Wednesday, July 29, 2009

RECURRING REMAINDER INCREASES PRACTICALITY









Recurring Remainders Leverage Exponentials.

Far-reaching Ranges Require Logarithms.

Motion Equations Turn Time
into Velocity into Distance.

Increase Particle Velocity
to Increase Range.










Conserve Mom: MShip × VShip=mExh × vExh
Let VShip = 9.80665 m/sec;
then, VShip /sec= 9.80665 m/s2= g
 Previous Work: MShip × gffSec × vExh
Light speed in vacuum: c = 299,792,458 m/s
Previous Work: MShip × gffSec × dc×c
Axiomatic: ffDay=86,400 sec/day×ffSec
=ffDay

 MShip
86,400 sec × ffSec

ffSec × dc × c / g 
=86,400 sec × g

dc × c
=.2826% GW

dc
=
.2826% GW

dc
EXAMPLE: Let particle exhaust speed = 56.5% light speed; then, dc = .565, and daily difference (∇) = 0.5% GW to produce g-force 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 - (1-2∇)t
RANGE
R = log(1-%TOGW)

log(1-2∇)

1) NEWTONIAN. G-force time for given distance to destination (d):  t = (2×d/g)  

2) ACCELERATE to halfway (d/2). Determine g-force 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 G-force solution: 2∇ = 1% GW/day
Daily GW remainder is: (1-2∇) = 1 - .01 = 99%.
Thus, every day's GW is 99% of previous day's GW.
Determine fuel consumed: 
initial GW (100%) minus expected GW(.99t).
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
Observedd

.5 AU/day²
1 - .99t
Range: R = log(1-%TOGW)/log(1-2∇)
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 (fExh) exceeds consumed fuel flow (ffsec).
Restate ffsec with n, mass multiple:    ffExh  = n × ffsec
ffExh = 
ffSec

(1 – dc2) 
Lorentz Transform (LT) quantifies relativistic mass increase of high speed particles.  EXAMPLE: Let one second of consumed fuel flow, ffsec, = 1 gm; let it achieve a relativistic velocity, dc×c =.700c = 70.0% light speed prior to exhaust;  
n
=
1

(1 - dc2)
=
1

(1 - .7002)
=1.4
then, 1.0 gm of ffsec will become 1.4 gm of exhaust fuel flow, ffExh,just before it exits vessel and contributes to forward momentum.

Thus, slightly change ship's mass  equation:
MShip =× ffSec  ×dc× c

g  
=dc

(1 - dc2)
 × 30.57×106 ffSec

However, TE still assumes daily consumption is
ffDay=86,400sec×ffSec
because fuel is consumed at rest.

Thus, re-accomplish daily diff. equation:
=ffDay

 MShip
=86,400sec×ffSec

ffSec×dc×c/(g×(1-dc2)
=86,400 sec × g/c

dc/(1-dc2)
=.2826% GW

n × dc
EXAMPLE: Let particle's exhaust speed (VExh) be .7c; then, light speed's decimal component, d, 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 g-force vessel's GW is ever decreasing at the daily rate of 2∇; thus, the quantity, 1-2∇, is a recurring remainder, and the exponential, t, an compute ship's gross weight (GWt) after t days of powered flight. EXAMPLE; Let accel time (t) be 30 days.
GW= GW0(1-2∇)t(1-.0058)30 GW0 = 84.1% GW0

If t, acceleration time, is known (Example: t = 60 days) ; then, determine fuel required:
t= GW0(1-(1-2∇)t) = (1-(.9942)30)GW0 = 15.9% TOGW
Under above conditions, vessel must expend 15.9% of initial GW to maintain g-force for 30 days (NOTE: Artificially estimates efficiency as 50%.)















Relativity Increases G-force Effect
ffExh × VExh / g = MShip = n × ffsec× dc × c / g
dcntProptVCrud
dec.
comp.
Mass
Mult.
per cent
Gross Wt.
Prop.
time
Accel
time
Cruise
Veloc.
Accel
Dist.
.3001.05.90% GW38.2 days9.6 days4.6 AU/dy28 AU
.4001.0965% GW53.2 days13.3 days6.4 AU/dy43 AU
.5001.1549% GW70.5 days17.6 days8.4 AU/dy75 AU
.6001.2538% GW91.6 days22.9 days10.9 AU/dy126 AU
.7001.40.29% GW119.9 days30.0 days14.1 AU/dy214 AU
.8001.67.21% GW
163.2 days
40.8 days18.9 AU/dy392 AU
.8662.00.16% GW212.7 days53.2 days24.2 AU/dy659 AU
Given1

(1 - dc2)
.2826%GW

n× dc
log(1-%TOGW)

log(1-ε∇)
tProp

P
c[1-Rct]
c×t +VCru

ln(Rc)
log(1-%TOGW)

log(1-ε∇)
Propulsion Time (tProp):  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)". EXAMPLE-1: 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.  EXAMPLE-2: If %TOGW is 50%; then, ship's minimum weight (fuel depleted) is 50%.  For convenience, this table assumes %TOGW = 50%.
--Daily Difference (): To maintain g-force 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 g-force 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 artifically assumes a dynamic efficiency model as described below.
tProp

P
Acceleration Time:  Typical interstellar flight profile can be divided into four phases; thus, P =4,  Phases are described below; For each phase,t = tProp/4.
Phase-1) Accelerate to midpoint. Fuel = GW0(1-(1-∇)t) = 15.9% TOGW
Phase-2) Decelerate to dest. Fuel = GW0(1-(1-∇)2t) = 29.3% TOGW
Phase-3) Accelerate back to midpoint. Fuel = GW0(1-(1-∇)3t) = 40.5% TOGW
Phase-4) Decelerate and return to dept.  Fuel = GW0(1-(1-∇)4t) = 50.0% TOGW
AXIOMATIC: All four phases require equal time/distance.
c[1-Rct]
Cruise Velocity: After t days of g-force propulsion, vessel accelerates to velocity which it will maintain throughout the cruise portion of flight to nearby star.
--Daily Light Speed Increment due to g-force (Δ):
 AXIOMATIC: After one day of g-force, 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 g-force day.
--Daily Remainder till light speed (Rc): Thus, on-board observer measures c(1-Δ) 0.997174 cremaining velocity until light speed, c for any g-force day. TE assumes that earth-bound observer can model ship's velocity as difference between c ( 299,792,458 m/s = 173.145 AU/day) and recurring remainder (1-Δ)which can be determined with exponential, t, as number of g-force days.
c×+VCru

ln(Rc)
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 super-heats 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.  
E' = loss/in = 250/1000 = 25%
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×.9n-2Thus,
E =  1 - E' = 1 - (.5×.9n-2)

ε
Efficiency Factor is Inverse of Efficiency: ε = 1/E.
If you need 1,000 high speed ions to move vessel at g-force 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:
ε =1

 1 - (.5×.9n-2)

Let one second of consumed fuel flow, ffsec,
be 1,000 gms (one kilogram);
 let it achieve a relativistic velocity, dc×c =.992c = 99.2% light speed;
then, 1.0 kg of ffsec will become 8.0 kgs of exhaust fuel flow, ffExh,
just before it exits vessel and 
contributes to forward momentum.
For convenience, note following identities:
n
=
1

(1 - dc2)
dc=  (n- 1)

n
n × dc = √(n- 1)
Increases G-force Efficiency
=ffDay

 MShip
=86,400sec×ffSec

n×ffSec×dc×c/g 
=.86,400×g/c

× dc
=.2826% GW

√(n2-1)
nΔεtVCrud
Mass
Mult.
%GWEfficiency
Factor
Accel
time
Cruise
Vel.
Accel
dist.
2.163%2.0053 days24.1AU/dy655 AU
3.100%1.8295 days40.9AU/dy 2,038AU
4.073% 1.68141days57.0AU/dy 4,295AU
5.058%1.57191days72.3AU/dy 7,510AU
6.048%1.49244days86.3AU/dy 11,709AU
7.041%1.42299days98.9AU/dy 16,870AU
8036%1.36357days110.1AU/dy 22,939AU
Given.2826%

√(n2-1)
1

1-(.5×.9n-2)
log(1-%TOGW)

4×log(1-εΔ)
c[1-Rct]
c×+VCru

ln(Rc)
Acceleration: After one year (365.25 days) of g-force 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 g-force velocity determined by: VCru=c[1 - Rct]
Distance: TE assumes g-force distance determined by: d = c[t + VCru/ln(Rc)]
Deceleration: To maintain g-force while decelerating from 64.43%c to essentially zero, deceleration requires same duration/distance as acceleration.
Interstellar Profile: Consistent Acceleration Time
DestinationAcceleration
Time
Accel.
Velocity
Accel.
distance
Cruise
distance
Cruise
time
Total
Time
StarDttDayVCruddCrutCrutrT
NameLYYrday% cLYLYYrsYrsYrs
α Cent4.3651.0365.364.43%0.383.615.603.347.60
Bar...Sta5.9701.0365.364.43%0.385.218.084.8310.08
Luh...166.6001.0365.364.43%0.385.859.075.4111.07
OBSERVEDGiven365.25tSee previous tables
to see how to compute
these values.

- 2d
dCru

VCru
tCru

n
tCru
+ 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:  dCru = 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.
tCru=dCru/VCru
Relativistic
Time
Due to relativistic effects,
 ship bound chronometer
measures:1.19 years

for each cruise LY.
tr=tCru×(1-VCru2/c2)
DestinationAcceleration
Time
Accel.
Velocity
Accel.
distance
Cruise
distance
Cruise
time
Total
Time
StarDttDayVCruddCrutCrutrT
NameLYYrday% cLYLYYrsYrsYrs
α Cent4.3650.8292.252.26%0.263.856.855.668.45
Bar...Sta5.9700.9328.760.56%0.315.348.827.0210.62
Luh...166.6001.0365.364.43%0.385.859.076.9411.07
Wolf..3597.7861.1401.867.92%0.46.9010.167.4612.36
Lal...211858.2911.2438.371.07%0.517.2710.227.1912.62
Sirius8.581.3474.873.91%0.587.4110.036.7612.63
OBSERVEDGiven365.25tSee previous tables
to see how to compute
these values.

- 2d
dCru

VCru
tCru

n
tCru
+ 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 trade-offs 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 in-flight 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.
ndcεAccel.
Time
(t)
Cruise
Vel. (
VCru)
Accel.
Dist. (d)
80.99220.036%GW1.36357 dy0.98Yr110.2AU/dy.636c 22,939AU0.36LY
90.99380.032%GW1.31417 dy1.14Yr120,0AU/dy.693c29,842AU0.47LY
100.99500.028%GW1.27479 dy1.31Yr128.5AU/dy.742c37,490AU0.59LY
110.99590.026%GW1.24542 dy1.48Yr135.8AU/dy.784c45,793AU0.72LY
120.99650.024%GW1.21605 dy1.66Yr141.9AU/dy.820c54,661AU0.86LY
130.99700.022%GW1.19670 dy1.83Yr147.2AU/dy.850c64,010AU1.01LY
140.99740.020%GW1.16735 dy2.01Yr151.5AU/dy.875c73,763AU1.17LY
150.99780.019%GW1.15801 dy2.19Yr155.2AU/dy.896c83,850AU1.33LY
Given(n2-1)

n
0.2826%

√(n2-1)
1

1-(.5×.9n-2)
log(1-%TOGW)

4×log(1-ε∇)
c[1 - Rct]
c×t +VCru

ln(Rc)
DestinationMass
Mult
.
Accel
time
 
Cruise
Vel. 
Accel.
dist
Cruise
dist.
Cruise
Time
Total
Time
StarDnt VCruddCrutCrutrT
NameLYYrdec. cLYLYYrYrYr
α Cent4.36580.980.64 c0.36 2.42 3.802.935.65
Bar...Sta5.97091.140.69 c0.473.69 5.333.847.61
Luh...166.600101.310.74 c0.593.99 5.373.607.99
Wolf..3597.786111.480.78 c0.724.836.163.829.13
Lal...211858.291121.660.82 c0.864.996.083.489.40
Sirius8.583131.830.85 c1.014.925.793.059.46
Luy.726-88.728142.010.88 c1.174.715.382.609.41
Ross 1549.681152.190.90 c1.335.305.922.6210.30
OBSERVEDSee green table in left panel.
-2d
dCru

VCru
tCru

nCru
tCru
+2t
CONCLUSION: Many ways to get there!!!




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



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