Friday, January 14, 2011

G-force Interstellar Range: Ship Mass Half Life

Thought Experiment
recognizes that even with particle accelerator propulsion systems, interstellar travel will take years (versus days for interplanetary). Thus, the interstellar portion of our thought experiment uses better computation methods to determine g-force ranges.

Interstellar Profile Assumptions

Constraint: g-force needed to maintain Earthlike conditions and best environment for human travellers.

Need g-force acceleration to attain sufficient portion of light speed to make interstellar flight duration reasonable. (few years).

Need g-force deceleration to slow down.

However, g-force propulsion requires fuel, and interstellar voyages take so long that a starship would easily consume well over 100% of its mass in fuel during the multi-year voyage.

Thus, interstellar flights also need cruise phase constant velocity. No fuel consumption, no g-force; thus, regain g-force through longitudinal spin, and crew/pax lives on inside of outer hull.

Fuel Consumption Assumptions

  1. For interplanetary flights, TE assumed that half the ship's mass could easily suffice(maybe water). For interstellar flights, TE might need to consider higher percentages of vessel mass to be available as fuel. More fuel increases potential propulsion time, tp , which can increase the durations of both acceleration/deceleration phases to further decrease the cruise phase duration:
    • cruise distance will be slightly shorter
    • cruise velocity will be slightly greater
  2. Higher performing particle accelerators could accelerate fuel particles to even greater exhaust speeds. The greater momentum would result in an even better energy exchange between exhaust particles and ship (less fuel could propel a bigger ship).startfuel

Ship's propulsion system accelerates a predetermined mass of ionized particles (plasma) to exit vehicle at relativistic speeds. Since these ions come from designated, onboard material, this material fuels the spaceship much like gasoline fuels our automobiles. For both vehicles, when we add fuel, the total gross weight (GW) increases; when fuel transforms into energy to produce motion, fuel is consumed, and GW decreases.

The fuel is consumed at a certain rate; thus, ship's gross weight (GW) decreases at this same rate. If we know the amount of fuel and the consumption rate; then, we can calculate the time required to consume this amount, the ship's range.

Additional Assumptions

  • Artificially assume 100% efficiency for fuel flow conversions for convenience. 100% Efficiency is very unlikely; however, for convenience:
    TE artificially assumes every single particle travels completely through the many kms of waveguides and exits spacecraft exhaust as intended.
    TE further assumes no additional energy is required for other energy requirements, such as electro-magnetic fields to guide the particles.
    Thus, TE artificially assumes 100% efficiency for convenience. Future work will adjust for less then perfect efficiency.
  • Marginal Fuel Requirements - it makes sense to carry "spare fuel"; it's always a good policy to plan the mission to end with some fuel remaining. Since this fuel is not planned for consumption, it might be better to consider marginal fuel as part of ship's infrastructure. Thus, thought experiment disregards spare fuel for computing range of spacecraft.
  • Assume continuous g-force acceleration due to constant particle exhaust speed.
    • TABLE-1 varies fuel loads given just one particle exhaust speed (vExh = .99c) to determine Vessel Ranges based on fuel load.
    • TABLE-2 varies particle exhaust speeds given just one fuel load (F0 = 50% GW0) to determine potential Vessel Ranges based on particle exhaust speed.

link to image
Previous work shows:
ffExh* vExh
Set particle exhaust speed such that, vExh = .99 c ; thus, dc = .99
= 7.09

Thus, Lorentz Transform (LT) shows that exhaust particles
will increase over 7 times greater due to relativity.
ffExh = mr ffsec = 7.09 ffsec

Substitute as shown:
mrffsec* dcc
7.09 ffsec* .99 c

Given g=9.8065 m/sec2 and c=299,792,458 m/sec, previous work leads to:

∇ = g * 86,400 sec/dy
c * √(mr2-1)
= 0.282%/dy

Thus, this example defines , Daily Difference in Ship's Gross Weight:
∇ = %TOGWDay = .04%/day


TE assumes following g-force values.
g =
9.80665 m/sec2
g =
.489 AU/day2
g =
.2826% c/day
Since first day of g-force flight takes ship to .2826% c, TE calls that value, Δ "daily difference".
Previous works shows that one year of g-force acceleration takes ship to
  • velocity of 64.43% c
  • distance of .38 LY.
Acceleration brings ship to a near light speed velocity which facilitates a reasonable cruise duration. After our spaceship accelerates at g-force for 1 year, it'll achieve a velocity of .6443 light speed (c) and will have expended perhaps 25% of available fuel. However, total distance for a year of g-force acceleration will still be only about .38 light year (LY), not even close to the Oort Cloud which borders our Solar System at about 1 LY from Sol.
  • Acceleration Time: Let tAcc = powered flight time to accelerate from departure until cruise phase begins; amount of time available for acceleration depends on vessel's range, propulsion time, tProp.
  • Acceleration Velocity ever increases as tAcc increases. Based on previous work, TE assumes final velocity can be computed as follows:

    v(tAcc ) = (1 - (1 - Δ)tAcc)c = vMax
    Max velocity is achieved at completion of acceleration phase.

  • Acceleration distance ever increases as tAcc increases. TE assumes this can be computed as follows:

d(tAcc) = c(tAcc) - [v(tAcc)/ln(1-Δ)]

Determine Propulsion Time, tProp

BACKGROUND: Thought Experiment (TE) assumes onboard particle accelerator increases particle speed to relativistic velocities and expels a predetermined mass of high speed particles out of ship's exhaust; resultant momentum propels ship for a velocity increase equal to g-force acceleration

Recall following points.

ffsec Fuel flow per second determines fuel consumption; thus, ever decreasing gross weight of g-force vessel. For every second of flight, our notional spaceship consumes a quantity of "at rest" fuel which we'll call fuel flow per sec (ffsec). This is the fuel's mass prior to conversion to plasma (ions, particles with charges). Fuel consumption is reckoned in terms of fuel at rest "ffsec" which is at zero velocity in relation to its host vessel. "At rest" fuel gives the daily difference in gross weight (∇, a certain gross weight percentage decrease per day).

ffExh Fuel exhaust mass provides propulsion for g-force vessel. For every second of flight, a particle accelerator uses a system of magnetic and electrical fields to accelerate these charged particles to large fraction of light speed. Relativism and a century of particle accelerator experience both tell us that ffsec will grow as particles accelerate to near light speed; we call this new quantity ffExh, fuel flow of exhaust particles. Collective exhaust particle momentum (near light speed velocity times relativistic mass) enables spaceship's propulsion. To account for relativistic growth, TE uses relativistic mass growth factor, mr; this is computed with Lorentz Transform (LT).

    To state this exhaust rate, TE borrows the symbol, ∇ ("nabla"), and uses it to represent percentage of ship's gross weight consumed each day, the "daily difference".

    • For discussion's sake, TE assumes exhaust particle speed, vExh, to be 99% light speed, c. Express particle exhaust speed (vExh) as decimal Light Speed (dcc) where:

      vExh/c = 99% = .99

    • Relativistic mass increase can be expressed. Lorentz Transform shows that relativity growth factor causes exhaust particle mass to grow about 7 times.
      mr =1 /√(1-dc2) = 7.0888
    • Determine Daily Difference in ship's Gross Weight (GW):
      Artificially assume 100% propulsion efficiency
      (no design flaws, no peripheral requirements);
      ∇ =(86,400 * g / c )
      = .04%/dy
      86,400 = secs/day
      g = 9.80665 m/sec2
      c = 299,792,458 m/sec
      86,400g/c = .002826 = .2826%/dy

    then, g-force fuel consumption =∇

    ∇= .04% GW/day

    If ship's initial gross weight (GW0) is 100 metric Tonnes (mT); then, ship will consume .04 mT of fuel on first day of powered flight.
    FDay = ∇ * GW0 = .04% * 100mT = .04 mT
    GW1 = GW0 - FDay = 100mT - .04mT = 99.96 mT


    About .38 LY from destination, spaceship must decelerate from 64.43% c to an operational velocity. This will take another year and another large portion of fuel. Reverse prop vector, equal prop time

    Deceleration phase values will essentially equal corresponding acceleration values.

    • Velocity
    • Distance
    • TimeTE assumes consistent g-force for all powered phases of flight; thus, max velocity, time and distance would be same for both acceleration and deceleration phases. {NOTE: If ship's crew choose an inconsistent g-force (differs from near Earth gravity), this would result in different values for deceleration distance and time.} Thus, TE assumes
      dAcc = dDec tAcc = tDec

    If ship's initial gross weight (GW0) is 100 metric Tonnes (mT); then, ship will consume .04 mT of fuel on first day of powered flight.

    FDay = ∇ * GW0 = .04% * 100mT = .04 mT
    GW1 = GW0 - FDay = 100mT - .04mT = 99.96 mT

    Assume we devote half of ship's mass to fuel; (%TOGW = 50%). Then, ship starts with 50 tons of fuel, and a we can initially approximate ship's propulsion range by dividing 50 mT by .04 mT (= 1,250 days). However, previous work reminds us this is only a rough approximation, and that we need more accurate methods. (For example, an upcoming "Half Life Method" uses logarithms to more accurately and quickly determine vessel range.

    The initial absolute consumption value of FDay ( = * GW0) is a good approximation for first day of flight. After the first day, we must consider that GW decreases as the fuel is consumed. Thus, Δ, as a daily dynamic percentage value, reflects the ever decreasing GW during powered flight.


    Interstellar flights will need a lengthy constant velocity duration to conserve fuel. This profile assumes a cruise phase at .6443c for several years.
    No prop vector, no prop time.
    • Cruise distance will be the remainder of the distance to destination after deducting acceleration distance and deceleration distance.

    dDest = dAcc + dCru + dDec
    dCru = dDest - dAcc - dDec

    • Cruise Velocity is max velocity achieved at completion of acceleration. With no propulsive force applied, ship maintains this velocity throughout cruise phase and consumes no fuel.
      vCru = v(tAcc) = vMax
    • Cruise Time is simply computed as
      tCru = dCru/vMax

    1st day of trip, GW1 = GW0 - first day's fuel consumption = 99.96 tons.
    2nd day, ship consumes .o4% of the remainder which will be less then .04 mT.

    On day, t, fuel consumption will be even less and so on as shown in following table.

    To daily decrement 1% of ship's gross weight (GWt), multiply original quantity (GWo), by (1-∇) as shown.
    TABLE-0: Consider Fuel Consumption
    0NoneGW0=TOGW =100 mT
    1100mT * .9996GW1=GW0(1-∇) =99.96 mT
    299.96 mT * .9996GW2=GW0(1-∇)2 =99.02 mT
    t100mT * .9996tGWt=GW0(1-∇)t =100mT * .9996t
    ...... ... ... ...... ...... ... ... ...... ... ...
    1250100mT * .99961250GW1250=
    ...... ... ... ...... ... ... ... ... ... ... ... ...
    1730100mT * .99961730GW1730 =
    =50.05 mT
    Ship's GW decrements by percent not by absolute weight.

    • Departure leg, the flight from our home sun, Sol, to a neighboring stellar system, will also consist of three phases: acceleration, cruise and deceleration. However, only the acceleration and deceleration phases use propulsive force and need to be included in range considerations.
    • Vessel's Range for Powered flight can be divided into four equal parts. Considering the acceleration time equals deceleration time for each leg, and each leg's time equals the other; max acceleration time equals .25 times the vessel's range.
    • Original Fuel Load: Thought Experiment now assumes no enroute refueling; thus, original fuel load, F0, at start of flight is all the fuel available to accomplish all the g-force travel for departure leg and return leg Thus, fuel consumption of F0 gives us our range.

    Range = TripTtl = DeptTtl + RetTtl

    • Requisite Flight Profile. We further assume time to travel departure leg is same as time for return leg:

      DeptTtl = RetTtl

    • Flight profile times can be further described:
    • Departure Leg's powered flight can be divided into two equal durations

    • DeptAcc: During this duration, spaceship attains an enormous velocity, vMax.

    • DeptDec: To accomplish the required slowdown and maintain g-force from midpoint to destination, spaceship decelerates by reversing propulsion vector for same duration as for acceleration. Thus:
    • DeptDec = DeptAcc = tAcc

    • Equal Halves: Consider it axiomatic for total departure propulsion time:
    Departure Powered Flight Time:
    tP-Dep = tAcc + tDec = 2 * tAcc


    Exponentials can readily approximate fuel consumption. Any inexpensive scientific calculator has an exponent function; thus, we can readily calculate numerous guesses as to how many days is required to reduce ship's original gross weight (GW0) by 50%.

    Consider that as fuel is consumed, GW decreases; thus, a lesser GW needs less fuel to propel it, and this cycle continues throughout powered flight.
    Thus, fuel consumption decreases throughout flight even though the GWDay percentage remains the same for consistent propulsion throughout powered flight.
    Recall that TE previously guessed a vessel range of 1,250 days of powered flight (because 50% divided by .04% gives us 1,250). However, a quick exponential calculation shows that 1,250 days of .04% consumption reduces ship's gross weight to only about 60% of ship's orginal gross weight (GW0).
    .99961,250 = .6065 = 60.65%

    One way of detemining better value for vessel range is try several more guesses with different values of t:
    .9996t = 50% GW0
    After trying 1800, 1700, 1750, 1720 ....;
    we eventually determine t integer which produces a value much closer to 50%.
    .99961730 = .5005

    Thus, we determine that at .04% GW consumption per day; 1,730 days takes us very close to 50% of original mass. Thus, an initial fuel load of TOGW = 50% and a ∇ of .04%/day gives our notional spaceship a range of about 1,700 days of powered flight.

    • Return leg: the flight from the destination back to Sol, can have a range of values which don't necessarily equal the phase values of the departure leg. However, for convenience, this chapter assumes corresponding phases to have equal values. Thus, the acceleration time of the Departure leg equals the deceleration time of the return leg and so on.

      • Return Leg's powered flight also has two equal durations:
        • RetAcc: Accelerating back from destination to maximum velocity at the beginning of the Return Leg's cruise phase.
        • Return time can differ from Departure time, but for convenience, TE assumes the two durations to be equal.
        • RetDec: From cruise phase, decelerate from max velocity back to trip's original departure point.
    • Return Fuel: from above we know fuel requirement for return leg:

    Return Powered Flight Time:
    tP-Ret = tAcc + tDec = 2 * tAcc

    • Acceleration Time, tAcc We've already determined range based on original fuel load, F0, (possible range in time (days) from %TOGW and GWFin; see above table).
      • Based on above, TE assumes all acceleration/deceleration durations are equal:

        tAcc = DeptAcc = DeptDec = RetAcc = RetDec

        4 tAcc = DeptAcc + DeptDec + RetAcc + RetDec

      • Thus, we can divide vessel's range (a.k.a. propulsion time, tProp) by four to determine:
        tAcc = tProp / 4

    There might be a more graceful way to determine this range.

    A more convenient method might use logarithms.
    A quick review follows:

    log10a = a

    For example,

    log(100) = log(102) = 2

    What if at = b, where a and b are known, but t is unknown? Exponential characteristics allow us to accomplish following:

    t * log(a)
    log(b) / log(a)

    For present case, use following values:

    .9996t =
    =log .5000 / log .9996
    -.301/ -0.0001745

    Thus, logarithms can quickly approximate range for different Fuel Loads.
    This is shown in Table-1.

    Original Fuel (F0)
    is fuel mass needed for powered spacecraft flight. This includes acceleration & deceleration phases (but not cruise phase).

    Since spare fuel is not planned to burn, TE considers it part of the ship's infrastructure and not part of F0.

    F0 can be expressed as a percentage of original gross weight (GW0).
    Gross Weight (GWt)
    is ship's gross weight (GW) after t days of powered flight.

    Original Gross Weight (GW0)
    is ship's initial gross weight (zero days of powered flight). It includes ship structure, payload, passengers, crew and original fuel, F0.
    Final Gross Weight (GW)
    is ship's gross weight (GW) after tp days of powered flight. (It is original GW less original fuel.)
    Vessel Range (tp)
    is total number of days of powered flight with original fuel load, F0. (Vessel Range could also be called powered flight time.)
    Time (tp)
    Planned days of g-force powered flight time includes durations for acceleration and deceleration but not cruise time.

    TABLE-1: Fuel Loads


    Half Life

    Term "half life" means a duration for which a specified shrinking substance decreases by 50%. Shrinking could be due to radioactivity like thorium or uranium; or it could be due to fuel consumption like automobiles, aircraft or g-force spacecraft.

    For example, radioactive thorium might decrease by 50% in 60 seconds.

    This could be described by Mtho(1-Δ)t = .5 * Mtho where Δ is % difference in mass per second and the t is time in seconds.

    Divide both sides by Mtho and set t = 60 sec; we get:

    (1-Δ)60 = .5

    Percent Initial Gross Weight (%GW0) is the amount of fuel required for entire mission. This value is expressed as a percentage of the vessels' initial gross weight, GW0.

    %GW0= F0/GW0

    Since current particle accelerators now accelerate particles to faster then .99c, it's certainly possible that we'll eventually design propulsion systems to accelerate particles that fast and even faster. We've previously shown that faster exhaust particles lead to much slower fuel consumption and consequently much greater range.

    TABLE-2. Exhaust Particle Speeds.

    This greatly extends range of g-force spacecraft.

    RelativisticGrowthDecimal cDailyDiff.










    67.33 =.39c
    6,337=.10 LY
    .40 LY
    84.86 =.49c
    11,224=.18 LY
    .72 LY
    99.28 =.57c
    17,038=.27 LY
    1.08 LY
    23,573=.37 LY
    1.48 LY
    30,905=.49 LY
    1.96 LY
    38,691=.62 LY
    2.48 LY

    Δ = .2826%
    ct +vMax

    LY = 63,241 AU
    4 * dAcc


    • mr is relativity growth factor computed from Lorentz Transform.
    • dc is decimal component of particle exhaust speed when expressed as decimal light speed.
    • ∇, "daily difference", is the amount of fuel needed for each day's propulsion, expressed as a percentage of ship's gross weight (%GW).
    • Δ , "daily delta", is percentage of light speed achieved after first day of g-force flight.
      Δ= 86,400 g / c = .2826%
    • %GWFin is the final gross weight of the spacecraft after the flight's required fuel (%TOGW) is consumed. %GWFin is expressed as a percent of orginal ship's mass and can be computed: 100% -%TOGW
    • tP is total powered flight time; this is also known as the "range".
    • tAcc is acceleration time.
    • vMaxis velocity achieved after g-force acceleration for tAcc days. c = 173.15 AU/day.
    • dAcc is acceleration distance. 1 light year (LY) = 63,241 AU.
    • dP is total distance for powered portion of flight for both destination and return flights. This includes distances for acceleration and deceleration (but not cruise).
      Powered Destination Distance = dAcc + dDec
      Powered Return Distance = dAcc + dDec
      TE assumes same propulsion distances for both destination and return flights. For consistent g-force, TE also assumes acceleration distance = deceleration distance; thus:
      dP =Destination (dAcc + dDec )+ Return (dAcc + dDec )= 4 * dAcc

    • TE assumes ever improving technology will lead to higher performance, higher efficiency g-force propulsion systems such that particle exhaust speed will keep increasings.
    • Ever increasing particle speed and system efficiencies will lead to ever shrinking ∇ s, daily increment of %GW needed to maintain g-force.
    • Leads to greater vessel range; i.e., available propulsion time, tP.
    • Half Life Method
      Radioactive Half LifeGW Remainder
      As duration increases, radioactivity decreases remaining mass of certain elements.As number of days of g-force flight increase, fuel consumption decreases ship's mass.
      ...easily computed with logarithms....easily computed with logarithms.

    • Portion of tP actually used for propulsion; that portion divided by 4 to determine acceleration time, tAcc .
    • Recall tAcc is g-force duration from beginning of flight to start of cruise phase. The longer the initial g-force duration, the higher velocity achieved, the shorter the cruise phase.
    • At completion of curise phase, an equal duration is required to decelerate to destination.
    • These three phases compose the departure leg (acceleration, cruise, deceleration); only the acceleration and deceleration phases required g-force power.
    • Similarly, return leg requires same three phases, and TE assumes their durations equal corresponding departure phase.
    • Thus, there are four g-force durations, each of them = initial tAcc.


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