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Old October 16th 09, 08:08 PM posted to rec.radio.amateur.homebrew
brian whatcott brian whatcott is offline
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First recorded activity by RadioBanter: Jul 2009
Posts: 48
Default The dish problem

Hi Mark,
I looked over the thread that your note kicked off.
Unusually, this was not very informative.
People were sticking with antenna measures, which weren't helping much
in this case.

I'll give it a try, introducing the missing concept: energy (and
power) DENSITY.

If you sit in the bright sun for 2 hours, you are likely to get a sunburn.

If you sit in a dark tent, with an opening for a magnifying glass of 2
inch diameter ( = 5 cm), you will get a small burn within 2 minutes.

Looking now at the numbers, you know that the effective power density
of the Sun at the surface is 700 watts per square meter. Your skin
represents say 1.5 sq meters, so you are getting a total power of say 1
kilowatt.
So we are suggesting, at a particular place it takes 1 kilowatt of solar
power for 2 hours to burn you all over.

Now lets look at that dark tent with a hole for a magnifying glass:
the power density at the glass is pi times 1 times 1 inches or pi X .025
X 0.025 square meters = 0.002 sq meters.
The Solar power at the glass is 700w X 0.002 m^2 = 1.4 watts

How the hell can 1.4 watts burn us in 2 minutes, but 1000 watts cannot
burn us in less than 2 hours?

Answer power density:
That 1.4 watts is focussed on a spot that is 1 mm diameter
( 40 / 1000 inch diameter) so its power density is
1.4 / pi X 0.001 X 0.001 watts per sq meter

....and THAT is 446 kilowatts /sq meter.

Oh look: the power density of the spot in the dark tent is about SIXTY
times as great as plain sunlight, so waddaya know, it burns us 60 times
as fast (but only over a LITTLE spot.)

And that's what this thread was all about - the effects of focussing
energy over a particular area.

I hope this helps.

Brian W

MarkAren wrote:
Hi All,

Modern engineering text indicates that for the same frequency, a large
fully illuminated dish will provide more gain than it’s smaller
equivalent. Why is this ?

Consider the following scenario:

======

Take a 10GHz RF source, and attach an appropriate feed horn to fully
illuminate a 2m dia dish.

Adjust focus so that the wave front out of the dish is parallel (2m
diameter ?)

At a distance, arrange the same setup, except the RF source is
replaced by a detector.

Make a note of received signal strength.

======

Repeat the same experiment at the same frequency, but with a
3mdiameter dish at each end, and appropriate feed horns .

Make a note of received signal strength.

======

Also, see http://www.idesignz.org/misc/dishproblem.pdf

In both cases all of the TX energy is transmitted in a parallel beam,
whose diameter is the same as the respective dish.

It would seem that if you can fully convert all of the TX RF into
parallel waves, at some point the diameter of the dishes should not
make any difference to the transfer efficiency of the system.

So, why would the received signal for the 3m + 3m dish be greater than
that of the 2m + 2m dish experiment ?

Thoughts please.

Mark

http://www.idesignz.org/AMPS/AMPS_BS.html – Experimental AMPS micro
Base Station project

http://www.idesignz.org/UAV/index.html – Early work on the four rotor
FlyingThingy