current returning to a different source

[crossman]

Winnie, I see your point although I need to put more thought in it.

For now, concerning the helicopter, I just cannot see that there could be any other way of looking at this than the diagram drawn below. I have broken it down to its most elementary components.

The helicopter cannot be looked at as some isolated body unaffected by the charge in the earth.

Now, if we want some semantic mish-mash of words that a capacitor does not "really and truely" complete a circuit for current flow, then I will toatlly agree with you. However, in most academic circles, I would bet that the capacitor is considered to be a completion of the circuit.

Rattus, expanding on your thought:

 

[crossman]

One other quick thing.... in the diagram, you will notice that when the helicopter probe begins arcing and then makes contact with the powerline, it is completing a circuit around a power source.

 

[winnie]

Crossman,

Your diagram is exactly how I intuit the situation.

IMHO the reason that it is important to understand the 'reservoir of charge' point of view is the rather interesting characteristics of the 'helicopter to ground' capacitor. Understanding this other way of looking at the situation is important if you want to actually have a chance of calculating the magnitude of the current flow.

The distance between the helicopter and ground is rather larger than the size of the helicopter. The normal capacitor equations which depend upon the distance between the plates being smaller than the size of the plates simply fall apart in this circumstance. IMHO a reasonable method for calculating the helicopter to ground capacitance is to calculate the capacitance of the helicopter as an isolated electrode, and the earth as an isolated electrode, and then calculate the series capacitance. To the extent that this calculation correctly describes the situation, if you double the height of the lines, thus doubling the altitude of the helicopter, the current will remain the same. _Not_ what you would expect if you double the spacing in a parallel plate capacitor.

This calculation looks more and more like the 'reservoir of charge' picture than the 'capacitor to ground' picture. They are, of course, the same...but sometimes it is useful to think of them as different in order to pick the calculation that best approximates the real world.

And I totally agree: it is purely a semantic distinction to say 'capacitive charging current closes the circuit', rather than saying 'a changing electric field permits current flow in an open circuit'. A capacitor doesn't carry _galvanic current_, but I'll happily use the term 'capacitive current'.

-Jon

 

[jghrist]

Oh, and for what its worth, the latest XKCD is right on target.
http://xkcd.com/386/

I've appreciated the discussion so far!

-Jon

Great! Now I have another place to waste time on the internet. Thanks a lot!

 

[crossman]

Crossman,

Your diagram is exactly how I intuit the situation.

IMHO the reason that it is important to understand the 'reservoir of charge' point of view is the rather interesting characteristics of the 'helicopter to ground' capacitor. Understanding this other way of looking at the situation is important if you want to actually have a chance of calculating the magnitude of the current flow.

Thanks Winnie. I'm glad to hear that we are on the same page again. I respect your opinion highly and when my thoughts disagree with yours, it definitely sends me running and my head spinning about what I am missing.

You know way more than I do about the mathamatical aspect of the capacitors. If you say there is an alternative way (or even a better way) to do the calcs, then I will just have to rely on your knowledge and accept it. Errr... or I could get off my lazy butt and learn the calcs myself.

 

[crossman]

But allow me to say one other thing for rick's benefit. Looking at that circuit above, the arcing from the helicoppter to the wire is ABSOLUTELY a completed circuit from one side of a source to the other.

 

[crossman]

Darn it, another brain spike.

Couldn't we or shouldn't we actually consider the capacitors as voltage sources?

We never looked at it from that aspect. A capacitor does not "complete a circuit" like a resitor does. It more closely resembles a voltage source. Do we even have to consdier that a capacitor completes a circuit?

 

[steve66]

Crossman:

I wouldn't consider the capacitor as a voltage source for an AC circuit. That's more for DC transient circuits.

Regarding the arc to the helicopter, I seem to recall something about the AC current in the line induces eddy currents in the metal of the helicopter. The arc to the helicopter has something to do with the energy flowing from the line to the helicopter, and being disipated in the helicopter. That's why the guy climbing on the wires wears a bunny suit of metal mesh. It keeps the eddy currents flowing around him. Not through him, which would be painful.

So you don't need the capacitor to ground to explain the current flow.

 

[rattus]

Crossman:

I wouldn't consider the capacitor as a voltage source for an AC circuit. That's more for DC transient circuits.

Regarding the arc to the helicopter, I seem to recall something about the AC current in the line induces eddy currents in the metal of the helicopter. The arc to the helicopter has something to do with the energy flowing from the line to the helicopter, and being disipated in the helicopter. That's why the guy climbing on the wires wears a bunny suit of metal mesh. It keeps the eddy currents flowing around him. Not through him, which would be painful.

So you don't need the capacitor to ground to explain the current flow.

Eddy currents result from alternating magnetic fields. They flow in closed paths in conductors such as Watt hour meter discs and transformer iron. I can't see any connection with this problem.

The capacitors are there, the voltage is there, therefore there will be current.

 

[rattus]

The straight scoop:

The straight scoop:

Winnie, I see your point although I need to put more thought in it.

For now, concerning the helicopter, I just cannot see that there could be any other way of looking at this than the diagram drawn below. I have broken it down to its most elementary components.

The helicopter cannot be looked at as some isolated body unaffected by the charge in the earth.

Now, if we want some semantic mish-mash of words that a capacitor does not "really and truely" complete a circuit for current flow, then I will toatlly agree with you. However, in most academic circles, I would bet that the capacitor is considered to be a completion of the circuit.

Rattus, expanding on your thought:

We have been told to look at the physics of this problem. Weeeeeell, okay!

Anyone who has passed Freshman Physics knows that a capacitor is formed by two conductive surfaces separated by a dielectric material. The value of a parallel-plate air-dielectric capacitor is provided by the formula,

C = e0 x A/d

This formula is difficult to apply in this case, but it can provide approximate values.

Clearly there exists capacitance between a transmission line cable and a nearby helicopter body. Let us call it C1.

Clearly there exists capacitance between said helicopter and many other objects in the vicinity, but let us consider only the capacitance between said helicopter and earth. Let us call it C2.

We now have two caps in series with a common plate as in Crossman?s simplified diagram.

One also learns that the current through a capacitor is computed with the differential calculus expression,

i(t) = C(dv/dt)

Clearly, with a sinusoidal voltage in the order of 230KV, dv/dt is not zero.

Therefore, there will be current through the series combination of C1 and C2.

And, somewhere we learned that air will ionize at some field strength and form an arc which can be roughly modeled as a resistor which carries current around C1 and then through C2 to ground.

Any notion that these capacitances and currents do not exist is not supported by any scientific principle. Such a notion is pure speculation and no more!

Now, steady state AC analysis is based on Physics and Calculus and is far easier to employ. That is the reason it has been developed. When it is all said and done, the results are the same.

Furthermore, I have never heard in all my decades of a circuit element called a ?charge reservoir?. If there is such an element, it looks a lot like a capacitor!

 

[crossman]

Myth busted. And the invisible dragon is dead!

 

[Smart $]

...
This is a crude representation of the electrostatic field which exists around the powerline when the powerline is negative in respect to earth.
...

Your missing at least two major factors for your diagram to even be a crude representation...

First, you have to account for the other two power lines. The only reason I mention them is you show positive charges along the tower when they are almost entirely negated by the e-fields of the other two lines.

Second, e-field strength is inversely proportional to the square of the distance. While your force lines of may be crudely representive of the vectors, they imply the e-field weakens linearly [actually somewhat less then linearly, but I'm giving you credit for drawing it and me not :wink: ). A more accurate depiction would be to start with a bunch of force lines at the power line and stop 3 of every 4 when the distance doubles. However, a simple mathmatical comparison will suffice... so let's say the power line and helicopter are 80' off the ground and the center of the conductive mass of the helicopter is 10' away from the power line. The e-field strength between power line and ground vs helicopter would be (10 ? 80)^2 = 0.0156 or 1.6%...!!!

 

[rattus]

Your missing at least two major factors for your diagram to even be a crude representation...

First, you have to account for the other two power lines. The only reason I mention them is you show positive charges along the tower when they are almost entirely negated by the e-fields of the other two lines.

Second, e-field strength is inversely proportional to the square of the distance. While your force lines of may be crudely representive of the vectors, they imply the e-field weakens linearly [actually somewhat less then linearly, but I'm giving you credit for drawing it and me not :wink: ). A more accurate depiction would be to start with a bunch of force lines at the power line and stop 3 of every 4 when the distance doubles. However, a simple mathmatical comparison will suffice... so let's say the power line is 80' off the ground and the center of the conductive mass of the helicopter is 10' away from the power line. The e-field strength at ground level would be (10 ? 80)^2 = 0.0156 or 1.6% of the field strength at the specified point of the helicopter...!!!

Yes, it is crude, but it proves the point. Crossman is keeping it simple.

Including the other phase lines provides even more paths for the current.

The field is inversely proportional to the distance--not the square of the distance.

 

[crossman]

A more accurate depiction would be to start with a bunch of force lines at the power line and stop 3 of every 4 when the distance doubles.

Electrostatic lines do not just "stop" or dissappear. They continue until they reach an oppositely charged body. Get a college level physics text and check it out. For example, page 435 of Fundamentals of Physics, Halliday and Resnick.

The reason the strength falls off at distance is because the field is spreading out in space and thereby has less density where the field is remote from a charged particle. But when that field approaches an oppositely charged particle, the field density increases back to its original level.

As for the transmission tower charge... in respect to the negative powerline, the tower is positive.

As for the other lines, this would simply add more capacitors into the diagram, but would in no way create current flow without paths.

We admitted that this was a elementary bare-bones diagram. However it does represent what is happening.

 

[crossman]

Samrt, look at it this way. There are field lines which are around the negative charges.

But.... there are just as many field lines at the positive charges on the ground/tower/earth complex.

 

[crossman]

While yall are chewing on that, here is another interesting tidbit which I think is true (but am open to criticism).

Take a 230 KV line phase to phase which is approx 135 kv to ground. (Yes, rough calculation and making assumptions, but ballpark.)

When the chopper is at a distance from the powerline which is about equal to its height above the ground, the voltage from ground to helicopter will be approximately equal to the voltage from helicopter to powerline. Line to chopper = 67 kv : chopper to ground = 67 kv

As the helicopter approaches the powerline, the voltage from helicopter to ground increases and the voltage from helicopter to powerline decreases. Just to put numbers, at some distance, line to chopper = 45 kv : chopper to ground = 95 kv

The closer the chopper gets to the powerline, the lower the voltage between them. At a closer distance, perhaps line to chopper is 20kv : chopper to ground is 115 kv.

When the chopper electrode probe makes contact with the powerline, then voltage from line to chopper is near zero and chopper to ground is 135kv.

Now, since the voltage between line and chopper is zero, then what voltage is causing the current to still flow between chopper and wire?

You guessed it! The full voltage on the capacitor from chopper to ground.

Very cool.

Edit: Don't freak out over the numbers. The concept is what is important. Think of elementary AC circuits. With series capacitors, which capacitor drops the most voltage?

 

[Smart $]

Yes, it is crude, but it proves the point.

Just so we're on the same page, I'm referring to his diagram of force lines, and not his representative circuit.

On that note, it doesn't prove any point to me... and I strongly advise others not to take its simplicity lightly.

Including the other phase lines provides even more paths for the current.

And that is exactly what negates most of the positive point charges on the tower.

The field is inversely proportional to the distance--not the square of the distance.

So you're saying the lesson on this web page is completely wrong in this respect? ...And what about the following parts of that Lesson?

 

[Smart $]

Electrostatic lines do not just "stop" or dissappear. They continue until they reach an oppositely charged body. Get a college level physics text and check it out. For example, page 435 of Fundamentals of Physics, Halliday and Resnick.

The reason the strength falls off at distance is because the field is spreading out in space and thereby has less density where the field is remote from a charged particle. But when that field approaches an oppositely charged particle, the field density increases back to its original level.

I concur with these statements. All I'm saying is that your diagram gives an impression that the force between the power line and earth is as strong as that between the power line and the helicopter. This is simply not the case.

As for the transmission tower charge... in respect to the negative powerline, the tower is positive.

But its not. Your reference is technically a neutral atom of the power line. We automatically assume earth to be of the same potential. If we agree on this premise, we then have to consider the tower is both conductive and grounded. So to say it is charged one way or the other relative to atomic neutral, we have to consider at least all major e-fields which affect it along with its conductivity to earth. What this means is, the same criteria by which you say the power line is negative has to be used on any point you say is positive. You cannot selectively abandon the criteria for the point of your argument.

As for the other lines, this would simply add more capacitors into the diagram, but would in no way create current flow without paths.

I feel in your using the capacitor model to describe the "system" you are forgetting that e-fields are vectorial, not scalar. The e-field between the power line and helcopter is, for the sake of discussion at or greater than 90? to that of the power lines' e-field to ground.

We admitted that this was a elementary bare-bones diagram. However it does represent what is happening.

Well, you get a star for the admission, but that's as far as I can go on positive notes :grin:

 

[crossman]

On that note, it doesn't prove any point to me... and I strongly advise others not to take its simplicity lightly.

A drawing of what you feel the electrostatic field should look like would be appreciated.

So you're saying the lesson on this web page is completely wrong in this respect? ...And what about the following parts of that Lesson?

You may not realize it, but the 1/d^2 formula actually proves that, when the helicopter is close to the wire, that the voltage from helicopter to ground is greater than the voltage from helicopter to wire. Winnie has already shown us this in a previous post. The further apart the charges are, the greater the voltage between the charges. 1/d^2 is actually an argument FOR what I am saying!

 

[crossman]

Re: Tower is positive with respect to the negative powerline

But its not. Your reference is technically a neutral atom of the power line.

No. My reference is the excess electrons on the surface of the wire due to the capacitance to ground. These excess electrons are negative in respect to the tower. The tower is positive in respect to the excess electrons accumulated in the capacitance of the wire to ground and other bodies in the area.

So to say it is charged one way or the other relative to atomic neutral

Has nothing to do with atomic neutral as mentioned above.

I feel in your using the capacitor model to describe the "system" you are forgetting that e-fields are vectorial, not scalar. The e-field between the power line and helcopter is, for the sake of discussion at or greater than 90? to that of the power lines' e-field to ground.

I do not understand what you mean, especially the part in bold. Why do the electrostatic fields have to be at 90 degree angles or greater?

Are you talking about the phase angle between voltage and current? That current leads voltage by 90 degrees in a capacitor? That doesn't have anything to do with it. Helicopter-wire and helicopter-ground are both capacitors. They both have the same phase angle.

I know of no science that requires electrostatic fields to be at 90 degree angles to each other.

If you are talking about "electric potential difference" between two points, then these lines cross the electrostatic lines at 90 degrees but again, I don't see what point you are trying to make. On that note, delving deeper into this aspect will also prove that when the helicopter is close to the wire, the potential difference from chopper to ground is greater than the potential difference from chopper to wire.

Well, you get a star for the admission, but that's as far as I can go on positive notes :grin:

I'll be waiting for your diagram which shows exactly what is going on.:grin:

Edit: added additional thought on potential difference