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Author Topic: Dummy question about how a Fan Dipole works?  (Read 1802 times)
WALTERB
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« on: July 08, 2011, 08:33:17 AM »

This is really a radio 101 for dummies quesiton, but thats why I'm asking.  Grin

If I broadcast on 40 meters on a 20 meter dipole without a tuner my SWR will be really high correct?

However on a fan dipole cut to 40, 30, 10, and 20 meter it won't.

So how do the electrons flowing through the wire "know" which one of these sets of wires to blossom out from?

in my mind it seems almost like one of those old coin sorters where the nickel rolls past the dime and penny slots because of its size and falls into the nickel slot.

I assume the size of the wave lengths are basically the same concept?

I told you it was dumb question.   Grin
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AE4RV
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« Reply #1 on: July 08, 2011, 08:39:16 AM »

The electrons are lazy and will seek the path of least resistance.
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KC9TNH
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« Reply #2 on: July 08, 2011, 08:49:39 AM »

The electrons are lazy and will seek the path of least resistance.
ROFL Grin
(That is classic sigline material.)

I believe this slovenly behavior applies to the 30,000 amp pretty blue bolt from the sky as well, no?

(now I've got to explain to the office what I'm laughing about)
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73
Wes -KC9TNH
"Don't get treed by a chihuahua." - Pete
AD4U
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« Reply #3 on: July 08, 2011, 09:09:00 AM »

The electrons are lazy and will seek the path of least resistance.

That is classic! 

Seriously in the fan dipole you speak of, MOST of the RF on which ever band you are transmitting on will tend to seek out (go into and radiate from) the section of the antenna that presents closest to a 50 ohm impedance.  Hence an acceptable SWR.  Conversely MOST of the RF will tend to ignore the other "elements" in the fan dipole that present something other than a 50 ohm impedance.

Dick  AD4U
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W6RMK
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Posts: 650




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« Reply #4 on: July 08, 2011, 10:05:54 AM »

Here's a bit more rigorous explanation...

Let's say you have a 20m dipole and a 15m dipole, each resonant on their own band.  The feedpoint impedance on 20m is around 70 ohms, ditto on 15m.  But the "out of band" dipole has a huge reactive impedance so not much current flows into it... here's some actual numbers.

What's the impedance of the "other" antenna that's in parallel..
The 20m dipole at 21 MHz is about 314+945j
the 15m dipole at 14 MHz is about 24-526j

So, at 21 MHz, all the current is going to go into the 15m dipole, and a little bit into the 20m dipole, which kind of looks like a series resistor and inductor across the 15m feed

And, same thing the other way.. at 14MHz, the 15m dipole looks like a capacitor across the feed of the 20m.  With not much current flowing (500 ohms vs 70)..

In practice, what you do is make the 20m dipole slightly longer (so it has an inductive reactance that just tunes out the capacitance of the 15m dipole.  You make the 15m dipole slightly shorter, so it's capacitive reactance tunes out the reactance of the inductor )..

In real life, the two will interact anyway, so you just fiddle with it until it works.

BTW, you can solve for 2 lengths that will do a nice two band fairly easily, but when you get to 3 or more, it gets very hairy (solving 3 equations with 3 unknowns, etc.)
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K0BG
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« Reply #5 on: July 08, 2011, 01:31:20 PM »

Don't pick on Mr. Noles, because his answer has a lot of truth behind it, albeit a bit of tongue in cheek.
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W5FYI
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« Reply #6 on: July 08, 2011, 01:43:27 PM »

W6RMK has a good explanation. Here's another way of looking at it; your 20-meter signal going to a 20-meter antenna should have an SWR of less than 2:1 (assume a 50-ohm coax connected to a 70-ohm resonant antenna, the SWR will equals 1.4:1). At 1.4:1, the antenna accepts around 97 percent of the power from the feed line. The same signal going to a 15-meter dipole sees 24 ohms of resistance along with 526 ohms of reactance, which is equivalent to an SWR of 233:1--a huge power roadblock.

In effect, the non-resonant legs of fan dipoles restrict transfer of power from the feed line, while only the resonant legs readily accept, and radiate, it.

Hope this helps.
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K0ZN
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« Reply #7 on: July 08, 2011, 02:24:08 PM »

 Home Run for AE4RV.... that is great !   ( chuckle, chuckle.....)

 Here is another "non-electrical" explanation !    YOU are a 20 meter signal happily walking down the corridor of a building headed out of the building.
 At the end of the hall, you come to three doors: Door # 1 is open and you can see it goes into a Biker bar, filled with very unhappy, irritated bikers, looking
for an opportunity to rough someone up and keep them out of "their bar"; clearly not a friendly environment....you don't want to go there.
That is the 40 M dipole section (High, nasty impedance). The middle door, door # 2, is open and leads to a nice park where you would love to
enter and take a long walk and see what is out there. (This is the 20 M dipole section....remember, you are a 20 M signal!).
The 3rd door is a very, very small door....looks like a "dog door"....you would have to get down on your hands an knees to try to crawl through it;
it would be very difficult and it doesn't look like there is much beyond the door even if you could get through the small door. That is the 10 M
dipole section !

 Bottomline: You will go where you are obviously welcome and it is easy to get there!

73,  K0ZN
« Last Edit: July 08, 2011, 02:46:00 PM by K0ZN » Logged
AE4RV
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« Reply #8 on: July 08, 2011, 03:59:04 PM »

I was just first, not original.

73,
Mr. Noles

Smiley
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G1YHE
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« Reply #9 on: July 08, 2011, 10:08:50 PM »

Door # 1 is open and you can see it goes into a Biker bar, filled with very unhappy, irritated bikers, looking
for an opportunity to rough someone up and keep them out of "their bar"; clearly not a friendly environment....you don't want to go there.

Easily spotted when you see the Megacycles parked outside.

 Smiley
« Last Edit: July 09, 2011, 02:42:11 AM by G1YHE » Logged
W8JI
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« Reply #10 on: July 09, 2011, 07:00:01 AM »

It is not correct to imply all of the current goes in the path of least resistance, or that maximum energy goes to the 50 ohm element. Clearly the SWR answers are incorrect. Current or power does not divide based on SWR.

What really happens are two things:

1.) The various element impedances are in parallel at the feedpoint, and current divides between the loads based on impedance. The LOWER the impedance, regardless of SWR, the more current flows in that path. It is really a division by impedance, not by SWR.

2.) A second effect is mutual coupling between elements, and transmission line effects between the wires. This is pretty complicated, but the addition wires can act like transmission lines and stubs. This is why it generally works better to space the wires apart.

Just my opinion or observation, but I really think saying "current seeks the path of lowest resistance" makes people think current just all goes one way.  I think it is clearer to say "current divides based on path impedance ratios".

73 Tom

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ND6P
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« Reply #11 on: July 09, 2011, 07:57:24 AM »

You have two tuning forks in front of a speaker.  One fork resonates at 1Khz and the other at 2 Khz.  You play a loud 1 Khz tone from the speaker for 10 seconds, then shut it off.  Which fork continues to emit sound, and at what frequency?

It's kind of like that.
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W8JI
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« Reply #12 on: July 09, 2011, 07:43:44 PM »

You have two tuning forks in front of a speaker.  One fork resonates at 1Khz and the other at 2 Khz.  You play a loud 1 Khz tone from the speaker for 10 seconds, then shut it off.  Which fork continues to emit sound, and at what frequency?

It's kind of like that.

Not at all. That's a totally invalid analogy because power is not mutually coupling to the elements through some indirect loosely coupled transfer.

Further proof of the invalidity is in an 80/40/20 fan dipole. The 80 meter element is resonant on 80 as a half wave and 40 as a full wave, yet the 80 meter element would have very little current on 40. The 40 meter element is resonant on 40 as a half wave and 20 as a full wave, yet the 40 meter element has very little current on 20.

This is strictly a parallel impedance problem. The lowest impedance, totally independent of anything else (like SWR or resonance), gets most of the current. Current divides by impedance ratios.  To confirm the truth of this, simply connect a six inch wire across the feedpoint. That wire has a terrible SWR, is not remotely close to resonance, yet will have virtually all of the current.

It's the impedance ratios at the coax connection point that matter.

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K0ZN
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« Reply #13 on: July 09, 2011, 08:17:05 PM »


To:  G1YHE:

That was good !!  Got a good chuckle out of that.......   This hobby would not be much fun if everybody took everything seriously.  Ya gotta lighten it up
 now and then.


73,  K0ZN
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ND6P
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« Reply #14 on: July 10, 2011, 08:54:30 AM »

You have two tuning forks in front of a speaker.  One fork resonates at 1Khz and the other at 2 Khz.  You play a loud 1 Khz tone from the speaker for 10 seconds, then shut it off.  Which fork continues to emit sound, and at what frequency?

It's kind of like that.

Not at all. That's a totally invalid analogy because power is not mutually coupling to the elements through some indirect loosely coupled transfer.

Further proof of the invalidity is in an 80/40/20 fan dipole. The 80 meter element is resonant on 80 as a half wave and 40 as a full wave, yet the 80 meter element would have very little current on 40. The 40 meter element is resonant on 40 as a half wave and 20 as a full wave, yet the 40 meter element has very little current on 20.

This is strictly a parallel impedance problem. The lowest impedance, totally independent of anything else (like SWR or resonance), gets most of the current. Current divides by impedance ratios.  To confirm the truth of this, simply connect a six inch wire across the feedpoint. That wire has a terrible SWR, is not remotely close to resonance, yet will have virtually all of the current.

It's the impedance ratios at the coax connection point that matter.



I think I need to ride my megacycle now Smiley
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