[Laser] EME Experiment

[Dave]

> Message: 1
> Date: Thu, 18 Jun 2009 02:41:13 -0700 (PDT)
> From: Tim Toast <toasty256 at yahoo.com>
> Subject: [Laser] Fw: Re:  EME Experiment
> To: laser at mailman.qth.net
> Message-ID: <873233.7801.qm at web37905.mail.mud.yahoo.com>
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> 
> to the point is good  :-) 

:-)

> I agree lasers are probably the way to go along with filters to block most of the background light on the receiver. Xenon has some hefty IR lines, they are not very narrow but any bit helps i imagine. 

I don't know that a laser will get you anything over a monochromatic
source.  By the time the light reaches the moon, it's going to be
noncoherent anyway due to atmospheric distortion of the beam.  And,
once reflected from the moon, it's certainly going to be noncoherent
(well, unless you're trying to bounce it off of one of the corner
reflectors that the Apollo missions left up there).

Using a laser does give you a fairly narrow angle of divergence
automatically, and that part is good.  However, you may be able to
achieve the same thing with careful optics and a monochromatic source.
The problem, though, is that there aren't that many high intensity
monochromatic sources readily available.

While Xenon does have some IR light output, some of it is broadbanded,
while another component is in bands.  Plus, the relative components
depend upon the discharge current.  Higher current densities tend to
produce broader banded output, with more in the red and infrared portion
of the spectrum, while lower current densities tend to favour the bluer
end of the spectrum and less broadbanded output.

> I had once figured i would need a fresnel at least a meter with 1 meter focal length (f1) with this particular strobe tube to get the beam small enough to fit inside 0.5 degrees without loosing any more power and actually gaining a bit due to faster f1 optics.
> An eclipse has a lot of indirect sunlight especially in the red and IR i imagine. The blue end of the spectrum from xenon has some strong lines. A few ppl had said they might try it with photo multipliers or blue sensitive pin's and apd's

If I'm remembering my physics correctly, sunlight peaks mostly in
the green portion of the spectrum (although this may have some
contribution by the Earth's atmosphere).

Of course, during an eclipse, the colour balance of the solar
illumination of the Moon will be skewed.

There are bands of absorption in the atmosphere, primarily due to water
vapour, but also due to other compounds.  Part of the trick will be to
pick a spectral band (or region) where that absorption is minimal.

> The time domain software is a "gate" of sorts but it's not discriminating between other synchronized signals that may be present at other phases in a "sweep", random noise at least is gated out. I guess both time domain and frequency domain software are gates. It's amazing they were able to predict exactly to a few nanoseconds when to open the gate for the original laser ranging experiments. I can imagine them sorting through thousands or millions of "channels" looking for the weak signal in the time domain "band" where it was likely to be. 
> 
> I'm not sure at all about the math when it comes to figuring how much power its going to take to actually see a given spot size on the moon. I had thought someone calculated that one time in here, a certain amount of power or so if they had a small telescope at least or binoculars?? But that's with an observer on the moon not on earth next to the transmitter.

The problem is that the Moon is, mostly, a diffuse reflector.  Thus, the
power to perform a one way transmission is tiny compared to the power
needed to perform the two way/reflected transmission.

I'm aware that there have been some government/university experiments
with bouncing a laser beam off of retroreflectors left on the moon.
That should give a starting estimate, although the fact that they were
bouncing off of a (specular) retro-reflector, rather than the (diffuse)
reflection of the surface makes a HUGE difference.

Also, there are some concerns (legalities?) involved with firing high
power laser beams into the air.

> I was really hoping there was something besides lack of transmitter power or lack of receiver sensitivity to explain why it couldn't be done optically. I guess current optical receivers are still many orders of magnitude less sensitive than the best RF receivers. But optical transmitters are now way beyond anything RF can do i assume as far as putting power on a spot.

It can be done.  However, there are several factors working against
amateurs doing it.  One is that it takes quite a bit of power.  And,
that power has to be directed/focused in the appropriate direction.

Then, you have to compete with the sun illuminating the moon, and that's
a LOT of energy to compete with.

You can, of course, increase the transmitter power, at least to a point.
But, sometimes, it's more practical to increase the effective power by
limiting the bandwidth of the transmitted signal.  That lets you put a
narrow band filter on the receiver, which can block out over 99 percent
of the solar energy you're competing with.

As for the receiver, I don't think sensitivity is the problem.
Photomultiplier tubes approach the level of being able to detect single
photons (e.g., 931A).  The problem, though, is that you're not trying
to detect a single photon.  You're trying to detect a single photon from
within a stream of billions upon billions of solar photons.  So, it's
not necessarily a matter of increasing the receiver sensitivity, but
rather increasing the receiver selectivity.

> Billions of joules sounds like a nice round number per square foot, that should do something... So far i've only spent about 5 dollars on a plastic lens and a few other odds and ends. I live in an electronics junk yard thankfully 8)

I'd start with a narrow band filter on the receiver that's tuned to a
monochromatic transmitter source, and go from there.  The trick, though,
may be finding an appropriate monochromatic transmitter source.

To give you some rough numbers to work with, the sun is pumping out
something like 1KW per square meter (rough, back of the envelope
calculation value, not exact!).  Figure the size of the moon, and it's
albedo, and that'll give you a starting value to work with (and, it'll
be a pretty HUGE number).  But, by using a narrow band optical filter,
you can exclude 99+ percent of that solar energy.  That improves the
situation by a factor of 100 right there.  Then, by carefully selecting
the time/phase of the Moon, you can improve the numbers even more.
It'll still be a challenge, but it can be done.

Dave