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

Determine proper independent variable for subsequent tables.

VExh km/sec dec. 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.	VExh dec. c km/sec 0.10c 30,000 0.20c 60,000 0.30c 90,000 0.40c 120,000 0.50c 150,000 0.60c 180,000 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.

Fuel Particle Size

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 ff_sec

• Recall that ff_sec is original fuel flow per second.

ff_sec 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 ff_sec varies throughout the flight.
In a general, theoretical manner, ff_sec can be determined as followed:

ffsec = 
g * MShip 
c * √(1 - n-2)

In a more practical way, realtime ff_sec 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.

VExh n * ffSec d km/sec n 0.10c 30,000 1.005 0.20c 60,000 1.021 0.30c 90,000 1.048 0.40c 120,000 1.091 0.50c 150,000 1.155 0.60c 180,000 1.250 dec. c 1 √(1 - d2) Per Lorentz Transform, fuel particle size strictly depends on fuel particle speed.	n * ffSec n d km/sec 1.1 0.42c 124,979 1.2 0.55c 165,831 1.3 0.64c 191,691 1.4 0.70c 209,956 1.5 0.75c 223,607 1.6 0.78c 234,187 1.7 0.81c 242,607 I.V. √(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.

%Ship Mass Converts to Energy

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.

ff_sec is fuel flow for one sec.

ff_i is daily fuel flow for ith day (= 86,400 * ff_sec)

GW = Fuel + Payload + Infrastructure

If GW₀ is initial gross weight,

then GW after first day of flight: GW₁=GW₀ - ff₁

---

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

∇ = 
ff1
GW1
=
ff2
GW2
=
ffi
GWi

∇ 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.

n * ffSec	dec. c	Exhaust	Daily Fuel
n	d	km/sec	∇
1.1	0.42c	124,979	0.63%
1.2	0.55c	165,831	0.43%
1.3	0.64c	191,691	0.35%
1.4	0.70c	209,956	0.29%
1.5	0.75c	223,607	0.26%
1.6	0.78c	234,187	0.23%
1.7	0.81c	242,607	0.21%
I.V.	√(1 - n-2)	d * 3x105km/sec	g * 86,400 c * √(n2 - 1)

Propulsion Time, t_p

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.

GW_i = (1- 2∇)ⁱ

Note that i = days of flight.

Find i such that

GW_i = 50% TOGW.

i = tp = 
log .5 
log(1-2∇) 

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

Range_Prac= log %TOGW /log(1-ε∇)

Instead of one ∇ to power spaceship for a day, it actually takes ε∇.

n * ffSec
n	d	km/sec	Δ	tp
1.1	0.42c	124,979	0.63%	54.8
1.2	0.55c	165,831	0.43%	79.5
1.3	0.64c	191,691	0.35%	99.6
1.4	0.70c	209,956	0.29%	117.6
1.5	0.75c	223,607	0.26%	134.2
1.6	0.78c	234,187	0.23%	150.0
1.7	0.81c	242,607	0.21%	165.1
I.V.	√(1 - n-2)	d * 3x105km/sec	.00288 √(n2-1)	log(%TOGW) log(1 - ε∇)

Interplanetary

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

Explanation

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.

Determine Range					Final	Cumm.	Examples
n * ffSec					Velocity	Distance	Dest.	Dist.
n	d	km/sec	∇	tp	AU/Day	AU		AU
1.01	0.14	42,111	2.03%	16.7	8.4 AU/day	69.8 AU	NEA	1 AU
1.02	0.20	59,117	1.43%	23.8	11.9 AU/day	142.1 AU	Mars	2 AU
1.03	0.24	71,877	1.17%	29.3	14.7 AU/day	215.3 AU	Asteroids	3 AU
1.04	0.27	82,401	1.01%	34.0	17.0 AU/day	289.5 AU	Jupiter	5 AU
1.05	0.30	91,473	0.90%	38.2	19.1 AU/day	364.4 AU	Saturn	10 AU
1.06	0.33	99,500	0.82%	42.0	21.0 AU/day	440.1 AU	Uranus	20 AU
1.07	0.36	106,726	0.76%	45.5	22.7 AU/day	516.7 AU	Neptune	30 AU
1.08	0.38	113,312	0.71%	48.7	24.4 AU/day	593.9 AU	Near Kuiper	40 AU
1.09	0.40	119,368	0.66%	51.8	25.9 AU/day	671.9 AU	Mid Kuiper	70 AU
1.10	0.42	124,979	0.63%	54.8	27.4 AU/day	750.7 AU	Far Kuiper	100 AU
I.V.	√(1 - n-2)	d * 3x105km/sec	.00288 √(n2-1)	log(50% TOGW) log(1 - 2∇)	g * t t = tp	g * t2/2 t = tp

Exoplanetary

Δ = .288% c /day
c = 172.8 AU/day

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, t_p. For remaining half of t_p, 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.

Determine Range				Acceleration Phase			Deceleration Phase
n * ffSec				Time	Distance	Velocity	Velocity	Distance	Time
n	Dec. c	∇	tp	tAcc	dAcc	vAcc	vDec	dDec	tDec
1.1	0.417	0.628%	54.8 Days	27.4 Days	182 AU	13.1 AU/Day	8% c	0.003 LY	0.08 Yr
1.2	0.553	0.434%	79.5 Days	39.7 Days	379 AU	18.7 AU/Day	11% c	0.006 LY	0.11 Yr
1.3	0.639	0.347%	99.6 Days	49.8 Days	590 AU	23.1 AU/Day	13% c	0.009 LY	0.14 Yr
1.4	0.700	0.294%	117.6 Days	58.8 Days	814 AU	26.9 AU/Day	16% c	0.013 LY	0.16 Yr
1.5	0.745	0.258%	134.2 Days	67.1 Days	1,053 AU	30.4 AU/Day	18% c	0.017 LY	0.18 Yr
1.6	0.781	0.231%	150.0 Days	75.0 Days	1,305 AU	33.6 AU/Day	19% c	0.021 LY	0.21 Yr
1.7	0.809	0.209%	165.1 Days	82.5 Days	1,571 AU	36.6 AU/Day	21% c	0.025 LY	0.23 Yr
1.8	0.831	0.192%	179.8 Days	89.9 Days	1,850 AU	39.5 AU/Day	23% c	0.029 LY	0.25 Yr
1.9	0.850	0.178%	194.1 Days	97.0 Days	2,142 AU	42.2 AU/Day	24% c	0.03 LY	0.27 Yr
2.0	0.866	0.166%	208.1 Days	104.0 Days	2,447 AU	44.8 AU/Day	26% c	0.04 LY	0.28 Yr
I.V.	√(1 - n-2)	.00288 √(n2-1)	log(%TOGW) log(1 - ε∇) %TOGW=.5 ε = 2.0	tp 2	ct- c[(1-Δ)t - 1] ln(1-Δ) t = tAcc c = 172.8 AU/day	c-c(1-Δ)t t = tAcc c = 172.8 AU/day	vDec = vAcc Different units; same value.	dDec=dAcc Different units; same value.	tDec = tAcc Different units; same value.

Interstellar

c = 172.8 AU/day

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.)

Considerations:

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.

Determine Range			Acceleration Phase			Cruise Phase		Deceleration Phase
n * ffSec			Time	Distance	Velocity	Distance	Time	Velocity	Distance	Time
n	∇	tp	tAcc	dAcc	vAcc	dCru	tCru	vDec	dDec	tDec
2.0	0.166%	208.1 Days	104.0 Days	2,447 AU	44.8 AU/Day	3.92 LY	15.13 Yr	26% c	0.04 LY	0.3 Yr
3.0	0.102%	340.0 Days	170.0 Days	6,157 AU	67.0 AU/Day	3.80 LY	9.82 Yr	39% c	0.10 LY	0.5 Yr
4.0	0.074%	465.7 Days	232.9 Days	10,934 AU	84.5 AU/Day	3.65 LY	7.47 Yr	49% c	0.17 LY	0.6 Yr
5.0	0.059%	589.2 Days	294.6 Days	16,609 AU	98.9 AU/Day	3.47 LY	6.07 Yr	57% c	0.26 LY	0.8 Yr
6.0	0.049%	711.6 Days	355.8 Days	23,039 AU	110.9 AU/Day	3.27 LY	5.10 Yr	64% c	0.37 LY	1.0 Yr
7.0	0.042%	833.4 Days	416.7 Days	30,104 AU	120.8 AU/Day	3.05 LY	4.36 Yr	70% c	0.48 LY	1.1 Yr
8.0	0.036%	954.8 Days	477.4 Days	37,701 AU	129.2 AU/Day	2.81 LY	3.75 Yr	75% c	0.60 LY	1.3 Yr
9.0	0.032%	1076.0 Days	538.0 Days	45,747 AU	136.2 AU/Day	2.55 LY	3.24 Yr	79% c	0.72 LY	1.5 Yr
10.0	0.029%	1197.0 Days	598.5 Days	54,170 AU	142.0 AU/Day	2.28 LY	2.78 Yr	82% c	0.86 LY	1.6 Yr
11.0	0.026%	1317.9 Days	658.9 Days	62,909 AU	147.0 AU/Day	2.01 LY	2.36 Yr	85% c	1.00 LY	1.8 Yr
12.0	0.024%	1438.7 Days	719.3 Days	71,914 AU	151.1 AU/Day	1.72 LY	1.97 Yr	87% c	1.14 LY	2.0 Yr
I.V.	.00288 √(n2-1)	log(%TOGW) log(1 - ε∇) %TOGW=.5 ε = 2.0	tp 2	ct- c[(1-Δ)t - 1] ln(1-Δ) t = tAcc c = 172.8 AU/day	c-c(1-Δ)t t = tAcc c = 172.8 AU/day	4LY - dAcc - dDec	dCru vDec	vDec=vAcc Different units; same value.	dDec=dAcc Different units; same value.	tDec=tAcc Different units; same value.