# Continuation of my 9/5 letter

```Hi Steve

[Note: Kevin too has suggested, on 9/4,

>How about a mission which has a constant beam power, the
>acceleration would drop off toward the turnaround point.  In
>this case, the crew would start off with earth-like gravity,
>and towards the middle of the trip, the gravity would be more
>lunar-like. ... The advantage would be simplified beaming
>requirements, and the disadvantage would be a slightly longer
>flight time.
>
>Questions:
>What would the top speed relative to Earth be?
>What is the total trip time. (crew time?)
>How much of this time is spent at less than 1/2 G?]

To pick up where I suspended my derivation--

....the desired description of the motion should be
theta = ln[(Pe * t'/M) + 1]   .

At the beginning of the flight, when thrust = To, acceleration
= ao and the received power, Pr, equals the emitted power, Pe,
To = M * ao = Pe/c   ,
ao = Pe/(c * M) --> Pe/M   (for c = 1 lt-yr/yr),
which makes the desired description of the motion
theta = ln(ao * t' + 1)   .

For one space dimension,
dx/dt' = u = sinh(theta)
= 0.5 * [exp(theta) - exp(-theta)]
= 0.5 * [(ao * t' + 1) - 1/(ao * t' + 1)]   .

Integrating this from x = 0 at t' = 0 gives
x = 0.5 * (0.5 * ao * t'^2 + t' - theta/ao)   .

The apparent (Earth) time, t, for the ship time, t', is
obtained from
dt/dt' = gamma = cosh(theta)
= 0.5 * [exp(theta) + exp(-theta)]
= 0.5 * [(ao * t' + 1) + 1/(ao * t' + 1)]   .

Integrating this from t = 0 at t' = 0 gives
t = 0.5 * (0.5 * ao * t'^2 + t' + theta/ao)   .

The Earth time of emission, te, of the energy that arrives at
the sail at t' is simply (for c = 1)
te = t - x   .

The proper velocity, u, is given by
u = sinh(theta)   .

The instantaneous proper acceleration, a, is given by the
velocity-parameter equation of motion--
a = c * d(theta)/dt'
= ao/(ao * t' + 1)       (for c = 1).

Putting these relations together in the Fortran program
CONOUTEM.FOR, which is appended, gives the following values of
theta, distance, proper velocity, instantaneous acceleration,
Earth time for t' and Earth time of emission for reception at
t', as a function of ship time, t', for an initial acceleration
of 1 g,

Tship    Theta     Dist   Prop Vel   Accel   TEarth    Temit
(yr)    (rad)   (lt-yr) (lt-yr/yr)   (g)     (yr)      (yr)
0.0   0.0000   0.0000    0.0000   1.0000   0.0000   0.0000
0.5   0.4162   0.1130    0.4283   0.6595   0.5161   0.4031
1.0   0.7092   0.4146    0.7702   0.4920   1.1016   0.6870
1.5   0.9355   0.8776    1.0781   0.3924   1.7838   0.9062
2.0   1.1200   1.4900    1.3693   0.3263   2.5748   1.0848
2.5   1.2756   2.2453    1.6509   0.2793   3.4809   1.2356
3.0   1.4103   3.1399    1.9266   0.2441   4.5059   1.3660
3.5   1.5290   4.1712    2.1983   0.2168   5.6522   1.4810
4.0   1.6350   5.3377    2.4673   0.1949   6.9215   1.5837
4.5   1.7309   6.6382    2.7343   0.1771   8.3148   1.6766
5.0   1.8184   8.0718    2.9999   0.1623   9.8332   1.7613

For an example trip to a star 4.4906 (= 2 * 2.2453) lt-yr from
Earth, the ship would accelerate for 2.5 ship years, reach a
maximum proper velocity of 1.6509 lt-yr/yr at an apparent
(Earth) time of 3.4809 yr, and at turnover receive power emit-
ted from Earth at 1.2356 yr after the departure date, giving an
instantaneous acceleration of 0.2793 g.

These results seem to support Kevin's intuitive inferences.

Please let me know if you find anything wrong or poorly stated
here or in my 9/5 letter.

We can address the deceleration phase (here assumed without
justification to be a mirror image of the acceleration phase)
in a subsequent dialogue.  Timothy may have already covered

Regards, Rex

PROGRAM CONOUTEM                              !9/7/96
101 FORMAT(2X, 21H Initial Accel (g) = )
102 FORMAT(1X, 6H Tship, 3X, 6H Theta, 5X, 5H Dist, 2X,
& 9H Prop Vel, 3X, 6H Accel, 2X, 7H TEarth, 2X, 6H Temit)
103 FORMAT(3X, F4.1, 3X, F6.4, 3X, F7.4, 4X, F7.4, 3X, F6.4,
& 3X, F6.4, 2X, F6.4)
1 CONTINUE
WRITE(*,101)
IF(AGO .EQ. 0.) GO TO 99
AO = 1.0324 * AGO
WRITE(*,102)
DO 10 IT = 1, 11
FT = IT - 1
TIM = 0.5 * FT
ARG = AO * TIM + 1.
THET = LOG(ARG)
DIST = 0.5 * (0.5 * AO * TIM*TIM + TIM - THET/AO)
PVEL = 0.5 * (ARG - 1./ARG)
ACC = AGO/ARG
TAPP = 0.5 * (0.5 * AO * TIM*TIM + TIM + THET/AO)
TEM = TAPP - DIST
WRITE(*,103) TIM, THET, DIST, PVEL, ACC, TAPP, TEM
10 CONTINUE
GO TO 1
99 STOP
END

```