Hello. I copied below a post to sci.astro about a possible means of communicating at high data rates from planetary missions using optical light transmissions.
A couple of questions about the communication from the Moon case. Based on the example of the "Iridium flares", would a 4mx4m size mirror be sufficient size to see flashes from a lunar distance with the naked-eye?
How large would be the size of the illuminated area on the Earth from the reflection of the Sun by the mirrors at the lunar distance?
Would parabolic mirrors increase the area illuminated or the intensity of the light as viewed at the Earth?
BTW, here's some info on how the intensity and the location of the "Iridium flares" can be predicted:

SeeSat-L Apr-98: Method for predicting flare.
http://satobs.org/seesat/Apr-1998/0175.html


Bob Clark

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From: Robert Clark
Date: Sat, 27 Jun 2009 23:54:14 -0700 (PDT)
Subject: High data rate space transmissions through visible light communication.
Newsgroups: sci.astro, rec.radio.amateur.space, rec.radio.amateur.antenna, sci.astro.seti, sci.physics

I had been thinking about methods of high data rate transmission in
regards to getting *video* transmissions from Mars orbiter missions. I
was irritated by the spotty coverage of the Mars surface at the best
resolutions so I wanted to send real-time *continuous* imaging back to
Earth receiving stations at the highest imaging resolutions. This
would require very high transmission rates, much higher than what is
currently used.
The idea would be to use light transmissions but only of the on-off
variety. You would use a large surface, many meters across, capable of
being alternatively lit up and darkened. There are computer chips of
course capable of operating at Ghz rates. This would determine if the
large surface was lit up or not electrically, possibly by using a
material whose reflective properties can be changed electrically.
I was worried though about the twinkling seen in point sources, which
this would appear to be, such as with stars due to atmospheric
effects. So this might require the telescope(s) to be in Earth orbit.
The question I had though was whether the atmospheric distortion would
cause an "on" signal to appear "off" and vice versa? My understanding
of atmospheric distortion is that it causes the point source to be
constantly apparently undergoing small shifts in position. But this
wouldn't be a problem if what you want to determine is whether it is
on or off. If that is the case then ground based telescopes would
work.
In the large reflecting surface, I actually wanted to use separate,
say, squares on the reflecting surface that could be put separately in
the on-off position to increase the information transmission rate. But
that would require being able to distinguish the squares from Earth
millions of kilometers away. This is why I wanted to use light rather
than radio for this since the larger wavelengths in radio would make
the reflecting surface impractically large for diffraction limited
resolution.
Even with light you couldn't do this with a single telescope. They
would have to be widely separated. Combining the signals from widely
separated scopes is common in radio astronomy but is not nearly as
successful in optical astronomy. That is because the light wavelengths
are so much smaller and you would have to have nanoscale accuracy in
positioning the widely separate mirrors in relationship to each other.
However, in the case of just detecting an on-off signal this shouldn't
be as big of a problem as you're not trying to form a usable image,
but only trying to see if a particular location is on or off. You
would need though highly accurate timing synchrony between the
separate scopes, within nanoseconds, to be sure they are detecting the
same on-off square. Note also here that the shifting in the image due
to atmospheric distortion very definitely would be bad for using
ground based scopes.

It occurred to me this might be a means of acquiring advertising
support for a Google Lunar X Prize entrant. I had also been trying to
come up with a method of having an illuminated image either on the
Moon or in lunar orbit that would be visible to the naked eye on
Earth. Such an idea was discussed here:

moon advertising.
put a billboard on the moon.
http://www.halfbakery.com/idea/moon_20advertising

I wouldn't be in favor of doing this in a way that would actually
advertise a product. But I was thinking about it as a way of sending a
message in favor of, for example, world peace. In this case you could
still have advertisers who could say in TV commercials for example
they supplied funding to support the mission and the message.
BTW, I would be in favor of advertisers who could pay to have
advertising signs set up at the rover landing site so that if anyone
who wanted to log on to the the rover transmissions or who watched a
TV program on the rover transmissions would see the ads. This to me is
something different than an ad that someone would be forced to see
just by looking up at the Moon.
In any case you would need something large enough so that with naked
eye resolution at the lunar distance it would still be
distinguishable. This page gives the naked eye resolution at the lunar
distance:

Purpose of Building Telescopes.
http://www.astronomy.org/astronomy-survival/telepur.htm

According to this page the resolution of the human eye at the lunar
distance would be about 22 miles. One single object clearly couldn't
do this. However, if you had separate illuminated landers or orbiters
at this large distance apart they could be used to send a message
visible to the naked eye on Earth.
It could work with orbiters by the example set of satellite formation
flying by the Cluster mission:

Cluster mission.
http://en.wikipedia.org/wiki/Cluster_mission

I also needed to find how large a brightly illuminated surface needed
to be at the lunar surface to be visible by the naked eye on Earth. I
thought of the example of the "Iridium flares":

Satellite flare.
http://en.wikipedia.org/wiki/Satellite_flare

The Iridium satellites have 3 antennas that happen to be also
reflective in visible light, totaling 4.8 m^2 in area. According to
the Wikipedia page, the flares can be up to -8 in apparent magnitude,
though typically at +6 magnitude, and are produced by an individual
antenna, so by one of area 1.6 m^2.
I'll assume the brightest flares are produced just by the orientation
the antennas happen to be in so we could make our reflective surfaces
be oriented with respect to the Sun to get the greatest brightness.
For the same size surface, the brightness would be lessened by the
greater distance to the Moon. The Iridium satellites are at about 780
km altitude so the Moon is about 500 times further. This would lower
the brightness by a factor of 500^2 = 250,000.
This page gives the apparent brightness commonly visible by the naked
eye in urban areas as +3:

Apparent magnitude.
http://en.wikipedia.org/wiki/Apparent_magnitude

The 250,000 times lesser brightness at the lunar distance for an
Iridium sized reflective surface would give it a +13.5 higher apparent
magnitude so up to +5.5 in apparent magnitude. To make our reflective
surface be at +3 apparent magnitude we could make the area be 10 times
larger, so at 16 m^2 area, or a square 4 meters across.
We would need a method for a flat reflecting surface of unfolding it
to this size. It might be easier instead to have the reflecting
surface be a balloon inflated by stored gas. Since this would be in a
vacuum, you wouldn't need much gas pressure or mass to accomplish
this.
Another consideration is that because of the brightness of the Moon
it could swamp out our illuminated surface. For the orbiter, this
could probably be alleviated by having the orbiter have a highly
elliptical orbit, (this also would be beneficial in minimizing the
required delta-v and fuel load) then it would be visible at the higher
distances from the Moon in its orbit. For the landers it might work
for them to land in the dark lunar maria.

To communicate the message though we would need a method to turn on
and off the reflecting surface. One possibility would be to have the
reflecting surface consist of very many small squares that could be
rotated to reflect toward the Earth or away. Another possibility might
be to have it covered with LCD's. Whichever method it would have to be
both lightweight and low power.
For our first attempts we probably would not want to send so many
orbiter or landers at once to form a naked-eye visible image. We would
first send just a single one to test it out. Note that this method
with a single vehicle could still be used to send high definition
video by having our single reflective surface be turned on and off at
the required rate, about 256,000 times per sec with compression.


Bob Clark
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 We might want to test this out in Earth orbit first before sending
to the Moon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

CubeSat.
http://en.wikipedia.org/wiki/CubeSat

Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
This should be a standard question in telescope optics if anyone knows
the answer. You have a mirror of a certain size. How large is the reflected
image according to distance to the imaging screen (the Earth in this case)?
For a flat mirror? For a parabolic mirror? For a bright source, how bright can
the image be at the imaging screen? Here's another way of looking at it:
the sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light that fell
on it to the Earth's surface it would still be only 10 watts. So this is saying a
10 watt light source at a distance of 780 km would be as bright as a -2.5
magnitude star.
Is this right even if you put behind it say a parabolic reflector as with a
flashlight so all the light was directed forward?


Bob Clark

   I suggest you read Glister's "Centauri Dreams".

  He discusses optical networks in the last chapters.  NASA is working on this concept now, and it is hoped that an optical system will replace the DSN.  In fact, it will likely have to.  The amount of data that the new probes/spacecraft are capable of sending, and the numbers of spacecraft, are increasing (Russia, China, India, Japan, the ESA and US all have active space programs that the DSN supports..)

  Optical systems can handle this load, and be almost entirely self-managed through its own operating system.  Uplinks/downlinks to Earth would be handled as well as collecting data streams from dozens of spacecraft.   The capacity/speed would allow video streaming from the moon, Mars, Jupiter, etc.. in real time (minus delay).

   The scopes themselves do not have to be that big, as the lasers being designed will focus beams far, far more efficently than radio (the point, I guess!)--an 8" - 12" mirror would be more than sufficient (I believe this was the size range Glister discussed...).  The laser, software, and CCD sensitivity are probably more critical, and these are improving rapidly.

  I hope I live to see it.  Maybe sometime early in the 2020s to coincide with the lunar program...  That would make sense.

 I found this page with the equation for concave mirrors:

The Mirror Equation - Concave Mirrors.
http://www.glenbrook.k12.il.us/gbssci/phys/class/refln/u13l3f.html

 The mirror equation gives the relationship between the focal length, the object distance, and the image distance:

1/f = 1/do + 1/di

 The object and image distances determine the magnification, which is the ratio of the image size to object size:

M = hi/ho = - di/do, where the minus sign indicates the image would be inverted.

 The object distance is the distance to the Sun at about 150,000,000 km. The image distance would be the distance from the Moon to the Earth at 380,000 km. Because the solar distance is so much larger than the lunar distance, the Earth would be quite close to the focus.
The size of the image is obtained from the equation hi/ho = - di/do, hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size of the Sun. So hi = -3,547 km, with an inverted image.
 What I'm still puzzling about, assuming the idea that a 4 m mirror at lunar distance would have an apparent magnitude of +3 is correct, is would this mean all the observers in that 3,547 km wide area would view it as being of +3 magnitude?


      Bob Clark

    Could you please state in 25 words or less what it is you are trying to figure out, and why?  I simply cannot follow your narrative!

 There are two things I'm trying to accomplish: 1.) send live *video* from Mars and other planetary missions, and 2.) send high definition TV from the Google Lunar X Prize mission.

  The reason for the 1st is that the large image size of the images using high resolution cameras on planetary missions means just individual photos can be taken at once, and just a few of the these can be sent in any one day. This means just widely separated spots on the Martian surface are imaged at the best resolution. Mars imaging has covered the entire surface at low resolution but these of course show little detail.

 What I want is to increase the data transmission rate so that dozens can be sent every *second*, essentially the same as sending live video, and at high resolution.

 For the second, sending high def video requires high data rates. This is of course being done now with satellite commercial TV services but this requires large satellites with large antennas and high power requirements. This wouldn't be possible with a low weight craft as is expected to be the case for an Google Lunar X Prize entrant.

 For the GXLP competition, there is a significant time constraint of it being accomplished by 2012 for the $20 million dollar prize, or by 2014 for the $15 million dollar prize.  Therefore I wanted to do the high def TV transmission requirement in a simple, low cost, low power way. Just reflecting sunlight from a mirror might satisfy this requirement.

 So I needed to know how large a mirror needed to be for the Sun reflection to be visible on the Earth at a lunar distance. If it turned out the mirror had to be, say, 1km across, clearly that wouldn't be feasible for a low cost, low weight mission. Because this mission was supposed to done by amateurs in order to extend the interest in spaceflight among the general public, it would also be ideal if in fact it could be visible with the naked eye, or at most with binoculars or a low power telescope.

  I'm fairly sure the turning on-and-off the reflection for the binary transmission can be done in a low cost, low power way (though I haven't worked this out in detail yet) so I'm just looking at the question now of getting the easily visible Sun reflection from the Moon. This should be an easily addressed question in geometrical optics.

  Here's a nice video on the Google Lunar X Prize competition:

Moon 2.0: Join the Revolution.
http://www.youtube.com/watch?v=9K4zosGUMBw

 

  Bob Clark

 (That's more than 25 words, but I think that's an easily understood summary.)

  Yes, thank you-- that was sufficiently dumbed- down for the peeps!

  So, you want to use the sun as a signal source.

  I wish you the best of luck in your future endeavors...