At the voltage levels of typical overhead transmission lines in the US, a bird can land on one and be just fine (as long as it doesn't do something like spread its wings and touch a tree or something else at lower electric potential).

However, what about a hypothetical powerline at much higher voltage (as in tens of megavolts). Could landing on such a powerline fatally-shock the bird even though it does not complete a circuit for sustained current? (Assume that the distance is long enough that electrical arc'ing is impossible.)

NOTE: My understanding of what happens when a bird flies from an earth object to a powerline (please correct me if I'm wrong) is that - upon contacting the wire - its electric potential changes from earth-potential to the powerline's potential. In order for this to happen, there is an initial transfer of electrical energy (i.e. flow of charge i.e. current) from the powerline to the bird which "equalizes" their electric potential, which happens nearly instantaneously. If this is correct, then my question can be restated more generally as "Can an 'equalization charge' such as this result in a fatal shock, if the potential difference that it's equalizing is high enough?"

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    \$\begingroup\$ This is going to depend on "what is the capacitance of an unladen swallow" \$\endgroup\$
    – pjc50
    Mar 8, 2015 at 16:52
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    \$\begingroup\$ Excellent question, and well formulated. I was wondering the same thing about humans. Just guessing, but I would say it compares to a RC circuit comprising the equivalent electrical capacity of the bird and the resistance of the contact HV wire-["center of charge" of bird]. This would give the i(t) characteristic, which we can assess (at least for humans) whether it is lethal or not in most cases. But even if I think we can estimate the capacity, I have no clue as to what the resistance can be. \$\endgroup\$ Mar 8, 2015 at 16:53
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    \$\begingroup\$ @pjc50 African or European? \$\endgroup\$
    – Majenko
    Mar 8, 2015 at 17:03
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    \$\begingroup\$ Worth a watch: youtube.com/watch?v=9tzga6qAaBA \$\endgroup\$
    – Majenko
    Mar 8, 2015 at 17:05
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    \$\begingroup\$ ... Especially to see how they have to equalize the potential of the helicopter all the time so they don't die. \$\endgroup\$
    – Majenko
    Mar 8, 2015 at 17:19

9 Answers 9


Assuming the bird still is at earth potential when entering in contact with the wire (say, it jumped right on it from the pole).

There are lots of unknowns in this problem but let's try to fill some gaps with data we kind of know in humans. So until an EE stackexchanger who is an ornithologist shows up with interesting data, let's assume humans can fly and like to chill out hanging from a high voltage cable.

All objects and living things have an equivalent electrical capacity. The Human Body Model is a convention which dictates humans are equivalent on that aspect to a 100pF capacitor (let's assume it doesn't reduce much from the ground to 23meters high, and call it a worst case scenario). Now, let's assume the contact resistance between the cable and wherever the geometric center of that capacitor is, is 3000Ohm - taken from the "Hand holding wire" case of the table in another thread - divided by two for a two hands contact. Then the total duration of the equilibrium current, taken as 5 times the time constant of the equivalent RC, is 0.75 microseconds.

Effects of currents through living things depend on the magnitude of the current and the duration. I have never seen any study showing any data below 10ms (e.g. the same study cited above), which is not surprising as apparently the response time of the cardiac tissue is 3ms. For 10ms, the current that generates irreversible effects is 0.5A, and it seems to have settled at that point (little dependent on the duration), certainly down to 3ms. Let's assume that past that point, the cardiac tissue behaves like an ineffective first order system, attenuating 20dB/decade. The required current for similar effects would be 20*4.25=90dB higher, or 15811A. For a contact resistance of 1500Ohms as used above, it means the voltage of the cable needs to be 23GV!

Burns solely depend on the energy transferred, so theoretically a high voltage could burn for such a small time. But how high? Well, "Electrical injuries: engineering, medical, and legal aspects", page 72, states:

The estimated lowest current that can produce noticeable first or second degree burns in a small area of the skin is 100A for 1s

Edit: Note that 100A is quite high, it is unclear how the author defines "first degree burns on small area of skin", but I would guess it would be for an area bigger than an inch, burning all epidermis and some of the dermis cells such that they peel away.

So for 750nanoseconds, that's 133MA required! If we use again the 1500Ohms resistance from above, that means the wire would need to be at 199GV, which is insane. Chances are there will be other nasty effects before those burns appear, but neither 23GV nor 199GV sound likely in the near future. Side note, as J... raised in the comments, a 23GV cable would spontaneously arc with anything at Earth potential within 7.6km and therefore would require an incredible amount of isolation.

As if it wasn't enough, you may have noticed that the above assume the maximum current is applied for the entire duration of the equilibrium current whereas in fact it is a decaying exponential... The average current over this duration is in fact 0.2 times the maximum, so these values should really be 115GV and 995GV!

Warning: This does not mean it is safe to jump on and hang from high voltage lines, this is a quick analysis with rough data estimates and modelling and shall not be considered a justification for your actions.

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    \$\begingroup\$ Excellent answer (+1). I will wait at least 24 hours before I "accept" anything, so others have time to post alternative theories and/or challenge the validity of other answers. \$\endgroup\$
    – etherice
    Mar 8, 2015 at 18:18
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    \$\begingroup\$ Great answer! Just one question, surely the current which will cause burns is 100 mA, not A? For example, you don't easily get 100 A from the wall socket, but you do get burns? \$\endgroup\$
    – tomnexus
    Mar 8, 2015 at 20:51
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    \$\begingroup\$ let's assume humans (...) like to chill out hanging from a high voltage cable. - until today, I thought everybody does it... now I feel alone again. \$\endgroup\$
    – user20088
    Mar 9, 2015 at 16:17
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    \$\begingroup\$ Also, just for curiosity, a cable at a potential of 23GV would spontaneously discharge (ie: make instant giant lightning) against anything at earth potential within at least 7.6km (~5 miles) (in normal atmosphere). Unless this was some sort of magic sky-cable, it would need an absurd amount of insulation just to keep it from arcing at anything and everything within that radius. \$\endgroup\$
    – J...
    Mar 10, 2015 at 11:51
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    \$\begingroup\$ The helicopter has a huge capacitance compared to the lineman, an equilibrium current of the helicopter going through the lineman could kill him, that's why they're equipped with Faraday cages/clothing. For your teacher, are you absolutely certain it was due to his own capacitance, and not because of an insufficient isolation to Earth? \$\endgroup\$ Mar 11, 2015 at 11:26

I mostly agree with Andy Aka explanation. I'll give a more detailed theory though (of course I might be overlooking something).

A body doesn't have a capacitance by itself, as it always needs the "second plate" of the capacitor. Humans relative to ground will have a given capacitance when standing (insulated) over then ground, and a different capacitance when flying (if able to) because then ground is farther.

A simple model of the bird could be the one in the next diagram:


simulate this circuit – Schematic created using CircuitLab

As the bird approaches the line C1 will increase and C2 will decrease. This is a capacitor divider and the potential of the bird will approach the High Voltage (HV) line one.

Let's assume, just to give some quick numbers, that C1 is 100 times C2 just before the bird's feets touch the line, the difference of potential between the bird and the HV line will then only be 1% of HV. Finally the bird's feets touches the line: C1 is "shorted" and the only capacitance to fill would be C2 (capacitance between bird and ground, which is very small as ground is far). Because body potential is already at 99% of HV, and it's capacitance to ground is very small, the current through the bird would be really small.

  • \$\begingroup\$ Great explanation, and it clarifies the point Andy Aka was making in his answer. +1 \$\endgroup\$
    – etherice
    Mar 8, 2015 at 22:02
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    \$\begingroup\$ It might be worth calculating estimates for those capacities, because I doubt C1 would be able to reach levels high enough for this to be significant (the cable doesn't have a flat surface which reduces the effective area, it's very small which reduces the overlap area with the body etc.). It is still a valid effect that I haven't covered in my answer, definitely worth a +1 (+1). \$\endgroup\$ Mar 9, 2015 at 12:27
  • \$\begingroup\$ @MisterMystère C1 is certainly small, but it only needs to be big relative to C2. In the other hand, the cable is not that small (typically it has at least 4 cm diameter) and it is very long. The complete picture would be a long cylinder for the wire, a sphere for the bird and an infinite plane for the ground. If anyone has time to work at this, it would be a nice contribution. I've found something with two spheres: iue.tuwien.ac.at/phd/wasshuber/node77.html but it is not quite the same! Thanks for the +1! – Roger C. 7 mins ago \$\endgroup\$
    – Roger C.
    Mar 9, 2015 at 16:50
  • \$\begingroup\$ I don't have the time to look into it but a plane is the limit of a sphere when the radius tends towards infinity, so both can actually be solved this way. \$\endgroup\$ Mar 9, 2015 at 17:59
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    \$\begingroup\$ All bodies have capacitance by themselves; their second plates lie on some distant stars ;) \$\endgroup\$ Mar 12, 2015 at 10:23

NOTE: My understanding of what happens when a bird flies from an earth object to a powerline (please correct me if I'm wrong) is that - upon contacting the wire - its electric potential changes from earth-potential to the powerline's potential

Here lies the crux of the matter. As the bird leaves the ground heading in the direction of the wire, it acquires a gradual change in potential. This is not an instantaneous change because if it were, the bird would experience a current jolt at that instant it landed.

So, no, it doesn't happen instantaneously and, bigger wire voltages = larger distance therefore a longer time period to reach said wire and, without going into the math, the small imperceptible current that the bird experiences will be the same.

Below is a picture of the way the voltage level changes with distance between ground and a "hot" wire: -

enter image description here

This is fairly classical electric field analysis. Emanating from the centre (assumed to be a point of high voltage) are black electric field lines. These exit in all directions from the wire and hit "ground" at right angles. If you chose any particular one of these E-field lines and "traveled" along it from the ground level by (say) 10% of its length, you would attain a voltage that is 10% of the hot wire.

If you did this thought experiment for all E-field lines at different percentages of the length you'd be able to plot all the lines of equipotential and that is what the red lines are.

As you should be able to see the potential that a small object can attain rising from ground to the "hot" wire is remarkably linear.

  • \$\begingroup\$ The wording I used in the question is "nearly instantaneous" to be concise. More importantly, the time it takes for the equalization to occur was not the question, but rather if the equalization could result in a fatal shock if the potential difference were large enough. \$\endgroup\$
    – etherice
    Mar 8, 2015 at 17:10
  • \$\begingroup\$ @etherice The point Andy is getting at is the situation you propose cannot happen. \$\endgroup\$
    – Matt Young
    Mar 8, 2015 at 17:16
  • \$\begingroup\$ I think he's arguing that it would not be instantaneous. \$\endgroup\$
    – pjc50
    Mar 8, 2015 at 17:41
  • \$\begingroup\$ +1 (your answer is much clearer after reading the explanation provided by Roger C) \$\endgroup\$
    – etherice
    Mar 8, 2015 at 22:56
  • \$\begingroup\$ Nice illustration. Only thing that I would object is that in a power transmission line there is 3 phases (120º apart, added voltage is 0) and not a single one. Because you have a superposition of the three electric fields one should not expect a linear potential increase from ground to one of the hot wires. The big potential increase step would happen when the bird really starts to approach one of the hot wires as its field would then become dominant. \$\endgroup\$
    – Roger C.
    Mar 11, 2015 at 20:48

It's a pity to see so many misinformed, high-ranking answers on this question - so I decided to finally open an account and contribute, after years of lurking :)

One way to see power transmission is current going through the wire - modeled as kinetic energy of particles (electrons) within. However, especially in AC installations, if one models electromagnetic energy (through Maxwell's equations), one sees the power carried in the space between and around the conductors.

So there is EM danger for anything even close to the lines. Its level depends, for a given bird-line system, on the overall power going through - tension and intensity!

This quantitative answer I found on https://van.physics.illinois.edu/qa/listing.php?id=1341 applies:

Q: Why don't birds feel a static shock? I understand that birds don't provide a path to ground, so they won't carry a steady-state current. But surely when they first land there is a current to charge up the bird's capacitance? I've read that static shocks are painful at around 10 kV. These power lines carry hundreds of kV, so wouldn't the static shock from a power line be very painful? Thanks, Ted - Ted (age 26) Stanford, CA, USA

A: Yes, It is not strictly true that there will not be any current at all. There are currents, but they are really small, and this not limited to landing only. Perhaps most negligible of all, the humid air is not a perfect insulator, so there will be losses from the body of the bird. But as you also point out, a bird can be considered a (roughly spherical) capacitor with second shell infinitely far away and at 0 potential. Therefore the bird will be charged and discharged at f=60Hz (50 Hz in Europe), because the power lines carry AC, not DC.

Lets make a rough calculation considering the bird as a sphere with 20cm diameter, the capacitance C should then be ~10pF. The rms current is then 2πfVrmsC f. Say there is 100kV on the wires, these parameters give about 400 µA for the rms current. For comparison, for a human being AC currents of around 10 mA start to become dangerous.(https://en.wikipedia.org/wiki/Electric_shock) For a bird, somewhat lower currents presumably can be dangerous. It sounds like even for the high voltage line, however, the purely capacitative current isn't quite a problem.

Tunc + Mike W.

  • \$\begingroup\$ Although the answer you quote seems sound, there is absolutely no electromagnetic danger. The electromagnetic field generated by a high voltage line is smaller than the Earth's (I can dig out sources if required). \$\endgroup\$ Mar 9, 2015 at 12:33
  • \$\begingroup\$ Ampere's law, in short: Electric current produces magnetic field. For a linear conductor B=miu * I / 2 * pi * d (en.wikipedia.org/wiki/Inductance#Inductance_of_a_coaxial_line). Where d is the distance from the wire and miu is on the order of 10^-6. Also, variable magnetic field -> electric field. Think of an induction heater, but of course power lines carry smaller currents ;) \$\endgroup\$ Mar 9, 2015 at 13:09
  • \$\begingroup\$ @VictorPopescu The excerpt that you have quoted states that it is not a danger.... \$\endgroup\$
    – scld
    Mar 9, 2015 at 13:14
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    \$\begingroup\$ You appear to be saying that power carried down a cable is current modelled as kinetic energy of the electrons. Do you have a proof of this? I'd love to see it. You also seem to say that the power (presumably that taken by the load possibly tens of miles away) is carried in the space around the conductors. I'd like to see some evidence of that too. \$\endgroup\$
    – Andy aka
    Mar 10, 2015 at 22:42
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    \$\begingroup\$ Andy, I'm happy you noticed the main point I was making. \$\endgroup\$ Mar 11, 2015 at 8:53

my understanding is that since the HV line is an AC line, the original potential of the bird is meaningless due to the fact that the potential of the wire is alternating above ground potential and below ground potential every 1/100 of a second in a 50hz situation. There is an equally likely chance that the potential relative to ground potential at the moment of contact of the bird's foot could be very near ground potential 1/100 of every second also.


I'm not an expert, but I think this is correct: The wire is a conductor; current is flowing through it. The bird won't be harmed. Current will flow up one leg and down the other, but the wire is a much better conductor, so that current will be miniscule. (On the other hand, if the bird landed on a very high-voltage source without current flowing, like a huge Van Der Graaf generator, then the electrostatic repulsion could blast its feathers off).


I know this is an old question and mostly hypothetical; however, it IS an engineering forum and, well, we all know how engineers are...

But the electrical field gradient will be a concern. Even a 250 kV line in the HV lab at my alma mater (way back when) was enough to light a fluorescent bulb strapped to a stepladder a few feet away from the line. Also, some lighting for aviation-related marking purposes is designed to hang on transmission lines with no other connection and allow the field to illuminate the lights. So even if the corona did not discourage the birds from landing on the lines, there would be some level at which the bird's size would bridge enough electric field differential to cause problems.

Note, too, that although corona can discourage birds at higher levels, even 138-kV can be host to our avian friends. When I worked at a major utility in the late 80's/ early 90's, there was a 138-kV line "Nuisance" tripping frequently enough to cause problems with manufacturing customers. Eventually we learned the cause was a high number of birds all flying away at the same time and causing the phases to gallup and slap together.


A hypothetical air-insulated HV line at 10s of megavolts does not exist because at those voltages, the power lost to corona discharge is greater than the power lost to wire resistance. As the voltage goes up, the current decreases in proportion, but past a certain point the corona discharge power loss is greater than the I squared R loss.

The "certain point" depends on the diameter of the conductor, which is one reason why all high voltage conductors (and particulary at 1KV+) have artificially inflated diameters: much of the "conductor" volume isn't conductor at all: it's steel.

Corona discharge occours when the voltage gradient is greater than the breakdown voltage gradient of air. This depends on humidity and air preasure (and pollution) and on the surface smoothness of the wire.

The equ-potential lines shown in the other answer are misleading. They should be much closer together near the wire, much farther apart near the ground. Here is an actual measured example: https://www.nms.org/Portals/0/Docs/FreeLessons/PHYS_Equipotential%20Lines%20and%20Electric%20Fields.pdf

Note the difference between the 8V-10V gap and the 4V-2V gap. Near a narrow wire, the field distribution is similar to that around an isolated point charge, where the voltage gradient rapidly approaches "infinity" for an "infinitely thin" wire.

I am unable to find actual figures for the electrical field gradient near a HV line. I would expect it to be less than 3.4MV/m under worse case conditions, or there would be failures. For comparison, humans will fail at about 0.01MV/m, and human skin will fail at about 500V. This suggests to me that there is not a lot of safety factor for a human being hanging from a HV line: you would be close enough to your ionization potential to start to worry.

Typical birds are much smaller/shorter than humans, and so would be exposed to much smaller voltage stress when landing on wires. Large birds might be comparible is size to humans, but do not normally perch on wires. Large birds normally perch on the transmission towers, not the wires, because the towers are always higher than the wire: I have no information on if large birds feel electrical voltage gradient discomfort when attempting to land on HV wires.

I am outside my area of experience, and welcome any corrections.


It's interesting to look at powerlines, and birds, and see what happens.

Birds tend to perch on low voltage powerlines, typically sub-100kV.

Birds tend not to perch on high voltage powerlines, typically >200kV.

The speculation (which I find entirely plausible) is that it's due to the corona which occurs on high voltage powerlines. This is why they tend to use bundles of wire, rather than single conductors, to reduce the gradient of the electric field round them. Any pointy thing sticking out of the smooth conductor will increase the corona loss.

A bird on a powerline acts like a 'bit sticking out', which worsens the corona discharge. Above some critical corona current, the bird finds this uncomfortable, and leaves. This will be felt by the bird flying in proximity to the line, even before it lands, the bird will distort the electric field and receive a corona current.


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