# starship-design: Motion of sail driven by constant-power beam

```Hi all

Steve has written, on 9/1 to me (and Timothy),

>I wonder what the worldline would look like for an object that
>is accelerated by a constant-output emitter?  In other words,
>the emitter would send constant output power, meaning the re-
>and acceleration as its proper time increases.

Kevin has written, on 9/4 to the Group,

>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?]

DERIVATION

For a power, Pe, sent out by an emitter, the power received by
a sail (ignoring inverse-square effects) that is receding at an
apparent velocity, beta lt-yr/yr, is
Pr = Pe * sqrt[(1 - beta)/(1 + beta)]
....Doppler shift
= Pe * gamma * (1 - beta)
....gamma = 1/sqrt(1 - beta^2)
= Pe * [cosh(theta) - sinh(theta)]
....gamma = cosh(theta); beta = tanh(theta)
(definition of velocity parameter, theta)
= Pe * exp(-theta)
....using exp forms of hyp functions.

(I believe this relation, with theta = a * t' for constant a, is
the source of the logarithmic time dependence introduced by Steve
in his email of 8/20 to the Group.)

The velocity-parameter equation of motion for a thrust, T = Pr/c,
applied to a mass, M, is
T = M * d(theta)/dt' = Pr/c = Pe * exp(-theta)
....c = 1 lt-yr/yr,
which gives the "differential equation"
exp(theta) * d(theta) = (Pe/M) * dt'   .

Integrating from theta = 0 at t' = 0, with Pe constant, gives
[exp(theta) - 1] = Pe * t'/M   ,

so the description of the motion of the sail, in terms of the
dependence of the velocity parameter, theta, on ship time, t',
is
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 description of the motion of the sail
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).

RESULTS

Putting these relations together in the Fortran program
COPOBM.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.
__________
*For a proper velocity, u, the apparent velocity, v, is given
by
v = u/sqrt(1 + u^2)   .
__________

These results seem to confirm Kevin's intuitive estimates regard-
ing acceleration levels.  They also substantiate Steve's conclu-
sion that the time of emission is limited (even for a constant-
output emitter); from the table above, the duration of emission
of the radiation accelerating a sail half way to a destination
more than 16 lt-yr away (the last entry) is only about a year
and three quarters.

The deceleration phase (here assumed without justification to
be a mirror image of the acceleration phase) needs to be
a lot of thought into it.

(Note: This exercise may turn out to be purely academic because
the inverse-square effects, without unforeseeable advances in
focusing abilities, would be much larger.)

Rex

PROGRAM COPOBM                              !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

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