Operational scale

Starting from rest, 1 gee acceleration for 24 hours yields a distance
traveled of 0.25 AU, pretty close--assuming I used the right formula
(d=1/2
a t^2 ), setting 1gee at 10m/sec^2; and assuming my arithmetic is right.

(I haven't worked out when you get to relativistic speeds).

So what's a reasonable scale for operational level (gravitic drives)? Is
1/4 AU per movement point about right?

> Laserlight wrote:

> Starting from rest, 1 gee acceleration for 24 hours yields a

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.

> So what's a reasonable scale for operational level (gravitic drives)?
Is
> 1/4 AU per movement point about right?

Seems ok to me...Have you thought about the weapons yet? Missles? FTL energy
beams?

> [quoted text omitted]

> Donald Hosford wrote:

> Nyrath the nearly wise wrote:

> Donald Hosford wrote:

So how close was I on the time taken?

> Donald Hosford wrote:

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

> Nyrath the nearly wise wrote:

> Donald Hosford wrote:

Thanks Perfesser! You are so nearly wise!

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

> Donald Hosford wrote:

> On Sat, 20 Mar 1999, Donald Hosford wrote:

> Laserlight wrote:

<kof kof> "reach light speed"? if you're using a relativistic system, you
can't reach light speed. that, i believe, is the point. i guess you mean
"reach 99% light speed", or something.

Tom

> Thomas Anderson wrote:

> On Sat, 20 Mar 1999, Donald Hosford wrote:

My simple minded 8-bit computer didn't know that...besides I was
curious. So 99% or so would be about 337 days. (to be sure, I would have to
write a new program to figure it out.)

> Nyrath the nearly wise wrote:
<SNIP>
> Earth Dist : Earth time speed ship time ship

Something is strange here, or maybe I just understand relativisitic motion.
With the ship that is constantly accelerating in this example, and using the
time and distance as measured by the ship, it takes it 1.28 years to travel
0.49 light years distiance. Then it would take at most twice that time, 2.56
years to travel twice that distance, 0.98 light years. But the table above
gives the distance at 2.98 years as 0.87 light years. In fact the distance the
ship travels ploted against time seems to be assymtoptic to one light years,
i.e. the distance travelled never reaches 1 light year. This doesn't make
sense to me.

Enjoy,