# Birds on power lines revisited
**URL:** https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321
**Category:** Factual Questions
**Created:** [April 4, 2014, 7:51am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321 "2014-04-04T07:51:58Z")
**Posts on this page:** 16
**Page:** 1
### Author:  [@Dog80](https://boards.straightdope.com/u/Dog80)
#### Post date: [April 4, 2014, 7:51am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/1 "2014-04-04T07:51:58Z")
Birds on power lines don’t get zapped because both feet are at the same electric potential.
But what would happen if power lines carried higher frequency voltage?
At 60Hz the wavelength is 5000 kilometers.
What if it carried voltage at 6Ghz? The wavelength at this frequency is 5cm, about same as the distance between a typical bird’s legs. When one leg is at a high peak, the other one is at a low peak of the sinusoid. If the line carries 20,000 volts rms, the peak to peak voltage is 56,000 volts
Wouldn’t that zap the bird?
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### Author:  [@aceplace57](https://boards.straightdope.com/u/aceplace57)
#### Post date: [April 4, 2014, 8:59am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/2 "2014-04-04T08:59:01Z")
Cecil’s column. He mentions one way the bird can get zapped.
> **[Why don’t birds on wires get electrocuted?](https://www.straightdope.com/21343084/why-don-t-birds-on-wires-get-electrocuted)**
>
> A STAFF REPORT FROM THE STRAIGHT DOPE SCIENCE ADVISORY BOARD
> [@](#):
>
> The bird’s resistance is much higher than the resistance of the wire between its legs, so no current flows through the bird. The key here though is that the bird’s legs and the wire are equally charged–they’re at the same voltage level, or “potential,” as the engineers say. Should the bird touch one of its legs, or anything else, to something that can suck up infinite numbers of electrons, e.g., the ground, this will no longer be the case. Suddenly, one leg will be charged and the other won’t. The electrons in the charged leg that repel each other now have somewhere to go, the uncharged leg. Current flows, heat is generated, and, well, you get the idea.
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### Author:  [@aceplace57](https://boards.straightdope.com/u/aceplace57)
#### Post date: [April 4, 2014, 9:51am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/3 "2014-04-04T09:51:35Z")
Basically the bird on the wire creates a parallel circuit. [like this](http://people.sinclair.edu/nickreeder/eet114/PageArt/CircuitParallelConnect.gif) Where R1 represents the resistance of the wire (say 1 ohm) and R2 is the bird (say 100k ohm)
kirchhoff’s law tells us the sum of the current going into the circuit will equal the sum going out. If you had two equal resistors (say 1k ohm each) half the current will go through each. say you had 1amp current. .5 amp will go through each branch of the parallel circuit. but the total is still 1amp.
with the bird the difference in resistance is so great that all the current goes through the wire. Electricity always takes the path of least resistance.
you can easily calculate the resistance of the parallel circuit
R=R1R2
…-------
…r1+r2
R=1\*100,000
…------------
…1+100,000
R= .999990 ohms
the bird didn’t change the resistance of the circuit in any significant way.
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### Author:  [@coffeecat](https://boards.straightdope.com/u/coffeecat)
#### Post date: [April 4, 2014, 9:54am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/4 "2014-04-04T09:54:41Z")
At 6GHz the wire would be radiating like an
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### Author:  [@coffeecat](https://boards.straightdope.com/u/coffeecat)
#### Post date: [April 4, 2014, 10:08am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/5 "2014-04-04T10:08:01Z")
Sorry, false start.
At 6GHz the wire would be radiating like anything. Practically all the energy going into the wire would be broadcast out as microwaves, leaving next to nothing to shock the bird.
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### Author:  [@aceplace57](https://boards.straightdope.com/u/aceplace57)
#### Post date: [April 4, 2014, 10:09am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/6 "2014-04-04T10:09:00Z")
The point I was trying to make with the parallel circuit is if the birds resistance was very low, say 100 ohms. Then the current would flow in both branches. A very bad scenario for the bird. 😉
you can calculate the current in each branch if R1 was 1 and R2 100. but lets just say the bird would be flambeed.
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### Author:  [@Dog80](https://boards.straightdope.com/u/Dog80)
#### Post date: [April 4, 2014, 10:24am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/7 "2014-04-04T10:24:15Z")
> [@aceplace57](#):
>
> Basically the bird on the wire creates a parallel circuit. [like this](http://people.sinclair.edu/nickreeder/eet114/PageArt/CircuitParallelConnect.gif) Where R1 represents the resistance of the wire (say 1 ohm) and R2 is the bird (say 100k ohm)
>
> kirchhoff’s law tells us the sum of the current going into the circuit will equal the sum going out. If you had two equal resistors (say 1k ohm each) half the current will go through each. say you had 1amp current. .5 amp will go through each branch of the parallel circuit. but the total is still 1amp.
>
> with the bird the difference in resistance is so great that all the current goes through the wire. Electricity always takes the path of least resistance.
>
> you can easily calculate the resistance of the parallel circuit
>
> R=R1R2
> …-------
> …r1+r2
>
> R=1\*100,000
> …------------
> …1+100,000
>
> R= .999990 ohms
>
> the bird didn’t change the resistance of the circuit in any significant way.
Actually a better representation, assuming the bird has some finite resistance, but the wire has zero resistance would be like this:
[http://i60.tinypic.com/579y0.png](http://i60.tinypic.com/579y0.png)
What I am asking though is that as the wave propagates along the wire, each leg will be at a slightly different potential. At 60Hz, the wave does a full cycle over a distance of 5000 kilometers. So the potential variation will be insignificant even over large stretches of cable.
But if the frequency was raised to some ridiculous number in the GHz range, the wavelength becomes small enough to have big potential variations even within centimeters of cable.
Unless I’m missing something, the bird would get fried in that case.
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### Author:  [@aceplace57](https://boards.straightdope.com/u/aceplace57)
#### Post date: [April 4, 2014, 11:13am UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/8 "2014-04-04T11:13:21Z")
kirchhoff’s law can be applied to AC circuits. You have to calculate the angular frequency.
I’ve been out of electronics school too long to do these equations now. But a large change in frequency does make a difference. A GHz range frequency will make induction and capacitance very important.
heres a pdf on AC calculations. Brings back a lot of memories. We worked our butts off doing these calculations in school.
[http://www.physics.ohio-state.edu/~gan/teaching/summer04/Lec3.pdf](http://www.physics.ohio-state.edu/~gan/teaching/summer04/Lec3.pdf)
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### Author:  [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)
#### Post date: [April 4, 2014, 12:05pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/9 "2014-04-04T12:05:35Z")
> [@aceplace57](#):
>
> Basically the bird on the wire creates a parallel circuit. [like this](http://people.sinclair.edu/nickreeder/eet114/PageArt/CircuitParallelConnect.gif) Where R1 represents the resistance of the wire (say 1 ohm) and R2 is the bird (say 100k ohm)…
At 6 GHz the apparent resistance of the wire is likely to be much higher. The [url-=[Skin effect - Wikipedia](http://en.wikipedia.org/wiki/Skin_effect#Examples)]skin effect compels AC current to flow near the surface of a conductor; the higher the frequency, the more tightly the current hugs the skin. At 6 GHz, the skin depth is less than a micron. So unless your conductor has a very large diameter (and therefore a lot of skin area), the resistance at this frequency is going to be very high, and we might indeed see a substantial voltage potential along its length (and therefore across the bird).
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### Author:  [@Dog80](https://boards.straightdope.com/u/Dog80)
#### Post date: [April 4, 2014, 12:28pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/10 "2014-04-04T12:28:13Z")
> [@Machine\_Elf](#):
>
> At 6 GHz the apparent resistance of the wire is likely to be much higher. The [url-=[Skin effect - Wikipedia](http://en.wikipedia.org/wiki/Skin_effect#Examples)]skin effect compels AC current to flow near the surface of a conductor; the higher the frequency, the more tightly the current hugs the skin. At 6 GHz, the skin depth is less than a micron. So unless your conductor has a very large diameter (and therefore a lot of skin area), the resistance at this frequency is going to be very high, and we might indeed see a substantial voltage potential along its length (and therefore across the bird).
Fair enough! But what about the wavelength effect? Is it a valid mode of electrocution? Aren’t different parts of the same wire at different potential?
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### Author:  [@Malacandra](https://boards.straightdope.com/u/Malacandra)
#### Post date: [April 4, 2014, 12:33pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/11 "2014-04-04T12:33:17Z")
> [@aceplace57](#):
>
> Basically the bird on the wire creates a parallel circuit. [like this](http://people.sinclair.edu/nickreeder/eet114/PageArt/CircuitParallelConnect.gif) Where R1 represents the resistance of the wire (say 1 ohm) and R2 is the bird (say 100k ohm)
>
> kirchhoff’s law tells us the sum of the current going into the circuit will equal the sum going out. If you had two equal resistors (say 1k ohm each) half the current will go through each. say you had 1amp current. .5 amp will go through each branch of the parallel circuit. but the total is still 1amp.
>
> with the bird the difference in resistance is so great that all the current goes through the wire. Electricity always takes the path of least resistance.
>
> you can easily calculate the resistance of the parallel circuit
>
> R=R1R2
> …-------
> …r1+r2
>
> R=1\*100,000
> …------------
> …1+100,000
>
> R= .999990 ohms
>
> the bird didn’t change the resistance of the circuit in any significant way.
No, “all” the current will not go through the bird. If the wire has a resistance in the milliohms range and the bird in the kilohms, a thousand volts across the bird’s feet will still pump of the order of an amp through it. It might be a rounding error as far as the circuit is concerned but that’s scant comfort to the bird.
What you do tend to get though at extreme frequencies is that the current will flow across the surface of the bird, not through its vitals.
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### Author:  [@Isilder](https://boards.straightdope.com/u/Isilder)
#### Post date: [April 4, 2014, 12:52pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/12 "2014-04-04T12:52:29Z")
> [@Dog80](#):
>
> But if the frequency was raised to some ridiculous number in the GHz range, the wavelength becomes small enough to have big potential variations even within centimeters of cable.
>
> Unless I’m missing something, the bird would get fried in that case.
Well kind of, the entire transmission line would fry before any bird got to sit on it.
IF the transmission line somehow survived operation at that frequency (and wasn’t so hot that it cooked the bird immediately…) then …
There is a reason transmission line efficiency is optimum right down low…
The inefficiency of moving to higher frequencies must result in heat… (well it can result in light, sound, explosions and so on… but mostly heat… )
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### Author:  [@johnpost](https://boards.straightdope.com/u/johnpost)
#### Post date: [April 4, 2014, 3:32pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/13 "2014-04-04T15:32:56Z")
instead of imagining a higher frequency to create shorter distance high potential differences what if we just imagined a bigger bird?
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### Author:  [@Dog80](https://boards.straightdope.com/u/Dog80)
#### Post date: [April 4, 2014, 4:56pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/14 "2014-04-04T16:56:30Z")
> [@johnpost](#):
>
> instead of imagining a higher frequency to create shorter distance high potential differences what if we just imagined a bigger bird?
Yes, that too!
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### Author:  [@aceplace57](https://boards.straightdope.com/u/aceplace57)
#### Post date: [April 4, 2014, 5:28pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/15 "2014-04-04T17:28:48Z")
A damaged and almost broken wire would have a much higher resistance at that spot. Bird lands and suddenly the path of least resistance is through the bird.
ouch.
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### Author:  [@Colophon](https://boards.straightdope.com/u/Colophon)
#### Post date: [April 4, 2014, 6:22pm UTC](https://boards.straightdope.com/t/birds-on-power-lines-revisited/685321/16 "2014-04-04T18:22:19Z")
> [@aceplace57](#):
>
> with the bird the difference in resistance is so great that all the current goes through the wire. Electricity always takes the path of least resistance.
Thus proving that the Borg were full of crap.