[Laser] Fw: Re: EME Experiment

[Tim Toast]

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 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

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.

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.

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)





--- On Wed, 6/17/09, mike1 at mgte.com <mike1 at mgte.com> wrote:

> From: mike1 at mgte.com <mike1 at mgte.com>
> Subject: Re: [Laser] EME Experiment
> To: "Tim Toast" <toasty256 at yahoo.com>, "Free Space LASER Communications" <laser at mailman.qth.net>
> Date: Wednesday, June 17, 2009, 10:34 PM
> 

Hi Tim,

I asked a friend of a friend via email various questions
regarding your experiments and here is his reply. He's quite to the point and spares no feelings. Hey, I'm just the messenger, so don't shoot me. :0)  Here's his analysis:

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>


(That is, the flash was on for 1 millisecond.)

> So how does that compare to the beam of a 20W CW laser?

1) The flashlamp pulse is broadband, whereas the laser puts out
a relatively narrow band of wavelengths. I get into the importance of this below, in #3.

2) I kinda doubt that you could see the spot from a 20W CW laser
on the moon; but if someone shows one to me I'll be happy to
change my mind.

3)  When people were first starting to use lasers to determine
the distance to the moon, they used an 8-joule laser pulse about
30 or 35 nanoseconds long, which they directed out through a
_very_ large telescope. This gave them a spot on the moon
that was relatively small. (I forget the exact size, but I think
there was an article in Scientific American about this, probably
some time around 1971-74 if I remember correctly, and they
give some numbers.) The pulse power from the laser was about
250 million watts -- 1 joule per nanosecond would be 1 billion
watts, but they had only about 1/4 of that.

On the moon was a special retroreflector, which bounced
the beam straight back toward the telescope, to within a very
narrow angle. Unfortunately, when the beam got back here
it was aimed wrong, because of the speed of light limitation --
the telescope had time to move a bit away from where it had
been when it sent out the pulse, so it was looking at the edge
of the return instead of the center.

I seem to recall that they detected 1 or 2 photons on an
average shot.

Got that? 8 joules out, millions of watts, carefully aimed into
a special reflector using a _big_ telescope, ...and they got
a couple of photons back.

It really helped that they used a Q-switched laser source that
put out nice short pulses, for two reasons --

1) They were able to filter out the light at other wavelengths,
and look only for photons with the correct energy. You
can't do that with a flashlamp.

2) Because they knew the timing of the outgoing pulse and
they knew the approximate distance to the moon, they were
able to gate their receiver so it was only looking for a return at roughly the correct time. Otherwise, they wouldn't have been
able to see anything at all.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

It is just crucially important to do the numbers on things like this before you try them. You don't have to know about the laser
project that I mention above; forget that, and just calculate the
amount of energy it would take. The other Mike doesn't seem
to have bothered to do this, so let's do it for him.

I'm going to assume that he has a nice telescope to look through,
with maybe 1 square foot of mirror area (that's probably a 14"
scope or thereabouts, pretty hefty for amateur stuff but not
out of the question), and I'm going to specify that he is _not_
sending the flash out through the scope. (We will, however,
allow the flash to be focused.) The moon has an albedo of
about 0.12, according to at least one Web page I've found.

The eye needs a bunch of photons to get enough contrast to see
anything when it is looking into anything other than blackness;
let's guess that it takes 10, though the actual number may be higher.

The surface area of a sphere is 4*pi*r^2. If we pretend that the moon is reflecting into half a sphere, there are 2*pi*r^2 square feet at our distance, where r is 238,000 miles * 5280 feet per mile; that is, something on the order of 10^19 square feet. This means that the moon has to reflect about 10^20 photons in order for him to be able to see anything through his handy backyard telescope, because 10 of those photons need to reach his particular square foot. The moon only reflects about 1/8 of the photons that hit it, though, so he'll need to put about 8x10^21 photons into it in order to see any when they get back here.

Now, if we assume that his flash unit is very tightly focused, and that it radiates into only 1/100 of a hemisphere, this means that there are only about 10^15 square feet at the distance of the moon (1/10,000 as many as there would be if it radiated into an entire hemisphere).

Thus, if it were actually radiating into just the moon, it would have to put out about 8x10^36 photons. Unfortunately, the moon is only 1/2 degree across, which I'm afraid is considerably less than 1/100 of a hemisphere, so the actual amount is a whole lot more because of the many photons that go past the moon instead of hitting it. Just for simplicity, however, I'm going to ignore that, and pretend that all of the photons that make it out through the atmosphere actually go to the moon.

The average energy of a visible photon is something on the order of 2 eV, so the energy in the outgoing light pulse from his flash unit will be about 1.6x10^37 eV. 1 eV is about 1.6x10^-19 J, so unless I have this wrong, the flashlamp needs to put out 10^18 Joules.

That's _1_billion_billion_ Joules, and that's the _output_; the 
electrical input to the lamp is higher, because the efficiency is less than 100%.

The bottom line is that even if my calculation is off by a factor of _100_billion_, this is a totally ridiculous waste of time and cycles.

(He _might_ be able to make a big enough flash for someone on
the moon to see with a nice telescope, if he had a _lot_ of money
to spend on a huge light source; ...but there isn't anybody up there to see it.)

>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>

So Tim, that's it in a nutshell. Hope there's no offense taken, he came right to the point for sure.

Best Regards,

Mike C.
Milton, FL