[Laser] Ramsey Kit range

[David D. Rea]

On Sun, 2003-11-09 at 15:20, TWOSIG@aol.com wrote:
> On the other hand, the area of beam, or the power in it, may not be descibed 
> by a mathematical relation to the square of the distance.  For "short 
> disatances" the intensity of the received beam may be linearly proportional to these 
> "short" distance numbers.  If that is the case then the power on the detector 
> would be proportional to the area of the lens, so that the useful range is 
> factored by the square of the diameter ratio of the lens to the detector.
> 
> Both ways of thinking about range seem to be oversimplification of the "real" 
> world, but is either a practical estimating tool?

Hi James-

You're fairly close to on-target with your theory. The received power
will fall off with the square of the distance, but this only occurs in
freespace. Keep in mind that you've got atmospheric nasties to deal with
as well - after you get above 1 Km or so, you'll start noticing the
effects of humidity and purturbations in the air; i.e. you'll lose power
due to absorption and dispersion of water mollecules, and you'll see a
"shimmering" effect as the beam traverses different thermal planes
between the transmitter and the receiver.

There has been quite a bit of work done in this subject, as I found out
when writing a paper on laser communication during an undergrad
independent study. You can get as deep in the math as you want; there's
no shortage of folks out there who have written PhD theses on this
stuff...

To add a layer of complexity, remember that no lens is perfect, either.
If you're dealing with a nice AR multi-coated glass lens, maybe you'll
throw away 5% of your signal (if you're lucky). But if you're using a
less expensive (but MUCH larger area) fresnel lens, plan on tossing
about 50% of the inbound light back toward the receiver. This is where
you compensate for reflection with sheer lens size...

Anyway - hope this helps a bit...

73 de Dave K2THZ