Thursday, March 28, 2013

EXTRAPLANETARY, Prep for the Stars

File:Kuiper oort.jpg
1-7 Day Velocities:
Simpler Newtonian 1-G velocities approximate the more accurate Einsteinian.
SIDEBAR:
Introduction

timeEinsteinian
Velocity
Newtonian
Vel/Dist
km/secAU/DayAU/DayAU
1 dy 847 0.49 0.49 0.24
2 dy 1,692 0.980.98 0.98
3 dy 2,534 1.461.47 2.20
4 dy 3,375 1.95 1.96 3.91
5 dy 4,207 2.43 2.456.11
6 dy 5,0422.912.93 8.80
7 dy 5,874 3.39 3.4212,0
tc (1 - (1 - Δ)t)g × tg × t² / 2
EINSTEINIAN and NEWTONIAN
velocities are close
for first few days
of g-force acceleration.

CONCLUSION:
Use simpler Newtonian method
to approximate velocities
to near planets.
c = 299,792.5 km

sec
= 173.1439 AU

day
= c
Different expressions
for lightspeed, c.
g = .2826%c

day
=847  kps

day
= 0.489 AU

day²
Different expressions for Earth gravity,
g= 9.8065 m/sec².
Interstellar Dress Rehearsal
Beyond Kuiper Belt (KB), we enter the vast space beyond the planets but well within the Solar System.

EXAMPLE: 12.5 days of g-force acceleration can get us to near parts of KB. However, an entire year of g-force only gets us to about one third of the distance to the Oort Cloud.

One year of g-force takes us to 23,800 AU. Compared to interplanetary (less than 100 AU), this distance is enormous; however, it is less than half way to the Oort Cloud, the physical border of the Solar System.
SIDEBAR:
Why go?????
10-70 Day Velocities:
Velocities computed via Newtonian 1-G acceleration diverge from Einsteinian 1-G
 Use Space Borne Resources

Eventually, various space borne enterprises will form strong collectives with considerable inter-commerce. They will eventually realize the enormous practicality of using space resources for space requirements; Kuiper Belt will have plenty of raw materials (i.e., construction), but Oort Cloud comets have the extremely useful fluids.

There will be an economic incentive to transport comets from Oort versus launching Earth's valuable water from Earth’s surface (bottom of Earth's gravity well).
Space Needs Best Met via Space Resources

timeEinsteinian
Velocity
Newtonian
Vel/Dist
%cAU/DayAU/DayAU
10 dy 3%4.834.8924.45
20 dy 5%9.529.7897.80
30 dy 8%14.08 14.67220.05
40 dy 11%18.5119.56391.20
50 dy 13%22.8224.45611.25
60 dy 16%27.0129.34880.20
70 dy 18%31.0834.231,198.1
tc (1 - (1 - Δ)t)g × tg × t² / 2
Daily Difference: Δ
Δ = 0.00282c
G-force vessel does a daily velocity increase toward c, light speed. Per Einstein, vessel observes same Δ everyday.
Daily Remainder: (1-Δ) 
(1-Δ) = 0.99718c
After each day of g-force acceleration, vessel velocity has considerable remainder till reaching c. Per Einstein,  (1-Δ) stays same for vessel observers..
Earthbound Observer
measures vessel's ever increasing velocity.

Thought Experiment (TE) assumes remainder exponential, (1-Δ)t, accurately predicts vessel velocities. 

Similar methods can compute radioactive half lives and certain probabilities.
100-700 Day Velocities:
Einsteinian 1-G Velocities continuously approach but never reach c.
SIDEBAR
How Go???

timeEinsteinian
Velocity
Newtonian
Velocity
AU/Day%c%c
100 dy 4325%c28%c
200 dy 7543%c56%c
300 dy 9957%c85%c
400 dy 11768%c113% c
500 dy 13176%c141% c
600 dy 14182%c169% c
700 dy 14986%c198% c
tc (1 - (1 - Δ)t)Impossible.
Thought Experiment assumes:
1. G-force acceleration daily increments ship's velocity, Δ = 0.283% c.
2. Paradoxically, all observers continue to measure light speed at constant value, c, throughout the voyage.
3. For any day's duration (86,400 sec), ship's observer measures velocity increase of .283% c with respect to the retreating Earth..
4. In like manner, Earth bound observer measures ship's ever increasing speed which approaches c but never reaches it.
Extended use of Newtonian has spaceship velocity exceeding c, an Einsteinian impossibility.
TE proposes exponential to compare ship's velocity against light speed, c.
Vt = c(1-(1-Δ)t
For first few days of g-force travel, the Newtonian (incremental) and Einsteinian (exponential) produce similar results. However, after 365 days, they produce much different results. Thought experiment assumes the Einsteinian to be more accurate.
PROPULSION
TE assumes ship's propulsion system uses a particle accelerator with powerful magnets to manipulate plasma streams. Accelerator magnets guide and focus ions into a tight beam which exit the vessel for exhaust momentum. Extremely high speeds of these tiny particles will collectively propel ship forward at g-force.



Accelerator magnets must be superconducting to consume much less energy then resistive magnets. Superconductor magnets will always be a necessary part of ship's on-board particle accelerator propulsion system. 



However, expanding range from interplanetary will increase trip time from days/weeks to months/years. This duration increase will need a significant design improvement.
SIDEBAR:
How go?????
100-700 Day Distances:
After 700 days of constant g-force, vessel finally gets to Oort Cloud.
SUPERCONDUCTORS

AXIOMATIC: Flights to Oort Cloud (about 65,000 AU) will last much longer then flights to Kuiper Belt (40-100 AU away).  Thus, energy efficiency will become much more important; even for superconductors.



On board particle accelerators will require highly intense electro-magnetic fields which require extremely powerful electro-magnets which require kilometers of precisely manufactured and carefully wound wire (also known as “windings”). Traditional (aka “resistive”) electromagnets would require enormous amounts of energy to produce the required magnetic fields.



Superconducting electromagnets are much more efficient and require much less energy; as a matter of fact, superconductor efficiencies are so significant that it’ll likely prove impossible to achieve g-force acceleration without them.

timeEinsteinian
Velocity (V)
Einsteinian
Distance (d)
AU/Day%cAULY
100 dy 4325%2,2290.04
200 dy 7543%8,1730.13
300 dy 9957%16,9170.27
400 dy 11768%27,7710.44
500 dy 13176%40,2160.64
600 dy 14182%53,8610.85
700 dy 14986%68,4081.08
tc (1 - (1 - Δ)t)c×t + V/ln(1-Δ)
0tv(t) dt = d(t) |t0
0tv(t) dt = ∫0tc (1-(1-Δ)t) dt

After integrating
and some rearrangement:

d = c × t + V

ln(1-Δ)
Einsteinian Distances
Thought Experiment presumes we need a different way to compute g-force distances for distant destinations. Use calculus to derive distance equation.

New Formula for Distance. Integrate velocity equation to determine the corresponding distance.
Two Types of Superconductor Magnets

Cable-in-conduit conductors (CICC) magnets can turn off / on relatively easily. For example, they can ramp up from 0 to 15 Tesla in minutes. When magnetic fields change intensity, they create electrical currents in adjacent conductors; with current comes heat. CICC magnets can handle this heat because the superconducting wires are bathed in liquid helium.
While CICC magnets are much more efficient then resistive magnets, they still use a small amount of power. For the short duration of interplanetary flights, this amount is almost inconsequential.
G-force accelerated trips to near planets will take days and at most weeks. Thus, CICC type magnets prove more than adequate. Their slight energy consumption will be relatively inconsequential compared to high momentum exchange from near light speed exhaust particles.
Wire-wound (WW) magnets can operate without power during most of long voyage's powered flight.  After initial power ramp up (a few weeks), wire-wound magnets disconnect from their power supply, because the superconducting current continues on its own.

For longer flights, the wire wound type will be even more important because the greatly reduced energy consumption will prove significant for an acceleration period of many months and possibly years.
In Summary, Note Key Differences
CICC quickly ramps up
but consumes some power throughout the trip.



Good for interplanetary trips inside KB.
Longer voyages need something more.
WW slowly ramps up 
but needs no additional power once there. This enormous gain in power efficiency would greatly benefit much longer voyages which will last months and perhaps years.



A ship with WW superconductors would need to accommodate this ramp up period prior to takeoff. Thus, preflight procedures might require a few weeks to ramp up; small price to pay for a trip which might take years.
Hybrid????
A ship with both types of superconductor magnet systems could start the voyage with CICC, then switch to WW after rampup process completes. PERHAPS, designers could implement a tradeoff between deleting preflight rampup period and adding considerable weight to facilitate a parallel magnet system.
Perhaps hydrids might help 
more advanced vessels.
To determine range capabilities of shorter duration flights,
intuition leads us to approximate range by dividing original fuel load by daily decrement, Δ (% GW0).
Inner Planetary Ranges (less than 100 AU from Sol)
 Express exhaust particle velocity
as decimal light speed, 

where dc is decimal component:
VExh = dc × c
Use Lorentz Transform to compute
relativistic mass growth (mrat very high speeds:

mr=1

(1 - dc²)
Express 1 sec of exhaust fuel flow (ffExh
as multiple of at rest fuel flow (ffsec) per sec:
ffExh=ffsec

(1 - dc²)
=mr × ffsec
When a propulsion system accelerates exhaust particles to significant fractions of light speed, resultant momentum can propel a large spaceship at g-force.
Use divisionR = %TOGW/Δ 
to quickly approximate, useful range durations.

Use Efficiency Factor: ε = 4,
a static estimate to conveniently assume some practicality.

Use Newtonian Motion Equation: d = g × t2/2
 to quickly approximate acceleration distances.

RELATIVISTIC GROWTH (mr)
At rest mass (ffSec) expands
to exhaust particle mass (ffExh).

MShip=ffExh × VExh

g
=mr×ffsec × dc× c

g
MShip=ffsec × mr

g
×(mr2-1) × c

mr
MShip=ffsec(mr2-1)×c

g
MShip=(mr2-1)×30.57×106×ffsec
Express ship size (MShip) as a mega multiple
of one second fuel consumed (ffsec)
SHIP SIZE (MShip)
capacity grows in correlation 
with particle exhaust velocity.
1) Greater velocity increases contribution to momentum exchange.
EXAMPLE: Conserve momentum and ignore relativity.Assume perfect efficiency. Let particle exhaust velocity be .866c,
   VExh = .866 × 299,792,500 m/s = ‭224,831,184 m/s
   MShip = ffSec × VExh /g = 22,790,794 ffSec
[i.e., about 23 million times one second fuel flow.  If ffSec is one gram of exhaust particles, then Ship Mass is about 22.8 metric Tonnes (mT)].
2) Relativistic growth increases exhaust masses to contribute even further.
EXAMPLE-2:  Conserve momentum and consider relativity.  Assume perfect efficiency. Let particle exhaust velocity by .866c.  Lorentz transform indicates exhaust particle mass doubles. (mr= 2.0).  
Previous work leads to following formula:
   MShip = ffSec(mShip²-1) × c / g = (4-1) × 30.57×106 × ffSec = 52,942,581 ffSec
(i.e., about 53 million times one second fuel flow. If ffSec is one gram of exhaust particles, then Ship Mass is about 52.9 metric Tonnes (mT). 
Of course, inevitable inefficiencies would decrease this amount.
DAILY DIFFERENCE (Δ) Fuel needed for one day of g-force.
Day = 3,600 secs/hr × 24 hr/Day=86,400 sec/day
ffDay=86,400sec × ffsec
ΔffDay

MShip
=86,400 sec × ffsec

ffSec(mr2-1) × c/g
Δg × 86,400sec

(mr2-1) × c
=.2826% GW

(mr2-1)
EXAMPLE: 
Let VExh be 86.6% light speed and arbitrarily assume perfect efficiency. At that speed, mass growth factor is 2.0 (i.e., mass doubles).
mr= 1/(1-dc2
m= 1/(1-.8662) = 2.00

Daily decrement of ship's mass is a small percentage of vessel's GW.
Δ = .2826%GW/(mr2-1)
Δ = 0.163%GW/Day
DIFFERENT RANGES (R) for nearby destinations.
Theoretical Range: Assume fuel to be 100% of ship's GW (entirely all fuel). 
For convenience, assume: 
RTheo ≈ 1/Δ = 100%/0.163%/day = 613.5 days

Feasible Range: assume ship's fuel between 10%-90% of GW.  For this example, let initial fuel be 40% of ship's initial GW. 
We could intuitively assume following:
RFeas ≈ 40%/Δ = 40%/0.163%/day  = 245.4 days
Practical Range: Assume much less than perfect efficiency (E=100%).  For inner planetary ranges, let E=25% and efficiency factor be 4 (ε = 1/E = 4).  Approximate:
RPrac ≈ 40%/(ε Δ) = 40%/(4 × 0.163%/day) = 61.35 days.
Recall 4 phases of a typical g-force voyage  
   1) Phase 1 is acceleration from departure to midpoint.
   2) Phase 2 is deceleration from midpoint to destination.
   3) Phase 3 is acceleration while returning back to midpoint.
   4) Phase 4 final phase is deceleration from midpt back to departure. 
 Since all 4 phases are equal, TE assumes one fourth of practical range i for Phase 1, acceleration time, tAcc.
DISTANCE (d) and TIME (t) to nearby destinations.
Phase I Acceleration Time (tAcc): Previous tables show value derived from Particle Exhaust velocity.  In turn, this value is basis for following terms.
Phase I Acceleration Distance (dAcc): ... can be closely approximated via Newtonian method: 
dAcc = g × (tAcc)²/2
Let g equal acceleration due to gravity,
g = 9.8065 m/sec² = 0.489 AU/day²
Phase II Deceleration Time (tDec):  that g-force deceleration time is same as acceleration time for same distance; thus, tDec = tAcc
Total Time Destination (tTtl):  
Add acceleration time and deceleration time.  
Since those times are equal, it doubles acceleration time:
 tTtl  = tAcc + tDec = 2 × tAcc

Total Distance Destination (dTtl): 
Since Phase I Accel takes vessel to midway point,
half the distance to destination; 
then, Phase II Decel distance must be remaining half.
Thus, total distance to destination is double the distance to midway.
Closer Destinations (VExh from .1c to .866c enable g-force ranges up to 30+ days.)
SIDEBAR:
Daily Decremental

NOTE: Exhaust Particle Velocity (VExh)= dc; such as .866c, 86.6% light speed.
Previous tables show how increasing exhaust particle velocity increases range of g-force vessel in both duration and distance.

Increasing particle velocity to .866c gives vessel a range to over 30 days of g-force to beyond Kuiper Belt.

Extraplanetary tables will use Einsteinian methods to show increased performance.
c = 299,792.5 km

sec
= 173.1439 AU

day
= c
Different expressions
for lightspeed, c.
g = .2826%c

day
=847  kps

day
= 0.489 AU

day2
Different expressions for gravity,
g= 9.8065 m/sec²
day × g

c
=86,400 sec × 9.8065m/sec²

299,792,500 m/sec
= 0.2826%
Daily increment(Δ): 1 day of g-force.
gains velocity equal to 0.2826%c.
Fuel Burn Reduces GW
As with trains, planes and automobiles, g-force vessel's GW decreases with fuel consumption. How do we compute ever decreasing Gross Weight (GW)?
Daily Decrement Model
Following intuitive model could approximate first few days of fuel consumption.
EXAMPLE-1: If fuel for one day of g-force propulsion (Δ) is 0.652% of vessel's initial Gross Weight (GW0); then, common sense suggests an arithmetic method of approximating fuel consumptions.
DayFuel
Burn (Δ)
Gross
Weight
00.000% GW0‭100.00% GW0
10.652% GW0‭99.348% GW0
21.304% GW0‭98.696% GW0
...................‬
tΔ×t100%GW0-Δ×t
EXAMPLE-2: If original fuel is 40% of ship's original GW, what is ship's range?
Let daily differential of ship's GW (Δ) continue to be 0.652% GW0. Intuition leads us to approximate range by dividing original fuel by Δ.  Thus, Approximate Range (RApp) is 40%/Δ = 40%/0.652% per day =61.35 days.
To determine range capabilities of longer duration flights,
use exponential methods; they are more accurate and more generous than division.

GWt =(1 - ε Δ)t GW0

GW is Gross Weight of vessel, 
units are in percentages of original GW.

GW0 is Original GW of vessel, 
prior to starting powered flight.

GWt is Daily GW of vessel, 
measured at start of day, t.
t is time, 
days of powered, g-force flight.

Δ is daily diff (%GW)
between any two consecutive days

ε is Efficiency Factor, inverse of E (ε =1/E);
ε can find additional fuel consumption 
needed to overcome inefficiency.
EXAMPLE-1 (Assume a very impractical 100% Efficiency): 
Previous work indicates exhaust particle speed of .866c;
theoretically, g-force vessel uses daily fuel consumption of
Δ = 0.163%GW
Thus, fuel consumption for 10 days of g-force
could theoretically reduce vessel's GW as shown:
GW10 = (1 - Δ)10 GW0 = (1 - .163%)10 GW0
GW10 = (0.99837)10 GW0 = 0.98382 GW0
EXAMPLE-2  (Assume more practical efficiency of E=25%)
For every unit contributing to propulsion, 
vessel must consume 4 units;
 thus: ε =1/E=1/.25= 4
Practical fuel consumption for 10 days of g-force
could reduce vessel's GW as shown:
GW10 = (1 - εΔ)10 GW0 = (1 - 4×.163%)10 GW0
GW10 = (0.99348)10 GW0 = 0.93668 GW0
NOTE: Practical Fuel (F) consumed after 10 days g-force: 
F10 =GW0 -GW10 = .06332 GW0
EXTRA-PLANETARY RANGES (need greater precision)
When a propulsion system accelerates exhaust particles to near light speeds,
resultant momentum can propel an even larger spaceship at g-force for a much longer duration.
For greater precision, Extra-Planetary Tables use more accurate methods than the Inner Planetary Tables.
Logarithms: 
=log(1-%TOGW)

log(1-Δ)
for more accurate, useful range durations.
Efficiency 
Factor:
ε = 41/(mr-1)
an arbitrary, dynamic model which assumes 
continuous performance improvement.
Einsteinian 
G-Force Equation:
d(t) =c×tAcc+V(t) 

ln(1-Δ)
for more accurate, useful distances.
RELATIVISTIC GROWTH (m r)
Mass consumed at rest  (ffSec) expands
to exhaust particle mass (ffExh).
ffExh = mr× ffsec
While it can be expressed 
as any rational number,
TE chooses to express mr 
as a series of integers. 
Each value of 
relativistic mass growth (mr
can map to corresponding 
particle velocity.
dc=(mr² - 1)

mr
DAILY DIFFERENCE (Δ)
Day=3,600s/hr×24hr/Day=86,400s/day
ffDay=86,400sec × ffsec
ΔffDay

MShip
=86,400 sec × ffsec

ffSec(mr2-1) × c/g
Δg × 86,400 sec

(mr2-1) × c
=.2826% GW

(mr2-1)
Determine fuel requirement
.... for one day of g-force.
EXAMPLE: Let relativity growth factor, mr, be 5 and arbitrarily assume perfect efficiency.

Daily decrement of ship's mass
is a small percentage of vessel's GW.
Δ = .2826%GW/(mr2-1) 
Δ = .2826%GW/(52-1)
Δ = .058%GW/Day
RANGE (R) to extra-planetary destination
Computed Feasible Range: Let initial fuel (%TOGW) be 40% of ship's initial GW.   Use logarithms to model range durations.  
EXAMPLE: Let relativity mass growth (mr) = 6.     Compute:
RFeas=log(1-%TOGW)/log(1-Δ)=log(1-40%)/log(1-048%)=1069 days
Efficiency Factor, ε, of ship's propulsion system. TE uses an arbitrary function to generate dynamic values which map to mr values.  Actual efficiency factors will eventually come from observations of actual propulsion systems. For now, arbitrarily compute:  ε = 41/(mr-1)
Computed Practical Range: TE assumes Efficiency (E) will ever approach but never reach perfection (E = 100%)..Thus, TE assumes efficiency improves as propulsion performance improves (increasing value of mr). Compute:
 RPrac = log(1-%TOGW)/log(1-εΔ) = log(1-40%)/log(1-1.32×.048%)=810 days
Phase 1 Acceleration Time (tAcc ):  Compute this value as one fourth of Practical Range:   tAcc  =  RPrac /4
TOTAL TIME (tTtl) and TOTAL DISTANCE (dTtl) to extra-planetary destinations
Phase I Acceleration Time (tAcc): Previously computed (see above.)
Einsteinian Velocity: Δ is daily progress of g-force vessel toward light speed, c.
Δ = day×g/c = 86,400sec×9.8065m/sec² /299,792,500m/sec = .2826%c
(1-Δ) is daily remainder of ship's velocity till reaching light speed. (1-Δ)=.997173c
Earth observer measures subsequent daily reminders per exponential: (1-Δ)t c
Earth observed vessel velocities:  c - (1-Δ)t c
Phase I Acceleration Distance (dAcc): ... Calculus gives us distance solution.
Phase II Deceleration Time (tDec):  To maintain g-force, deceleration duration must be same as for acceleration (Phase I).
Total Time Destination (tTtl):  Double acceleration duration from Phase I.  This includes acceleration time from departure to midpoint and deceleration from midway to destination.
Total Distance Destination (dTtl): Double acceleration distance from Phase I which sums acceleration distance plus deceleration distance..  Note this distance is much less than distance achieved by g-force acceleration for same time
SIDEBAR:
Daily Exponential
Oort Cloud Destinations
G-force performance can be measured by relativistic mass growth (mr). For Oort Cloud,  mr values 2 to 10 generate propulsion ranges exceeding 52,000 AUs.
Fuel consumption reduces ship's Gross Weight (GW).  Let daily decrement (Δ) be fuel consumed for one day of g-force propulsion.
EXAMPLE:  if Δ = .163% GW; then, vessel's will decrease by .163% in one day.  For subsequent days of powered flight, vessel's GW will continue to shrink.
Exponential Model
For longer duration voyages, use exponentials to readily determine vessel's gross weight for t days of g-force flight
GWt = (1-Δ)t × GW0
For corresponding fuel requirements.
Ft = GW0 - GWt
For the lengthier voyages, the precise exponential model yields much more encouraging results than the daily decremental model.


CONCLUSION

After travel to/from Oort Cloud 
becomes routine,
interstellar travel is next.
VOLUME 0: ELEVATIONAL
VOLUME I: ASTEROIDAL
VOLUME II: INTERPLANETARY
VOLUME III: INTERSTELLAR

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