00100		We shall derive formulas for the time required to  accomplish
00200	an interstellar journey of distance %s% on the basis of the following
00300	technology: energy from a nuclear reactor (fission or fusion) is used
00400	to  expel  mass at a velocity that is varied during the mission in an
00500	optimal way.  We shall assume that  the  performance  of  the  system
00600	(reactor  +  rocket)  is characacterized by a number %p% equal to the
00700	power the  system  can  handle  per  unit  mass  of  apparatus.   The
00800	plausible  values  of  %p%  are  between  one  watt/kilogram and 1000
00900	watts/kilogram.  Since the time turns out  proportional  to  %p-1/3%,
01000	this  will  only  mean  a  factor  of  ten  in  the  time required to
01100	accomplish a given journey.
01200	
01300		We introduce symbols as follows:
01400	
01500		s = length of journey
01600	
01700		T = time of journey
01800	
01900		t is a time variable
02000	
02100		M = initial mass of the system
02200	
02300		m0 = final mass of the system
02400	
02500		m is a mass variable
02600	
02700		α = M/m0 = the mass ratio
02800	
02900		w is an exhaust velocity variable
03000	
03100		a(t) is the acceleration
03200	
03300		p = power available for unit mass of system
03400	
03500		P is a power variable.
03600	
03700		Conservation of momoentum and conservation of energy give the
03800	following equations:
03900	
04000	         .
04100	1)	-mw = ma
04200	                 .  2
04300	2)	P = -1/2 m w .
04400	
04500	                                                                   .
04600		Solving for %w% in 1), substituting in 2) and solving for %m%
04700	gives
04800	
04900	3)
05000	
05100	
05200		We  now  distinguish  two cases: in a single stage rocket, we
05300	have
05400	
05500	4)
05600	
05700	expressing  the  fact that the power available is proportional to the
05800	final mass of the system.
05900	
06000		In a continuously staged rocket, we have
06100	
06200	4')	P = pm,
06300	
06400	expressing the fact that the power available is proportional to the
06500	current mass.  (We suppose that every so often a nuclear reactor or
06600	a rocket is taken out of service, vaporized and expelled as working
06700	fluid).
06800	
06900		In the two cases, we get the equations
07000	
07100	
07200	5)
07300	
07400	
07500	and
07600	
07700	
07800	5')
07900	
08000	
08100		Taking  into  account the initial and final masses we get the
08200	following results by integration:
08300	
08400	
08500	6)
08600	
08700	
08800	and
08900	
09000	
09100	6')
09200	
09300	
09400		Using %α%=M/m % and setting for the two cases
09500	
09600	7)	q = p(1 - 1/α)
09700	
09800	and
09900	
10000	7') 	q = p log α,
10100	
10200	we get the following equation valid in both cases:
10300	
10400	
10500	8)
10600	
10700	
10800		Assuming that the journey begins and ends at rest we have
10900	
11000	
11100	9)
11200	
11300	
11400	The final distance is given by
11500	
11600	
11700	10)	s =
11800	
11900	
12000	from which
12100	
12200	
12300	11)	s =
12400	
12500	
12600	follows by integration by parts.
12700	
12800		Our  goal is now to determine the acceleration profile %a(t)%
12900	satisfying equations 8),9) and 11) so that %T%  is  minimized  for  a
13000	given %s.  Before doing this, however, we shall treat the simple case
13100	in which we use a constant magnitude acceleration reversed in sign at
13200	the midpoint of the journey.  This assumptions gives from 8) and 11)
13300	
13400	
13500	12)
13600	
13700	and
13800	
13900	13)	s =
14000	
14100	
14200		Solving for %T% and %a% gives
14300	
14400	
14500	14)
14600	
14700	and
14800	
14900	
15000	15)	a =
15100	
15200		We shall now labor mightily to optimize %a(t), but the  eager
15300	reader  is  warned  that  this  only  changes  the  co-efficient 2 in
15400	equation 14) to 1.817 which might not be considered worth either  the
15500	mathematics or the engineering.  Well, onward!
15600	
15700		Instead of holding %s% fixed and minimizing %T, we  take  the
15800	equivalent   but  simpler  problem  of  maximizing  %s%  holding  %T%
15900	constant.  Using Lagrange multipliers and taking variations gives
16000	
16100	
16200	16)
16300	
16400	
16500	
16600	
16700	
16800	and since this must hold for arbitrary variations %%a(t), we have
16900	
17000	17)
17100	
17200		Combining  17) with 8), 9), and 11) gives (if we have finally
17300	gotten the algebra right)
17400	
17500	
17600	18)
17700	
17800	
17900	and
18000	
18100	
18200	19)	T =
18300	
18400	
18500		Thus  if we optimize acceleration and use continuous staging,
18600	we get
18700	
18800	
18900	20)	T =
19000	
19100	
19200		As  a  numerical  example,  we let %s%=%10 ~ 100 light years,
19300	p%=%1000, and α%=%e%%~%3000.  We then get
19400	
19500		T = .9  10   sec = 3000 years.
19600	
19700	Remarks:  The  zero  in  acceleration  at the midpoint means that the
19800	calculation is invalid at that time because with constant power, that
19900	corresponds to infinite exhaust velocity.  The actual optimum profile
20000	involves emitting only the waste products of the nuclear reactor near
20100	the midpoint.
20200	
20300		Since  the  time  is  proportional  to  the  2/3 power of the
20400	distance, clearly the formula is incorrect for  long  distances.   It
20500	becomes  incorrect when it recommends exhaust velocities greater that
20600	corresponding to emitting only the  waste  products  of  the  nuclear
20700	reaction.