ANNUAL SNOWBALLS

If one resupply vessel is great,

multiple resupplies must be even greater!

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SNOWBALL FROM OORT introduces the concept of a 7G cargo vessel launching from Sol and overtaking a previously launched 1G vessel, much slower with passengers (pax).

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SNOWBALL FROM OORT shows how a cargo vessel can overtake the prior pax vessel exactly one year into its cruise phase.

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MORE SNOWBALLS extend this profile with two variations.

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First, suppose that mission planners decide to up the game and send an annual resupply vessel throughout the pax vessel's cruise period.

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Second, consider a variable 1G vessel profile with different cruise velocities achieved by different durations of 1G acceleration (explained further below).

RENDEZVOUS: SNOWBALL JOINS PAX VESSEL
PREPARE FOR RENDEZVOUS.  When the 7G snowball (habitat encased in ice) starts deceleration, it is 1,104 AUs (= 6,484 AU - 5,380 AU) behind the 1G vessel which rotates during cruise phase to simulate gravity.	DESIRED END STATE:  Throughout the 48¼ day deceleration duration, this gap gradually decreases till they rendezvous at the end of this duration. Thus, 7G vessel precisely matches 1G vessel's cruise velocity at each Interception Point.
7G CRUISE START/STOP POINTS
MATCH 1G CRUISE SPEED.  At end of each snowball's cruise phase (Points: ⓐ  ⓑ  ⓒ ), it must decelerate for 48¼ days to match speed of Pax vessel (111.8 AU/day).	DESIRED END STATE:  Resupply (7G) vehicle matches velocity of Pax vessel exactly upon interception. For rest of flight, vessels fly in tandem.  CONCLUSION:  Such maneuvers can happen multiple times during the flight.
7G ACCELERATIONS
7G CRUISE SPEED.   After 7G acceleration for 100 days and 9,802 AUs, snowball attains cruise velocity (149.9 AU/dy). For annual intercepts, assume that snowballs A, B, and C must start cruse at 360.2 days (①),453.7 days (②) and 547.3 days (③) respectively after pax vessel launches at 1G (zero days).	7G LAUNCH LAGS 1G LAUNCH Thus, first snowball's launch time, A, lags pax launch by 260.2 days (= ① -100 days). Similarly, B lags by 353.7 days, and C lags by 447.3 days after Pax vessel launch.
Achieve Different 1G Cruise Speeds by adjusting duration of 1G accelerations for Passenger vessel.
FASTER CRUISE SPEEDS come from longer accelerations. EXAMPLE: 1G propulsion for 1.1 year (401.7 days) takes vessel to a velocity of 117.9 AU/day over a distance of 28,072 AU. BASELINE: By comparison, 1G propulsion for 1.0 year takes vessel to 111.8 AU/day and 23,876 AUs. SLOWER CRUISE SPEEDS come from shorter accelerations. EXAMPLE:  1G propulsion for .9 year (328.7days) takes vessel to a 105.1 AUs/day over 19,913 AUs.
Acceleration: 1 × g Time (t) Velocity (Vt) Distance (dt) Years Days %c AU/day LY AU Faster 1.1 Y  401.7 d 67.63%c 117.1AU/d    0.458 LY 28,972AU Baseline 1.0 Y 365¼ d 64.43%c 111.9AU/d    0.377 LY 23,842AU Slower 0.9 Y 328.7 d  60.70%c 105.1AU/d    0.315 LY 19,913AU Given Vt = c × (1-(1-Δ)t) Δ = .2826% dt = c × t + Vt ln(R) R = 1-Δ = 0.997174	For convenience, arbitrarily define BASELINE 1G acceleration profile with a duration of 1 year; thus, start acceleration at day 0 and end after 365.25 days.  Resultant Velocity becomes vessel's cruise velocity.  See dashed line on above diagram. NOTE: To readily compare different cruise phases (following frames), arrange acceleration start times for all cruises to start at a common time at 365.25 days. For faster cruise, increase baseline acceleration duration by 10% from 365.25 days to 401.7.  Thus, start faster acceleration at -36.45 days (before day zero, BASELINE start.) For slower cruise, decrease baseline acceleration duration by 10% from 365.25 days to 328.7 days.  Thus, start slower acceleration at +36.55 days (after day zero, BASELINE start.)
	EXAMPLE:  Consider a slower than baseline trip to our nearest stellar neighbor, Alpha Centauri, with total distance (dTtl) from Sol of 4.3 Light Years (LYs). Typical interstellar flight profile includes: Acceleration Distance (dAcc). Accelerate at 1G from Sol for 0.9 years for reasonable cruise speed (.60c) and distance of .31 LYs. Deceleration Distance (dDec).Decelerate at 1G for 0.9 year (same duration as acceleration) starting .31 LY prior to destination. Cruise Distance (dCru). Between acceleration and deceleration, cruise for several years at constant velocity, 60% Light Speed, c. Compute cruise distance as follows: dCru = dTtl       - dAcc     - dDec  dCru = 4.3 LY - .31 LY - .31 LY = 3.68 LY  Earth bound observer would measure cruise duration as 6.13 years (3.68 LY ÷ .60 LY/yr).
For each year (as observed from Earth) of each vessel's cruise phase, one could plan for each 1G vessel to intercept a 7G resupply vessel. PREPLANNED ANNUAL INTERCEPTS Cruise Start (after 1 yr Baseline 1G) Cruise Distances (at annual interviews) Dist. (d0) Vel. (V) 2 yr 3 yr 4yr Faster ➀ 28,972 AU 117.9 AU/d ⓐ 72,036AU ⓑ 115,098AU ⓒ 158,161 AU Baseline ➁ 23,842 AU 111.8 AU/d ⓓ 64,678AU ⓔ  105,511.9 AU ⓕ 146,347AU Slower ➂ 19,913 AU 105.1 AU/d ⓘ 58,301AU ⓙ 96,689AU ⓚ 135,076AU SeePrev. Table d0 + V × (t-1) First three years are shown in above table and associated diagram (to the right.)
	Snowballs Decelerate to Match Different Pax Cruise Speeds NOTE:  See 7G deceleration values from following link: (7G Dec)
	MATCH SPEED OF 1G's FASTER CRUISE (117.9 AU/day): To match this velocity, 7G vessel must decelerate from its cruise speed (149.9 AU/day) for 43 days over a distance of 5,843.2 AUs.
	BASELINE CRUISE speed is arbitrarily chosen as vessel velocity (111.8 AU/day) attained after 1 year of 1G propulsion:  Thus,  7G vessel must decelerate for 48.25 days over a distance of 6,444.9 AUs.
	MATCH SPEED OF 1G's SLOWER CRUISE (105.1 AUs/day): Due to greater velocity differential, 7G vessel must decelerate for a longer duration, 53.4 days, and a greater distance, 6,974.1 AUs.
Consider Selected Annual Intercepts Determine "start point" of 7G Decel. Annual Intercept 7G Deceleration Interval 7G Deceleration  Start Point time (tAnn) distance (dAnn) time (tInt) dist (dInt) time dist Faster 2 Yr  ⓐ 72,036 AU 43 days 5,843.2 AU 687.5 days 66,192.8 AU Baseline 3 Yr ⓔ  105,511.9 AU 48.25 days 6,444.9 AU  1,047.5 days 99,067 AU Slower 4 Yr ⓚ 135,076 AU 53.4 days  6,974.1 AU 1,407.6 days 133,668.4 AU Previous Tables 7G Dec  tAnn - tInt  Assume measurements accomplished by Earth bound observers.	After slight 7G deceleration interval, resupply vessel matches velocity of pax vessel at exact time/distance (Selected examples include points: ⓐ , ⓔ,  ⓚ.) Thereafter, both vessels travel in tandem.
	End point of 7G cruise segment; coincides with start point of 7G Deceleration 7G Cruise Segments Cruise End Point Cruise Start Point Time (t1) Dist. (d1) Time (t0) Dist. (d0) Faster 687.5 days 66,193 AU   312 days 9,841 AU Baseline 1,047.5 days 99,067 AU   452 days  9,841 AU Slower 1,407.6 days  133,668 AU   582 days  9,841 AU See Prev. Table t0 = t1 - (d1 - d0)/V Cruise start point of 7G cruise segment; coincides with endpoint of 7G Acceleration.
	Jan Hendrik Oort "GREAT OAK OF ASTRONOMY" (28 April 1900 - 5 November 1992)   Subrahmanyan Chandrasekhar: "The great oak of Astronomy has been felled, and we are lost without its shadow".
Jan Hendrik Oort was a Dutch astronomer; best known for Oort Cloud of comets which bears his name. Oort Cloud is the de facto boundary of our Solar System. In 1935, Dr. Oort became professor at the observatory of the University of Leiden; Ejnar Hertzsprung was the director. (Dr. Hertzsprung is well known for the Hertzsprung-Russell diagram, an essential tool in the study of stars.) Fascinated by radio waves from the universe, Dr. Oort pioneered radio astronomy with an old radar antenna confiscated from the Germans after WWII. Radio interferometry was suggested by Oort well prior to their experimental tests by others (Ryle in Cambridge and Pawsey in Sydney).   In the 1950s, he raised funds for a new radio telescope in Dwingeloo, in the east part of the Netherlands, to research the center of the galaxy. In 1950, Dr. Oort observed most comet origins to be within the Solar System.  HYPOTHESIS:  SOL IS SURROUNDED BY BILLIONS OF COMETS. While very few enter the inner Solar System, and even fewer are observed; an analysis of these few indicate the existence of the Oort Cloud.	Summarize 7G Resupply Solution 1) Determine Annual Intercepts of 1G Cruises. Plan for resupply vessels to join passenger vessels at specific time/distance points.  For convenience, arbitrarily choose these points at one year intervals (as measured by Earth observers). 2) Determine Start Point of 7G Deceleration.  7G vessel must decrease its cruise velocity (from 149.9 AU/day) to precisely match velocity of its target pax vessel (1G cruise velocity of about 110 AU/Day.) 3) Determine Start Point of 7G Cruise. Upon completion of 7G acceleration (100 days), Resupply Vessel starts the cruise portion of its journey at 149.9 AU/day. 4) Determine Start Point 7G Accelerations at 100 days prior to corresponding end-times.  (Recall these end-times coincide with start-points of 7G cruise segments.)
COMPARE INTERSTELLAR  FUEL CONSUMPTIONS: Consider different 1G profiles due to different 1G acceleration durations Following three examples consider 1G acceleration durations of .9 year, 1.0 year and finally 1.1 years.  HYPOTHESIS: Increased durations increase fuel consumptions.	Oort's other discoveries include: In 1924, Oort discovered the galactic halo, a group of stars orbiting the Milky Way but outside the main disk. In 1927, he calculated that the center of the Milky Way was 5,900 parsecs (19,200 light years) from the Earth in the direction of the constellation Sagittarius Mass of Milky Way exceeds 100 billion Solar masses.  Light from Crab Nebula is produced by synchrotron emission.

1G Vessel: Least 1G Duration (0.9 Yr)  Daily Difference (Δ) Previous work estimates that under certain conditions, a vessel consumes .0468% of its gross weight (GW or "mass") for one day of 1G-force propulsion.  Thus, Δ = .000468 GW Thus, compute daily remainder: RΔ = (1 - Δ)GW = 0.999532 GW Similarly, assume 7G daily diff is seven times greater; thus, for 7G propulsion: 7×Δ = .00328 GW R7Δ = (1-7×Δ)GW = .99672 GW. Flight Phase Passenger Vessel Resupply Vessel n×G time (t) Phase Rem. Cum. Rem. Cum. Fuel Consumed n×G time (t) Phase Rem. Cum. Rem. Cum. Fuel Consumed Accel 1×G 329days RΦ=.8573GW RC=.8573GW0 F=.1427GW0  7×G 100 days RΦ=.7118GW RC=.7118GW0 F=.2882GW0  Cruise RΦ  =  (1-.000468)t F = 1 - RC = 1 - P(RΦ) RΦ  = (1-.00328)t Decel 7×G 53.4days RΦ=.8342GW RC=.5938GW0  F=.4062GW0  Cruise F = 1 - RC = 1 - P(RΦ) Decel 1×G 329days RΦ=.8573GW RC=.735GW0 F=.2651GW0  1×G 329days RΦ=.8573GW RC=.509 GW0 F=.491GW0
1G Vessel: Baseline Duration (1 Yr)
Flight Phase Passenger Vessel Phase Remainder (RΦ) 1G profile has 2 propulsion phases (i.e. "burns"): 1) 1G acceleration to cruise; 2) 1G deceleration to dest. 7G profile has 3 burns: 1) 7G accel to cruise; 2) 7G decel to match 1G vel.; 3) Finally, 7G profile's 3rd phase is same as 1G decel burn. Phase Remainder (RΦ) is vessel mass remaining after a vessel completes a propulsion phase. For each 1G propulsion, compute RΦ: RΦ = (1-Δ)t= (.999532)t For each 7G burn, compute RΦ as follows: RΦ = (1-7×Δ)t= (.99672)t Resupply Vessel n×G time (t) Phase Rem. Cum. Rem. Cum. Fuel Consumed n×G time (t) Phase Rem. Cum. Rem. Cum. Fuel Consumed Accel 1×G 365¼days RΦ=.8377GW RC=.8377GW0 F=.1623GW0  7×G 100 days RΦ=.7118GW RC=.7118GW0 F=.2882GW0  Cruise RΦ = (1-.000468)t F = 1 - RC = 1 - P(RΦ) RΦ  = (1-.00339)t Decel 7×G 48¼days RΦ=.8540GW RC=.608GW0  F=.392GW0  Cruise F = 1 - RC = 1 - P(RΦ) Decel 1×G 365¼days RΦ=.8377GW RC=.7017GW0 F=.2983GW0  1×G 365¼days RΦ=.8377GW RC=.5092GW0 F=.491GW0
1G Vessel: Greatest Duration (1.1 Yr) Flight Phase Passenger Vessel Resupply Vessel Cumulative Fuel Consumption increases as cumulative remainder (RC) decreases.  RC is product of previous phases: RC = P(RΦ) For 1 G profile: RC= R1G-Acc × R1G-Dec For 7 G profile: RC=R7G-Acc×R7G-Dec ×R1G-Dec For all profiles: F = (1-RC) GW0 n×G time (t) Phase Rem. Cum. Rem. Cum. Fuel Consumed n×G time (t) Phase Rem. Cum. Rem. Cum. Fuel Consumed Accel 1×G 401days RΦ=.8289GW RC=.8289GW0 F=.1711GW0  7×G 100 days RΦ=.7118GW RC=.7118GW0 F=.2882GW0  Cruise RΦ = (1-.000468)t F = 1 - RC = 1 - P(RΦ) RΦ = (1-.00339)t Decel 7×G 43 days RΦ=.8641GW RC=.6150GW0  F=.385GW0  Cruise F = 1 - RC = 1 - P(RΦ) Decel 1×G 401days RΦ=.8289GW RC=.6870GW0 F=.3130GW0  1×G 365¼days RΦ=.8289GW RC=.5098GW0 F=.490GW0
CONCLUSION: More Propulsion Time Requires More Fuel. AXIOMATIC: Most interstellar voyages require same energy for deceleration as for acceleration.
For 1G profile, following three variations assume: deceleration energy = acceleration energy Variation Acceleration Time (tAcc) Deceleration Time (tDec) Propulsion Time (t) 1G Fuel 1) Slower 0.9 year 328.73 days 656.46 days 26.49% GW0 2) Baseline 1.0 year 365.25 days 730.50 days 28.96% GW0 3) Faster 1.1 year 401.78 days 803.56 days  31.35% GW0 Given Given 365.25 days/yr × tAcc tAcc+ tDec  (1 - (1 - Δ)t) GW0 CONCLUDE:  1G fuel consumed increases as propulsion time increases.
For all 3 1G variations, resupply vessels use 100 days of 7G acceleration to high speed cruise (149.9 AU/day).  Thus for all 3 variations: t7G-Acc = 100 days f7G-Acc = (1 - (1 - 7×Δ)100) GW0 = .2882 GW0 7G Propulsions 1G Propulsions Total Variation (Cruise Speed) 7G Decel. Time(t7GD) 7G Prop Time (t7G) 7G Fuel Consumed (f7G) 1G Decel. Time (t1GD) 1G Decel Fuel (f1GD) Ttl Resupply Fuel (fResup) 1) Slower     (105.1AU/day) 53.4 days 153.4 days 39.55% 328.73 days 14.26% GW0 53.81% GW0 2) Baseline (111.8 AU/day) 48¼ days 148¼ days 38.52%  365.25 days 15.72% GW0 54.24% GW0 3) Faster    (117.9 AU/day) 43 days 143.0 days 37.45% 401.78 days 17.14% GW0 54.59% GW0 Given Numerical Methods (t7GD + 100) dy (1 - (1 - 7×Δ)t) GW0 See Prev. Table (1 - (1 - Δ)t) GW0 (f7G + f1GD) GW0 CONCLUDE:  Fuel consumed still increases as 1G propulsion time increases.

SLIDESHOW

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VOLUME 0: ELEVATIONAL
VOLUME I: ASTEROIDAL
VOLUME II: INTERPLANETARY
VOLUME III: INTERSTELLAR

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