Sunday, October 22, 2006

The Planets, Initial Thoughts


LIST OF TABLES
Typical Distances
Sol Earth
(AU) (AU)
Sun
0.00
1.00
Mercury
0.39
1.39
Venus
0.72
1.72
Earth
1.00
2.00
Mars
1.50
2.50
Jupiter
5.20
6.20
Saturn
9.50
10.50
Uranus
19.20
20.20
Neptune
30.10
31.10
Kuiper Belt
44.00
45.00
dSol d = dSol + 1
In considering travel distances from Earth to neighboring planning consider the following.

Intuitive "straight line" distance don't consider orbiting considerations nor necessary detours around the Sun.
In spite of this, maximum straightline distances can  be easily computed as shown in the table (just add one to distances from Sol, second column).

This simple change does increase distance from Earth to Earth from from zero AU to two AUs. There might be a need for a spacecraft to fly to opposite side of Earth's orbit, perhaps a test flight or "shakedown cruise".

Conjunction and Opposition
 Consider Mars in two positions

Opposition: Mars and Sol are on opposite sides of Earth (Terra). Thus, Earthly observers easily see Mars.  Conjunction: Mars hides behind the Sun (for Earth bound observers).
Both positions occur annually.




Conjunction shows maximum straight line distance from Earth to Mars. Thus,  the conjunction distance to another planet is a good "typical" distance to approximate travel distance from Earth to a position on the far side of the other planet's orbit.


Light Time

Distance Travel Time

to Earth for photon

(AU) (secs) (hrs)
Sun 1 500 0.14
Mercury 1.39 695 0.19
Venus 1.72 860 0.24
Earth 2 1,000 0.28
Mars 2.5 1,250 0.35
Jupiter 6.2 3,100 0.86
Saturn 10.5 5,250 1.46
Uranus 20.2 10,100 2.81
Neptune31.1 15,550 4.32
Kuiper Belt 4522,500 6.25
d t=d * 500 tHr=t/3600

Table adds a column to describe travel time of a photon traveling at c, speed of light, during maximum distance to Earth. Purpose is mainly academic as it puts an lower limit on travel time; this is the absolute quickest time anything (even electronic communications) can travel those distances.
Determining number of seconds required for photon to travel from Sun to Earth is very simple. Recall that c = 300,000 km/sec , and AU = 150,000,000 km. Since time = distance/velocity = t = d/v, photon travel time = AU/c = 500 sec for one AU. Since all the Max Dist's are expressed in AUs, they can all be converted to "light seconds" (ls) by simply multiplying by 500.
However, this method will quickly produce very large, cumbersome numbers of ls. A more convenient number would be "light hours" (lh), distance traveled by light in one hour. This conversion is easily done. Recall that 1 hr = 60 min * 60 sec/min = 3600 sec. Thus, dividing number of ls by 3600 will give us lh's for all the max dist's.
For all practical purposes, this upper limit on speed, c, (or lower limit on travel time, distance divided by light speed, c) is unachieveable except for electronic communication signals. Perhaps we can find a more useful purpose for these values.

Communications

Distance Comm Time

to Earth Light Time Comm Delay

(AU) (hrs) (hrs)
Sun 1 0.14 0.28
Mercury 1.39 0.19 0.39
Venus 1.72 0.24 0.48
Earth 2 0.28 0.56
Mars 2.5 0.35 0.69
Jupiter 6.2 0.86 1.72
EXAMPLE7.21.002.00
Saturn 10.5 1.46 2.92
Uranus 20.2 2.81 5.61
Neptune31.1 4.32 8.64
Kuiper Belt 45 6.25 12.50
d
t =d * 500



3,600
tCom= 2* t
Signal travel is limited by speed of light. Let's assume an interplanetary traveler is using radio waves to maintain constant contact with his/her colleagues on Earth. In practice, Terran sponsors want a constant flow of data from their expensive enterprise vessel; on the other hand, the traveler wants a constant flow of data from Earth for various reasons: guidance, security, entertainment, news, views, etc. Thus, a vessel would likely have a constant flow of two way traffic. However, these two flows of information would likely be independent; actual communication would have to undergo significant delays as shown by following example.
Assume Interplanetary Traveler (IT) to be somewhere between the orbits of Jupiter and Saturn such that light has to travel an hour to reach Earth. (If 1 AU takes 500 seconds for light to travel, then 3600/500 AU = 7.2 AU would take light a full hour.) Next, assume IT wants to conduct a quick test of the communication system by sending "1234567890 THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG'S BACK. ACK" First part is known as Quick Brown Fox (QBF) used in the early days of teletype (when all caps, upper case, was the norm for then expensive global communications) to test most characters. ACK just means acknowledge receipt by sending immediate reply. Thus, we can further assume zero processing time by the receivers on Earth. Of course, this results in a total communications delay of one hour to Earth and another hour from Earch for a total of two hours. We obviously picked 7.2 AU distance to make this thought problem more convenient. Point being, any delay over a few seconds will quickly surpass a typical human threshold for patience. Certainly any delay over a few minutes will undoubtedly do this. Today's space missions travel much slower and they encounter this problem. Accelerating at g will quickly achieve sufficient distance to create noticeable and then unbearably comm delay; thus, interactive communications between Earth and space travelers will quickly cease to be possible, mandating considerable autonomy to the space vessel occupants as well as necessary authority to the vessel's "captain".
Computations: communication time = signal travel from sender to receiver + process time by receiver + response signal travel from receiver to sender. Since both send and receive signals both travel at speed of light over the interplanetary distances and assume process time to be essentially zero, min comm time can be set equal to twice the light distance. Thus, all the values are doubled, or 2 * Dist * 500/3600 = MaxDist * 5/18. (Note: We previously assumed zero processing for data; this simple calculation also disregards slight distance traveled by spacecraft during period of comm delay.)

On the spot interactivity will become an artifact of terrestrial presence as stark distances will greatly prolong times of electronic signals between communicators. Thus, current email practices will quickly become the norm because noone will have the time not to mention the patience to wait for replies. Audio and video messages will be in the constant stream of data from Earth, but immediate responses (not to mention conferences) won't happen because simple physics prevent it. Thus, circumstances will force onboard crews to become much more autonomous; the ship's captain will really be in charge.
Accelerate to Destination

Distance Comm Delay At Destination
to Earth fm Light Travel Travel Time Final VelocityEscape Velocity
(AU) (hrs) (Days) (km/sec) (km/sec)
Sun 1 0.28 2.00 1,728
Mercury
1.39
0.39
2.36
2,039

4.25 Venus
1.72
0.48
2.62
2,263

10.4 Earth
2
0.56
2.83
2,445

11.2 Mars
2.5
0.69
3.16
2,730

5.02 Jupiter
6.2
1.72
4.98
4,302

59.5 Saturn
10.5
2.92
6.48
5,599

35.5 Uranus
20.2
5.61
8.99
7,767

21.3 Neptune
31.1
8.64
11.15
9,634

23.7 Kuiper Belt
45
12.50
13.42
11,595

d

CD=2*d/c

Vfin=t*g
Observed





(g=864 km/sec









 day



)

Interplanetary Flight Profile

Destination
Distance
Accelerate
Max Velocity
Decelerate
Accel + Decel
to Earth
Dept to Midway
(at midway)
Midway to Dest
Total Travel Time
(AU)
(days)
(km per sec)
(days)
(days)
Sun
1
1.41
1,218
1.41
2.83
Mercury
1.39
1.67
1,443
1.67
3.33
Venus
1.72
1.85
1,598
1.85
3.71
Earth
2
2.00
1,728
2.00
4.00
Mars
2.5
2.24
1,935
2.24
4.47
Jupiter
6.2
3.52
3,041
3.52
7.04
Saturn
10.5
4.58
3,957
4.58
9.17
Uranus
20.2
6.36
5,495
6.36
12.71
Neptune
31.1
7.89
6,817
7.89
15.77
Kuiper Belt
45
9.49
8,199
9.49
18.97
d





VMax=tACCEL*g
(g=864km/sec/day)









distGiven(d)40 AU40 AU range puts us within Kuiper Belt with lots of resources (asteroids for building materials; comets for fuel).
time4√(d/g)(t)


35.78 days
Formula, 4√(d/g), approximates time for two way trip. Recall that g-force acceleration for d/2 (= 20 AU) will take √(d/g) (= 8.94 days). Double this to complete the 1 way trip takes 17.9 days; double again to include return trip: 35.8 days.
vExh mr%TOGWDay%TOGWPercent Take Off Gross Weight is portion of ship's original mass required for fuel.










.866c20.16%






5.85%
Accelerated particles double in mass when they reach .866 light speed. Thus, only one sixth of a percent of ship's mass is needed to maintain g-force. After about 36 days of g-force flight, only 6% of ship's mass needs to convert to kinetic energy (assuming 100% efficiency).










(dc*c)


1



√(1-dc2)








0.283% 









√(mr2-1)




t * %TOGWDay






 











 





    

    

    
    

      posted by JimODell at 3:56 AM
      

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Monday, October 09, 2006

  

  
     
  
  

         
    

	 
	 HUBS
	 
    

    

	         

	

      

















  
CONTENT

    
8 OCTANTS

    
X, Y, Z COORDs

    
DISTANCES TWEEN STARS

    
TRAVEL TIMES

    
SELECT HUB

    
SUMMARY

 


















SELECT A HUB FOR EACH
OF THE 8 OCTANTS.




Thus far, humanity has discovered about
51 stellar systems within 15 LYs of Sol.

  







DETERMINE CARTESIAN (X, Y, Z) COORDINATES


Consider notional (X, Y, Z) coordinates in Octant I

O = (0, 0, 0) Origin

P1 = (1, 2, 2); P2 = (2, 4, 4); P3 = (3, 6, 2)



Pythagorean Theorem

Compute 3-D distances from origin in following manner:

D = √(X2 + Y2 + Z2)

D0,1 = √(12 + 22 + 22) = √(1 + 4 + 4) = √(9) = 3

D0,2=√(22+42+42)=6 ; D0,3=√(32+62+22)=7



Furthermore,quickly determine 3-D distances between any two points by further leveraging P-Theorem on the coordinate differences.

EXAMPLE: compute 3-D distances from P1 in following manner.

Da,b = √[(Xb-Xa)2 + (Yb-Ya)2 + (Zb-Za)2]

to P2: D1,2 = √[(2-1)2 + (4-2)2 + (4-2)2] =√[1+4 +4]=3

to P3: D1,3= √[(3-1)2 + (6-2)2 + (2-2)2] =√[4+16 +0]=4.47





DETERMINE DISTANCES BETWEEN ADJACENT STARS









Observed Astrometrics

Consider Groombridge 34 (G..34) and Teegarden's Star (T..St),



Sol's neighbors in Octant One.






StellarRADecDist.


Systemαδd


G..344.6°44°11.62 LY


T..St43°17°12.51 LY







Right Ascension (RA=α) and Declination (Dec=δ)

are readily obtained in  decimal degrees.

Observed distance are traditionally obtained for nearby stars

by carefully measuring brightness and parallax.





  



Compute 3-D Coordinates

Seq.StardCEPxyz


i =0Sol0  LY0  LY0  LY0  LY


i =1G..348.36 LY8.3 LY0.7 LY8.1 LY


i =2 T..St11.6 LY8.7 LY8.2 LY3.6 LY


Givend×cos(δ)dCEP×cos(α)dCEP×sin(α)d×sin(δ)



Use trigonometric functions 

to convert astrometrics

to three dimension coordinates.

Compute Leg Distances

Determine  two legs of voyage to interstellar destination.

EXAMPLE: Leg-1 distance from Sol to Hub, G..34;

and Leg-2 distance is from hub to destination, T-Star.






Seq.StardLegΔdXΔdYΔdZ 


i =0Sol0  LY0  LY0  LY0  LY


i =1G..3411.62 LY8.3 LY0.7 LY8.1 LY


i =2 T..St8.76 LY-0.4 LY-7.5 LY-4.5 LY


GivenSee belowXi - Xi-1Yi - Yi-1Zi - Zi-1



Use Pythogorean Theorem to obtain leg distances..

 dleg = √[(ΔdX)2 + (ΔdY)2 + (ΔdZ)2]





EXAMPLE: dleg2 = √[(0.4)2 + (7.5)2 + (-4.5)2] = 8.76 LY 

  


TOTAL TRAVEL TIMES FROM SOL TO HUB TO STARS

Consider Hub Concept
Instead of direct to all stars, use a well situated star as stop over "hub" for its neighbors. Thus, interstellar voyages could transit this hub enroute to other destinations.

Likely criteria:

1) PROXIMITY:  Closeness to Sol reduces flight time.

2) WELL SITUATED:  Position among other stars is very useful.

3) WELL PROVISIONED:  In situ materials (comets and asteroids) could resupply transit vessels.








Convert interstellar distances (LYs) to time (Yrs)

Previous TE work assumes following model

for insterstellar g-force travel times (Yrs).






Choose to G-force Accelerate for 1 Year:

 tAcc = 1 year; distance = dAcc= .38 LY;

achieve cruise velocity; vCru = 6443c


G-force Decelerate for same duration as acceleration:

tDec = 1 year; distance = dDec= .38 LY


Cruise at Constant Velocity with no Propulsion:





Distance:dCru = dW - dAcc - dDec = 7.8 LY-38 LY- 38 LY =7.04LY



Time:



tCru =  dCru / vCru = 7.04 LY / (.644c) = 10.93 years








Travel Time

tTtl =  tAcc +  tCru  + tDec = 1.0 yr + 10.93 yr + 1.0 yr = 12.93 yrs







Stopover Flights



  EXAMPLE: Consider Wolf 359 (W359) as Hub;
  for spokes, try Ross 129 (R129) and Lalande 21185 (L185)
  

DeptLeg-1HubLeg-2DestTtl-Time


DistTimeDistTime


Sol7.8 LY12.9 YrsW359 3.8 LY6.7 YrsR12819.6 Yrs


Sol7.8 LY12.9 YrsW3594.1 LY7.2 YrsL18520.1 Yrs

















One long voyage can divide into two shorter flights; use a well placed star as a hub between Sol and subsequent destinations. Thus, a  stopover flight profile might prove more tolerable and probably much more useful than direct flights to the more distant neighbors. Stopover flights make even more sense if the "hub" has plentiful "in situ" resources (comets/asteroids) to build and provision other star ships. 


SELECT OCTANT'S MOST SUITABLE STAR AS HUB



Put it all together:

From each octant, select most suitable star as a hub.



Express traditional observed coordinates:
  • Right Ascension (α) in degrees (°)
  • Declination (δ) in degrees (°)
  • Distance in decimal Light Years (LY)



Transform into three dimensional (3D): 
  • Cartesian Coordinates: X, Y and Z in LY.
  • 
Assume Common Ecliptic Plane (CEP) as base for X-Y plane.



Use typical Interstellar Profile to determine travel time:
  • 
given cruise velocity of 0.6443 c,
  • 
with one year acceleration (accomplished during initial 0.37 LY from Sol),
  • 
with 1 year deceleration (accomplished during final distance of 0.37 LY prior to destination),
  • BEARINGS to nearby stars help determine when to transition from cruise to deceleration.






IIIIIIIVVVIVIIVIIIAll 8 Octants






  


OCTANTS  
  Example

Hub		RAdec
  
  Dist3-D Coord.Time

  

αδdxyzt

  
  

OriginSoln/an/a00000

  
  

OCT-IGroombridge344.6°44°11.628.30.78.118.9

  
  

OCT-II
  
  Wolf 359
  
  164°7°7.78-7.42.10.912.9

  

OCT-IIIBarnard's Star270°5°5.96-0.1-5.90.510.1

  

OCT-IVRoss 248356°44°10.327.34-0.67.216.9

  
  

OCT-VLuyten 726-8 25°-18°8.797.53.5-2.714.4

  
  

OCT-VISirius101°-17°8.58-1.68.2-2.614.4

  
  

OCT-VIIα Centauri217°-63°4.37-1.6-1.4-3.87.6

  
  

OCT-VIIIRoss 154283°-23°9.71.9-8.7-3.815.9

  
OBSERVED:  α;  δ;  ddCEP  = d × cos(δ)


x = dCEP×cos(α)y = dCEP×sin(α)z = d×sin(δ)



INTERSTELLAR PROFILE TRAVEL TIME:
t = 1 yr + (d-.76 LY)/.644c + 1 yr














SUMMARY: HUBS



Hub stars are selected to leverage as much as possible the three hub criteria of 



    1) proximity to Sol, 



    2) situation to other stars, 



    3) provision of on site resources.


  
 
  
  
  
 
  
 
  
 
  
 
  

  

  



Hub's stellar names are placed to conveniently designate parent octants; they are not placed to show exact location within each octant.












    

    

    
    

      posted by JimODell at 6:03 AM
      

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