IntroduceFlight profile (transient)
Relativistic Mass Increase  

 Apply Lorentz Transform to determine how much exhaust particle grows from original (ff_{sec}).  
Particle Size Maps to Particle Speed  
We get a more pleasing picture if we list ff_{Exh} as integer multiples of ff_{sec}. Thus, we restate relationship between exhaust particle's speed vs. mass. 
 
Size & Speed to Ship Mass  
 New ship size (M_{ship}) column lists values as "megaff_{sec}". Thus, if fuel flow values are in grams, then M_{ship} would be metric Tonnes. Further, if ff_{sec} values are in kilograms, then M_{ship}values would be in kilo metric Tonnes (kmT)...  
Determine Daily Diff, Δ  
The daily portion of ship's mass that must convert to kinetic energy is a very small percentage. Furthermore, this percentage gets even smaller as exhaust particle speed increases. 
 Departure Leg  Return Leg  

ff_{Exh} (multiff_{sec} )  Practical Range (Rge_{Prac})  Phase I Time (t_{Half})  Phase II Time  One Way Time (t_{1way})  Phase III Time  Phase IV Time  Dest Dist. (d_{1way})  
* ff_{sec}  Days  Days  Days  Days  Days  Days  AU  
2  208  52  52  104  52  52  1,353  
3  340  85  85  170  85  85  3,613  
4  455  114  113  228  113  113  6,778  
5  589  147  147  295  147  147  10,848  
6  712  178  178  356  178  178  15,823  
7  833  208  208  417  208  208  21,704  
8  955  239  239  477  239  239  28,489  
9  1,076  269  269  538  269  269  36,180  
10  1,197  299  299  599  299  299  44,775  
11  1,318  329  329  659  329  329  54,276  
IV  DV  Rge_{Prac}/4  Rge_{Prac}/4  =t_{PhI}+t_{PhII}  Rge_{Prac}/4  Rge_{Prac}/4  g * t^{2}_{1way} 
 Departure Leg  Return Leg  

ff_{Exh}  Practical Range  Phase I  Phase II  One Way  Phase III  Phase IV  Two Way  
n*ff_{sec}  (Rge_{Prac})  t  d  t  d  t_{1way}  d_{1way}  d  t  d  t  (t_{2way})  
n  Days  Days  AUs  Days  AUs  Days  AUs  Days  AUs  Days  AUs  Days  AU  
2  208  52  676  52  676  104  1,353  52  676  52  676  
3  340  85  676  85  676  170  3,613  85  676  85  676  
4  455  113  676  113  676  226  6,778  113  676  113  676  
5  589  147  676  147  676  295  10,848  147  676  147  676  
6  712  178  676  178  676  356  15,823  178  676  178  676  
7  833  208  676  208  676  417  21,704  208  676  208  676  
8  955  239  676  239  676  477  28,489  239  676  239  676  
9  1,076  269  676  269  676  538  36,180  269  676  269  676  
10  1,197  299  676  299  676  599  44,775  299  676  299  676  
11  1,318  329  676  329  676  659  54,276  329  676  329  676  
IV  DV  Rge_{Prac}/4  Rge_{Prac}/4  =t_{PhI}+t_{PhII} 
Fuel per second  Ship propulsion  Per day  

Original Mass  Exhaust Velocity  Exhaust Mass  Ship Velocity  Ship Mass  Daily %TOGW 
ff_{sec}  v_{Exh}  ff_{Exh}  g  M_{Ship}  %TOGW_{Day} 
kgm/sec  %c  kgm/sec  m/s^{2}  mT  % M_{Ship} 
0.48  86.6  0.96  10  25,000  0.17% 
0.96  86.6  1.92  10  50,000  0.17% 
1.73  86.6  2.88  10  75,000  0.17% 
1.92  86.6  3.84  10  100,000  0.17% 
ff_{Exh}  Given  M_{Ship }* %TOGW_{Day}  Con.  Given  Given 
Since the Earthlike gravity condition requires extremely stable incremental velocity increase throughout the flight, any increased momentum from increased exhaust massvelocity must go to increase amount of mass being propelled; this is size of spaceship, MShip(= ffExh* vExh / g).
Discussion when planning a mission, it's likely that planners will specify payload requirements or perhaps payload capacity and fuel flow will be adjusted to satisfy requirements. Thus, it makes sense to make table such that one can determine required fuel requirements to satisfy mission.
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Due to Lorentz Transform (LT), mass of exhaust particle (ff_{Exh}) is double that of original (ff_{sec}) in this specific case when exhaust particle's speed is 86.6% light speed.
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After one day of gforce acceleration, spacecraft achieves 864 km/sec.
Compute: v_{Final} = t * g ;
where t = time in seconds and
g = 10 meters/sec per sec = 10 m/sec^{2}
v_{Final} = 86,400 secs * 10 m/s^{2} = 864,000 m/s = 864 km/sec
This is an extremely large value compared with velocity required to escape Earth's gravity, e = 11.5 km/sec. Of course, a vessel wishing to orbit Earth would have to decrease its velocity to less then e.
For above destinations, max velocity happens a midway between dept and dest. In a previous chapter, we have determined that two way travel time (go and return) for a flight profile where ship must accelerate to midway, then decelerate from midway to destination for one way. Thus max velocity is acheived at midway. To compute time to midway: t_{Half} = SQRT(d/g)
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1)  ff_{Exh}  =  ff_{sec} 

2)  ff_{sec}  =  ff_{Exh * }SQRT(1v^{2}_{Exh}/c^{2}) 
3)  ff_{day}  =  ff_{sec}*86,400 sec/day 
4)  ff_{day}  =  ff_{Exh * }SQRT(1v^{2}_{Exh}/c^{2})*86,400 sec/day 
5)  %TOGW_{Day}  =  ff_{day} 
6)  %TOGW_{Day}  =  ff_{Exh * }SQRT(1v^{2}_{Exh}/c^{2})*86,400 sec/day 
7)  ff_{Exh}  =  g * M_{Ship} 
8)  %TOGW_{Day}  =  g * M_{Ship}_{ * }SQRT(1v^{2}_{Exh}/c^{2})*86,400 s/dy 
9)  %TOGW_{Day}  =  g*SQRT(1  v^{2}_{Exh}/c^{2})*86,400 s/dy 
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PERSPECTIVE from bottom row of values: After we've expended 100% of the ship's mass, we accomplished following profile of go and return spaceflight.
PHASEI. ACCELERATE TO HALFWAY. We have departed from vicinity of Earth and accelerated at g for half the one way distance (5,406 AUs = 10,812AUs/2). At this distance, constant gforce acceleration for 147 days causes our notional ship to attain 127,051 kms/sec which 42% light speed.
PHASEII. DECELERATE TO DESTINATION. Our ship then decelerates for 147 days and another 5,406 AUs. Then, our velocity decreases to essentially zero, and we've completed the initial one way distance, 10,812 AUs. One LY = 63,115 AU; so, this takes us far less then 1/4 distance to Oort Cloud (appox 1 LY from Sol).
PHASE III. ACCELERATE BACK TO HALFWAY. With 50% theoretical mass left, notional spaceship then reverses course for return trip and accelerates for 147 days and about 5,400 AUs toward Earth.
PHASEIV. DECELERATE BACK TO START POINT. Finally, the spaceship decelerates for 147 days and remaining 5,406 AUs for a safe orbital velocity in Earth's vicinity.
1) BAD NEWS!! After we've accomplished the tremendous, remarkable feat of accelerating enormous amount of mass to 86.6% speed of light for a 2way trip of almost 600 straight days, we've still barely moved out of the center of our Solar System.
1a) Even Badder News!!!! We can't use 100% of ship's mass as fuel. Let's arbitrarily pick a max of 50% as a reasonable limit for subsequent discussion.
2) GOOD NEWS!! Range of 50% TOGW is better then we might think. For example, if daily %TOGW is 1% and our ship weighs 100 tons, then first day, ship will consume 1 ton of fuel. A crude approx, 50 tons of fuel, thus, range is 50 days. However, range is really better then this.
The %TOGW_{Day} is really only good for first day. After the first day, we must really use percent Gross Weight (%GW) because the GW decreases as the fuel is consumed.
2nd day of trip, GW = %TOGW  first day's fuel = 99 tons. 2nd day, ship consumes 1% of 99 tons which is slightly less then 1 ton,
3rd day, fuel consumption will be even less and so on as shown in following table.
Days  Assoc'd Calculation  Remaining  

0  None  100 mT  
1  100mT * .99  99 mT  
2  99 mT * .99  98.81 mT  
3  100mT * .99 * .99 * .99  97.03 mT  
4  100mT * .99^{4}  96.06 mT  
5  100mT * .99^{5}  95.10 mT  
...  ... ... ... ... ...  ... mT  
n  100mT * .99^{n}  100*.99^{n} mT 
.99^{50} = .6050; too high.
.99^{100} = .3660; too low.
.99^{60} = .5472; too high.
.99^{70} = .4948; too low, but close.
.99^{69} = .4998; about right.
0.9983^{50}  =  0.9184  
0.9983^{100}  =  0.8435  
0.9983^{200}  =  0.7112  
0.9983^{400}  =  0.5063  
.Similiar, exercise for fuel consumption rate of .17% /day which we calculate from an exhaust particle speed of .866c and we get table to right.
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Perhaps a more convenient method might involve logarithms.
Recall log10^{10} = x
For example, log10^{100} = 2.
What if a^{x} = b, where a and b are known, but x is unknown.
Due to exponent characteristics, x = log b / log a;
in our case, .9983^{x} = .5000;
then x = log .5000 / log .9983 = 407.4.
Thus, exponentials and logarithms give a more accurate approximation of how many days 50% TOGW will take us if we're expending .17% GW per day.
EVEN BETTER NEWS!! It's very likely that we'll eventually design propulsion systems to accelerate particles much faster then .866c which means a much less fuel consumption rate and consequently much greater range.
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