[extropy-chat] unidirectional thrust

[Hal Finney]

Now I will analyze this thrusting device using relativistic rather than
Newtonian physics.  A quick review.

With uniformly accelerated motion, let the proper time as measured by
the accelerating observer be t' ("t prime").  Let the acceleration by
"a", and for convenience let us use geometric units where c is 1.
Then velocities are dimensionless values which are interpreted as
fractions of the speed of light, and accelerations have units of 1/time.

The standard formulas are:

x = 1/a * (cosh(a*t') - 1)
t = 1/a * sinh(a*t')
v = tanh(a*t')
E = m * cosh(a*t')

cosh, sinh and tanh are the standard hyperbolic functions.  x, t, v, and
E are position, time, velocity and energy as measured in the rest frame.
The E term is in geometric units where c=1.  It includes the rest energy
of E=m (commonly written as E=mc^2 when using non-geometric units), so to
get the kinetic energy we would subtract that:

KE = m * (cosh(a*t') - 1)

It remains to estimate the acceleration "a".  Mike several times has
pointed to a document,
<http://www.geocities.com/ekpworld/doc/EKP_satellite_maneuvering.doc>,
which is the only one I have found which actually claimed to have measured
lifter thrust in air and in vacuum.  The author reported a thrust of 2.38
mN in atmosphere and 0.31 mN in vacuum, with his setup (much smaller than
the one used by Naudin).  That would imply that lifter vacuum thrust is
0.13 times that in atmosphere.

So, instead of my previous analysis using Naudin's lifter which gave
0.3 Newtons of force, I will scale that down by this factor of 0.13
and take the force in vacuum as 0.039 N.  With a 20 kg device that is
an acceleration of .00195 m/s/s.  To go to geometric units, we divide
that acceleration by c and get an acceleration of 6.5E-12/sec.

Let's see how things are cooking after 60 days of acceleration.  This is
5.184 million seconds.  Our device has been drawing 132.9 Watts, times
5.184 million seconds is about 690 MJ.  That's how much power we've used.

Meanwhile our velocity, from the formula above with a = 6.5E-12 and
t' = 5.184 million, is 3.37E-5 in geometric units, meaning it is that
fraction of the speed of light.  Multiply by c to get the speed in regular
units and it is 10.1 km/sec, a very modest speed, not even Earth escape
velocity.  And the distance travelled, x, is 87.34 from the formula,
which in these geometric units is light-seconds.  Multiply by c and get
26 million km, so you could stay near the sun and keep the power for
your solar panels.  In fact you could safely head towards the sun the
whole time, it's 150 million km away.

What we really want is the kinetic energy.  From the KE formula above,
it is 1.14E-8 kilograms, which we have to multiply by c^2 to return to
regular units: 1.02 GJ, just over a gigajoule.

So there you go.  Energy in is 690 MJ, energy out is 1.02 GJ, after 60
days of acceleration.  Over unity.  You get out more energy than you
put in.  You're only going 10.1 km/sec at the end, having travelled 26
million km.  Calculation done with relativistic, Lorentzian formulas.
Using a figure from a document Mike pointed to as authoritative, for
the power output from a lifter in a vacuum.  Okay?

Hal