Sunday, July 20, 2014

7G TABLE: Decelerate for 48 ¼ Days





















To effectively accomplish an interstellar resupply mission, an Artificial Intelligent (AI) vessel could use 7G propulsion to quickly intercept the primary 1G vessel.  However,  the 7G vessel must slow down prior to intercept in order to match the 1G vessel cruise velocity. This table describes daily progress throughout the deceleration duration.
7G Deceleration for 48 ¼ Days 

Previous table, 7G Acceleration for 100 Days, describes daily progress of vessel accelerating from zero velocity at 7g.
After 100 days of 7g acceleration, a vessel travels total distance of 9,842 AUs (.155 Light Year, LY).  It gained velocity of about 150 AU/day (.866 c, 86.6% light speed).  
Assume 7G vessel cruises at this velocity until the optimal time/distance to start decelerating at 7g.
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
0 days149.93 AU/dy86.59% c9,811.1 AU0.1551 LY0.0 AU9,811.1 AU27.974% GW0
1 days149.46 AU/dy86.32% c9,661.7 AU0.1528 LY149.4 AU9,960.5 AU28.210% GW0
2 days148.98 AU/dy86.04% c9,512.8 AU0.1504 LY148.9 AU10,109.4 AU28.445% GW0
3 days148.48 AU/dy85.76% c9,364.4 AU0.1481 LY148.4 AU10,257.9 AU28.679% GW0
4 days147.98 AU/dy85.46% c9,216.5 AU0.1457 LY147.9 AU10,405.8 AU28.913% GW0
5 days147.46 AU/dy85.17% c9,069.1 AU0.1434 LY147.4 AU10,553.2 AU29.146% GW0
6 days146.94 AU/dy84.86% c8,922.2 AU0.1411 LY146.9 AU10,700.1 AU29.378% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
For success, the 7G resupply vessel must slow down to match the 1G vessel's cruise velocity (.644c = 111.5 AU/day).  This table describes daily progress throughout the 48¼ days of 7G deceleration.
Interstellar 1G Vessel
starts its one year of acceleration at day 0.

After this year, vessel stops acceleration and cruises at 111.5 Astronomical Units per day  (.644 c).

One year of 1G acceleration has taken vessel to 23,841 AUs,  .377 Light Year (LY).
Interstellar 7G Snowball
In this example, 7G snowball (traditional habitat encased in large volume of ice) starts its 100 day acceleration at day 260 of 1G vessel’s acceleration.

On day 360, 7G vessel stops propulsion, and cruises at 149.9 AUs per day (.865c).

100 days of 7G acceleration takes vessel to 9,842 AU, (= .155 LY).
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
7 days146.40 AU/dy84.56% c8,775.8 AU0.1388 LY146.4 AU10,846.5 AU29.609% GW0
8 days145.86 AU/dy84.24% c8,630.0 AU0.1365 LY145.8 AU10,992.3 AU29.840% GW0
9 days145.30 AU/dy83.92% c8,484.7 AU0.1342 LY145.3 AU11,137.6 AU30.070% GW0
10 days144.73 AU/dy83.59% c8,340.0 AU0.1319 LY144.7 AU11,282.3 AU30.299% GW0
11 days144.15 AU/dy83.26% c8,195.8 AU0.1296 LY144.1 AU11,426.4 AU30.527% GW0
12 days143.56 AU/dy82.91% c8,052.3 AU0.1273 LY143.6 AU11,570.0 AU30.755% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
Points thru map out following events:
After 360 days, resupply vessel ends 100 days of 7G propulsion and starts cruise at 149.9 AU/day.
After 365¼ days of 1G acceleration, primary vessel cruises at 111.5 AU/day.
After 400 days (about 35 days of constant velocity), resupply vessel is about 12,000 AU behind primary vessel.
After 500 days (about 135 days of constant velocity),, resupply vessel closes the gap to slightly over 8,000 AU.
 600 days (100 more days of constant velocity), gap closes even more to just over 4,000 AU.
 684 days, gap shrinks to about 1,000 AU in prep for 7G vessel’s deceleration duration of 48¼ days needed to match speed of primary vessel.
Describe primary vessel’s cruise distance by linear equation:
d = 111.5 t - 16,884 AU
Resupply vessel’s cruise distance:
d= 149.9 t - 44,152 AU
By inspection, we expect a pending intercept of these two cruise tracks about one LY and two years from start of primary vessel’s track. HOWEVER, the resupply vessel track must become nonlinear at about 684 days for 48¼ day deceleration maneuver.
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
13 days142.96 AU/dy82.57% c7,909.3 AU0.1251 LY143.0 AU11,712.9 AU30.982% GW0
14 days142.34 AU/dy82.21% c7,767.0 AU0.1228 LY142.3 AU11,855.3 AU31.208% GW0
15 days141.72 AU/dy81.85% c7,625.3 AU0.1206 LY141.7 AU11,997.0 AU31.433% GW0
16 days141.08 AU/dy81.48% c7,484.2 AU0.1183 LY141.1 AU12,138.1 AU31.658% GW0
17 days140.42 AU/dy81.10% c7,343.7 AU0.1161 LY140.4 AU12,278.6 AU31.881% GW0
18 days139.76 AU/dy80.72% c7,203.9 AU0.1139 LY139.8 AU12,418.3 AU32.105% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
19 days139.08 AU/dy80.32% c7,064.8 AU0.1117 LY139.1 AU12,557.5 AU32.327% GW0
20 days138.38 AU/dy79.92% c6,926.4 AU0.1095 LY138.4 AU12,695.9 AU32.549% GW0
21 days137.67 AU/dy79.51% c6,788.7 AU0.1073 LY137.7 AU12,833.6 AU32.770% GW0
22 days136.95 AU/dy79.10% c6,651.7 AU0.1052 LY137.0 AU12,970.6 AU32.990% GW0
23 days136.21 AU/dy78.67% c6,515.4 AU0.1030 LY136.3 AU13,106.9 AU33.210% GW0
24 days135.46 AU/dy78.24% c6,379.9 AU0.1009 LY135.5 AU13,242.4 AU33.428% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
After 684 days of mission time, resupply vessel flight profile used 7G acceleration and high speed cruise to reach a position 58,340 AU from Sol.
At this point, 7G vessel is only 1,076 AU behind the 1G vessel.  
The 7G vessel must start the 48¼ days of  7G deceleration required to slow down from 149.9 AU/day to 111.5 AU/day, cruise velocity of the primary vessel.

For functional rendezvous, both the primary mission vessel and the resupply vessel must match velocities  at the same position at the same time.  
In this example, both vessels simultaneously reach 64,796 AU (about one LY) along the track from Sol to neighboring star.
Determine Intersection Point
Assume the 1G pax vessel accelerates at 1G for one year; then, cruises at constant velocity. Thus, determine distance traveled per following linear equation:
d1G = -.267 LY + .644 c × t
Assume resupply vessel starts 7G accelerates at 265.25 days after mission vessels begins.  100 days later (365.25 days into mission), 7G vessels stops acceleration and starts cruise. Thus, determine total distance traveled via following linear equation:  
 d7G = -.711 LY + .866 c × t
To determine intersection point of the two linear equations, set them equal as shown:     
d1G = -.267 LY + .644c × t = -.711 LY + .866c × t  = d7G
For convenience, substitute following terms: LY = 63,241 AU and c = 173.15 AU/day:     
-16,885 AU + 111.5 AU/day × t = -44,964 AU + 149.9 AU/day × t
thus, solve for: tInt = 731.2 days and dInt= 64,644 AUs.  
These linear equations enable us to determine that the two cruising vessels will meet at time equals 731.2 days (2 years) since day zero and distance equals 64,644 AU (about 1 LY) from 1G pax vessel's starting point. However, the two vessels cannot gracefully rendezvous because they still have a huge velocity differential of about .22 c.  A following section proposes a deceleration method to synchronize velocity in addition to time and distance.
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
25 days134.70 AU/dy77.79% c6,245.1 AU0.0988 LY134.8 AU13,377.2 AU33.646% GW0
26 days133.91 AU/dy77.34% c6,111.1 AU0.0966 LY134.0 AU13,511.2 AU33.864% GW0
27 days133.12 AU/dy76.88% c5,977.9 AU0.0945 LY133.2 AU13,644.4 AU34.080% GW0
28 days132.30 AU/dy76.41% c5,845.5 AU0.0924 LY132.4 AU13,776.8 AU34.296% GW0
29 days131.47 AU/dy75.93% c5,713.9 AU0.0904 LY131.6 AU13,908.4 AU34.512% GW0
30 days130.62 AU/dy75.44% c5,583.1 AU0.0883 LY130.7 AU14,039.1 AU34.726% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
31 days129.76 AU/dy74.94% c5,453.2 AU0.0862 LY129.9 AU14,169.0 AU34.940% GW0
32 days128.88 AU/dy74.43% c5,324.2 AU0.0842 LY129.0 AU14,298.1 AU35.153% GW0
33 days127.98 AU/dy73.91% c5,196.1 AU0.0822 LY128.1 AU14,426.2 AU35.366% GW0
34 days127.06 AU/dy73.38% c5,068.9 AU0.0802 LY127.2 AU14,553.4 AU35.577% GW0
35 days126.13 AU/dy72.84% c4,942.6 AU0.0782 LY126.3 AU14,679.7 AU35.788% GW0
36 days125.17 AU/dy72.29% c4,817.3 AU0.0762 LY125.3 AU14,805.0 AU35.999% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
37 days124.20 AU/dy71.73% c4,692.9 AU0.0742 LY124.4 AU14,929.4 AU36.208% GW0
38 days123.20 AU/dy71.15% c4,569.5 AU0.0723 LY123.4 AU15,052.8 AU36.417% GW0
39 days122.19 AU/dy70.57% c4,447.1 AU0.0703 LY122.4 AU15,175.2 AU36.626% GW0
40 days121.15 AU/dy69.97% c4,325.7 AU0.0684 LY121.4 AU15,296.6 AU36.833% GW0
41 days120.10 AU/dy69.36% c4,205.4 AU0.0665 LY120.3 AU15,416.9 AU37.040% GW0
42 days119.02 AU/dy68.74% c4,086.1 AU0.0646 LY119.3 AU15,536.1 AU37.246% GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
43 days117.92 AU/dy68.11% c3,968.0 AU0.0627 LY118.2 AU15,654.3 AU37.452%GW0
44 days116.80 AU/dy67.46% c3,850.9 AU0.0609LY117.1 AU15,771.4 AU37.657%GW0
45 days115.66 AU/dy66.80%c3,735.0 AU0.0591 LY115.9 AU15,887.3 AU37.861%GW0
46 days114.49 AU/dy66.12% c3,620.2 AU0.0572 LY114.8 AU16,002.1 AU38.065%GW0
47 days113.30 AU/dy65.44% c3,506.6 AU0.0554 LY113.6 AU16,115.7 AU38.268%GW0
48 days112.09 AU/dy64.74% c3,394.2 AU0.0537 LY112.4 AU16,228.0AU38.470%GW0
Resupply Vessel Matches Cruise Velocity of Baseline "Pax" Vessel (1.0 yr of 1G Accel.)
at exactly the time/distance of the intercept.
48¼days111.78 AU/dy64.56% c3,366.3 AU0.0532 LY27.9 AU16,256.0 AU38.520%GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t
Decel.
Time(t)
Spot
Velocity (Vt)
Spot
Distance (dt)
Daily
Dist. (dΔ)
7G
Dist.(d7G)
Total
Fuel (F)
52 days106.98AU/dy61.79%c2,966.6 AU 0.2636LY107.3AU16,667.7AU39.272%GW0
53 days105.64AU/dy61.01%c2,860.5AU 0.2652LY106.0AU16,773.7AU39.471%GW0
Resupply Vessel Matches Cruise Velocity of Slower "Pax" Vessel (0.9 yr of 1G Accel.)
at exactly the time/distance of the intercept.     
53.4 days105.10AU/dy60.70%c2,818.4AU 0.2659LY42.0 AU16,815.7AU39.551%GW0
54 days104.27AU/dy60.22%c2,755.7AU 0.2662LY62.6 AU16,836.3AU 39.669%GW0
Given
(1 - (1-Δ)100-t) × c
c×(100-t)+ Vt

ln(1-Δ)
dt-1 - dt dΔ + ΣdΔ 1-(1-ε∇)100+t

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