Saturday, November 30, 2013

Determine Travel Time to ν = 90° (Transient 4.5.2014)

Determine Travel Time to ν = 90°

When object reaches ν = 90°,
the positional vector, R, will be the semi-latus rectum, ℓ.
Some calculations for T= 2 years and ν = 90°.
aecbR90
AU AUAUAUAUdeg
1.591/30.531.50 1.411.4170.5°
T2/3Givena × e(a2-c2)b2

a
 

1+e×cos(90°)
tan-1b

c
First, determine area of E, a circular sector.
AE = π a2 × E/360° = 1.555 AU2
We also know: AE =  A + A 
Next, compute: A= 0.5×b×c=0.5×1.499 AU×0.53 AU=0.397AU2
Finally, compute: A= AE - A=(1.555-0.397)AU2=1.158AU2
Calculate edge area (A) of EA Sector.
EAEA A
AUAU AU degAU2AU2AU2
1.591.4990.5370.5°1.560.401.16
 (2 yr)2/3 Given(a2-b2
tan-1b

c
π a2 E

360°
b × c

2
AE - A
When object's True Anomaly (TA) is 90 degrees (ν = 90°), we can redefine internal triangle for special case when the TA area equals the edge area (AΔ). Thus, TA area can be determined through proportionality shown in figure to right. Further, we can now determine travel via Kepler's Second Law which implies Time/Area proportionality.
Calculations to determine travel time (Δt).
A AOrbAΔΔt
1.16 AU27.49 AU21.41 AU1.09 AU2106.4 days
AE-Aπ a bb2

a
×A

b
2 yrs×AΔ

AOrb
SLR Q-II. Semiorbit time was previously computed as half the orbital period, T. Thus, 2nd SLR quadrant travel time easily computed as difference between t1 and t2.
SLR Q-I. Orbit from t0 to directly above Sun, t1. Table shows area/time calculations to give us asteroid travel time, 106.6 days.
SLR Q-III. Due to orbit's symmetry about major axis (q to Q), area and travel time mirror from Q-2 to 3.
SLR Q-IV. Final, fourth quadrant mirrors from quadrant 1. It's also difference between t3 and t4.

 
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.
730.5 days =T= 2 yrs a =1.59 AU b = 1.499 AU c = 0.53 AU e = 1/3
EE'YEXEYTXTrν'ν
degdegAUAUAUAUAUdegdeg
90901.590.001.500.531.5970.5109.5
105751.540.411.450.941.7357.0123.0
120601.380.801.301.331.8544.4135.6
135451.121.121.061.651.9632.6147.4
150300.801.380.751.912.0521.5158.5
165150.411.540.392.072.1010.6169.4
18000.001.590.002.122.120.0180.0
Given180-Ea*sin(E')a*cos(E')b*sin(E')a*cos(E')+c√(xT2 + yT2)tan-1(yT/xT)180° - ν'
yE = a*sin(E')
xE =a*cos(E')
yT = b*sin(E')
xT =a*cos(E') + ae
next row
xxxxxxxxx
Step 4. For safety, humanity should construct two co-orbiting 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.
Positional vectors, r.
νcos(ν)e * cos(ν)1 + e * cos(ν)r
degAU
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
Givencos(ν)e * cos(ν)1 + e * cos(ν)

1+e*cos(ν)
730.5 days = T = 2 yrs a =1.59AU b =1.499AU c =0.53AU e = 1/3 l =1.41 AU

Compute Partial Areas of TA Triangular Area from apex, AT∠.
νrxTyTAT∠AT∇AT
30°
1.10 AU
0.95 AU
0.55 AU
0.26  AU2
? AU2
0.26AU2 +?
45°
1.14AU0.81 AU0.81 AU0.33 AU2? AU20.33AU2 + ?
60°1.21AU0.61 AU1.05 AU0.32 AU2? AU20.32 AU2 + ?
90°1.41AU0.00 AU1.41  AU0.00 AU2? AU20.00 AU2 + ?
120°1.70 AU-0.85 AU1.47  AU0.62 AU2? AU20.62 AU2 + ?
150°1.99 AU-1.72 AU0.99 AU0.85 AU2? AU20.85 AU2 + ?
Given

1+e*cos(ν)
r × cos(ν) r × sin(ν) 0.5×XT×YTUnkAT∠+AT∇


Note trig identify:cosν sinν = 0.5 [sin2ν]Thus:0.5 r cosν rsinν = 0.25 r2[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.

Determine Eccentric Anomaly and Corresponding Areas

 True
Anomaly
Radius Object's
Y value
 Object's
AC Y value
Eccentric
Anomaly
 Total
E Area
 Triangle
Area
Edge
Area
νrYTYACEAEAE∠A
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

1+e×cos(ν)
r×Sin(ν)YT×a

b
Sin-1(YAC

a
)
πa2×E

360°
a2×Sin(2E)

4
AE-AE∠

Compare Edge Areas, AE∇ & A

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.TotalTriangleEdgeEdgeTriangleTotal
νrν'EAEAE∠AE∇AAT∠AT
DegAU DegDegAU 2AU 2AU 2AU 2AU 2AU 2
30°1.1030°21.5°0.4730.4300.0430.0410.2600.30
45°1.1445°32.6°0.7200.5740.1460.1380.3270.46
60°1.2160°44.4°0.9800.6320.3480.3280.3180.65
90°1.4190°70.5°1.5560.3971.1581.0920.0001.09
120°1.7060°78.5°1.7310.2481.4831.3980.6232.02
135°1.8545°60.7°1.3400.5390.8000.7540.8551.61
150°1.9930°41.5°0.9160.6270.2880.2720.8551.13
Given

1+e*cos(ν)
Acute ν'
Sin-1(r×Sin(ν')


b
)
π a2E


360
a×Sin(2E)

4
AE -AE∠
AE∇b


a
r2 × Sin(2ν')

4
AT∠+A
e=0.33 = 1.41 AUIf ν>90°a =1.59 AUb =1.499 AUc = 0.53 AUAOrb=7.484AU2T=730.5Days
e=c/a = b2/aν'=180°AE = AE∠ + AE∇AOrb= π a bT= 2π(a3/μ)
xxxxxxxxxxxxxx

Compare Edge Areas, AE∇ & A

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.TotalTriangleEdgeEdgeTriangleTotal
νrν'EAEAE∠AE∇AAT∠AT
DegAU DegDegAU 2AU 2AU 2AU 2AU 2AU 2
30°1.1030°21.5°0.4730.4300.0430.0410.2600.30
45°1.1445°32.6°0.7200.5740.1460.1380.3270.46
60°1.2160°44.4°0.9800.6320.3480.3280.3180.65
90°1.4190°70.5°1.5560.3971.1581.0920.0001.09
120°1.7060°78.5°1.7310.2481.4831.3980.6232.02
135°1.8545°60.7°1.3400.5390.8000.7540.8551.61
150°1.9930°41.5°0.9160.6270.2880.2720.8551.13
Given

1+e*cos(ν)
Acute ν'
Sin-1(r×Sin(ν')


b
)
π a2E


360
a×Sin(2E)

4
AE -AE∠
AE∇b


a
r2 × Sin(2ν')

4
AT∠+A
e=0.33 = 1.41 AUIf ν>90°a =1.59 AUb =1.499 AUc = 0.53 AUAOrb=7.484AU2T=730.5Days
e=c/a = b2/aν'=180°AE = AE∠ + AE∇AOrb= π a bT= 2π(a3/μ)

Compare Edge Areas, AE∇ and A

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.AngleEdgeEdgeTriangleTotal
νrν'EAE∇AAT∠AT
30°1.10 AU30°21.5°0.043AU20.041AU20.260AU20.30AU2
45°1.41 AU45°32.6°0.146AU20.138AU20.327AU20.46AU2
60°1.21 AU60°44.4°0.348AU20.328AU20.318AU20.65AU2
90°1.41 AU90°70.5°1.158AU21.092AU20.000AU21.09AU2
120°1.70 AU60°78.5°1.483AU21.398AU20.623AU22.02AU2
135°1.85 AU45°60.7°0.800AU20.754AU20.855AU21.61AU2
150°1.99 AU30°41.5°0.288AU20.272AU20.855AU21.13AU2
Given

1+e×cos(ν)
Acute ν'
Sin-1(r×Sin(ν')

b
)
E×πa2

360°
-.5a2Sin(E)
AE∇b

a
r2 ×Sin(2ν')

4
AT∠+A
e=0.33
e=c/a
= 1.41 AU
= b2/a
If ν>90°
ν'=180°- ν
b = 1.499 AUxxxxxxxxxxxxxxxxxxxxxxxxx

Closing thot: Relate six orbital elements with cycler pathes.
Recall alpha/omega, which are notional coorbiting habitats at Earth-Sun 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 g-force 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 r2[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.
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: semi-minor 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
AE: area enclosed by Eccentric Anomaly and orbit .
A : Subset of AE, area enclosed by triangular portion.
A: Subset of AE, area enclosed by edge portion.
AOrb 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.'
AΔ ,subset of AT: 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(YAC

a
)
x=
νEAATAtTA
30°21.46°.011 AU2.301 AU229.36 dy
45°32.65°.038 AU2.465 AU245.27 dy
60°44.41°.095 AU2.646 AU263.00 dy
x
νθ Eθ  
120°60° 78.45°  
135°45° 60.72°  
150°30° 41.50°  
x
AYθ ν Y-AC
.646 AU2 60°  78.45°
.465 AU2 45°  60.72°
.301 AU2 30°  41.50°
x
A
x
Y1 < 2 × A1 / Base1  = 2 × A1 / Q = 2 × A1 /(a + c) = 2 × A1 /(a(1 + e))
x
θ = Asin(
Y2 < 2 × A2 / Base2  = 2 × A2 / q = 2 × A /(a - c) = 2 × A2 /(a(1 - e))
If θ = 1°, then ν= 180° - θ = 179°;  and yθ  = Rν  × Sin(θ)= .0369 AU.
Insert EdgeAQH
.464 AU2 2.12 AU.4377 AU
  .
Givena + c.2A/Q

θν RνyθY 
 1° 179°2.115AU.0369AU 12.205°
 11.86° 168.14°   
Given 180-θ
 

1+e×cos(ν)
 Rν×Sinθ
.4505 AU

yθ
xxxxxxx
θASect=0.464 AU2= νASect
AE%AEdgeATriQyθ
1.06%0.00492 AU20.4591 AU22.12 AU0.4331 AU
1.07%0.00496 AU20.4590 AU22.12 AU0.4331 AU
1.08%0.00501 AU20.4590 AU22.12 AU0.4330 AU
% of ASectASect×AE%ASect - AEdgea + c2×ATri

Q
x
xx
x
True Anomaly Eccentric Anomaly
TAE%TAEdgeyθEθEAEdgeTAEdge
1.050.004870.4331016.7950.00530.00498
1.060.004920.4330516.7930.00530.00498
1.070.004960.4330116.7920.00530.00498
1.080.005010.4329616.7900.00530.00498
1.090.005060.4329216.7880.00530.00498
1.10.005100.4324816.7870.00530.00497
1.20.005570.4320516.7690.00530.00496
1.30.006030.4316116.7520.00520.00494
1.40.006500.4311716.7340.00520.00493
1.50.006960.4307316.7170.00520.00491
1.60.007420.4302916.6990.00520.00490
1.70.007890.4298616.6820.00520.00488
1.80.008350.4294216.6640.00520.00487
1.90.008820.4289816.6470.00510.00485
20.009280.4285416.6290.00510.00484
2.10.009740.4281116.6120.00510.00482
2.20.010210.4540916.5940.00510.00481
GivenTASect×TAE%(TASect-TATri)/Qsin-1(yθ/b)a2×(πEθ/360⁰-.5yθ/b)b×EAEdge/a
x
1.073%0.00498 AU20.459 AU22.12 AU0.4330AU
% of ASectASect×AE%ASect - AEdgea + c2×ATri

Q
x
x
8. Travel Time Mirrors across Major Axis
For θ, angle from Q,
compute area and time.
See sample travel times:
νθ Eθ  
120°60° 78.45°  
135°45° 60.72°  
150°30° 41.50°  
From ν = 0° to any acute angle,
compute time per previous steps.
See sample travel times:
νEAATAtTA
30°21.46°.011 AU2.301 AU229.36 dy
45°32.65°.038 AU2.465 AU245.27 dy
60°44.41°.095 AU2.646 AU263.00 dy
For ν = 180° to 315°, Δt = 320 days
Time reflects from top half.
For ν = 0° to 315°, t=685.25 days
Others:
For ν=315° to 360°, Δt=45.27 days
Time reflects from top half.
For ν = 0° to 360°, t=730.5 days
Others:

xxx

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