Monday, December 11, 2006

INTERSTELLAR SUPER G

If g-force is great, is super g-force even greater?

Expectations continually adjust to ever higher levels. For example, it's very possible that someone will become a dot com billionaire; you'd think she'd be happy forever. However, relevant experience indicates such happiness lasts for a much shorter time.

Our billionairess will soon learn that a billion dollars is not infinite, and materialistic desires will quickly grow to accommodate that finite amount. For instance, she can fly her private jet; but she now dreams of space travel.

In similar fashion, consistent g-force propulsion will prove a great boon to interstellar travelers as g-force propulsion will reduce interstellar voyage durations from centuries to years.  However, humans will soon want even faster transport for much quicker cargo deliveries. Interstellar shipments of essential materials could be greatly expedited if they could fly at super-g.


Near Earth's surface, a free falling object will accelerate at rate g, approx. 9.8 m/sec2.

Thus, an acceleration of 10 g
would be about 98.1 m/sec2.

Using Newtonian motion equations, we previously determined the first second of 1-g free fall will move object 4.9 meters. Extend  to one day of constant g-force acceleration to move vessel to .245 Astronomical Units (AUs).
Expand our thought experiment's envelope to exceed g-force acceleration for selected vessels. Assume no humans, but Artificial Intelligence (AI) devices will guide a vessel with essential cargo.
Express 10g differently.
10g98.065 m

sec2
×(86,400 sec)2

day2
×1 AU

149,597,870,700 m
10g ≈4.893 AU

day2
If double g-force does not damage ship or contents; then, much shorter travel time would be an enormous benefit.  
d=a×t2

2
4.893 AU

day2
×(1 day)2

2
=2.45 AU
let a = 10g = 4.893 AU/day²
NOTE: 1 sec of 10-g acceleration moves an object 49.1 m; thus, extend to 1 day (86,400 sec) for distance of 2.45 AU.
First Day of i×G-force Propulsions
Describe greater accelerations as integer (i) times g, gravity force experienced on Earth's surface.
DISTANCE EXAMPLE:  After one day of 10g-force acceleration, object travels a great distance:
dAccel = .5 × a × t2 = .5 × 10g × t2 = .5 × 10(.245AU/day2) × day2 = 2.447 AU
Attain Great Speeds with 1-G Propulsion
Re-express light speed, c
c=299,792.458 km

sec   
= c ×AU

149,597,871 km



×86,400 sec

day



=173.145 AU

day
Daily Difference:
Δ1-G=VDay

c



.489 AU/day

173.145 AU/day 
0.283% c
Daily Remainder: (1 - Δ)

R1-G=c (1 - Δ1-G)  =c(1-.00283)=.99717 c
Previous work shows Vt = c(1 - Rt)
Time
Einsteinian
Velocity
Newtonian
Velocity
Days
% Light Speed
AU/Day
AU/Day
1
0.283% c
0.490
0.490
5
1.407% c
2.436
2450
10
2.794% c
4.838
4.900
t
c (1-Rt)
Convert
t × a1-G
Show Einsteinian velocities as both percent light speed (%c)
and equivalent AUs per day.
For lesser velocities, Einstein and Newton velocities are fairly close.
Much Greater Speeds with 10 G Propulsion
Daily Difference for 10g propulsion
Δ10G= 10× 0.283%c = 2.83%c
Daily Remainder for 10g propulsion
R10G=c (1 - Δ10G)  =c(1-.0283)
=.9717 c
For 10g propulsion, let R = .9717.
Time
Einsteinian
Velocity
Newtonian
Velocity
Days
% Light Speed
AU/Day
AU/Day
1
2.830% c
4.838
4.838
5
13.371% c
23.152
24.190
10
24.955% c
43.208
48.380
t
c (1-Rt)
Convert
t × a10G
After 10 days of 10-g propulsion,
Einsteinian velocity significantly differs from Newtonian velocity.
TE assumes Einsteinian velocity is more accurate.  See next table.

Velocities for Different Super-G's

Accel.a = 1 × ga = 4 × ga = 7 × ga = 10 × g
Daily Diff.Δ = .2826%cΔ = 1.1305%cΔ = 1.9784%cΔ = 2.826%c
Daily Rem.R = 99.7174%R = 98.8695%R = 98.0216%R = 97.1738%
Time (t)EinsteinNewtonEinsteinNewtonEinsteinNewtonEinsteinNewton
AU/dayAU/dayAU/dayAU/dayAU/dayAU/dayAU/dayAU/day
1 day
0.489
0.4891.9571.9573.4253.4254.8934.893
10 days4.8324.89318.60819.57431.36034.25443.15848.935
100 days42.6848.93117.60195.74149.67342.54163.70489.35
200 days 74.8497.87 155.33391.48169.96685.09172.58 978.69
300 days99.07 146.80 167.43587.22 172.71 1,027.63173.11  1,468.04
365¼ days 111.56 178.73170.42 714.92  173.03 1,251.14173.141,787.34
400 days 117.33195.74171.31782.96 173.091,370.17173.141,957.39
Given
Vt = c ×(1 - Rt)173.145 AU/day

c
Einsteinian Velocities
Vt = t × a = t × i×g =t × i ×0.489 AU

day²
Newtonian Velocities
For even faster interstellar cargo transport, consider super G-forces even greater than 1g.
With acceleration as i × g,  table considers integer "i" values as high as 10.
Newtonian method computes velocities greater than light speed (c=173.145 AU/day) ;
IMPOSSIBLE!!!!
Thus, Thought Experiment proposes Einsteinian Velocities for subsequent use.

Interstellar Distances for Different Super-G's

Accel.a = 1 × ga = 4 × ga = 7 × ga = 10 × g
Daily Diff.Δ = .2826%cΔ = 1.1305%cΔ = 1.9784%cΔ = 2.826%c
Daily Rem.R = 99.7174%R = 98.8695%R = 98.0216%R = 97.1738%
Time (t)VelocityDistanceVelocityDistanceVelocityDistanceVelocityDistance
%cLY%cLY%cLY%cLY
1 day
0.28%
0.00000391.13%0.00001551.98%0.000032.83%0.00004
10 days2.79%0.0003810.75%0.0015018.11%0.0025624.93%0.00358
100 days24.65%0.03567.92%0.11086.44%0.15594.31%0.184
200 days43.22%0.12989.71%0.33298.16%0.41399.68%0.452
300 days57.22%0.26889.71%0.58999.75%0.68599.98%0.726
365¼ days64.43%0.37798.43%0.763 99.93%0.86399.9972%0.905
400 days67.76%0.44098.94%0.85799.97%0.95899.9990%0.999999
Given
Vt = (1 - Rt)× c
Einsteinian Velocities
dt =c × t+Vt

ln(1-Δ)
Einsteinian Distances
For even faster interstellar cargo transport, consider super G-forces even greater than 1g.
With acceleration as i × g,  table considers integer "i" values as high as 10.
Thought Experiment proposes Einsteinian Velocities and Distances.


Interstellar Fuel Flow

Relativistic fuel particles grow in mass
which further increases momentum of exhaust flow.
EXAMPLE: Let one second of consumed fuel flow, ffsec, be 1,000 gms of at rest water particles.  Accelerate these particles to a relativistic exhaust velocity ( VExh) of .992 light speed.
VExh = dc×c =.992c = 99.2% light speed
As shown in the right hand panel, 1.0 kg of ffsec will grow to become 8.0 kgs of exhaust fuel flow, ffExh,  just before it exits vessel and contributes to forward momentum.  NOTE:  Consume fuel particles at rest; expel at relativistic speeds.
ffExh = n × ffsec
For convenience, note following identities:
Growth Factor
n
=
1

(1 - dc2)
Growth Factor (n)
Uses Lorentz Transform (LT) as shown.
Decimal Component (dc)
If particle speed is expressed as decimal light speed; then, dc is the decimal portion.
Decimal Component
dc     = (n- 1)

n
LT can be rewritten
to solve for d (see table at right). 
Thus, conveniently generate below table.
Growth
Factor (n)
12345678
Fuel Exhaust
Speed(dc×c)
0c.866c.942c.968c.980c.986c.990c.992c
For convenience, n is expressed as series of integers,
but n could also be a rational number greater than 1.



To discuss following tables,
TE arbitrarily assumes a relativistic growth factor of 8.

Theoretical Daily Diff: (∇E=1)

Consistent acceleration ever increases daily velocities and daily distances; it also requires daily fuel,
best expressed as a percentage of ship's gross weight (%GW).
First, consider daily diff with 100% Efficiency (E=1),
a misconception, soon corrected.

E=1=ffDay

 MShip
Determine ratio
of daily fuel
to Ship's Mass
E=1=86,400 × ffSec

 MShip
Daily fuel =
86,400 seconds

of at-rest fuel consumption
 To determine ship's mass:
Recall Momentum Exchange
Huge mass of ship × small velocity increase =
tiny collective mass of fuel exhaust particles × huge relativistic velocity.
MShip × VShip = MExh × VExh 
Divide each side by one second for useful restatement.
MShip × VShip

sec
= MExh × VExh

sec
  • Ship's velocity per second (VShip/sec) can be expressed as ship's acceleration (AShip).
  • Exhaust mass per second (MExh/sec) can be expressed as ship's exhaust fuel flow (ffExh).

Substitute and re-arrange:
Equal Forces
MShip × AShip = ffExh × (dc × c) 

State ship's acceleration (AShip) as multiple of g, Earth's gravity:
Equal Forces Restated
MShip × (i × g )  = (n ×ffsec) × (dc × c)

E=1=ffDay

Mship
=86,400 × ffSec

ffSec×dc×c/(i×g)
Solve for Ship's mass (Mship) and substitute:

E=1 =ffDay

Mship
=86,400 ×i×g/c

n × dc

Further
simplify:

n ×dc =n ×
(n- 1)

n
=(n- 1)
Recall decimal component in terms of n, growth factor:

E=1=i×.2826% GW

(n2-1)
Make following substitutions:
Light speed, c = 299,792,458  m/sec
Earth gravity, g = 9.8065 m/sec²

E=1=i × .0356% GW
Make following substitutions:
Growth factor, n = 8
(n2-1) = 7.937

Gravity
Mult (i)
12345678
Exhaust
Flow (∇)
0.0356%0.0712%0.1068%0.1424%0.1780%0.2136%0.2492%0.2848%
If perfectly efficient vessels were possible; then, a 7g-force vessel could use 1% of it's GW for 4 days propulsion. However, we must also consider
INEVITABLE INEFFICIENCIES!!
TE proposes a coefficient {efficiency factor (ε)} to increase fuel consumption (∇ ) to cover both the Exhaust Flow (ffExh=n×ffsec) and inevitable particle losses: design flaws and peripheral systems.
Synchrotronic Propulsion System
will have inevitable inefficiencies.
However, they will improve over time.
 
Inefficiency  (E'= loss/input) reflects the inevitable loss of consumed fuel particles which will not exit the vessel to contribute to ship's velocity increase.  This loss may be intentional such as particles diverted for other services.  Loss may be unintentional due to design flaws.  TE assumes that subsequent designs will decrease particle loss.
Interplanetary Inefficiency: TE assumes  50%.
Interstellar Efficiency must be much better than interplanetary; thus, TE assumes constant tech improvement correlating to n, growth factor. (For following examples, let n = 8.)
E' = (.5 × .9n-2) = .2657 

Efficiency, E, can be stated as complement of Inefficiency, E'.
E =  1 - E' = 1 - (.5×.9n-2) = .7347 

Efficiency Factor (ε) is a necessary coefficient of daily exhaust flow (∇).  As the reciprocal of Efficiency, ε always exceeds one:
ε =1

1 - (.5×.9n-2)
= 1.362

Interstellar Daily Difference

Efficiency Factor (ε) will increase fuel consumption to overcome inevitable inefficiencies.
Gravity
Mult (i)
12345678910
Daily
Diff. (ε×∇)
0.0468%0.0936%0.1404%0.1872%0.1872%0.2808%0.3276%0.3744%0.4212%0.4680%
Given 7g acceleration, 1% ship's gross weight will provide about 3 day's propulsion.

Interstellar Fuel Consumptions for Different Super-G's

Accel.a = 1 × ga = 7 × g
Daily Diff.ε∇ = .0468% GWε∇ = .339% GW
Daily Rem.(1-ε∇) = 99.9532% GW(1-ε∇) = 99.61% GW
Time (t)Velocity (Vt)Distance (dt)Fuel (ft)Velocity (Vt)Distance (dt) Fuel (ft)
percent cLight Yearpercent GW % light speedLight Year%Gross Wt.
1 day0.28%c0.000 003 9 LY 0.048% GW 1.98%c0.000 03LY 0.339% GW
10 days 2.79%c0.000 38 LY 0.48% GW 18.11%c0.002 56LY3.34%  GW
100 days 24.65%c0.035 LY 4.73%GW86.44%c0.155LY28.82%GW
200 days 43.22%c0.129 LY9.24%GW98.16%c0.413LY49.34%GW
300 days 57.22%c0.268 LY13.54%GW99.75%c0.685LY63.94%GW
365¼ days  64.43%c0.377 LY 16.23 %GW99.93%c0.863LY71.11%GW
400 days  67.76%c0.440 LY  17.63%GW99.97%c0.958LY74.33%GW
Given
Vt = c ×
( 1 - (1-Δ)t )
Einsteinian Velocities
dt =c × t+Vt

ln(1-Δ)
Einsteinian Distances
ft=
1 - (1-ε∇)t
Fuel Consumed
For even faster interstellar cargo transport, consider 7g acceleration.
Assume fuel exhaust particles have velocity, VExh = 99.2%c, with a corresponding growth factor, n = 7.
Thought Experiment proposes Einsteinian motions and exponential fuels.
7g acceleration greatly increases distance and speed much quicker than 1g acceleration;
DOWNSIDE:  7g propulsion requires MUCH MORE FUEL!!!
For more, see "Snowball from Oort".



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




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