Determine Travel Time to ν = 90° (Transient 4.5.2014)
Step 1. Basics of Earth orbit. Recall that the ordinary annual calendar is an excellent guide to where Earth is in its orbit. Describe Aries as the reference point.  

Step 2. Discuss resonance. Resonant asteroids have orbits which are exact multiples of one year. Discuss has a, semimajor axis, relates to orbit's period. Discuss how asteroids can be useful, cyclers, if they consistently come near Earth's orbit at same day every year, every other year, or some other multiple. Show resonance for same point but different period cycler orbits.  

Step 3. Discuss resonance again, but contrast with step 2 by showing  how utility is increased by evenly spacing out cyclers throughout Earth's orbit.  

E  E'  Y_{E}  X_{E}  Y_{T}  X_{T}  r  ν'  ν  deg  deg  AU  AU  AU  AU  AU  deg  deg  90  90  1.59  0.00  1.50  0.53  1.59  70.5  109.5  105  75  1.54  0.41  1.45  0.94  1.73  57.0  123.0  120  60  1.38  0.80  1.30  1.33  1.85  44.4  135.6  135  45  1.12  1.12  1.06  1.65  1.96  32.6  147.4  150  30  0.80  1.38  0.75  1.91  2.05  21.5  158.5  165  15  0.41  1.54  0.39  2.07  2.10  10.6  169.4  180  0  0.00  1.59  0.00  2.12  2.12  0.0  180.0  Given  180E  a*sin(E')  a*cos(E')  b*sin(E')  a*cos(E')+c  √(x_{T}^{2} + y_{T}^{2})  tan^{1}(y_{T}/x_{T})  180°  ν' 

Step 4. For safety, humanity should construct two coorbiting habitats which lead and lag Earth in its orbit about the sun. Previously discussed as Alpha and Omega.
Orbit Multiples. This is by far the simplest method. One can readily predict orbital positions for one, two or more periods in the future (or past). For example, assume that the period, T, for Apollo's orbit is exactly 650 days. (Apollo is a Near Earth Asteroid, NEA). Then, position of Apollo for 650 days from now will be at the same orbital position.
Recall knowing a, semimajor axis is sufficient to know T, orbit period. You don't have to know e, eccentricity, or any other value about the specific orbit under study.
If we assume that Apollo passes q, its perihelion, on April 11, 2009, then we can safely predict that it will come back to q 650 days later on Jan 20, 2011.
Another example: Consider another point in Apollo's orbit. To arrive at 45° past q will take Apollo 21 more days (arriving March 1, 2009); from that point, we can safely predict another arrival at this new point in another orbital cycle. Thus, we predict that at Mar 1, 20o9 plus T (plus 650 days = Feb 10, 2011), Apollo will once again arrive at 45° past q.
This leads us to a simple brute force method where observations can be made daily throughout an asteroid's orbit; then, position predictions can be made accurately for any of those positions and interpolated for times between those exact daily times. For example, we could make 650 daily observations of Apollo from start to stop of its orbit, then forecast future positions accordingly.
Add blurb from Lewis, Mining Asteroids, where cyclers are constructed to have exact orbits of two years. Comments.
The downside of orbital multiples, there are an infinity of positions impossible to make observations to accommodate all of them. A more elegant method would involve a procedure to calculate positions for any time, then use observations to confirm.
730.5 days = T = 2 yrs a =1.59AU b =1.499AU c =0.53AU e = 1/3 l =1.41 AU
_{}
Note trig identify:cosν sinν = 0.5 [sin2ν]Thus:0.5 r cosν rsinν = 0.25 r^{2}[sin2ν] Still, need to compute edge area which is unknown due to variable range of position vector.
Thus, determining True Anomaly Area must wait on this value.
xxxxxxxxxxxxxx
Closing thot: Relate six orbital elements with cycler pathes.
Recall alpha/omega, which are notional coorbiting habitats at EarthSun L4/L5 Lagrange points in Earth's Solar orbit (lead/lag Earth by 60°).
Initial cyclers will hopefully be developed before accelerator spaceships which can constantly accelerate at gforce for several days. Thus, initial cyclers will by necessity have to orbit such that cycler's perihelion will take it to a resonant point near Earth's orbit. HOWEVER, safety and sanity will eventually prevail to decision makers to move these resonance points away from Earth toward Alpha/Omega co orbiters. Recall that 60° away from Earth is a full AU in straight line distance, a considerable distance for today's space traveling technology; however, our notional spaceships could fly there in 2.8 days (using standard flight profile).
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ν  cos(ν)  e * cos(ν)  1 + e * cos(ν)  r 

deg  AU  
30

0.866

0.29

1.29

1.10

45

0.707

0.24

1.24

1.14

60

0.500

0.17

1.17

1.21

90

0.000

0.00

1.00

1.41

120

0.500

0.17

0.83

1.70

135

0.707

0.24

0.76

1.85

150

0.866

0.29

0.71

1.99

180

1.000

0.33

0.67

2.12

Given  cos(ν)  e * cos(ν)  1 + e * cos(ν)  ℓ 1+e*cos(ν) 
ν  r  x_{T}  y_{T}  A_{T∠}  A_{T∇}  A_{T} 

30°

1.10 AU

0.95 AU

0.55 AU

0.26 AU^{2}

? AU^{2}

0.26AU^{2 }+?

45°
 1.14AU  0.81 AU  0.81 AU  0.33 AU^{2}  ? AU^{2}  0.33AU^{2 }+ ? 
60°  1.21AU  0.61 AU  1.05 AU  0.32 AU^{2}  ? AU^{2}  0.32 AU^{2 }+ ? 
90°  1.41AU  0.00 AU  1.41 AU  0.00 AU^{2}  ? AU^{2}  0.00 AU^{2 }+ ? 
120°  1.70 AU  0.85 AU  1.47 AU  0.62 AU^{2}  ? AU^{2}  0.62 AU^{2 }+ ? 
150°  1.99 AU  1.72 AU  0.99 AU  0.85 AU^{2}  ? AU^{2}  0.85 AU^{2 }+ ? 
Given  ℓ 1+e*cos(ν)  r × cos(ν)  r × sin(ν)  0.5×X_{T}×Y_{T}  Unk  A_{T∠}+A_{T∇} 

_{}
Note trig identify:cosν sinν = 0.5 [sin2ν]Thus:0.5 r cosν rsinν = 0.25 r^{2}[sin2ν] Still, need to compute edge area which is unknown due to variable range of position vector.
Thus, determining True Anomaly Area must wait on this value.
True Anomaly  Radius  _{} Object's Y value  _{}_{} Object's AC Y value  Eccentric Anomaly  Total E Area  _{} Triangle Area  Edge Area  

ν  r  Y_{T}  Y_{AC}  E  A_{E}  A_{E∠}  A_{EΔ}  
30°

1.10

0.55

0.58

21.5°

0.473

0.430

0.043
 
45°

1.14

0.81

0.86

32.6°

0.720

0.574

0.146
 
60°

1.21

1.05

1.11

44.4°

0.980

0.632

0.348
 
90°

1.41

1.41

1.50

70.5°

1.556

0.397

1.158
 
120°

1.70

1.47

1.56

78.5°

1.731

0.248

1.483
 
135°

1.85

1.31

1.39

60.7°

1.340

0.539

0.800
 
150°

1.99

0.99

1.05

41.5°

0.916

0.627

0.288
 
Given  ℓ 1+e×cos(ν)  r×Sin(ν)  Y_{T}×a b 
 πa^{2}×E 360°  a^{2}×Sin(2E) 4  A_{E}A_{E∠} 
Need ever increasing True Anomaly (ν) to get variable range of positional vector, r.  Eccentric Anomaly Areas  True Anomaly Areas  

Need acute angle, ν', to obtain triangle adjacent to major axis.  Total  Triangle  Edge  Edge  Triangle  Total  
ν  r  ν'  E  A_{E}  A_{E∠}  A_{E∇}  A_{TΔ}  A_{T∠}  A_{T}  
Deg  AU  Deg  Deg  AU ^{2}  AU ^{2}  AU ^{2}  AU ^{2}  AU ^{2}  AU ^{2}  
30°  1.10  30°  21.5°  0.473  0.430  0.043  0.041  0.260  0.30  
45°  1.14  45°  32.6°  0.720  0.574  0.146  0.138  0.327  0.46  
60°  1.21  60°  44.4°  0.980  0.632  0.348  0.328  0.318  0.65  
90°  1.41  90°  70.5°  1.556  0.397  1.158  1.092  0.000  1.09  
120°  1.70  60°  78.5°  1.731  0.248  1.483  1.398  0.623  2.02  
135°  1.85  45°  60.7°  1.340  0.539  0.800  0.754  0.855  1.61  
150°  1.99  30°  41.5°  0.916  0.627  0.288  0.272  0.855  1.13  
Given  ℓ 1+e*cos(ν)  Acute ν' 

 a^{2 }×Sin(2E) 4  A_{E }A_{E∠} 
 r^{2 }×^{ }Sin(2ν') 4  A_{T∠}+A_{TΔ}  
e=0.33  ℓ = 1.41 AU  If ν>90°  a =1.59 AU  b =1.499 AU  c = 0.53 AU  A_{Orb}=7.484AU^{2}  T=730.5Days  
e=c/a  ℓ = b^{2}/a  ν'=180°  A_{E} = A_{E∠} + A_{E∇}  A_{Orb}= π a b  T= 2π√(a^{3}/μ) 
Need ever increasing True Anomaly (ν) to get variable range of positional vector, r.  Eccentric Anomaly Areas  True Anomaly Areas  

Need acute angle, ν', to obtain triangle adjacent to major axis.  Total  Triangle  Edge  Edge  Triangle  Total  
ν  r  ν'  E  A_{E}  A_{E∠}  A_{E∇}  A_{TΔ}  A_{T∠}  A_{T}  
Deg  AU  Deg  Deg  AU ^{2}  AU ^{2}  AU ^{2}  AU ^{2}  AU ^{2}  AU ^{2}  
30°  1.10  30°  21.5°  0.473  0.430  0.043  0.041  0.260  0.30  
45°  1.14  45°  32.6°  0.720  0.574  0.146  0.138  0.327  0.46  
60°  1.21  60°  44.4°  0.980  0.632  0.348  0.328  0.318  0.65  
90°  1.41  90°  70.5°  1.556  0.397  1.158  1.092  0.000  1.09  
120°  1.70  60°  78.5°  1.731  0.248  1.483  1.398  0.623  2.02  
135°  1.85  45°  60.7°  1.340  0.539  0.800  0.754  0.855  1.61  
150°  1.99  30°  41.5°  0.916  0.627  0.288  0.272  0.855  1.13  
Given  ℓ 1+e*cos(ν)  Acute ν' 

 a^{2 }×Sin(2E) 4  A_{E }A_{E∠} 
 r^{2 }×^{ }Sin(2ν') 4  A_{T∠}+A_{TΔ}  
e=0.33  ℓ = 1.41 AU  If ν>90°  a =1.59 AU  b =1.499 AU  c = 0.53 AU  A_{Orb}=7.484AU^{2}  T=730.5Days  
e=c/a  ℓ = b^{2}/a  ν'=180°  A_{E} = A_{E∠} + A_{E∇}  A_{Orb}= π a b  T= 2π√(a^{3}/μ) 
Need ever increasing True Anomaly (ν) to get variable range of positional vector, r.  Eccentric Anomaly and Edge Area  True Anomaly Areas  

Need acute angle, ν', to obtain triangle adjacent to major axis.  Angle  Edge  Edge  Triangle  Total  
ν  r  ν'  E  A_{E∇}  A_{TΔ}  A_{T∠}  A_{T}  
30°  1.10 AU  30°  21.5°  0.043AU^{2}  0.041AU^{2}  0.260AU^{2}  0.30AU^{2}  
45°  1.41 AU  45°  32.6°  0.146AU^{2}  0.138AU^{2}  0.327AU^{2}  0.46AU^{2}  
60°  1.21 AU  60°  44.4°  0.348AU^{2}  0.328AU^{2}  0.318AU^{2}  0.65AU^{2}  
90°  1.41 AU  90°  70.5°  1.158AU^{2}  1.092AU^{2}  0.000AU^{2}  1.09AU^{2}  
120°  1.70 AU  60°  78.5°  1.483AU^{2}  1.398AU^{2}  0.623AU^{2}  2.02AU^{2}  
135°  1.85 AU  45°  60.7°  0.800AU^{2}  0.754AU^{2}  0.855AU^{2}  1.61AU^{2}  
150°  1.99 AU  30°  41.5°  0.288AU^{2}  0.272AU^{2}  0.855AU^{2}  1.13AU^{2}  
Given  ℓ 1+e×cos(ν)  Acute ν' 


 r^{2 }×Sin(2ν') 4  A_{T∠}+A_{TΔ}  
e=0.33 e=c/a  ℓ = 1.41 AU ℓ = b^{2}/a  If ν>90° ν'=180° ν  b = 1.499 AU  xxxxxxx  xxxxxxxx  xxxxx  xxxxx 
Closing thot: Relate six orbital elements with cycler pathes.
Recall alpha/omega, which are notional coorbiting habitats at EarthSun L4/L5 Lagrange points in Earth's Solar orbit (lead/lag Earth by 60°).
Initial cyclers will hopefully be developed before accelerator spaceships which can constantly accelerate at gforce for several days. Thus, initial cyclers will by necessity have to orbit such that cycler's perihelion will take it to a resonant point near Earth's orbit. HOWEVER, safety and sanity will eventually prevail to decision makers to move these resonance points away from Earth toward Alpha/Omega co orbiters. Recall that 60° away from Earth is a full AU in straight line distance, a considerable distance for today's space traveling technology; however, our notional spaceships could fly there in 2.8 days (using standard flight profile).
Note trig identify:cosE sinE = 0.5 [sin2E]Thus:0.5 r cosν rsinν = 0.25 r^{2}[sin2ν] Still, need to compute edge area which is unknown due to variable range of position vector.
Thus, determining True Anomaly Area must wait on this value.
Thus, determining True Anomaly Area must wait on this value.
T, Period of orbit:The orbital period is the time taken for a given object to make one complete orbit about another object 
eccentricity: The fifth Keplerian orbital element, specifying the ratio, (from 0 to 1) of the distance between the foci and the length of the major axis of the satellite's orbit. 
b, semiminor axis: semiminor axis of an ellipse is one half of the minor axis, running from the center, halfway between and perpendicular to the line running between the foci, and to the edge of the ellipse. The minor axis is the longest line that runs perpendicular to the major axis. 
c, focus: distance from the center of the ellipse to a focus is designated c, so the distance between the two foci is 2c. 
E, Eccentric Anomaly: The angle between the periapsis of an orbit and a given point on a circle around the orbit, as seen from the centre of the orbit (see diagram). The point concerned is found by drawing a line perpendicular to the major axis through the actual position of the orbiting body until it cuts the circle around the orbit, as in the diagram. The diameter of the circle is equal to the major axis of the orbit, and its centre is also the orbit's centre. The angle of eccentric anomaly is measured in the direction of orbital motion 
A_{E}: area enclosed by Eccentric Anomaly and orbit . 
A _{∠}: Subset of A_{E}, area enclosed by triangular portion. 
A_{∇}: Subset of A_{E}, area enclosed by edge portion. 
A_{Orb} Orbit area:The area of the ellipse is π a b. 
l, semilatus rectum: The chord through a focus parallel to the conic section directrix of a conic section is called the latus rectum, and half this length is called the semilatus rectum (Coxeter 1969). "Semilatus rectum" is a compound of the Latin semi, meaning half, latus, meaning 'side,' and rectum, meaning 'straight.' 
A_{Δ ,}subset of A_{T}: area enclosed by edge portion. In this special case, where True Anomaly, ν = 90°, the edge portion is the entire TA area. 
TravTime: travel time of asteroid from q to present position directly above Sun. Sun and asteroid will be at opposite sides of l, semilatus rectum. 
Sin^{1}(  Y_{AC} a  ) 

x=
x
ν  E  A_{∇}  A_{TA}  t_{TA} 

30°  21.46°  .011 AU^{2}  .301 AU^{2}  29.36 dy 
45°  32.65°  .038 AU^{2}  .465 AU^{2}  45.27 dy 
60°  44.41°  .095 AU^{2}  .646 AU^{2}  63.00 dy 
x
x
A_{∠}
x
Y_{1} < 2 × A_{1} / Base_{1} = 2 × A_{1} / Q = 2 × A_{1} /(a + c) = 2 × A_{1} /(a(1 + e))
x
θ = Asin(
Y_{2} < 2 × A_{2} / Base_{2} = 2 × A_{2} / q = 2 × A /(a  c) = 2 × A_{2} /(a(1  e))
θA_{Sect}=0.464 AU^{2}= νA_{Sect}  

A_{E%}  A_{Edge}  A_{Tri}  Q  y_{θ} 
1.06%  0.00492 AU^{2}  0.4591 AU^{2}  2.12 AU  0.4331 AU 
1.07%  0.00496 AU^{2}  0.4590 AU^{2}  2.12 AU  0.4331 AU 
1.08%  0.00501 AU^{2}  0.4590 AU^{2}  2.12 AU  0.4330 AU 
% of A_{Sect}  A_{Sect}×A_{E%}  A_{Sect}  A_{Edge}  a + c  2×A_{Tri} Q 
xx
x
True Anomaly  Eccentric Anomaly  
TA_{E%}  TA_{Edge}  y_{θ}  E_{θ}  EA_{Edge}  TA_{Edge} 
1.05  0.00487  0.43310  16.795  0.0053  0.00498 
1.06  0.00492  0.43305  16.793  0.0053  0.00498 
1.07  0.00496  0.43301  16.792  0.0053  0.00498 
1.08  0.00501  0.43296  16.790  0.0053  0.00498 
1.09  0.00506  0.43292  16.788  0.0053  0.00498 
1.1  0.00510  0.43248  16.787  0.0053  0.00497 
1.2  0.00557  0.43205  16.769  0.0053  0.00496 
1.3  0.00603  0.43161  16.752  0.0052  0.00494 
1.4  0.00650  0.43117  16.734  0.0052  0.00493 
1.5  0.00696  0.43073  16.717  0.0052  0.00491 
1.6  0.00742  0.43029  16.699  0.0052  0.00490 
1.7  0.00789  0.42986  16.682  0.0052  0.00488 
1.8  0.00835  0.42942  16.664  0.0052  0.00487 
1.9  0.00882  0.42898  16.647  0.0051  0.00485 
2  0.00928  0.42854  16.629  0.0051  0.00484 
2.1  0.00974  0.42811  16.612  0.0051  0.00482 
2.2  0.01021  0.45409  16.594  0.0051  0.00481 
Given  TA_{Sect}×TA_{E%}  (TA_{Sect}TA_{Tri})/Q  sin^{1}(y_{θ}/b)  a^{2}×(πE_{θ}/360⁰.5y_{θ}/b)  b×EA_{Edge}/a 
1.073%  0.00498 AU^{2}  0.459 AU^{2}  2.12 AU  0.4330AU 
% of A_{Sect}  A_{Sect}×A_{E%}  A_{Sect}  A_{Edge}  a + c  2×A_{Tri} Q 

x

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