[Laser] pictures from the Moon
Art
[email protected]
Mon, 23 Feb 2004 10:55:06 -0500
What a difference a factor of 173,000 can make. Thanks Stewart.
I'm not so sure that the problem of ambient light can be managed, even with
the best optical passband filters, which are +/- 1 nm. Obviously, a plain
old laser diode won't do for a transmitter, they shift frequency as they
age and with temperature changes. A He-Ne would be practical.
I'm not sure the doppler shift would allow such tight optical passband
filters to be used though.
In any event, it would still be a technical challenge.
Keep dem lasers lit.
Art
At 07:51 AM 2/23/04, you wrote:
>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
> > _______________________________________________
> > Laser mailing list
> > [email protected]
> > http://mailman.qth.net/mailman/listinfo/laser
> >
>
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