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# Re: Operational scale

From: Nyrath the nearly wise <nyrath@c...>
Date: Sun, 21 Mar 1999 13:21:38 -0500
Subject: Re: Operational scale
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Donald Hosford wrote:
>
> Nyrath the nearly wise wrote:
>
> > Donald Hosford wrote:
> > > I figured it out on a computer a long time ago. (back when they
were only 8
> > > bit...)
> > > At 1 gee acceleration, it takes something like 340 earth days to
reach light
> > > speed.  I have forgotten how far you would travel by the time you
reached light
> > > speed.
> >
> >	    IIRC about half a light year.
>
> So how close was I on the time taken?

Well, the article I read just said "one year".
(A STEP FARTHER OUT by Jerry Pournelle)

Failing that, herein follows all that I know
about relativisitic motion, cribbed from various sources:

RELATIVISITIC MOTION

gamma = 1 / sqrt ( 1- ((v^2) / (c^2)))

For two inertial (unaccelerated) frames of reference, if frame S' is
moving
with respect to frame S with velocity V, in the positive direction along
the
x-axis during time t, then:

x' = gamma * (x - (V * t))
x  = gamma * (x' + (V * t))
t' = gamma * (t - ((V * x)/(c^2)))
t  = gamma * (t' + ((V * x')/(c^2)))

For a rocket moving with constant acceleration "a", due to thrusting in
its
proper frame, then the total elapsed proper time deltaT' (as time is
measured
on the rocket) over distance S:

deltaT' = (c/a) * cosh^-1 (1 + (a * S)/(c^2))

where  cosh^-1(x) = inverse hyperbolic cosine of x

As (a*S)/(c^2) approaches 1.0, the equation becomes

deltaT' = (c/a) * 1n((2 * a * S)/(c^2))

The vehicle's velocity after accelerating for deltaT' and reaching
distance S is:

V = c * sqrt(1 - (1 / (1 + ((a * S) / (c^2)))^2))

If the rocket accelerates at "a" up to the midpoint, then deaccelerates
at "-a" to destination:

deltaT' = ((2 * c)/a)cosh^-1(1 + (a/(2 * c^2)) * S )

As (a*S)/(c^2) approaches 1.0, the equation becomes

deltaT' =((2 * c)/a)ln( (a/(c^2))S )

Velocity at turnover is

Vturnover = c*sqrt( 1 - ( 1 + (a/(2 * c^2)) * S )^(-2) )

Erik Max Francis max@alcyone.com says:

deltaV = u 1n lambda/sqrt[1 + (u^2/c^2) 1n^2 lambda]

where deltaV is the deltavee, and lambda is the mass ratio, or the ratio

of the initial to the final mass.

v = c t/sqrt[c^2/a'^2 + t^2]
r = c [(c^2/a'^2 + t^2) - c/a']
t' = (c/a') ln [(1 + a'^2 t^2/c^2)^(1/2) + a' t/c]
t = c v/a'/sqrt(c^2 - v^2)

a' = subjective acceleration
v = objective velocity
r = objective displacement
t = objective elapsed time
t' = subjective elapsed time
"objective" = from the rest frame (at rest relative to the departure
point)
"subjective" = from the ship frame
Subjective (not objective) acceleration is constant;
acceleration is all in one direction only.

Example:
t = c v/a'/sqrt(c^2 - v^2)

if v = 0.995 c and a' = 5000 gee
then t = 8.61e4 sec = 23.9 hours
Now plug t into
t' = (c/a') ln [(1 + a'^2 t^2/c^2)^(1/2) + a' t/c]
and get 2.04e4 sec = 5.67 hours
Plug t into
r = c [(c^2/a'^2 + t^2) - c/a']
to get objective displacement of 2.4e13 m (about 160 au)

Bill Woods <wwoods@ix.netcom.com>:
Assuming a magical stardrive which allows you to accelerate
continuously at constant acceleration a, as measured onboard the ship,

a  : ship acceleration
tau: ship time (proper time)
d  : ship distance

T: Earth time
D: Earth distance
A: Earth acceleration

Mo: initial mass
M : mass of ship

theta(tau) = (a/c)tau	     : velocity parameter
beta   = v/c  = tanh(theta)
= tanh((a/c)tau)
gamma  = 1/sqrt[ 1 - beta^2 ]

v(tau) = c*tanh[(a/c)tau]
D(tau) = (c^2/a)*( cosh[(a/c)tau] - 1 )
tau(D) = (c/a)arccosh[ (a/c^2)D + 1 ]
d(tau) = D/cosh(theta) = (c^2/a)*( 1 - sech[(a/c)tau] ) -> c^2/a
d ~ c^2/a for tau > 6c/a

T(tau) = (c/a)sinh((a/c)tau)	    (a/c)T   =	  sinh( (a/c)tau )
tau(T) = (c/a)arcsinh((a/c)T)	    (a/c)tau = arcsinh( (a/c)T	 )

Alternately, in the frame of a stationary observer,

A = a / gamma^3

A(v) = a*sqrt( 1 - (v/c)^2 )^3
D(T) = (c^2/a)*( sqrt[1 + ((a/c)T)^2] - 1 )
T(D) = (c/a)sqrt[ ( (a/c^2)D + 1 )^2 - 1 ]
v(T) = a*T / sqrt[ 1 + ((a/c)T)^2 ]
= c / sqrt[ 1 + (c/aT)^2 ]
beta(T) = v(T)/c = 1 / sqrt[ 1 + (c/aT)^2 ]

tau(T) = (c/a)ln[ (a/c)T + sqrt( 1 + ((a/c)T)^2 ) ]

A(T) = a / sqrt( 1 + ((a/c)T)^2 )^3

For acceleration at 10 m/s^2, the time taken to reach various distances
is:

Earth Dist :   Earth time	speed	      ship time   ship distance
__________     __________	_____	      _________   _____________

.06 ly :	  0.34 yr      0.34	c      0.34 yr	   0.06 ly
0.25 ly :	  0.73 yr      0.61	c      0.67 yr	   0.20 ly
0.50 ly :	  1.10 yr      0.755	c      0.94 yr	   0.33 ly
1 ly :	  1.70 yr      0.873	c      1.28 yr	   0.49 ly
2 ly :	  2.79 yr      0.9467	c      1.71 yr	   0.64 ly
4 ly :	  4.86 yr      0.9814	c      2.22 yr	   0.77 ly
10 ly :	 10.91 yr      0.99622	c      2.98 yr	   0.87 ly
25 ly :	 25.93 yr      0.99932	c      3.80 yr	   0.92 ly
50 ly :	 50.94 yr      0.99982	c      4.44 yr	   0.93 ly
100 ly :	100.95 yr      0.999947 c      5.09 yr	   0.94 ly
1000 ly :     1000.95 yr      0.999991 c      7.27 yr	   0.95 ly
10000 ly :    10000.98 yr      0.999992 c      9.46 yr	   0.95 ly
d -> 0.9500 ly

For a trip which goes from standing start to standing finish, calculate
the
time to cover half the distance, then double the T and tau variables.

DistAlphaCen	= 4.3 ly = 41 Pm = 41e15 m
1/2 DAC	= 20.5e15 m
1/2 tauAC	= 55.7e6 sec
1/2 TAC	= 93.7e6 sec
TauToAlphCen	= 111e6 sec = 3.5 yr
TimeToAlphaCen	= 187e6 sec = 5.9 yr

For a perfectly efficient photon rocket,

theta = ln(Mo/M) , so	     M(tau) = Mo*e^[-(a/c)tau]

or more conveniently,	     theta(Tau1/2) = ln(2) = 0.7

so the rocket's halflife is  Tau1/2 = 0.7c/a

for instance, for a = 1 kgal (= 1000 cm/s^2 ~ 1 "gee")
Tau1/2 = 21e6 s ~ 8 months

TauToAlphaCen = 111e6 s = 3.5 years ~ 5.3 Tau1/2
so initially the rocket must be more than 31/32 fuel.

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