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°. a e c b ℓ R90 E  AU   AU AU AU AU deg 1.59 1/3 0.53 1.50  1.41 1.41 70.5° T2/3 Given a × e √(a2-c2) b2 a  ℓ 1+e×cos(90°) tan-1 b 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. a  b  c  E AE A ∠ A ∇ AU AU  AU  deg AU2 AU2 AU2 1.59 1.499 0.53 70.5° 1.56 0.40 1.16  (2 yr)2/3  Given √(a2-b2)  tan-1 b 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 ∇ AOrb ℓ AΔ Δt 1.16 AU2 7.49 AU2 1.41 AU 1.09 AU2 106.4 days AE-A∠ π a b b2 a ℓ×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 E E' YE XE YT XT 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 180-E a*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

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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.

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)

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, A_T∠.

ν	r	xT	yT	AT∠	AT∇	AT
30°	1.10 AU	0.95 AU	0.55 AU	0.26  AU2	? AU2	0.26AU2 +?

45°

1.14AU

0.81 AU

0.81 AU

0.33 AU2

? AU2

0.33AU2 + ?

60°

1.21AU

0.61 AU

1.05 AU

0.32 AU2

? AU2

0.32 AU2 + ?

90°

1.41AU

0.00 AU

1.41  AU

0.00 AU2

? AU2

0.00 AU2 + ?

120°

1.70 AU

-0.85 AU

1.47  AU

0.62 AU2

? AU2

0.62 AU2 + ?

150°

1.99 AU

-1.72 AU

0.99 AU

0.85 AU2

? AU2

0.85 AU2 + ?

Given	ℓ 1+e*cos(ν)	r × cos(ν)	r × sin(ν)	0.5×XT×YT	Unk	AT∠+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
ν	r	YT	YAC	E	AE	AE∠	AEΔ
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)	YT×a b	Sin-1( YAC a )	πa2×E 360°	a2×Sin(2E) 4	AE-AE∠

Compare Edge Areas, A_E∇ & A_TΔ

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	AE	AE∠	AE∇	ATΔ	AT∠	AT
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'	Sin-1( r×Sin(ν') b )	π a2 E 360	a2  ×Sin(2E) 4	AE -AE∠	AE∇ b a	r2 × Sin(2ν') 4	AT∠+ATΔ
e=0.33	ℓ = 1.41 AU	If ν>90°	a =1.59 AU	b =1.499 AU	c = 0.53 AU	AOrb=7.484AU2		T=730.5Days
e=c/a	ℓ = b2/a	ν'=180°		AE = AE∠ + AE∇		AOrb= π a b		T= 2π√(a3/μ)

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Compare Edge Areas, A_E∇ & A_TΔ

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	AE	AE∠	AE∇	ATΔ	AT∠	AT
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'	Sin-1( r×Sin(ν') b )	π a2 E 360	a2  ×Sin(2E) 4	AE -AE∠	AE∇ b a	r2 × Sin(2ν') 4	AT∠+ATΔ
e=0.33	ℓ = 1.41 AU	If ν>90°	a =1.59 AU	b =1.499 AU	c = 0.53 AU	AOrb=7.484AU2		T=730.5Days
e=c/a	ℓ = b2/a	ν'=180°		AE = AE∠ + AE∇		AOrb= π a b		T= 2π√(a3/μ)

Compare Edge Areas, A_E∇ and A_TΔ

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	AE∇	ATΔ	AT∠	AT
30°	1.10 AU	30°	21.5°	0.043AU2	0.041AU2	0.260AU2	0.30AU2
45°	1.41 AU	45°	32.6°	0.146AU2	0.138AU2	0.327AU2	0.46AU2
60°	1.21 AU	60°	44.4°	0.348AU2	0.328AU2	0.318AU2	0.65AU2
90°	1.41 AU	90°	70.5°	1.158AU2	1.092AU2	0.000AU2	1.09AU2
120°	1.70 AU	60°	78.5°	1.483AU2	1.398AU2	0.623AU2	2.02AU2
135°	1.85 AU	45°	60.7°	0.800AU2	0.754AU2	0.855AU2	1.61AU2
150°	1.99 AU	30°	41.5°	0.288AU2	0.272AU2	0.855AU2	1.13AU2
Given	ℓ 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∠+ATΔ
e=0.33 e=c/a	ℓ = 1.41 AU ℓ = b2/a	If ν>90° ν'=180°- ν	b = 1.499 AU	xxxxxxx	xxxxxxxx	xxxxx	xxxxx

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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=

ν	E	A∇	ATA	tTA
30°	21.46°	.011 AU2	.301 AU2	29.36 dy
45°	32.65°	.038 AU2	.465 AU2	45.27 dy
60°	44.41°	.095 AU2	.646 AU2	63.00 dy

x

ν	θ	Eθ
120°	60°	78.45°
135°	45°	60.72°
150°	30°	41.50°

x

A∠	Y	θ	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 y_θ  = R_ν  × Sin(θ)= .0369 AU.

Insert EdgeA	Q	H
.464 AU2	2.12 AU	.4377 AU
		.
Given	a + c	.2A/Q

θ	ν	Rν	yθ	Y∠
1°	179°	2.115AU	.0369AU	12.205°
11.86°	168.14°
Given	180-θ	ℓ 1+e×cos(ν)	Rν×Sinθ	.4505 AU yθ

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ν Sector Area Mirrors to θ Sector Area			θASect=0.464 AU2= νASect
AE%	AEdge	ATri	Q	yθ
1.06%	0.00492 AU2	0.4591 AU2	2.12 AU	0.4331 AU
1.07%	0.00496 AU2	0.4590 AU2	2.12 AU	0.4331 AU
1.08%	0.00501 AU2	0.4590 AU2	2.12 AU	0.4330 AU
% of ASect	ASect×AE%	ASect - AEdge	a + c	2×ATri Q

x
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x

True Anomaly			Eccentric Anomaly
TAE%	TAEdge	yθ	Eθ	EAEdge	TAEdge
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	TASect×TAE%	(TASect-TATri)/Q	sin-1(yθ/b)	a2×(πEθ/360⁰-.5yθ/b)	b×EAEdge/a

x

1.073%	0.00498 AU2	0.459 AU2	2.12 AU	0.4330AU
% of ASect	ASect×AE%	ASect - AEdge	a + c	2×ATri Q

x

x

8. Travel Time Mirrors across Major Axis For θ, angle from Q, compute area and time. See sample travel times: ν θ  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: ν E A∇ ATA tTA 30° 21.46° .011 AU2 .301 AU2 29.36 dy 45° 32.65° .038 AU2 .465 AU2 45.27 dy 60° 44.41° .095 AU2 .646 AU2 63.00 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:

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