Intro to Volume Three

ASTEROIDAL Vast majority of space travel will take place in asteroidal habitats. Asteroidal orbits Asteroid materials Over the next century, mankind will become adept of constructing cylindrical habitats from material already in space (mostly asteroids, but comets have their place). Just like transport of most goods is accomplished today not by subsonic jet liners but by huge, very cost effective, cargo ships (well over 100,000 vessels greater then 100 tons traverse the oceans today). We'll gain plenty of experience building, moving and dwelling in asteroidal habitats. Most interplanetary space travel in the Solar System will happen not in the quick spaceships with particle accelerator propulsion systems, but on these much more comfortable habitats. Cyclers. Some habitats will cycle between Earth's orbit and orbits of other planets. The trips will take months, but the travel conditions will be excellent (like living at home or better). Habitats will be so big that the spin around the longitudinal axis won't be noticed, but this spin will produce an Earth like gravity. They'll have their own fields, forests, jungles, rivers or any environment that enterprise perceives a market for. A few habitats will be "towed" by g-force vessels directly to their, perhas to orbit around Jupiter or another giant gas planet. A few habitats will even share Earth's orbit around the Sun. Cruise Phase of Interstellar Trips Interstellar vessel must cruise at a constant velocity for most of their voyage due to inherent fuel limits on g-force propulsion.  Without g-force, how will the vessel simulate earth like gravity? During cruise, gravity comes from angular momentum, constant rotation of axial rotation about the longitudinal axis (i.e., asteroidal habitat). Occupants will transition from walking on the decks and considering direction "up" as being the direction of travel (toward forward end of vessel which is pointing toward destination); they will now walk on the inside of the outer hull of the vessel and consider "up" as toward the center of the vessel (center being an imaginary longintudinal axis running throughout the vessel length). Eventually, the vessel will come to the distance where it needs to decelerate. During this phase, "up" will once again be toward forward end of vessel (now pointing to departure star) and people will start walking on decks again. I suppose one could call such vehicles interstellar "cruisers". Conclusion: When interstellar flight is a new, novel experience, humans should have acquired lots of experience living in large, rotating habitats in space. Humans should get plenty of experience with asteroids to fulfill interplanetary missions because economies of scale will need large safe vessels with lots of capacity for passengers and cargo to accomodate the large distances in our Solar System. These same type vessels will prove indispensable to accommodate the much greater distances between stellar systems. Divided into three sections: @@@Thinking About It@@@ @@@Getting Ready@@@ @@@Getting There@@@ Near Earth Objects (NEOs) present much more opportunity then risk. Near Earth Objects (NEOs) present much more opportunity then risk. Most NEOs present absolutely no risk at all because their trajectories do not impact the Earth ever. These NEOs are pure opportunity to harvest at will; the only risk is that we'll refuse to harvest them at all. For the very few NEOs which will impact the Earth; they present an imperative for ASAP harvesting before they present any danger.  Asteroids will provide most material for habitat construction. Compare orbits for Earth and Apollo, a NEO with great opportunity. The paths of habitats will essentially be those of asteroids; thus, TE computes orbits of asteroids, objects orbiting the Sun. Recall that Kepler proved that all orbits are elliptical.

Intro to Volume Two

INTERSTELLAR INTERSTELLAR flights will take years; thanks to G-force. Thought Experiment's vessel would reach 60% light speed (.6c) after 323 days of g-force acceleration (distance is about .3 LY); thus, it would need another 323 days (another .3 LY) to slow down just prior to arrival at destination star. Thus, this much g-force (646 days) enables the vessel to cruise the interim distance between acceleration and slowdown at a constant velocity of .6c. For example, a trip to a nearby stellar system, Alpha Centauri, would only take about 8 years (as observed from Earth). INTERSTELLAR G-force consumes considerable fuel. If 0.1% of the vessel's mass can fuel one day of constant g-force acceleration; then, G-force flight for 646 days would require 46% (=100%×(1-.999646)) of ship's original gross weight (GW).  Since 646 days of g-force only covers a distance of .6 LY, common sense compels us to conclude that g-force propulsion over most interstellar distances (>4 LYs) would require much more then 100% of vessel's GW (more fuel then ship, not practical).  Thus, TE assumes that a vessel must g-force accelerate for about a year to gain feasible velocity (significant portion of c); and TE further assumes that vessel must cruise at this velocity for several years before deceleration and arrival. INTERSTELLAR travelers must maintain awareness. Constant comm stream with Mother Earth. It's very likely that one of the many "chores" for interstellar crews will be to deploy and recover comm pods, deployed to assure constant communications with home station at Earth as well as the many other interstellar vessels along same route. Not only is this desirable for many reasons, consider this overwhelming factor - if this ship encounters unexpected hazards (large debris such as ejected comets and asteroids, rogue planets, brown dwarfts, singularities, even comm modules from previous flights), the time that communications terminate will alert listeners to exercise caution during subsequent trips to that stellar destination. Three Sections: +++Thinking About It+++ +++Getting Ready+++ +++Getting There+++ Practical Interstellar Scenario G-force acceleration/deceleration plus a lengthy cruise make interstellar trips practical. There are many other concepts for interstellar travel; most are infeasible and/or inpractical.  For example, warp drives, worm holes and other theoretical mind exercises are not feasible.  "Sailships" and generational space trips are not practical.  However, a few concepts, such as the Bussard Ramjet might prove useful during the long cruise portion of an interstellar voyage. Regardless of g-force duration, light speed remainder stays the same. Recall Einstein: Regardless of relative velocity, all observers measure same value for c. At start of voyage, two obervers, Earth bound Alan Ein (Al) and ship board Bertrand Stein (Bert), are at relative rest, and they both measure light speed, c, as 299,792,458 meters per sec (m/s). After Bert g-force accelerates for one day, Al observes Bert's velocity to be .283% c; thus, the remaining velocity, 99.717%c still remains to be achieved.  However, Bert still measures c as precisely 299,792,458 m/s. On second day, Bert continues g-force acceleration, and he again plans to achieve .283% c; thus, remaining velocity will still be 99.717%c.  Again, both Al and Bert still measure c as precisely 299,792,458 m/sec. Everyday, Bert continues to calculate .00283c achieved with a constant remainder of .99717c.  Fortunately, Bert does not despair because he notices that home star, Sol, is receding at ever greater speeds; also, destination star is approaching at ever greater speeds. Thus, light speed remains constant, but g-force changes ship's relative velocity compared to departure star as well as destination star. Back on Earth, Al continues to measure a daily increase in Bert's velocity. Intuitively, one would think that a constant daily increase of .283% c would cause a g-force vessel to achieve light speed after 354 days (354 days × .283%c/day = 100.182%c). However, Einsteinian concepts clearly state that cannot happen. Thus, Thought Experiment (TE) assumes that following exponential equation is a better model of g-force velocities ("t" is time in days).  Einsteinian: vt = c [1- Rt]= c[1 - .99717t] TE assumes: After 354 days of g-force, Al measures Bert's velocity, v354 , as 63.331% c.

Intro to Volume One

INTERPLANETARY Spaceflights to neighboring planets will become practical and eventually routine (much like airline travel today) because future propulsion methods will reduce travel time from months and years to weeks and perhaps even day. Furthermore, spacecraft will maintain comfortable Earthlike conditions (gravity, atmosphere, comfortable billets, entertainment, etc) throughout the flight. Equivalence Principle. Consider Einstein's thought experiment about an accelerating elevator. If the elevator accelerates at same rate as free falling objects near Earth's surface, then occupants will feel equivalent g-force as if they're static on Earth's surface. Instead of Einstein's elevator, our Thought Experiment notionalizes a high performance spaceship to accelerate at rate, g, to produce gravity like force (g-force). However, our TE assumes that g-force propulsion can be achieved with slight enhancements to current technology. Three sections: ***Thinking About It*** ***Getting Ready*** ***Getting There*** Different Expressions for g (acceleration of freely falling object near Earth's surface). After one day of g-force acceleration, vessel would achieve a velocity of  864 km/sec which equals 0.5 AU/day or 0.3% of light speed, c. 10 m/sec2 = 864 km/sec/day = g  = .5 AU/day2 = 0.3% c/day "Galileo's famous demonstration at the Leaning Tower of Pisa showed that heavier objects fall at the same rate as lighter objects. In fact, he did numerous experiments in a more practical fashion, rolling balls down sloping troughes at different angles. He discovered that an object's freefall velocity increases with time, not with mass. In fact, numerous observations have determined that a freely falling object accelerates per the duration of the fall not the mass of the object." G-force to other planets is more practical then current method of transfer orbits. Transfer orbits between planets now take months and years. This might work for robots and other AI devices; it won't work so well for humans and other biologics. By contrast, g-force propulsion would reduce interplanetary flight time to days. On the other hand, g-force has its concerns. For example, g-force to Jupiter could increase ship's velocity to 2,700 km/sec at the midpoint. To decrease speed and still simulate gravity (thus, the term, "g-force"), spaceship reverses direction of fuel exhaust and decelerates at g for remaining days of travel. Momentum Exchange: Expel small mass of high velocity gas in one direction, and much larger mass (rocket) slowly increases speed in opposite direction. Mship × Vship = mfuel × vfuel Let large space vessel's speed increase by 10 m/s for every second of powered flight. Thus, let Vship be a constant 10 m/s; then, change it into a rate by dividing both sides of equation by a second. (Recall: g, acceleration due to gravity near Earth’s surface, is approximately 10 m/sec2). Thus, TE assumes spaceship can expel enough high speed particles to provide 1-g force throughout the flight. This propulsion will bring ship to high velocity and simulate Earth gravity for ship and contents. Mship × g = (mfuel × vfuel)/sec

*GETTING READY*

*******************GETTING READY*******************

Enabling Technologies

Contents6. Time to Upgrade7. Plasma Particles and Momentum8. Accelerators - Storage Rings9. Payloads from Earth10. Lunar Launch Platform

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We already have the pieces to the puzzle ...

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We just need to put them together.

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LIST OF TABLES Select from following:Table 1. Likely DestinationsTable 2: Typical TO DistancesTable 3: Typical TO VelocitiesTable 4. Transfer Orbital Periods.Table 5. Compare Travel TimesTable 6. Compare Travel VelocitiesTable 7. ⇑Pros and ⇓Cons of Two Space Travel ConceptsTable 8. Fuel to the PlanetsTable 9. More Fuel to the PlanetsTable 10. G-force Fuel & Return 6. Time to Upgrade Transfer Orbits vs. G-force Profiles Typical transfer takes a semi-orbit (SO); thus, total distance traveled would be one half the distance of the relevant ellipse. At orbital speeds, this takes a while. Most destinations take years. Destination Transfer Orbit (TO) G-force Profile SemiMajor Axis Typical LOS SO Dist Travel Times Travel Times aD d CT/2 TY Td tAcc tDec tTtl Uranus 19.18 AU 20 AU 34.58 AU 16.03 Yr 5,854 dy 6.32 dy 6.32 dy 12.65 dy Observed Assumed π√(aT2+bT2) √2 (1+aD)3/2 5.656 365.26TY √(2(d/2)) √g √d √g 2√d √g Line of Sight (LOS) is a straight line from Earth to destination, LOS distance is much less then SO distance. G-force acceleration would enable ships to approximate a straight line path at much greater speeds; this would greatly decrease travel time. Typical destinations would take days. It's time to upgrade from transfer orbits to g-force profiles.

7. Plasma Particles: G-force Fuel High speed particles make momentum happen. Mship= 30.57 *dc*ffsec Non-Relativistic Mship dc ffsec mT dec. c gm 3.057 .1 c 1.0 To Mars Mship= 30,570 *dc*ffExh Low Relativistic Mship dc ffExh mT dec. c kg 6,240 .2 c 1.021 To Ceres Mship= 30,570 *dc*mr*ffsec Mid Relativistic Mship dc mr mT dec. c 22,928 .6 c 1.25 To Uranus Ms=30,570√(mr2-1) ffsec High Relativistic Mship mr ffsec mT kg 52,949 2 1.0 Exoplanetary

7a. Close Look at the Particle Stream As fuel burns, gross weight decreases. As vessel weight decreases, burn rate also decreases. To consider ion quantity requirements, TE constructs following three tables. Table-1: Every Day Differs. Fuel consumption remains a consistent percentage of current GW; thus, GW is ever decreasing due to fuel burn. TE uses an exponential method {(1-Δt)t}to model daily fuel requirements. Table-2: Any day: 86,400 Unique Seconds TE arbitrarily chooses Day 20 as an example; Day 20's fuel requirement is 222.91 mTs of water. Simple division approximates an average burn rate of .00258 mT (= 2,580 grams) per second. Table-3: Pulse requirements During each second, PA will create and expend many plasma packets. Each will contain small quantity of water but a large number of particles. TE arbitrarily assumes a Packet Repetition Frequency (PRF) of 10,000 per second.

Particle Accelerator components: Select from following:1. Plasma Source - charged particles. 2. Injector - ions into wave guide.3. Wave Guide - particles travels in a tube.4. Klystrons - generate microwaves. 5. Superconducting Magnets6. Detectors - examines colliding particles.7. Vacuum System - clears wave guide.8. Cooling System - removes heat. 9. Storage Rings - stores particles.10. Extractor - ions into exhaust 8. Accelerators - Storage Rings Storage Rings keep momentum happening. Summary Component Accelerator Purpose Leverage Space Environ Plasma Source Provides charged particles for acceleration. Maybe water will work: Plenty of water in the oceans plenty of comets in space plenty of He3 in the giants Injector Puts ions into wave guide. Rate determined by ship's mass. Wave Guide vacuum tube provides path for particle beam. Design determined by size/shape of ship. Klystrons generate microwaves for the particles to ride. Text coming Magnets focus and steer the particle beam. Required for particle acceleration. Superconductors Save enormous energy Required for magnet practicality. Detectors examines radiation from particles colliding with each other and with walls of wave guide. Needed for efficiency as well as safety concerns. Not easy to repair wave guide, even harder during powered flight. Vacuum System keeps wave guide completely clear. Plenty of vacuum in deep space. Cooling System removes heat generated by equipment. Plenty of cold in space (4° K); can make liquid hydrogen from water supplies. Storage Rings store particle beams temporarily when not in use. Key design factor. Extractor Puts ions into exhaust ports. Rate determined by efficiency.

************CONTENTS************ Select from following: Space Elevators Carbon composite, nanotube ribbon" into space.Counterweight will keep the ribbon taut.Mechanical lifters will take payloads to space.Power for upward bound lifters will come from laser beams.Risk mitigation will take several forms.Operational Analysis helps determine true utility. 9. Payloads from Earth Materials from Earth start the process. One Thousands Lifts per Mission. Terran topsoil, oceanic water and many other Earth bound resources must elevate to orbit to support humanity's space borne mission. If a "lifter" can transport 10 mTs per lift, it will take 300 cycles to export sufficient water for one mission. If there is an equivalent requirement for Terran top soil, then another 300 cycles. Presume another 400+ cycles for other payload requirements then one g-force mission might take 1,000 lifts. With one strand of ribbon and one lifter; then, elevator capacity is 1 load per one 15 day cycle. At that rate, each g-force mission needs 15,000 days (about 41 years) to completely supply. With multiple strands and multiple lifters, capacity could conceivably increase to 1 load per day. This reduces supply time to 1,000 days or about 3 years. Conveyor Belt Configuration. TE assumes greatly increased ribbon durability and ribbon redundancy to enable elevator to adopt a conveyor belt configuration. If this enables 1 lifter per hour, then, supply time decreases to about 50 days which would be a reasonable duration to be included into a mission's Work Breakdown Schedule (WBS). Improvements could further reduce load time.

CONTENT Select from following: Lunar ResourcesLunar Launch Lunar Transformation 10. Lunar Launch Platform Moon makes final contribution. SUMMARY Filler for TE SpaceshipOne ton of helium-3 requires processing 14,000 tons of lunar regolith. This resource could be used to help build enormous infrastructure of g-force spacecraft. Home Berth G-force vehicle orbits around Luna when not traveling to/from interplanetary destinations. Orbiting g-force vehicle can leverage Lunar Transformations Energy source: He-3 Reactors. Terran Topsoil Terran Ocean Water

Close Look at the Particle Stream

Carefully consider  quantities of water ions

required for an effective particle stream.

Readily transformed into steam, superheated steam and then plasma, water molecules will become the ions needed to enter the ship's particle accelerator (PA) to gain near light speed velocity then exit from the spacecraft. (TE assumes "equivalent water molecules": H2O broken into various ions, OH-, H- and O-.) Consequent momentum exchange imparts a slight velocity increase to the ship in the opposite direction. To maintain g-force, spaceships will need plenty of water. Life support requirements of an ample water supply are obvious; however, an even more compelling requirement is to sustain g-force throughout the flight. Fortunately, there is plenty of water in Earth's oceans for the first few g-force ships. For longer term, there are plenty of water bearing comets throughout interplanetary space to provision the fleet of g-force ships needed to service likely travel requirements. Previous work causes us to conclude that g-force spaceship's Gross Weight (GW) is everdecreasing due to fuel consumption. In turn, decreasing GW results in decreased fuel requirements; thus, each day's fuel burn will be slightly less then previous day.

As fuel burns, gross weight decreases. As vessel weight decreases, burn rate also decreases. To consider ion quantity requirements, TE constructs following three tables. Table-1: Every Day Differs. Fuel consumption remains a consistent percentage of current GW; thus, GW is ever decreasing due to fuel burn. TE uses an exponential method {(1-Δt)t}to model daily fuel requirements. Table-2: Any day: 86,400 Unique Seconds TE arbitrarily chooses Day 20 as an example; Day 20's fuel requirement is 222.91 mTs of water. Simple division approximates an average burn rate of .00258 mT (= 2,580 grams) per second. Table-3: Pulse requirements During each second, PA will create and expend many plasma packets.  Each will contain small quantity of water but a large number of particles. TE arbitrarily assumes a Packet Repetition Frequency (PRF) of 10,000 per second.

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Table-1: Everyday is different. t GWt Ft fft-sec Δt days metric Tonnes mT/day daily ave gm/sec daily ave difference 0 100,000 mT  n/a n/a n/a 1 99,767 mT 233.00 2,697 gm/sec n/a 2 99,530 mT 232.46 2,690 gm/sec  6.90 gm/sec ... . . . . . . . . . . . . 19 95,664 mT 223.45 2,586 gm/sec  6.044 gm/sec 20 95,442 mT 222.91 2,580 gm/sec  6.027 gm/sec 21 95,219 mT 222.39 2,574 gm/sec  6.010 gm/sec ... . . . . . . . . . . . . 39 91,304 mT 213.24 2,468 gm/sec 5.76 gm/sec 40 91,091 mT 212.74 2,462 gm/sec 5.75 gm/sec Given %GWt*TOGW GWt-1*ΔDay Ft 86,400 s/d fft-1 - fft Δt= 0.233% / day    %GWt = (1-ΔDay)t TOGW = GW0= 100,000 mT

TE work concludes that a vessel could maintain g-force acceleration if --exhaust particles achieve .866c --daily fuel consumption is .233% of its gross weight.  Table 1 arbitrarily assumes ship's initial gross weight (GW0) to be 100,000 mTs.  With fuel consumption of .233%/day, ship will consume 233 mT of water during day 1, and ship's gross weight will decrease by 233 mTs. GW1 = 99,767 mT Day 2's fuel consumption will be .233% of 99,767 mT which further decreases ship's gross weight and so on.  Subsequent calculations (expedite with exponentials) eventually lead to fuel consumptions shown on Table 1 rows for mid-flight: days 19, 20, and 21. Since a metric Tonne (mT) equals 1,000,000 grams (gm), and one day equals 86,400 seconds, we can easily compute a daily per second average as shown in Table-1a. Table-1a: Daily Per Second Average t Ft fft-sec fft-sec = Ft 86,400 s/d day grams/day ave gm/s 19 223,454,000 2,586.27 20 222,912,000 2,580.00 21 222,385,000 2,573.09

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Total fuel requirement

for a 40 day, g-force mission is about 9,000 mTs of water.

(NOTE:  One Olympic sized swimming pool contains about 3,000 mTs of water.)

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Assume each day's "average value" occurs exactly at midday.

Table-1b. Three Days of Mid-Flight t sec fft-sec Δt day of flight precise second  at mid-day assume fuel flow at precise sec difference 19 43,200thsec 2,586.03gm/sec Δ19=+6.028gms 20 43,200thsec 2,580.00gm/sec Δ20=0.000 gms 21 43,200thsec 2,573.99gm/sec Δ21=- 6.010gms Arbitrarily select Day 20 as a reference; set Δ20 to zero.  Thus, burn rate for previous day (19) is a little greater (+), and burn rate for following day (21) is a little less (-).

Since vessel's GW and fuel burn rate gradually decrease throughout each day, common sense compels us to conclude the per second burn rate will range from slightly above the daily average to slightly below. TE artificially assumes each day's average per second fuel consumption value to be the "exact" value for the 43,200th second of the day (exactly halfway through the day's 86,400 seconds). This midday value will differ between succeeding middays and can be readily discerned.  For example, Table-1b shows the19th midday value to be 6.028 grams greater then value for 20th midday. In turn, the 21st midday value decreases by 6.01 grams.

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Avagadro, the Number The name "Avogadro's Number" (NA) is an honorary name attached to the calculated value of the number of atoms, molecules, etc. in a gram molecular weight of any chemical substance. To determine number of particles in a given mass, use Avagadro's Number (NA ) approximately 6.0221415×1023.   NA denotes quantity of molecules in one mole, material's atomic weight in grams.  A standard mole is defined as the value of quantity of atoms in 12 grams of pure carbon-12 (12C), carbon's primary isotope with atomic weight, 12. Any substance's mean molecular weight expressed in grams has same number of molecules. For example, the mean molecular weight of natural water is about 18.015; so, one mole of water is about 18.015 grams  with 6.0221415×1023 molecules, same quantity as 12 grams (one mole) of Carbon-12. The mole proves to be a convenient way for chemists to express the amounts of reagents and products of chemical reactions. For example, the chemical equation: 2 H2+ O2 → 2 H2O states that 2 mol of dihydrogen and 1 mol of dioxygen react to form 2 mols of water. Divide a mole by water's molecular weight  to determine quantity of molecules in  one gram of water. Water: 1 gm  = 6.0221415×1023  molecules / 18.015 1 gm = 0.334285×1023 molecules (of H2O) If one gram of liquid water displaces one cubic centimeter (cc = cm3); then, 33.428×1021 water molecules Rel. 1 cm3 Cube root both sides of this relation; then, TE presumes 3.2213×107 molecules (of liquid water) take up 1 cm in length. Thus, average distance between liquid water molecules could be about: 10-2 meter / 3.2213×107 = 3.1×10-10 m = .31 nanometers.

Focus on Day 20 To more closely approximate per second fuel flows for Day 20, use different Δs for first and second half of the day. Table-1c. Day 20: Start to Finish t sec fft-sec Δt-sec day_of flight precise second assume fuel flow at precise sec difference 20 00,001stsec 2,583.014gm/sec Δ20AM=+3.014 gms 20 43,200thsec 2,580.000gm/sec Δ20Mid=0.000 gms 20 86,400thsec 2,576.995gm/sec Δ20PM=-3.005 gms Determine morning  Δ  by halving previous's day's difference.  Δ20AM = Δ19 ÷ 2=+3.014 gms Determine afternoon Δ  by halving following day's difference. Δ20PM = Δ21 ÷ 2 =-3.005 gms For convenience, TE assumes linearity for determining precise and unique fuel consumption values for each of the 43,200 seconds in each half day.  A linear model requires a per second decrement (let i represent quantity of elapsed seconds): ffi= FFDay - i*Δ Decrement (Δ) can be expressed as mass or as quantity of potential particles (molecules not yet ionized). At exactly midday of Day 20, 43,200th sec, vessel consumes 2,580 grams or 862.4549×1023 water molecules. 12 hours earlier, Day 20, first sec; fuel consumption is 3.014 grams more; 3.014 gms of water contain 1.007 534 99×1023  molecules.  To determine per second decrement, divide by 43,200. Δ = -2.332 257×1018  = -.000 023 253×1023  12 hours later, Day 20, 86,400th sec; fuel consumption is 3.005 grams less; 3.005 gms of water contain 1.004 524 92×1023  molecules.  To determine per second decrement, divide by 43,200. Δ = -2.325 289×1018  = -.000 023 253×1023

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Table-2. Day 20: 86,400 Unique Seconds. t ffsec Nsec Remarks sec estimated grams estimated H2O molecules 1 gram of H2O contains 0.334 284 5×1023  molecules (or equivalent) 1 2,583.015 gm  863.461 878×1023 First half day: 3.015 grams of H2O contain 1.007 868 81×1023  molecules (or equivalent) Determine average decrement per second, divide by 43,200. Δsec = .000 023 33×1023  molecules 2 ≈ same  863.461 854×1023 ... Δ << 1 gm Δ≈.000 023 33×1023 43,199 ≈ same 862.454 033×1023 Δsec= +.000 023 33×1023  43,200 2,580.000 gm 862.454 010×1023 Reference Point: Exact Midday: Δ= 0.0 43,201 ≈ same  862.453 987×1023 Δsec= -.000 023 25×1023  ... Δ << 1 gm Δ≈ 000 023 25×1023 Second half day: 3.005 grams of H2O contain 1.004 524 92×1023  molecules (or equivalent) Determine average decrement per second, divide by 43,200. Δsec = .000 023 25×1023  molecules 86,399 ≈ same 861.449 508×1023 86,400 2,576.995 gm 861.449 485×1023 Given Ft 86,400 s/d NA*ffsec mole Assume day 20 of 40 day voyage. Fuel consumption is 222.91 mT for average fuel flow of  2,580 gm/sec

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Avogadro, the Method In 1811, Avogadro first proposed his hypotheis: volume of a gas (at a given pressure and temperature) is proportional to the number of atoms or molecules regardless of the nature of the gas. The term "Avogadro Number" was first used in a 1909 paper by Jean Baptiste Jean Perrin (1870-1942) entitled "Brownian Movement and Molecular Reality." (Translated from French to English  in "Annales De Chimie et de Physique" by Fredric Soddy.) In 1909, French physicist, Jean Perrin, proposed naming the constant in honor of Avogadro. Perrin: "The invariable number N is a universal constant, which may be appropriately designated 'Avogadro's Constant'." (now known as NA) Perrin, won the 1926 Nobel Laureate in Physics for his work on the discontinuous structure of matter and the discovery of sedimentation equilibrium. A large part of his work determined the Avogadro constant by several different methods. Determinating Avogadro's number required accurate and precise measurement of a single quantity on both the atomic and macroscopic scales. In 1834, Michael Faraday's works on electrolysis contained value for electrical charge of a mole of electrons ("Faraday constant"). In 1910, Robert Millikan, an American physicist,  measured the charge on a single electron.  Divide the charge on a mole of electrons by the charge of a single electron to determine Avogadro's number.  Since 1910, newer methods  more accurately determined the values for Faraday's constant, the elementary charge, and NA.

Thus far, TE has considered various water quantities required to g-force propel 100,000 mT vessel: For the entire trip, the 40 day requirement is about 10,000 metric Tonnes, enough water to fill a large pond or a small lake. For one day, the daily requirement will be about 250 mTs, a small swimming pool. Each day has 86,400 seconds. For each second, fuel consumption is a few liters, typical daily ration for drinking, cooking, etc. During each second, PA could eject 10,000 particle packets. Each packet could contain about a quarter gram of ionized water (1/4 gram liquid water is about 5 raindrops). If the Packet Repetition Frequency (PRF) is 10,000 per second, the associated Period, P, is 100 µseconds.  Thus, each packet has a 100 µsec window.  If the particles travel at .866c (259,620,268 meters/sec), PRF's associated wavelength is about 26,000 m.  Thus, a packet could travel 26 km in one packet period, more then sufficient to travel TE's projected PA guidepath of one km. Lay literature states the PA particle beam is "pencil thin".  Thus, TE arbitrarily assumes a circular cross section with max radius of one millimeter. TE further assumes following packet dimensions: A= π r2 = 3.14 (.1 cm)2   = .0314 cm2   L = 10 m = 10 × 100 cm = 1,000 cm V = A × L = .0314 cm2 × 1,000 cm  = 31.4 cm3 Since 30 cubic centimeters (cc) can contain 30 grams of liquid water,    TE assumes 31.4 cc is enough volume for one quarter gram of water ions.

TE assumes onboard PA uses "storage tubes" where particles will orbit at .866c for lengthy periods before being diverted into exhaust tube, final 1 km "guidepath" just prior to exit from spacecraft. This assumption justifies constant speed considerations reflected in diagram below.  However, 25 extra kilometers leaves plenty of margin for adjustment.  For example, if particles had to accelerate from zero to .866c within one km guidepath, this would reduce the average speed to .433 c, enough velocity to travel 13 km during the 100 µsec period; 12 km to spare.

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Divide each second into 10,000 equal periods.

 During each 100 µsec period, allocate one packet.

For each each 100 µsec period, the packet occupies the one km guidepath for only 4 µsecs; plenty of time to prepare for next packet.
At only one FLoating OPeration (FLOP) per nanosecond, each period has enough time for 100,000 FLOPs; TE assumes this to be sufficient computing power to make needed minor adjustments per packet.

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Due to repulsion of like charged particles, plasma particles naturally tend to spread out into available volume.  However,TE assumes that well designed PA uses superconducting magnets (sextupoles and quadrapoles) to focus each packet both longitudially and radially.

≈0	.008×1021	0.181×1021	1.17×1021	2.94×1021	2.94×1021	1.17×1021	0.181×1021	.008×1021	≈ 0
<<.001NP	.001NP	.021NP	.136NP	.341NP	.341NP	.136NP	.021NP	.001NP	<<.001NP

 N_P = 8.624 540 100×10²¹ water molecules (ionized into particles)

≈0

.008×10²¹

0.181×10²¹

1.17×10²¹

2.94×10²¹

2.94×10²¹

1.17×10²¹

0.181×10²¹

.008×10²¹

≈ 0

<<.001N_P

.001N_P

.021N_P

.136N_P

.341N_P

.341N_P

.136N_P

.021N_P

.001N_P

<<.001N_P

Longitudinal cross section shows particle density throughout the total packet length (10 meters).  TE assumes normal distribution of particles throughout the 10m length; this results in a elongated bell curve with highest density at the mean (μ) length (5m) from start of packet. For convenience, TE assumes standard deviation (σ) of one meter; thus, 68.2% of particles are within one meter of μ; 95.4% of particles are within 2 meters of μ; and so on.

Radial cross sections at various lengths along packet show particles filling up circular areas at various radii from the center. Radial focusing moves particles toward packet's center line. Particle density is greatest at the mean length of the packet, but particle density decreases as packet length differs from the mean. Thus, the lesser particle quantity is more affected by the magnetic radial focus which drives them closer to packet center line.

As described above, TE assumes a packet length of 10 m (packet volume > 30 cc) can easily contain .25 gms of water ions (recall 1 gm of liquid water is about 1 cubic cm). However, this model leaves plenty of margin for adjustment, and the packet length can easily be increased as required. Other adjustable parameters include: PRF, packet size, guidepath length, cross sectional radius and no doubt others.

Table-3a:Every day differs. t fft-sec Pgm Ppar days daily ave gm/sec ave gram per pkt ave particles per pkt 1  2,697 gm/sec .270 gm 9.026×1021 ... . . . . . . . . . 20  2,580 gm/sec .258 gm 8.624×1021 ... . . . . . . . . . 40  2,462 gm/sec .246 gm 8.223×1021 Given See Tables 1 & 2. fft-sec PRF K * Pgm PRF= 10,000 packets per second Conversion Factor: K = 33.428×1021 (Qty of water molecules/gm) From Tables One and Two, determine average fuel flow per second for any day during the voyage.  TE arbitrarily shows values for first, mid and last days of the 40 day trip. For each day's 86,400 seconds, TE assumes a range of 86,400 decreasing, unique values. Thus, fuel flow values will range from slightly above average to slightly below. TE assumes each day's average fuel flow is the precise fuel flow for the 43,200th second of that day (precise midday). For each precise midday, TE takes assumed value and determines average packet size by dividing by PRF (assume: 10,000 packets per second).  Like values for seconds of the day, values for packets of the second will range from slightly above to slightly below. TE assumes: At the precise midsecond, the 5,000th packet contains precisely that second's average particles per packet.

Avogadro, the Man ************* Amedeo Carlo Avogadro was born in Turin in 1776 to a noble family of Piedmont, Italy. He dedicated himself to physics and mathematics (then called positive philosophy).  In 1809, he started teaching at a liceo (high school) in Vercelli, where his family had property. In 1811, he published an article ("...Determining the Relative Masses of the Elementary Molecules of Bodies and the Proportions by Which They Enter These Combinations"), in a French journal, (Journal of Physics, Chemistry and Natural History); so, his major work "Avogadro's Hypothesis" was first written in French, even though Avogadro was Italian. In 1820, he became Professor of Physics at the University of Turin.  Due to regional politics, he lost his chair during 1823; however,  Avogadro was recalled to the university in Turin in 1833, and he taught for another twenty years. Though active in politics in younger life, Avogadro's private life was mostly sober and religious. He married Felicita Mazzé and had six children. Professionally, Avogadro was very active; he held posts dealing with statistics, meteorology, and weights and measures (he introduced the metric system into Piedmont) and was a member of the Royal Superior Council on Public Instruction.

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TE uses different forms of Scientific Notation: a×10b (operand, a, times power of ten) Form TE Example Why this form? Normalized   1 ≤ a < 10 1 mole = 6.0221415×1023 molecules This normalized SN value is the the textbook definition of the mole, standard number of molecules in atomic weight (in grams) regardless of substance. Non-normalized   a < 1 1 gm (of  H2O)=1 mole/18.015  =  0.334285×1023 molecules Non-normalized value of water particles per gram readily compares with above value of particles per mole by keeping same value for exponent. Non-normalized   a  > 1 typical ffsec = 2,580 gm (of  ionized H2O)/sec = 862.454 010×1023 equivalent molecules/sec Non-normalized value of ionized water particles per second readily compares with above values of particles per mole and particles per  gram. Change to Normalized 862.454010×1023=8.62454010×1025  Equivalent value of particles per second can be expressed in Normalized SN by --decrease magnitude of operand (move decimal point left 2 digits) --increase power of  10 (add 2 to exponent) Reduce Magnitude 8.624 540 10×1021 particles per packet At 10,000 packets per second, approximate particles per packet by dividing by 104. Due to nature of exponents, this is done by simply subtracting 4 from 25. Engineering Notation 10b;  b is mult, of 3 33.428×1021 molecules  (of liquid water)  displace 1 cm3 in volume Returning to quantity of H2O molecules in one gram, TE chooses to change from Scientific Notation (SN) to Engineering Notation (EN).  3.3428×1022 = 33.428×1021  Cube Root  3.2213×107 molecules (of liquid water)  take up 1 cm  in length Cube root volume to find cube's edge length in both cm and molecules.  Cube root readily done when power of ten is multiple of 3. (33.428×1021)1/3  = 3.2213×107 Negative Exponents Intermolecular  Distance = 3.1×10-10 m 1 cm = 10-2  m 10-2  m/ 3.2213×107 = 3.1×10-10 = .31 nanometers.

Table-3b. Mid-day 20 of G-force Flight For each second of powered flight, average particles per packet is readily computed as shown in Table 2. TE assumes this value to also be actual value of packet which transits the guidepath during precise midsecond (500,000 µsec shown below as decimal ".500").   t NP Δsec Δpkt Decimal Seconds Ave Particles per Packet Per Second Diff Particles/Second Per Packet Diff Particles/Packet 43,199.500 sec 8.624 540 33×1021 Δsec= +2.333×1015  Δpkt= +2.333×1011  43,200.500 sec 8.624 540 10×1021 Reference: Δsec= 0.0 Reference: Δpkt= 0.0 43,201.500 sec 8.624 539 87×1021 Δsec= -2.325×1015   Δpkt= -2.325×1011   Mid-day See Table-2 See Table-2 Δsec / 104 Consider three seconds at the exact middle of 20th day of powered flight (41,199; 42,000; 42,001).  Let exact middle of 42,000th second be the reference; then, compare values with one second before reference (midpoint of 41,199s) and with one second after reference point (midpoint of 42,001s).

Table-3c. Precise Middle Second of Day 20 First half second is slightly greater. First Half Second: packet size decreases by 1.1665×1015  particles.. Determine per packet decrement (divide by 5,000):  Δpkt = Δ / 5,000 = 2.333×1011  particles/packet P Timing NP Remarks Packet µsec µsec Particles per Packet Semi-Second Diff in Particles/Packet 0001 000,001 000,100 8.624 540 22×1021 Δ = +1.1665×1015  5,000 499,901 500,000 8.624 540 10×1021 Reference: Δ= 0.0 10,000 999,901 1,000,000  8.624 539 98×1021 Δ = -1.1625×1015   Mid-day Packet's first µsec Packet's last µsec See Table-2 See Table-2 Last half second: Δpkt = 1.1625×1015/5,000 = 2.325×1011  particles/packet Let exact middle of 42,000th second be the reference; then, compare values with 1/2 second before reference (start of 43,200s) and with 1/2 second after reference (end of 43,200s).

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Table-3. 10,000 Unique Values for 10,000 Packets. Particle Quantity: First to Last Packet of Second 43,200 P t ffP NP Remarks Packet µsec estimated grams estimated H2O molecules 1 1-100 .2583 gm 8.624 540 220 0000×1021 First half second: First to Mid: packet size decreases by 1.1665×1015  particles. Divide by 5,000 for per packet decrement:  2.333×1011  part. 2 101-200 ≈ same 8.624 540 219 7667×1021 ... ... ... ... 4,999 499,801-499,900 ≈ same 8.624 540 100 2333×1021  Δpkt = .000 000 000 2333×1021  molecules 5,000 499,901-500,000 .2580 gm 8.624 540 100 0000×1021 Reference Point: Exact Midday: Δ= 0.0 5,001 500,001-500,100 ≈ same 8.624 540 099 7675×1021  Δpkt = .000 000 000 2325×1021  molecules ... ... ... ... Second half second: Mid to Last: packet size decreases by 1.1625×1015  particles. Divide by 5,000 for per packet decrement:  2.325×1011  part. 9,999 999,801-999,900 ≈ same 8.624 539 980 2325×1021 10,000 999,901-1,000,000 .2577 gm 8.624 539 980 0000×1021 Given 1 packet passes through the PA every 100 µsec,  10,000 packets per second. NA*ffsec mole Assume day 20 of 40 day voyage. Fuel consumption is 222.91 mT for average fuel flow of  2,580 gm/sec

OTHER CONSIDERATIONS  

Plasma Source. To continuously generate plasma, TE assumes significant PA design such that significant portion of particle stream diverts to impact a designated "pool" of water. At near light speed, this particle impact would impart sufficient energy to "superheat" the solid/liquid water and ionize the water molecules.

Real Efficiency. Hence, one of the many reasons for particle stream inefficiencies or less then100% conversion of plasma ions to propulsion particles. Thus, Thought Experiment arbitrarily assumed an efficiency of 70%; thus, best case scenario 30% of PA particles diverted to superheat water and create more plasma ions for more PA cycles.

Model Adjustments. Very likely that not all consumed fuel will exit (due to inherent inefficiencies); however, if some particles are put to other uses other then propulsion (whether by design or defect) can those consumed particles correctly be modeled as GW decrement.  Smarter modelers will improve TE's current model; however, real data from actual flights will have the final word.

Particle Flow Precision. Finally, is it necessary or even possible to adjust particle quantities for individual packets??? Table 3 (above) approximates a theoretical per packet decrement of .23 parts per billion for subsequent packets.  Even with greatly enhanced future technology, that sounds impractical if not impossible.  A more practical method might be distribute weight scales throughout the vessel; these weight sensors could stream data into a centralized systems of servos/computers/etc to constantly make practical adjustments to PA's plasma flow.  The objective: keep vessel's acceleration very close to g-force as perceived by humans.

Afterthought! At 5 drops per packet of particles;  fuel flow is about 50,000 drops per second.