the first day, GW decreases due to fuel consumption. Thus, Δ is a dynamic daily percentage value, and it reflects the ever decreasing GW during powered flight. For convenience, let daily decrement be ten percent of ship's Gross Weight (10% = .10).
2nd day, ship consumes 10% of GW_{1}; Δ = 10% ×90 mT = 9 mT.
Thus, 2nd day's fuel consumption is less then 1st day's. Furthermore, GW at end of day 2 (GW_{2}) is less than GW_{1} at end of day 1.
3rd day, fuel consumption will be even less and so on as shown in following table.
Day t  GW_{0 }× (1Δ)^{t}  =  GW_{t}  =  100mT × .9^{t}  <<  100 mT 
To determine ship's gross weight (GW) for any given day "t" (GW_{t}), multiply original quantity (GW_{o}), by initial daily decrement "(1Δ)" raised to the power "t".
 
Exponentials
We see that exponentials can readily approximate fuel consumption. Any inexpensive scientific calculator has an exponent function; thus, we can readily calculate numerous guesses as to how many days is required to reduce ship's original gross weight (GW_{0}) by 50%.Our initial intuituive guess was 5 days; recall it seemed that 50 mTs divided by 1 mT/day might give us 5 days of powered flight.
Divide %TOGW by first day's Δ: (50 mT/10 mT/day = 5 days)
However, our guess fails to consider that ship's GW decrements by percent not by absolute weight. Consider that as fuel is consumed, GW declines; a lesser GW needs less fuel to propel it, and this cycle continues throughout powered flight (as shown by above table). Thus, mass of fuel consumed decreases throughout flight even though the percentage of GW_{Day} remains the same.
Let Δ = 10%; then, 1Δ = 90% = .9
A quick exponential calculation shows that 5 days of 10% consumption only reduces ship's gross weight to about 60% GW_{0}.
.9^{5} = .5905 = 59.05%
However, we want to determine time (t) for GW_{t} = 50%;
thus, we might test several t values:
.9^{t} = 50% GW_{0}
to eventually determine a t value for a GW near 50%.
Thus, we determine that at 10% GW consumption per day; 6.6 days takes us very close to 50% of
original mass.
Thus, a daily difference (Δ) of 10%/day gives our notional spaceship a range of about 6 and a half days of powered flight.
There might be a more graceful way to determine this range. 
a^{t}  =  b 

log(a^{t})  =  log(b) 
t × log(a)  =  log(b) 
t  =  log(b) / log(a) 
For present case, substitute following terms:
For present case, use following values:
Daily Decrement: Δ = 10%; thus, 1Δ= .90.
Percent TakeOff Gross Weight is 50%; thus, 1%TOGW = .5000
Thus, logarithms can quickly approximate ranges for different values of %TOGW.
(1  Δ)^{t}  =  1  %TOGW 

log(1  Δ)^{t}  =  log(1  %TOGW) 
t × log(1  Δ)  =  log(1  %TOGW) 
Eq9: 


Daily Decrement: Δ = 10%; thus, 1Δ= .90.
Percent TakeOff Gross Weight is 50%; thus, 1%TOGW = .5000
.90^{t}  =  .5000 

t  =  log .5000 / log .90 
t  =  .301/ (.04576) 
t  =  6.58 
Original
Fuel Load 
Final
Gross
Weight

Decimal
Light
Speed**

Relativistic
Mass
Increase**

Gross Wt
Daily
Difference**

Gforce
Vessel
Range


F_{0}  GW_{Fin}  d_{c}  m_{r}  Δ  R_{Days} 
(%TOGW_{0})  (%GW_{0})  (d_{c }c)  (%GW)  (days)  
10%  90%  .27 c  1.04  10%  1.00 
20%  80%  .27 c  1.04  10%  2.12 
30%  70%  .27 c  1.04  10%  3.39 
...  ...  ...  ...  ...  ... 
40%  60%  .27 c  1.04  10%  4.85 
50%  50%  .27 c  1.04  10%  6.58 
60%  40%  .27 c  1.04  10%  8.70 
70%  30%  .27 c  1.04  10%  11.43 
80%  20%  .27 c  1.04  10%  15.28 
90%  10%  .27 c  1.04  10%  21.85 
Indep. Variable  1%F_{0}  Given  1 √(1d_{c}^{2})  ε × .282% √(m_{r}^{2}1)  log(%GW_{Fin}) log(1Δ) 
GW_{Fin} = 
(GW_{0 } F_{0}) ÷ GW_{0}


GW_{Fin} = 
1  %GW_{0}

Daily Difference (Δ)
is percentage amount that ship's GW decreases per day.
Efficiency Factor (ε) helps account for design flaws and peripheral power needs. Δ is expressed as percentage of gross weight, GW.  
Vessel Range (R_{Days})
is total days of powered flight available from original fuel load. Following four phases of interplanetary travel must fit within. 
Acceleration Time, t_{Acc}
We've already determined range based on original fuel load, F_{0}, (possible range in time (days) from %TOGW and GW_{Fin}; see above table).
Thus, we can divide vessel's Range (R_{Day}) by four to determine maximum acceleration time for Phase I:
t_{Acc }= R_{Day} / 4
Increasing particle exhaust speed, V_{Exh}, to .866c
doubles mass of exhaust particles (ff_{Exh }= 2 × ff_{sec}) to greatly expand spaceship's range.
Common sense compels us to continue assuming painful but practical efficiency factor, ε = .1. Time will tell, but for now, TE assumes that for every 100 ff_{sec }particles consumed, only 10 actually become effective propulsion particles (ff_{Exh}). Thus, we must consume ten times more ff_{sec }than we'd otherwise expect.
Daily Velocity Increase. Due to constant gforce acceleration, interplanetary gforce flights will take days; even this brief duration will increase spaceship's velocity to enormous values. Recall:
Thus, spaceship attains: 864 km/sec in one day, 1,728 km/sec after two days, etc.
where t = time in days.
Flight Profile Summary. Slowdown is required for both legs of journey: departure and return. This is described in TABLE2. 




Range
Acceleration
Velocity
5%
95%
3.14
0.79
678 kps
10%
90%
6.45
1.61
1,393 kps
20%
80%
13.7
3.43
2,959 kps
30%
70%
21.8
5.45
4,709 kps
40%
60%
31.0
7.75
6,696 kps
50%
50%
42.1
10.53
9,094 kps
GW_{0}
100%%TOGW
log(1Δ)
R_{Day}
4
4
sec
Current particle accelerators now accelerate particles to velocities faster then .99c; thus, it's very likely that we'll eventually design propulsion systems to accelerate particles much faster then .866c. Thus, daily fuel consumption will decrease, and range will increase. Consider following points.
 ff_{sec }For every second of flight, our notional spaceship consumes an ever decreasing quantity of "at rest" fuel which we'll call fuel flow per sec (ff_{sec}). This is the fuel's mass prior to conversion to plasma (ions, particles with charges). Thus, fuel consumption is reckoned in terms of fuel at rest "ff_{sec}" at zero velocity in relation to ship.
 ff_{Exh }For every second of flight, a particle accelerator uses a system of magnetic and electrical fields to accelerate these charged particles to large fraction of light speed. A century of particle accelerator experience confirms that ions grow as they accelerate to near light speed; thus, the consumed quantity of particles (ff_{sec}) transform to an increased quantity ff_{Exh}, fuel flow of exhaust particles. Collective exhaust particle momentum (near light speed velocity times relativistic mass) enables spaceship's propulsion.
 Determine Acceleration Distance. Once we determine acceleration time, t_{Acc}, from one fourth of the practical range (see previous work); then, we can determine acceleration distance, d_{Acc}, as shown below.
d_{Acc}= g × (t²_{Acc})
2  Departure Distance. Determine total distance for departure leg by multiplying by 2 because deceleration distance equals acceleration distance, d_{Dec} = d_{Acc}. This was previously discussed in Table2.
d_{Dept} = 2 × d_{Acc}  100% Efficiency is impossible for following reasons.
 We must artificially assume every single particle makes it all the way through the many kilometers of waveguides and exits spacecraft exhaust as intended.
 Furthermore, we must incorrectly assume no additional energy is required for other energy requirements, such as electromagnetic fields to guide the particles.
Thus, thought experiment arbitrarily assumes 10% efficiency for interplanetary flights because relativistic momentum provides so much capability that realistic efficiencies can be overcome. (NOTE: Interstellar flights will require much greater fuel efficiency; this will be discussed in a subsequent volume.)
TABLE3. Even Faster
Add columns to cover increasing relativistic growth factors and associated speeds.This table uses a constant final gross weight (GW_{Fin}) of 50% of original GW.
Growth
Difference
Range
Time
Time
Acceleration
Time
Time
Acceleration
Distance
Distance
Total
Distance
Distance
1.04
2
...
3
4
5
m_{r}
√(m_{r}^{2}1)
log(1Δ)
4
2
Definitions:
 m_{r} is relativistic growth factor computed from Lorentz Transform.
 d_{c} is decimal component of particle exhaust speed when expressed as decimal light speed.
 ε is an Efficiency Factor introduced to assure practicality of fuel requirements. All we now know is that fuel flow efficiency will start out much less than 100%, and it will improve over time. For convenience, Thought Experiment assumes an efficiency of 10%; thus, ε = 10.
 Δ is the amount of at rest fuel consumed for each day's propulsion, expressed as a percentage of ship's gross weight.
 %GW_{Fin }is the final gross weight of the spacecraft after the original fuel load fuel (%TOGW) is consumed. %GW_{Fin }is expressed as a percent of original ship's mass and can be computed: 100% %TOGW. Above (Table3) uses a constant %GW_{Fin }of 50%.
 t is total time from F_{0}; this capacity is the range.
 t_{Acc} is acceleration time.
 d_{Acc} is acceleration distance.
 d is total distance for departure leg. Note that this table only uses half the planned fuel for departure leg; other half is for return leg.
SUMMARY: Good News/Bad News
Good News After gforce spaceships can accelerate exhaust particles to 98% c; then 50% TOGW spaceflight profiles will have capability to accelerate for 30 days then decelerate for another 30 days to travel about 451 AUs.
Bad News In spite of this impressive feat, we've still well within our Solar System. 451 AUs is less than 1% of a light year (LY= 63,241 AU); however, the Solar System is bordered by the Oort Cloud which is almost a full LY away from the Sun.)
More Bad News Thus, it follows that we've only traveled a tiny fraction of the distance to our nearest stellar neighbor, Alpha Centauri, about 4 LYs away.
v_{Fin }limitation (can't exceed c)  There is a physical limit on spacecraft speed; it can't exceed light speed. This doesn't limit time of acceleration, but it does mean a more accurate model needs us to compute resulting spacecraft velocity in a different way. After a certain point, incremental increases in acceleration time can no longer be modeled by a corresponding linear increase in spacecraft velocity.
Father of Nuclear Physics
Born 1871 in Nelson, New Zealand, Ernest Rutherford entered Nelson Collegiate School at age 16; in 1889, he was awarded a University scholarship to University of New Zealand, Wellington where he entered Canterbury College. He graduated M.A. in 1893 with a double first in Math and Physical Science; and he continued research work at the College for a short time.
In 1851 (well prior to Lord Rutherford's birth), Prince Albert, husband of Queen Victoria, set up a scholarship fund during the 1851 Crystal Palace Exhibition. Prince Albert specifically wanted to increase opportunities for deserving students outside of England. In 1894, Rutherford was awarded this scholarship; he learned of this while working in his parent's garden and said "this is the last potato that I'll ever dig". Thus, he attended Trinity College, Cambridge as a research student at the Cavendish Laboratory.
His initial experience was not as friendly as he would have liked. As one of the first two nonCambridge graduates to enter their graduate program, he was subjected to some ridicule just for being a provincial. However, his genius quickly asserted itself, and his brilliance could not be denied for very long. For example, he was invited to demonstrate one of his experiments to an evening audience of students, faculty and guests, and Ernest Rutherford did very well. It also helped that he was taken under the wing of the then Cavendish Director, Dr. J. J. Thompson, a brilliant physicist in his own right. (Dr. Thompson discovered the electron.)
In 1898, Dr. Rutherford identified alpha and beta rays in uranium radiation and indicated some of their properties. During this time, he was described by his colleagues as "force of nature". He remained a strong influence on physics throughout his lifetime. Later that same year, he went to Canada to take the Chair of Physics at McGill University, Montreal. There, he discovered new gas, thoron, isotope of radon. With his colleague, Dr. Frederick Soddy, a chemistry professor, he discovered "halflife" of radioactivity.
In 1908, Rutherford was awarded Nobel Prize for Chemistry.
Lord Rutherford died at age 66; his ashes were buried in the nave of Westminster Abbey near the graves of Sir Isaac Newton and Lord Kelvin.
 
Half Life
Term "half life" means a duration for which a certain substance declines by 50%.
For example, radioactive thorium might decrease by 50% in 60 seconds.
This could be described by
M_{tho}(1Δ)^{t} = .5 × M_{tho}
where Δ is % difference in mass per second and the t is time in seconds.
Divide both sides by M_{tho} and set t = 60 sec; we get:
(1Δ)^{60} = .5
After some manipulations with logarithms, we get:
Now, we determine amount of time for any amount of radioactive decay.
For example, how much time (in seconds) to decay to 10% of original?

0 Comments:
Post a Comment
Links to this post:
Create a Link
<< Home