Re: Laser ranging to moon begs questions

On 22 Oct 2007 23:52:45 -0400, Craig Markwardt
<craigmnet@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


John C. Polasek <jpolasek@xxxxxxxxxx> writes:

On 21 Oct 2007 22:09:28 -0400, Craig Markwardt
<craigmnet@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


John C. Polasek <jpolasek@xxxxxxxxxx> writes:

On 21 Oct 2007 14:21:10 -0400, Craig Markwardt
<craigmnet@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


John C. Polasek <jpolasek@xxxxxxxxxx> writes:
On 20 Oct 2007 08:32:03 -0400, Craig Markwardt
<craigmnet@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:


Yep, for example:
http://ilrs.gsfc.nasa.gov/stations/sitelist/MDOL_sitelog.html

Pulsed ND Yag laser, pulsed at 10 Hz with a 200 ps duty cycle. Those
parameters appear to be pretty standard (though the peak power varies
from one station to another).

Also, the beam-width of most laser systems reported on that page
appear to be in the 10's of arcsecond range, not the 0.075 microrad
values previously reported.

Craig
This spec says divergence is adjustable from 0-20 seconds. It also
says "final beam diameter .75m ". This latter is probably as adjusted
out of the telescope which may have a final mirror of .75m.

snip

This is really valuable information that straightens things out
considerably:
From Dickey et al. 1994,
Ranging to the moon is technically challenging. An outgoing pulse
of laser light transmitted from the a collimating telescope with a
beam divergence of 3 to 4 s of arc, consistent with atmospheric
seeing, spreads to an area of approximately 7 km in diameter on the
lunar surface. ...

[Corner cube ] arrays intercept only 10^{-9} of the area of the
impinging light beam. The angular spread of the returning pulse is
set by diffraction, polarization properties, and irregularities of
the array's individual corner cubes. In the case of the the 3.8 cm
diameter Apollo corner cube, the spread is approximately 10 sec of
arc. Thus, the diameter of the spot produced on the Earth is
approximately 20 km. A 1 m diameter receiving telescope would
collect only 2 x 10^{-9} of the returning photons. A variety of
practical matters, such as quantum efficiencey, mirror reflectance,
optical performance under thermal stress, and velocity aberration
(which slightly shifts the center of the returning beam from the
location of the transmitting and receiving telescope), make the
product of the transmitting, lunar retroreflecting, and receiving
efficiences considerably less than unity. The overall signal loss
of approximately 10^{-21} puts a premium on the detailed design of
ground stations to minimize their losses.
...


With 20 seconds of beamwidth, (which I doubt, since, if it's
adjustable, what would be the point?) we reduce the photon count from
6.43e22 to 4.02e16 photons/s/m^2. The field strength would go down by
sqrt(1.6e6) or1265 to 1.9millivolts/meter.

According to Shelus et al, approx 3 x 10^{17} photons are launched per
pulse, which is reduced by the losses to significantly less than one
returned photon per pulse (approx 1/3rd of a return per minute, at a
10 Hz pulse rate). I have read elsewhere that under the best
conditions (rather than average), the McDonald station can return 1
photon per minute.

On the other hand, newer stations like APOLLO (Apache Point) can
return several photons per pulse.

CM


References

Dickey et al., "Lunar Laser Ranging: A Continuing Legacy of the Apollo
Program," Science, 1994, Vol. 265. no. 5171, pp. 482 - 490

Shelus et al. "McDonald Observatory Lunar Laser Ranging: Beginning the
Second 25 Years," IAU Symp, 1996, v.172, p. 409
Thank you Craig for getting some real data.

Using their figures, the attenuation, as I figure it, is 7.7x10^-28
yielding a "robust" 15 photons per pulse m/l in line with their
"several photons per pulse".



You are mistaken.


from a 3 second divergence and not microradians.

The area of a *single reflector* is 3.8 cm
diameter, but the assembly has multiple reflectors. However, the 3.8
cm number is relevant for the returned diffraction pattern.


They reckon the
return patch 21km vs the first patch of 7 km arose from divergence
angle tripling from 3 sec to 10 seconds on the way back. Twice would
be better.
My calculations:
WL = 532nm
Ep = hc/WL = 3.7e-19J per photon
Ebeam = 1.75J Pwide = 200ps

N = photons/pulse = Ebeam/Ep = 4.02e18 photons/pulse Shelus:3e17
(Shelus figure is 13 times lower (wd make 15ppp go to 1ppp). I will
argue for my figure: 1.75joules, 3.73-19J/photon.

You of course are assuming peak power output which appears to be
incorrect.
No. I am using the stated *energy* per pulse of 1.75J, and dividing by
the classical energy per photon which has to be
Ep = h*c/532nm = 3.7e-19J per photon giving
N = 1.75J/3.7e-19JP = 4.02e18 photon/pulse as above.

N' from article = 3e17, fewer by 13x
Why the 13 times discrepancy? 4 numbers, all givens, are involved.

The difference is likely to be that the energy consumed by the laser
is not all directed at the target. This is one of the inefficiency
factors involved.
Yes, 8% efficiency would cover 13x. But how separate out, just from a
reading of input power, the 532 nm energy draw from that of the
1065nm energy "that is not used"?

Phorate = N/Pwide = 2e28 ph/sec 1 pulse during pulse.
This is photons per second during the pulse.
Calculating the photon rate "per pulse" is not relevant here, only the
total number of photons per pulse.
I see you calculating the seconds per pulse below. That is a rate.
Total number is my N = 4.02e18.
In fact I make direct use of Phorate down below in my paragraph which
you had the bad taste to cut out.
Calculate attenuations starting from the .75m scope, their 7km patch
At1 = bexit/patch1 = (.75m/7km)^2 = 1.15e-8 7km patch not 28m
At2 = mirror/patch1 = (3.8cm/7km)^2 = 2.95e-11 3.8cm not 50
At3 = receiver/patch2 =( 1m/21km)^2 = 2.6e-9 1m vs .1m
Atoveral = 7.67e-28

I can't believe you cut this out, so here it is again:
Net photon rate 1 pulse = Phorate x Atoverall =15.4 ph/sec/pulse
This translates to 15 photons per pulse which is considerably more
than their newer stations record of "several photons per pulse".

Your value (and units) are non-sensical. If you *really* mean 15.4
photons per pulse *per second of laser on-time* -- per above, you
described as "ph/sec ... during pulse" -- then to find photons per
second, you should multiply by 10 pulses per second, times 200
picoseconds per pulse. The result is an absolutely miniscule number
(3e-8 ph/sec).

But that is because you made multiple errors in calculating the
attenuation.
(a) your value for the area of the retroreflector array was
incorrect (as noted previously);

I used their literal value of 3.8cm, and lost track of previous
reference describing the array size.

(b) you "double-counted" attenuation by computing (.../7 km)^2 twice;

No, I was working power density, so A1 is density thinning. A2 is
geometric "catch" ratio for the mirror. A3 is the ground geometric
ratio.
(c) you did not account for other non-geometric losses, as detailed
by Dickey above.

I dont think they were detailed.

Even if one were to assume pure geometric calculations (which are
incorrect), your calculations are still incorrect. The number of
photons received at the retroreflector are,
You evidently determined a new array size of .33m whereas I took their
.038m

Apparently you didn't bother to read the note above. The diameter of
a *single* corner cube is 3.8 cm, but there are 100 cubes.

I don't see a reference to 100 cubes; you just introduced it.

N_retro = N_transmitted * (A_retro/A_lunar_spot)
= N_transmitted * (1100 cm^2 / 38 km^2)
= N_transmitted * (~2 x 10^{-9})

which is in line with the Dickey quotation. The number received at
earth is,

N_received = N_retro * (A_receiver / A_earth_spot)
= N_retro * (0.44 m^2 / 346 km^2)
= N_retro * (~1 x 10^{-9})


again comparable to the Dickey estimates. However, what you *don't*
account for by the geometry-only approach is the other loss factors
described above. The true attenuation is around 10^{-21}, but this
actual value depends on observing angles and other factors.

So for 3e17 photons launched per pulse, about 0.0003 photons are
received. Thus, it takes about 300 sec to receive one photon (my
previous estimate was incorrect).

What is the use of all the labored precision if your attenuation,
which is 2.42e-18, which,when multiplied by 3e17 photons comes out as
.726 photons returned, and you "round it off" with a 2400x adjustment
to 0.0003 photons????

Apparently you didn't bother to read about the other system losses as
described by the Dickey et al paper.


The Apache Point observatory (which is *different* than McDonald) is
much more capable because of the much larger aperture area on the
earth.

CM
I think I've had enough of this; it's too time consuming. New
parameters are being brought up from time to time to change the
results.

Lunar laser ranging is a complicated system. You are pretending to be
able to solve/refute it with a few unsubstantiated and erroneous
calculations.

Someone might want to recheck on an apparent conspicuous error in
the photons per pulse using 1.75J and dividing by E = Hc/lambda at
532nm and see why I get 4.02e18 photons and the study cites 3e17.

The Dickey paper describes geometric losses of about 10^{-18}, but
total system losses of 10^{-21}. I am willing to admit that the laser
might consume as much as 1500 mJ per pulse, but only a ~tenth of that
may end up directed through the telescope and to the moon (hence the
3e17 photons).

Still, in that best case scenario, using your emission value, the
total returned signal is (4e18 x 1e-21 = 0.004 photons / pulse). In
other words, the returned signal for the MRLS system is still
miniscule. [*]

CM

[*] - As mentioned before, the new Apache Point system has quite a bit
better throughput, but Apache is not MLRS.

Why don't we start a new thread detailing how it is possible to range
the moon to within 1 cm if, as you say, there is only 1 photon every
300 seconds. The said photon could have originated anywhere in a 60
meter span: 3e8*200*e-9 = 60 meters.
Question begging continues, ameliorated only by a sense of tedium.
John Polasek
.

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