Vol 1. Tables, A Summary
Recall momentum exchange: M * V = m * v
Apply to fast moving fuel particles and a slow moving spaceship: Mship * Vship = mfuel * vfuel
Rearrange to solve for spaceship size: Mship = mfuel * vfuel / Vship
Per above text, give terms following values:


 1,000,000 gms = 1,000 kgms = 1.0 metric Tonnes = 1 mT 
Reconsider: mfuel * vfuel = Mship * Vship. Divide both sides of original equation by one sec. (mfuel * vfuel)/1sec = (Mship * Vship)/1sec
Rearrange: (mfuel /1sec) * vfuel = Mship * (Vship/1sec)
Recall that we set Vship equal to 10 m/s; thus, Vship/ 1sec = 10m/s/s = g, acceleration due to gravity near Earth’s surface.
We introduce new term, ff (fuel flow), which we define as mfuel/1sec.
Thus, we’ll set ff equal to 1.0 gm/sec.
ff * vfuel = Mship * g
To continue expressing vfuel as decimal c and Mship in metric Tonnes, we'll continue use of conversion constant, k1 = 300/c.
This gives equation at top of table: Mship = 300/c (ff * vfuel / g)
Table2. Lorentz Transform.
To quantify mass increases due to relativistic (near light speed) velocities of exhaust particles, use the Lorentz Transform (LT). This mass increase also contributes to the momentum exchange which propels the ship.
 Original fuel flow (ff). Convert one gram of matter to plasma every second; accelerate all those particles to near light speed; and expel them from the spacecraft. (See T1. Momentum Exchange.)
 Relativistic fuel flow (ffLT). Use LT to adjust original fuel flow to account for relativistic mass increase. (See following table.)

Transition from m_{r} equation to ff_{LT} equation
Terms in m_{r} equation
 m_{o}, original mass
 c, light speed, is a constant.
 v, velocity achieved by particles at time of exhaust from spaceship.
 m_{r}, relativistic mass, which increases as v increases.
Rearrange Some Expressions
 Express v as decimal c, then
m_{r} = m_{o} /SQRT(1v^{2}/ c^{2})
reduces to
m_{r} = m_{o}/SQRT(1v^{2}_{ratio}).  Divide both sides by one second.
m_{r}/1sec = (m_{o}/1sec) /SQRT(1v^{2}_{ fuel}).
Terms in m_{r} equation
 m_{o}/sec becomes ff (fuel flow), grams per second. This is used to determine fuel consumption.
 m_{r}/sec becomes ff_{LT}, fuel flow's mass increased by relativistic velocity. This is used to determine momentum exchange.
Table3. Percent Take Off Gross Weight (%TOGW).
 
 
Note: Determine equation for %TOGW which strictly depends on vfuel and not directly on Mship Table4. Spaceship Range (in days of flight).
Academically, it makes senses to see how much ship can be be propelled by one gram of particles ejected at very high speeds. Above tables show that one gram at .1 c can propel 3 mT, one gram at .2 c can propel slightly more then 6 mT, and so on. That’s an impressive fuel/mass ratio, but note following two things. 3 metric Tonnes is about the mass of my Ford Explorer; our spaceship will have to be much bigger. If we increased fuel consumption to 1 kgm per second, that could propel 3,000 mT to accelerate another 10 m/s during one second. 3,000 mT is much better then just 3 mT, but it still may not be enough to launch a scientific expedition. Furthermore, using fuel flow to determine ship size is really putting cart before the horse. It would be more appropriate to use ship size determine required fuel flow. Therefore, let’s readjust terms so that we keep ship size constant (though substantial) and determine required fuel flows needed at incremental propulsion particle speeds. Thus, we’ll arbitrarily pick a nice round number, 100,000 mT as the mass of our ship, determine required fuel flow as we uniformly increase particle propulsion speed as shown by the table 
g  ff  v_{fuel}  M_{ship}  ff_{day}  %TOGW_{day}  ***Range 
con  DV  IV  con  DV  DV  DV 
m/s/s  kgm/sec  dec c  mT  mT/day  %/day  days burn 
10  33.2  0.1  100,000  2,866  2.87%  8.7 
10  16.3  0.2  100,000  1,411  1.41%  17.7 
10  10.6  0.3  100,000  916  0.92%  27.3 
10  7.6  0.4  100,000  660  0.66%  37.9 
10  5.8  0.5  100,000  499  0.50%  50.1 
10  4.4  0.6  100,000  384  0.38%  65.1 
10  3.4  0.7  100,000  294  0.29%  85.1 
10  2.5  0.8  100,000  216  0.22%  115.7 
***Assumptions about range of spaceship  
Cap_{fuel}  Capacity of fuel likely limited to one half mass of M_{ship}  
Eff_{fuel}  Efficiency of fuel is 50% due to auxilliary energy requirements.  
Disregard  Exponential function which considers declining fuel consumption due to decreasing mass of M_{ship.} 
 Force of gravity will accelerate a free falling object (near Earth's surface) 10 m/s for every second of free fall. After one second, object is falling 10 m/s. After two seconds, object is falling 20 m/s. If spaceship self propels at force of gravity for one entire day (86,400 seconds), it will have achieved final velocity of 864,000 m/s. If spaceship (or falling object) starts with initial velocity of 0 m/s (zero) and maintains constant acceleration throughout duration, then average velocity equals one half of final velocity (vave = .5 vfin). Thus, after one day of gforce accelerated flight, our space ship has averaged 432,000 m/s. Distance traveled equals time times average velocity (d = t * vave). Thus, our spaceship has traveled a very large distance of 432,000 m/s times 86,400 seconds. 
If units of meters and seconds give us huge numbers inconvenient for long durations of acceleration, let's try days and AUs.
Meters to AUs. Accelerating for one second at g will move an object 5 meters, accelerating for one day at g will move an object how far?? Basic acceleration calculations give us about 37.5 billion meters (= 37.5 million kms) distance for one day of gforce acceleration. Since 1 AU equals 150 million kms, one day acceleration distance equals 1/4 AU. 
 We now have values t and d in days and AUs for one day's gforce acceleration. What about correspondings values for Vave, Vfin, and g? Vave = distance/time = d/t = .25 AU /1 day = .25 AU/dy Vfin = twice average speed = 2 * Vave = 2 * .25 AU/dy = .5 AU/dy g = velocity increase per duration = Vfin/t = (.5 AU/dy) / (1 day) = .5 AU/dy/dy 
Using the constant, g (= .5 AU/dy/dy), we can determine all inflight speeds, times, and distances with the variables: time and distance (t & d).
d = 0.5 * g * t**2 On the other hand, we can rearrange this equation to make distance the Independent Variable (IV) and time the Dependent Variable (DV). t = SQRT(2*d/g) 

Table6. Enroute to Mars: Distances and Flight Times.
Recall our thot experiment has spaceships which use particle accelerators to constantly accelerate throughout their spaceflights anywhere in our Solar System. Our notional spaceships can easily reach the orbit of Mars within a few days. The table shows several approximate distances which vary from .5 AU (absolute closest with Mars and Earth on same side of Sun) to 2.5 AU (Mars and Earth separated by Sun). (See orbit diagram.) 
Useful flights must adopt a flight profile as follows:
1. Accelerate from rest to the midpoint between departure and destination. Occupants feel simulated gravity due to force from constant acceleration; this enables comfort and optimal productivity.
2. Decelerate from midway to destination by using same gforce in opposite direction. I conjecture this would be achieved by turning the space vessel 180 degrees and pointing stream of accelerated particles toward the destination.
Examining table, we can see that only one half day of gforce acceleration would bring the ship to 432 km/sec, much higher then Mars's escape velocity of 5 km/sec (ship would have to be slight slower to orbit around Mars) .
Using acceleration/deceleration profile, we can then determine flight times for various distances.
Earth  Mars  g = 10 m/s/s = 0.5 AU/day/day  
mean distance from Sol  1  1.52  After 1 day, V_{fin} = 864 km/sec  
escape velocity (km/sec)  11.18  5.02  g = 0.5AU/dy / dy = 864km/sec / day  
Profile to Midpoint (Acceleration)  Final V at Midpoint  Profile fm Midpoint (Deceleration)  Totals  
Dist (AU)  Time (day)  Velocity (km/sec)  Dist (AU)  Time (day)  Total Dist (AU)  Total Time (Day) 
d_{1}  t_{1}  V_{Mid}  d_{2}  t_{2}  D  T 
0.25  0.50  432  0.25  0.50  0.5  1.00 
0.50  0.71  611  0.50  0.71  1.0  1.41 
0.75  0.87  748  0.75  0.87  1.5  1.73 
1.00  1.00  864  1.00  1.00  2.0  2.00 
1.25  1.12  966  1.25  1.12  2.5  2.24 
D / 2  SQRT(2*d_{1}/g)  864 * t_{1}  d_{2} = d_{1}  SQRT(2*d_{2}/g)  Given  t_{1} + t_{2} 
1. Even the shortest flight (.5 AU when Mars and Earth are at their closest) for a constant gforce accelerating spacecraft achieves an enormous speed at the midpoint of the flight. Since spacecraft needs the force from constant acceleration to simulate Earth gravity, then the spacecraft must reverse direction of propulsion flow of particles to apply same force in opposite direction. This would continue gravity conditions for second half of flight and would decelerate the spacecraft.
2. For example, a vessel would have to start from Earth at least as fast as the indicated escape velocity, 11.2 km/sec. After accelerating for one day, the ship will have achieved 864 km/sec, 77 times greater. To operate in the Martian planetary system, the spacecraft must reduce speed to about 5 km/sec, less then 1% of VMid . For convenience, next table will approximate departure and destination velocities as zero.
3. Finally, we can simply the table by changing relevant equations as follows:
D/2 = d1 = d2
T = t1 + t2 = SQRT(2d1/g) + SQRT(2d2/g)
T = SQRT(2 * D/2 /g) + SQRT(2 * D/2 /g) = 2 * SQRT(D/g)
Solar System Destination  Mars  Jupiter  Saturn  Uranus  Neptune  Kuiper Belt  
Typical Distance from Earth (AU)  1  5  10  20  30  40  
Flight Time (Days) = 2 * SQRT(D/g)  2.83  6.32  8.94  12.65  15.49  17.89 
Solar System Destination  Mars  Jupiter  Saturn  Uranus  Neptune  Kuiper Belt  
Typical Distance from Earth (AU)  1  5  10  20  30  40  
2 way Flight Time (Days) = 4 * SQRT(D/g) = t  5.66  12.65  17.89  25.30  30.98  35.78  
%TOGW_{Day}  **Range (Days)  v_{fuel} (dec. c)  %TOGW  %TOGW  %TOGW  %TOGW  %TOGW  %TOGW  
2.87%  8.7  0.10  16.21%  36.25%  51.26%  72.49%  88.79%  102.52%  
1.41%  17.7  0.20  7.98%  17.85%  25.24%  35.69%  43.72%  50.48%  
0.92%  27.3  0.30  5.18%  11.58%  16.38%  23.17%  28.37%  32.76%  
0.66%  37.9  0.40  3.73%  8.35%  11.80%  16.69%  20.45%  23.61%  
0.50%  50.1  0.50  2.82%  6.31%  8.92%  12.62%  15.46%  17.85%  
0.38%  65.1  0.60  2.17%  4.86%  6.87%  9.71%  11.90%  13.74%  
Increment  
g = 0.5 AU/dy/dy  **Range Limit1: Assume fuel is limited to one half mass of spaceship. This limits %TOGW to 50%.  
**Range Limit2: Assume fuel efficiency of 50%. This further limits %TOGW to 25%. 
 
Why Approximate?Approximate values are much easier to work with. 300 million m/s is commonly used for c, speed of light; so, we'll use that value. Similarly, 10 m/s^{2} is also commonly used for g, gravity near Earth; so, we'll use that as well. Reaccomplishing above equation with new values: M_{ship} = m_{fuel} * v_{fuel}/ V_{ship} M_{ship} = 1.0 gm * 30,000,000 m/s / 10 m/s M_{ship} = 3,000,000 gm NOTE: ratio of .1 c to g for one second (29,979,246 / 9.80665 )is closely approximated by 30,000,000 / 10 because 29.98 million is just under 30 million and 9.8 is just under 10. Even with approximation, 3 million gms and 30 million m/s both have a lot of zeros. For further convenience, use conversion constants to mix units.  

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