Saturday, June 06, 2009

Ref Mat'l: Review Rate and Range

PURPOSE: Discuss why thought experiment assumes spacecraft range relates to exhaust speed of fuel particles.

First discuss how we can most conveniently express these values.

Assume Fuel Particle Speed

km/secdec. c
10,000 0.03c
20,000 0.07c
30,000 0.10c
40,000 0.13c
50,000 0.17c
60,000 0.20c

Intuitively, take a nice round number (i.e., 10,000 kms/sec) and use as increment. This becomes the Independent Variable (IV). Above table compares absolute speed with decimal light speed.

dec. ckm/sec

Alternatively, we could independently vary speed as decimal light speed values. An inherent benefit, decimal c implicitly increases precision, because c by definition is exact speed of light. On the other hand, c = 300 million m/sec is a gross approximation.

Determine proper independent variable for subsequent tables.

Fuel Particle Size

VExhn * ffSec
dec. c
√(1 - d2)

Per Lorentz Transform, fuel particle size strictly depends on fuel particle speed.

n * ffSec
n dkm/sec
√(1 - n-2)
d * 3x108m/sec

We could even make fuel particle size the IV. In above table, n is multiple of original particle size at rest.

Per Einstein's Theory of Relativity, particle size maps to particle speed. For example, if a particle increases its speed to 60% c, it increases mass to 1.25 times size of original mass (at rest). Thus, if we use n as the IV, a value of 1.25 gives us two useful facts:

  1. Fuel particle mass increases 25%.
  2. Fuel particle speed increases to .6 c (due to LT mapping).
  • Momentum of fuel particle becomes a product of the two

.6 c * 1.25 ffsec

  • Recall that ffsec is original fuel flow per second.

ffsec is strictly defined as the amount of fuel required to increase ship's speed another 10m/sec for that one second of flight. Since this increase is exactly the acceleration experienced by objects in free fall near Earth's surface, it stands to reason this acceleration produces a propulsive force on the ship's contents such that near Earth gravity is simulated. (Recall Einstein's thought experiment.)

Since this consumes a slight amount of fuel, ship's weight decreases slightly and subsequent seconds will require slightly less fuel flow. Thus ffsec varies throughout the flight.
In a general, theoretical manner, ffsec can be determined as followed:

ffsec = g * MShip
c * √(1 - n-2)

In a more practical way, realtime ffsec will likely be closely controlled by sophisticated servos which monitor perceived "weight" of control objects and subsequently adjust fuel flow. Thus, fuel flow will constantly adjust in accordance with how closely propulsion force resembles near Earth gravity.

%Ship Mass Converts to Energy

n * ffSecdec. cExhaustDaily Fuel
n dkm/sec
√(1 - n-2)
d * 3x105km/sec
g * 86,400
c * √(n2 - 1)

Explanation of Terms

g : acceleration due to near Earth gravity, thought experiment rounds this value to 10 meters per second per second (10 m/sec2)

86,400: number of seconds per day (= 3,600 sec/hr * 24 hr/day)

c: speed of light, thought experiment rounds this value to 300 million meters per second.

n: measure growth of fuel particles due to relativistic speed.

∇: Daily difference of ship's gross weight (GW) due to fuel particles exiting propulsion system.

ffsec is fuel flow for one sec.
ffi is daily fuel flow for ith day (= 86,400 * ffsec)
GW = Fuel + Payload + Infrastructure
If GW0 is initial gross weight,
then GW after first day of flight: GW1=GW0 - ff1

ffi decreases daily because it takes less energy to move lesser GW. However, the ratio of daily fuel flow (ffi) to daily gross weight (GWi) should be consistent.

∇ = ff1

∇ can be expressed as a percentage as well as a ratio.
Example: ratio of 1:100 = 1%

Thus, the daily mass of fuel consumed is ever decreasing, but the ratio of daily fuel consumed vs. current ship's GW stays the same if the ship's acceleration remains constant.

Propulsion Time, tp

n * ffSec
n dkm/secΔtp
√(1 - n-2)
d * 3x105km/sec
log(1 - ε∇)

Explanation of Terms

%TOGW: Percent Take Off Gross Weight meaning the percentage of ship's initial mass which needs to be devoted to fuel for entire voyage. One way of computing could be by summing up every day's daily fuel for total days of planned flight. Another way could be by determining how much weight to be devoted to fuel, then determing range. For example, if designers decide to allocate 50% TOGW to fuel; then, we must determine "How many days of powered flight are available?". That's what above table does.

∇: Daily Difference, daily percentage of ship's mass which must convert to kinetic energy to propel ship at g-force acceleration. Previous table discussed this in detail.

ε: Efficiency coefficient. Efficiency will be far from perfect; thus, we must adjust the ∇ value to reflect fuel consumption for energy in addition to the fuel particles which actually exit via propulsion system.

Spaceship must actually consume more then one ∇ per day because additional fuel will be needed for additional energy requirements (see below). For convenience, assume an additional ∇ will accommodate additional energy requirements. If true, then mass quantity 2∇ will accomodate both propulsion and additional requirements. Actual spaceflight experience will determine down the exact amount. Of course, ship designers will work to lower the ε coefficient.

Examples of non-propulsion energy requirements:

  • --Peripheral needs: life support, comm, nav, enter, etc.
  • --Design flaws.
    --Propulsion Auxiliary Power Needs. Acceleration magnets need power.
    --Margin. Good judgment would plan for some spare fuel to remain.
GWi = (1- 2∇)i

Note that i = days of flight.

Find i such that

GWi = 50% TOGW.

i = tp = log .5

If practicality considers these efficiency factors; then, above "i" can be the definition of "Practical Range"

RangePrac= log %TOGW /log(1-ε∇)
Instead of one ∇ to power spaceship for a day, it actually takes ε∇.


10 m/sec = g = .5 AU/day

Determine Range

n * ffSecVelocityDistanceDest.Dist.
n dkm/sectpAU/DayAUAU
1.010.1442,1112.03%16.78.4 AU/day69.8 AUNEA1 AU
1.020.2059,1171.43%23.811.9 AU/day142.1 AUMars2 AU
1.030.2471,8771.17%29.314.7 AU/day215.3 AUAsteroids3 AU
1.040.2782,4011.01%34.017.0 AU/day289.5 AUJupiter5 AU
1.050.3091,4730.90%38.219.1 AU/day364.4 AUSaturn10 AU
1.060.3399,5000.82%42.021.0 AU/day440.1 AUUranus20 AU
1.070.36106,7260.76%45.522.7 AU/day516.7 AUNeptune30 AU
1.080.38113,3120.71%48.724.4 AU/day593.9 AUNear Kuiper40 AU
1.090.40119,3680.66%51.825.9 AU/day671.9 AUMid Kuiper70 AU
1.100.42124,9790.63%54.827.4 AU/day750.7 AUFar Kuiper100 AU
√(1 - n-2)
d * 3x105km/sec
log(50% TOGW)
log(1 - 2∇)
g * t

t = tp

g * t2/2

t = tp


Even the smallest "n" value (1/100 increase in mass) gives the ship more then 16 days of propulsion time (given 50% of ship's initial mass used as fuel). 16.7 days of g-force acceleration enables the ship to attain a final velocity of 8.4 AU/day (approx 5% c, speed of light) for a total distance of almost 70 AU.

Comparing this distance to a typical Near Earth Object (NEO), Mars, or even Jupiter, we see that a relatively low performance from our notional particle accelerators can offer tremendous benefits for traveling between nearby planets.

Since this tables propulsion times are fairly short, we're using the Newtonian formulas to approximate final velocities and cummulative distances. Following tables (Exoplanetary and Interstellar) use much longer propulsion times; thus, they use Einsteinian formulas.


Δ = .288% c /day
c = 172.8 AU/day

Determine Range

Acceleration Phase

Deceleration Phase

n * ffSecTimeDistanceVelocityVelocityDistanceTime
n Dec. ctptAccdAccvAccvDecdDectDec
1.10.4170.628%54.8 Days27.4 Days182 AU13.1 AU/Day8% c0.003 LY0.08 Yr
1.20.5530.434%79.5 Days39.7 Days379 AU18.7 AU/Day11% c0.006 LY0.11 Yr
1.30.6390.347%99.6 Days49.8 Days590 AU23.1 AU/Day13% c0.009 LY0.14 Yr
1.40.7000.294%117.6 Days58.8 Days814 AU26.9 AU/Day16% c0.013 LY0.16 Yr
1.50.7450.258%134.2 Days67.1 Days1,053 AU30.4 AU/Day18% c0.017 LY0.18 Yr
1.60.7810.231%150.0 Days75.0 Days1,305 AU33.6 AU/Day19% c0.021 LY0.21 Yr
1.70.8090.209%165.1 Days82.5 Days1,571 AU36.6 AU/Day21% c0.025 LY0.23 Yr
1.80.8310.192%179.8 Days89.9 Days1,850 AU39.5 AU/Day23% c0.029 LY0.25 Yr
1.90.8500.178%194.1 Days97.0 Days2,142 AU42.2 AU/Day24% c0.03 LY0.27 Yr
2.00.8660.166%208.1 Days104.0 Days2,447 AU44.8 AU/Day26% c0.04 LY0.28 Yr
√(1 - n-2)
log(1 - ε∇)

ε = 2.0
ct-c[(1-Δ)t - 1]

t = tAcc
c = 172.8 AU/day
t = tAcc
c = 172.8 AU/day
vDec = vAcc
Different units; same value.
Different units; same value.
tDec = tAcc
Different units; same value.

Above distances are well past the planets but well short of the Oort Cloud (approx 1 LY away).

Above table's profile shows total fuel load as 1/2 ship's mass. This fuel is expended such that for half the time, fuel particles propel craft to increase ship's speed for half the propulsion time, tp. For remaining half of tp, fuel particles propel craft in opposite direction to decelerate craft. At destination, crew needs to refuel from nearby accessible sources (presumably comets), then accomplish same profile to return.


c = 172.8 AU/day

Determine Range

Acceleration Phase

Cruise Phase

Deceleration Phase

n * ffSecTimeDistanceVelocityDistanceTimeVelocityDistanceTime
n tptAccdAccvAccdCrutCruvDecdDectDec
2.00.166%208.1 Days104.0 Days2,447 AU44.8 AU/Day3.92 LY15.13 Yr26% c0.04 LY0.3 Yr
3.00.102%340.0 Days170.0 Days6,157 AU67.0 AU/Day3.80 LY9.82 Yr39% c0.10 LY0.5 Yr
4.00.074%465.7 Days232.9 Days10,934 AU84.5 AU/Day3.65 LY7.47 Yr49% c0.17 LY0.6 Yr
5.00.059%589.2 Days294.6 Days16,609 AU98.9 AU/Day3.47 LY6.07 Yr57% c0.26 LY0.8 Yr
6.00.049%711.6 Days355.8 Days23,039 AU110.9 AU/Day3.27 LY5.10 Yr64% c0.37 LY1.0 Yr
7.00.042%833.4 Days416.7 Days30,104 AU120.8 AU/Day3.05 LY4.36 Yr70% c0.48 LY1.1 Yr
8.00.036%954.8 Days477.4 Days37,701 AU129.2 AU/Day2.81 LY3.75 Yr75% c0.60 LY1.3 Yr
9.00.032%1076.0 Days538.0 Days45,747 AU136.2 AU/Day2.55 LY3.24 Yr79% c0.72 LY1.5 Yr
10.00.029%1197.0 Days598.5 Days54,170 AU142.0 AU/Day2.28 LY2.78 Yr82% c0.86 LY1.6 Yr
11.00.026%1317.9 Days658.9 Days62,909 AU147.0 AU/Day2.01 LY2.36 Yr85% c1.00 LY1.8 Yr
12.00.024%1438.7 Days719.3 Days71,914 AU151.1 AU/Day1.72 LY1.97 Yr87% c1.14 LY2.0 Yr
log(1 - ε∇)

ε = 2.0
ct-c[(1-Δ)t - 1]

t = tAcc
c = 172.8 AU/day
t = tAcc
c = 172.8 AU/day
4LY - dAcc - dDec

Different units;
same value.

Different units; same value.


Different units; same value.

Explain Cruise Phase. Thot experiment assumes impractical to accelerate continuously to interstellar destination. However, it might be practical to accelerate starship to a significant portion of c, then cruise for a few years, then slow down to essentially zero speed at destination stellar system. For example, if we assume the Alpha Centauri system to be 4 LYs away, then Earth bound systems could observe the starship accelerating for a year to attain .64c; cruising for about 5 years; then, decelerating for another year. Of course, these times would be the times observed by Earth bound observer, and total time would be 7 years. (Time dilation would make the times shorter for the starship occupants.)


1. How does ship simulate gravity for the 5 year cruise phase.
A. Rotate the vessel at a predetermined angular velocity; resultant centrifugal force will "press" objects against the outer hull to simulate near Earth gravity. By the time, humans accomplish interstellar flights, asteroidal habitats will have simulated near Earth gravity for many decades if not centuries.

2. How does ship obtain required energy for the several years well away from Sol's "biosphere", volume of space where Solar energy is sufficient to support life. Thought experiment assumes biosphere distance to extend to perhaps Mars. By the time, we do interstellar, asteroidal habitats will have long since been using He-3 (mined from the "gas giants": J, S, U, N). Thought experiment assumes that He-3 will also work to provide same time of energy needs for starship during Cruise Phase. It's possible that propulsion system might be able to "bleed off" some power for these "peripheral needs" during powered flight during the acceleration and deceleration phases.


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