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Old October 21st 09, 12:17 AM posted to rec.radio.amateur.homebrew
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First recorded activity by RadioBanter: Jul 2008
Posts: 242
Default The dish problem

Hey OM

If I look at the node of a source of 20mW and connect 10 wires to that
node with equal loads, I will have 2mW going down each wire. No matter
how I try and add them up I can't get more than 20 mW.

Now I take a 20 mW isotropic emitter, now your'e saying I feed a dish
with that 20mW emitter through a feed horn and now all of a sudden I
have gain, I have more than 20 mW? Because the dish and feed horn are
able to add the particles and waves, where as with 10 wires I cannot
add the waves and particles?

The only difference in both these are the shape of the dish and the
medium used to convey the wave, particles.
And you can add the waves particles until you run out of dish.


73 OM
de n8zu



On Oct 18, 8:40 pm, brian whatcott wrote:
terry wrote:

...

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.


....
That 700 watts per sq. metre is an interesting number? Bright sun

at midday at the equator perhaps?
But I doubt that here at around 48 degrees north and with our climate
we get anything close to that?
Also now wondering how efficient a solar panel is at converting that
700 watts into electric power (for say home use). Which may be why
some 'solar' collectors are both photo-voltaic and heat collecting.


The scientific name for this measure is Solar Irradiance. If you hold up
a square frame 1 meter on a side, perpendicular to the Sun's direction,
in space near Earth's orbital distance, about 1400 watts passes the
frame - and twice as far from the Sun, that frame only frames a quarter
of that amount (of course!)

That number 700 W/m^2 takes into account the absorption of power by
stuff in the atmosphere, more at slant angles of course.

This URL can show the kind of Solar power you can expect at various
geographical coordinates.http://re.jrc.ec.europa.eu/pvgis/apps/radday.php

You'll see a peak value around noon - and this can be more than 700
W/m^2 at that 48N latitude.

Brian W


 
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