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Author Topic: Law of Inverse Square question  (Read 1206 times)
KC5AOS
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Posts: 54




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« on: January 24, 2011, 01:24:59 PM »

Ok, I know the Law of Inverse Square can be applied to both an isotropic radiator and toroidal radiation pattern such as from a 1/2 wave dipole.  From the law, we can deduce that in order to double the distance and maintain the signal strength at the receiver, we must multiply the power by 4.

The problem I am having is this; . . .   if you take a three dimensional model of a toroidal radiation pattern, cut it in half and compare it to a toroidal radiation pattern that is twice the diameter. . ie twice the radius or twice the distance, then compare the volumes of the two toroids, you find out the volume of the toroid that is needed to send the signal twice as far actually has 8x the volume of the smaller toroid.  This is also true for the isotropic radiator.

I am at a loss trying to understand this.  Given the volumes of the toroidal patterns, the power needed to increase the distance should be 8x if you relate the volume of the radiation lobes to the output power, but the law of inverse square says it should only be 4x.


Help me, my brain is frying.  Some math is fine, but no calculus please.
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WX7G
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« Reply #1 on: January 24, 2011, 01:57:46 PM »

It's not the volume that matters, it's the surface area.
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W1BVV
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« Reply #2 on: January 25, 2011, 07:17:18 AM »

The inverse square law only applies to fields radiating from point sources, uniformly radiating in all directions.  The basis unfortunately is from calculus, basically integrating in 3 dimensions with respect to r (radius) around the source.

Field strength from an infinitely long wire would be proportional to 1/r,  field strength from an infinite plane is constant (hence the field in a capacitor).

Antennas (other than a theoretical isotropic radiator) have complex radiation patterns making close integration solutions impossible and therefore the r dependence of radiation is very complicated, always falling somewhere between 1/r^2 and 1/r depending on pattern. 

Hope this helps.

Dave, W1BVV
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KC5AOS
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« Reply #3 on: January 27, 2011, 09:14:52 AM »

Welllll. . . . I think I'll accept the laws of physics as is and stop killing my self over it. Wink

Just would have been nice to fit the power output (or equate it) directly into the toroidal  radiation pattern.

Damon
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WX7G
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« Reply #4 on: January 28, 2011, 07:04:12 AM »

You are fitting the power into the toroidal volume. Fit the power into the toroidal surface area and you will understand.
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G8HQP
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« Reply #5 on: January 28, 2011, 08:41:26 AM »

Once you are sufficiently far away, any real antenna (i.e. non-infinite) looks like a point source. Therefore in free space you will always get 1/r^2 behaviour in the radiation zone. You can get other laws if you are not in free space (e.g. over plane or curved earth), or too near the antenna.

As others have said, area is the issue (not volume). Think of the power from the antenna flowing through the surface of a sphere. Double the radius, and you have to spread the same power over four times the area.
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KC5AOS
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« Reply #6 on: January 31, 2011, 06:53:50 AM »

Thanks guys.  ~4x to double distance.  I thought I had an Einstein moment, but I'll just leave the laws of physics alone.
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