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Re: [amsat-bb] Laser Satellite Comms

Bob Bruninga wrote:
> Great!  I just love these little "Fermi" problems and back of the envelope
> calculations.  Your starting point of the 2 inch spot back down, and Tom's
> calculation of the divergence of a cheap laser, then suggests that all of
> the transmitted energy will hit the receiver, but reduced by the capture
> area of the 1 inch mirror compared to Tom's 150 meter uplink beam.
> I calculate this loss at 75 dB?

At the transmit end, this is correct, assuming the mirror reflects with
100% efficiency and that the spot has uniform intensity over the 150 meter

> So wouldnt a "2 inch" optical receiver capture this 230 usec burst with
> only a loss of 75 dB?  Ah!  But now what is the chance of the 2 inch spot
> hitting "your" 2 inch receiver?  Well... Real Small!  How small?

No, the "spot" on the ground is diffraction limited. In my first 
posting I stated "Therefore you will have a return beam that has
a size ~1/40000 radians = 5 arc seconds. On the ground, the "spot
size" will be about 10 meters in diameter.". 

If you have a ~10 cm (4") receiving telescope, then you intercept 
only ~(10cm/10m)**2 = -40 dB of the energy when you are in the 
center of the beam, and even less when off the peak. So your path 
loss is more like 120 dB.

> [snip]
> Lets assume there are 60 mirrors around the circumference, thus there are
> many "sweeping" beams every 6 degrees across the ground.  From 250 miles
> away, that is every 25 miles.  Thus your chances are 8 inches in 25 miles
> of seeing the spot at any instant.(.0005 %)  But at 1 RPM, you get a new
> sweep every second.  And during a 5 minute pass, you get 300 such chances.
> So your chances are 0.15% per pass?  At 2 good passes a day and a 6 month
> mission, this results maybe a 50% chance at success in receiving ONE
> packet at a megabit per second...

But its a two dimensional problem. Consider the plane formed by the 
line from you to the satellite and the line normal to (perpendicular to) 
the mirror in question at a particular instant, and let these two
lines form an angle alpha. Then the reflection point will be in the 
same plane (on the other side) at an angle = alpha. You now have to
consider precisely where the mirror is on the surface, and precisely
where the spin axis is with respect to both the transmitter and 
receiver in order to know if the spot will even be seen. So, using 
your numbers, the spots are 25 miles = 40 km apart in two dimensions.
Don't know where you came up with the 8" figure, since the spot
size on the ground (again from my earlier posting) is ~10 meters.
Therefore the spot fills ~(10 meters/40 km)**2 = (0.00025)**2 =
0.0000006 = 0.00006% = -72 dB. If I transmit at 1Mbit/sec solidly
for a 10 minute pass, and I can keep the transmitter pointed 
perfectly for the entire pass, and my laser is strong enough to 
give an echo above the noise, then statistically I will get back
225 bits sometime during the pass using this "shooting goldfish
with a shotgun in a reservoir" approach. 

David Forham also mused about the use of big mylar satellites, a la
Echo. Unfortunately these don't work very well either. If you fly
it at low altitude, than residual atmospheric drag makes its orbit
decay very rapidly -- witness the problems keeping MIR in orbit.
The decay time relates to the satellite's ballistic coefficient --
the ratio of surface area to mass. A cannonball (little area, lots
of mass) will survive much longer than a light balloon.

It's hard to keep a balloon inflated in orbit, since the ballon 
is easily punctured (like by a micrometeorite or other space junk). 
Once its mechanical integrity is violated, it loses its rigidity. 
At that time, then the atmospheric drag and radiation pressure
from solar photons will cause it to distort away from its spherical

In a communications sense, a balloon is just a passive reflector.
It intercepts the signal incident upon it and then relects it back
into half the sky (2*pi steradians in physicsgeek terminology).
Since the energy it intercepts varies as the square of the distance
from the transmitter, and the reflected signal also varies as
the square of the distance, this makes the signal go down rapidly
as the 4th power of the distance. This is why EME ain't easy!
If you want the balloon to work well you want it to be close to
you (and hence it appears big on the sky), but this means the
satellite will be in low orbit and it will "Chicken Little" fall
from the sky easily.

The solution to several of these problems is to make the satellite
be an active "repeater" with the signal amplified and re-transmitted.
Which is why AMSAT has been in the active satellite game for 30
years now [Y'all come to the 30th Birthday Party on Mar.13th!!!].

73, Tom

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