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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,
your acceleration is measured as:
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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