[Laser] pictures from the Moon

Mon, 23 Feb 2004 04:51:19 -0800

Hi James and all,

It's the numbers that are "fishy".

173,000 square miles is right, but there are 5280 * 5280 square feet
per square mile, so it's "only" 173000 * 5280 * 5280 = 4.8e12 square feet.
A telescope with one square foot capture area would receive 1 / 4.8e12 =
about 2e-13 watts.  If there were no interfering light, that's a lot
of power.  A photon at 700 nm has an energy of about 2.8e-19 Joules,
so we're talking about 700,000 photons per second.  The human eye,
when adapted to darkness, can detect light as weak as a few photons
per second.

Saying that the NEP of a PMT is "x" watts is only meaningful when a
bandwidth is specified.  A good PMT has an NEP of about 1e-16 watts
per square-root-Hz, so its noise would equal our 200-femptowatt signal
in a bandwidth of 4 MHz.  The major impairment would be quantum
noise, limiting the data rate to perhaps 100 kilobits per second.

Of course, in the real world, there is lots of interfering light.
When the region with the laser is in daylight, communication is
clearly impossible.  But even earthshine is bright, when compared
with these power levels.  One could mitigate the interference
with a narrow optical filter, and/or by using a pulsed laser and gated
receiver.  However, those calculations are beyond my capabilities.

EME communication on 1296 MHz with 3-meter dishes normally needs
several hundred watts for conventional CW, or several tens of watts
for advanced modes such as JT65.  In either case, it takes at least
a few minutes to exchange call signs and signal reports.  The
effective data rate is about one bit per second, hardly useful for
sending images.

Hope that makes you feel better.

73,

Stewart KK7KA

----- Original Message ----- 
From: <[email protected]>
To: <[email protected]>
Sent: Sunday, February 22, 2004 8:58 PM
Subject: Re: [Laser] pictures from the Moon

> If Art's math is right, then  to receive a one watt laser beam from the Moon
> with a 14 inch diameter amateur telescope (which has an area of about 1 square
> foot), I need to narrow the beam by the ratio of about 470 miles divided by
> the square root of 125 square feet.
>
> Thats about 19 miles.  The beam divergence of 19 / 239000 is about one
> thirteen of a milliradian.  The 1 meter telescope that the Lunar range
experiment
> uses sends a more powerful pulse of light to the Moon that expands to about
2000
> meters.  The technology exists, to send the beam down, but it is not in the
> realm of amateurs.
>
> Well, shucks!  If radio waves are the way to downlink, then I guess I cannot
> use this to promote thousands of light communications experimenters.  The idea
> of picture from the Moon is still good, I just wasn't trying to promote
> interest in radio communications.
>
> Ok.  If those are the numbers for a 1 watt signal sent from the Moon to the
> Earth in a beam that is 2 mR wide, then those would also be the numbers for a
1
> watt radio signal signal with the same beam width.  Seems to me that the
> telemetry antennas used on the Moon had beam widths of more than 5 degrees (87
mR)
> or more.  Receive antenna were more than 50 feet in diameter.
>
> I don't understand.   If a 1296 MHz radio contact is possible with 5 watts of
> power sent up to the Moon and then bounced back to the Earth, then why can't
> I see a laser beam from the Moon in a 6 inch amateur telescope?  If I can see
> it, then why can't I detect it electronically?   (Grumble.  Grumble.
> Grumble..............)
>
> James
> N5GUI
>
>
> #########################################
>
> In a message dated 2/22/2004 12:52:08 PM Central Standard Time,
> [email protected] writes:
> At 12:07 PM 2/22/04, you wrote:
> >On Sun, 22 Feb 2004 11:30:24 EST, you wrote:
> >
> > >The Moon is tide locked
> > >to the Earth so that once aligned, neither the camera nor the downlink
> > >antennas would need to be adjusted.
> >
> >If I remember right, there's some wobble to it. I wonder how the
> >wobble compares to the spread of the laser beam?
> >
> >Of course, the trick would be to make the laser beam wide enough that
> >the variation in the moon's orientation w.r.t. the earth wouldn't
> >matter.
>
> The wobble is called libration. It happens because the Moon wants to turn,
> but the Earth's gravity yanks it back-there is a proverbial tug of war.
>
> The Earth has libration as well, which is influenced by the Sun, other
> planets and the Moon itself.
>
> I doubt that a laser beam that is spread out over 470 miles diameter would
> be readable on earth for 2 reasons....
>
> #1      The Moon itself has libration, which causes the ambient light level
> from the Moons reflection to vary in phase and amplitude. This variation is
> much stronger than the laser is and would be interference.
>
> #2      The field strength of a laser spread over a 470 mile diameter would
> be nil, you just wouldn't be able to have a light collector large enough to
> recover any real amount of laser power. Let's say the laser was 1 watt and
> that there was no loss in the receiver optics or the collimation optics on
> the transmitter. Let's also say there is no attenuation or scattering from
> the Earth's atmosphere. Let's also assume our receiver has a 10 femtowatt
> NEP, which is about what a PM tube has. Let's also assume there was no
> interference from the ambient light reflected from the Moon (which is a
> combination of Sunshine and Earthshine).
>
> The area of the footprint on the Earth is Pi X R X R, which is 3.14 X 235 X
> 235 (in miles), which is about 173,000 square miles, which is also about 8
> X 10e17 square feet.
>
> The total area of the footprint on the earth is 8 X 10e17 square feet.
>
> 10 femtowatts divided by the output power of the laser is 1watt/10 X
> 10e-15= .1 X 10e15 or 1 X 10e14.
>
> The area of our lens (in square feet) needs to be 1 X 10e14/8 X 10e17=125
> square feet.
>
> Thus, we would need a 125 square foot lens to recover 10 femtowatts of the
> original signal.
>
> My math is rusty, perhaps I made an error?
>
> It seems to me that a simple fixed laser pointed towards the Earth is not
> going to be useful for any sort of real data transfer.
>
> I think the laser must be collimated to a very narrow beam and that it has
> to be a tracking laser in order to keep it pointed at the same spot on Earth.
>
> I would think that a radio link would be much more practical.
>
> Regards,
>
> Art
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