[Laser] Re: Lunar Eclipse #3

Wed Feb 27 02:24:04 EST 2008

Hi Tim,

Some more thoughts ...
If you have light leaving the earth and illuminating the whole side of the 
moon facing us, then the path difference between a ray to the center of that 
face and one to the edge is twice the radius of the moon, one radius longer 
going and one radius longer coming back.  Pretty much a pulse would be 
lengthened by the time of flight of two moon radii.  The main difference 
will be in amplitudes ... a ray illuminating the center area reflects back 
to us normal to the surface.  A ray going to the edge has its energy spread 
over a large area and we are way off to the side from nornal.  Of course, 
objects with surface roughness of even a few mils scatter light almost 
omni-directionally.  The is not quite like 2 meters where 1 foot roughness 
is "flat".

John Matz KB9II

----- Original Message ----- 
From: "Tim Toast" <toasty256 at yahoo.com>
To: "laser mailinglist" <laser at mailman.qth.net>
Sent: Tuesday, February 26, 2008 2:42 AM
Subject: [Laser] Re: Lunar Eclipse #3

> here are some of the numbers without crunching them too
> much:
>
> speed of light 299792.4 km/second
>
> wavelength of 120 Hz = 2498.27 km
> wavelength of 360 Hz = 832.75 km
> wavelength of 100 Hz = 2997.92 km
> wavelength of 300 Hz = 999.30 km
>
> radius of moon 1738 km
> radius of earth 6378 km
>
> TOF 1738 km = 5.7973 x 10-3 seconds (5.7973 milliseconds)
> TOF 6378 km = 2.1274 x 10-2 seconds (21.274 milliseconds)
>
> pulse of 5.7973 ms = half cycle of 86.247 Hz sinewave
> pulse of 21.274 ms = half cycle of 23.502 Hz sinewave
> pulse of 1.0000 ms = half cycle of 500 Hz sinewave
>
> length of 1 ms pulse = 299.79 km
>
> While pondering things, one thing that struck me was this
> broadening of a pulse by the reflection off the curved
> surface of the moon. And this is also assuming you are
> illuminating the whole surface of the moon (0.5 degree beam
> or larger). The moon's radius of 1738 km seems to lead to a
> minimum pulse length of 5.79 milliseconds regardless of how
> short the initial pulse was.
>
> So, if you are using the entire visible surface of the moon
> on your photodiode, any pulse reflected back to you will be
> 5.79 milliseconds long plus the initial pulse length. Is
> this correct?
>
> For example, if you send a 1 millisecond pulse, the
> reflection would be 5.79 + 1.000 or 6.79 ms long. I assume
> this is some sort of constant well known to the RF moon
> bouncers, and is related to the fastest modulation that can
> be sent without being garbled by the pulse lengthening that
> occurs with wide FOV's.
>
> This 5.79 ms pulse length can also be thought of as equal
> to a frequency of 86.247 Hz. The 5.79 ms pulse in this case
> represents a half cycle of a sinewave of 86.247 Hz
> frequency.
>
> How this relates to the frequencies of 120 Hz or 100 Hz i'm
> not sure. Would modulations above 86 Hz be diminished when
> using the entire moon on your photodiode?
>
> If your FOV is LESS than the entire hemisphere of the moon,
> then it seems like higher modulation frequencies would be
> OK. There's probably a formula for how the pulse
> lengthening relates to field of view. But it seems like,
> the smaller your field of view, the higher the frequency
> you can see (ungarbled by pulse lengthening).
>
> I was thinking, if your field of view is half the moon's
> surface (0.25 degrees), then you might have twice the
> modulation resolution. Meaning instead of 5.79 ms pulses,
> you would have 2.895 ms reflected pulses and 172 Hz
> bandwidth capability. Does this sound like it could be
> correct? I'm sure this translates badly, sorry.
>
> A picture is worth a thousand words 8)
>
> This concept of circular bands of light moving from center
> to limb was mentioned in here before but i'm not sure who
> brought it up.... Anyway, imagine a single light pulse sent
> toward the moon. When the resulting plane wave front
> arrives, it strikes near the center portion of the moon
> first. Then radiates in a circular pattern outward toward
> the edges (limb). The time it takes to go from the center
> to the limb is about 5.79 milliseconds. (TOF of Moon's
> radius)
> If you send out pulses at a rate of one every 5.79
> milliseconds, then just as one pulse has radiated outward
> to the edge, another one starts at the center and moves
> outward. If you space out the pulses to about twice this or
> 11.58 ms, then you have a nice 50/50 duty cycle pulse train
> being reflected back to you from the moon. That is the
> 86.247 Hz signal that appears to be an upper limit for
> undistorted modulation when using the entire moon's surface
> on the photodiode.
>
> I hope this calls to attention anything that might be of
> value for anyone pondering these problems.
>
>
> ------------------------------------
>
>>I need more  time for this, and I need
>>paint two pictures for ilustration of  situation.
>>I hope, that my teory evoke interesting  discussion.
>>shortly:
>>1. wavelenght of 120Hz is 2500km! after  phaseshift of
> 3phase powerline is
>>833km!
>>2 Diferent between short  and long way of bounced light is
> minimal 4800km!
>>3. Lamp is powered from  long distance from few electric
> power station
>>....this is next phase  shift.
>>4. more milions sources of different phase!
>>5. phase shift  of 120Hz earth surface lamp is interlaced
> up to DC level.
>>Is it any wrong  at this?
> Robo
>
>
> ...
>>With the  synchronic networks, the chaotic light sources
>>repartition could
>>build some  beats with by constructive or deconstructive
>>adding's.
>>This effect could  create circular interferences from the
>>heath and moon
>>curvatures.
>>It is  possible a fraction of percent of modulation can
>>occur in favourable
>>circumstances.
>>The city light pollution is equivalent to hundreds
>>megawatts  optical power.
>>The FFTDSP are so sensitive it seems to me something could
>>be  detected in
>>very favourable circumstances.
> ...
>>73 Yves F1AVY
>
>
>
>
>
>
> 
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