Re: Operational scale

Nyrath the nearly wise wrote:

> 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.

Thanks Perfesser!  You are so nearly wise!

(Wow!  Do you think about this stuff all the time?)

Donald Hosford