lightning protection

[walt]

Phil, please check out this reference

http://www.marinelightning.com/Information/GroundingConcepts.htm

Take a look at sections 4.2.1, 4.2.2, 4.2.3

The corners of the tank compress the equipotential surfaces just outside it and result in a local enhancement of the electric field. As a consequence the inside upper corner is likely to form a sideflash to the lightning conductor, and the outside lower corner to the water.

I'm not sure what you are implying by saying my mast is not an insulated conductor - some point I missed (the mast on my trailer sailboat is an insulated conductor). But assume in this case a conductor of length 20 foot and it is just off the ground (which is at ground potential). Charge is brought over the ground by a cloud such that an electric field is induced. In the free space away from the mast, lets say the potential at 20 feet is 200KV (ie, field gradient is 10KV per foot). The mast mid point is at 10 feet.

Phil, what do you believe the potential on the conductive mast would be? If its not uniform across the surface of the mast, ohms law says current would have to continually flow. But where would it flow to (there is no circuit path)?

If the mast does take on a single potential (and I think it will take on the single potential of its mid point which in this case would be 100 KV assuming no charge flow Since the mast is all at one potential, the the top of the mast is also at 100KV. But the free space potential at this height would have been 200KV. So at the top of the conductor, the potential goes from 100KV to 200KV in a compressed distance compared to free space. So the conductor did influence the electric field in the vicinity of the conductor- which is also what I believe is shown on Dr Thomson web site I linked to.

 

[T37 Phil]

Phil, please check out this reference

http://www.marinelightning.com/Information/GroundingConcepts.htm

Take a look at sections 4.2.1, 4.2.2, 4.2.3

Perfect conductor (you'll recall I asked). No interior field, as I suspected.

I'm not sure what you are implying by saying my mast is not an insulated conductor - some point I missed (the mast on my trailer sailboat is an insulated conductor).

Of what material is it made? Air, at least not perfectly dry air, is not a perfect insulator. There is no insulating material applied to your mast, is there? So it's not the case that current may flow only from the top and bottom of your mast, or some such picture. This relates more to the pre-bolt field than to where the current of the bolt travels. Perhaps you meant insulated in that there is no connection to ground and I simply misunderstood your point.

Phil, what do you believe the potential on the conductive mast would be? If its not uniform across the surface of the mast, ohms law says current would have to continually flow. But where would it flow to (there is no circuit path)?

Current is charge redistribution and the asymmetry of charge would alter the field, particularly near the mast. However, that possibility doesn't seem right to me either. I have not had the time yet to dig out my notes and come to a more firm conclusion.

On a weird note, I noticed this morning that when I walked into the dark bathroom and reached out for the switch, the fluorescent bulbs nearby glowed faintly. No closed circuit - something was surely 'flowing' somewhere, wasn't it?

I agree that a conductor will take on a single potential, no problem. I'm much less certain about what that potential will be. Your link confirms the former, but doesn't address the latter. I also have no problem with the observation that the presence of a conductor in an electric field will alter the field. However, if you'll note on your linked site, it depends on the geometry of the object too. It's not at all clear to me that a tall, narrow pole normal to the equipotential lines of a uniform field will do what you are saying they will do.

I'm sorry for dragging this out. I really need to check a few things and then either agree with you or give a detailed reason for not doing so.

 

[T37 Phil]

Walt - your site also mentions "induced flow of charge into the surrounding air when there is no direct connection ...", which tells me that an uninsulated conductor, like a metal mast, can exchange charge with the air around it. Well, maybe no exchange, but 'communicate' charge movement and experience mutual influence. That changes things a bit, don't you think?

Edit: I like the site, now that I'm reading it. I think I've also found the source of the confusion, perhaps. In Fig 4-2, a 2d potential is shown with two conductors running in and out of the page. The 'added' conductor (note the care to mention no excess charge) takes on a potential depending on where it is located within the field generated by the first conductor. If the second conductor is thin, no problem. If it's relatively thick, then it distorts the field and it's isopotential is something like the mean that you propose.

However, the mast situation is not that depicted in Fig 4-2. The atmosphere in that 3d case can be describe as stacked up isosurfaces, where the potential depends only on the height above the water. In that case the first conductor is no long a thin line running in and out of the page. Rather, it's more like a capacitor, two infinite sheets separated by a distance. The second conductor is still a line with some thickness, but it now passes through many different potential surfaces. If you will, Fig 4-2 is a slice through a picture which could be continued infinitely by stacking up identical pictures (forgetting about end effects of the finite mast length). The geometries are different. It may be that Fig 4-2 is a model of what happens during the bolt, if you consider the current path of the bolt to be a [infinite] line charge.

So, I'm still left pondering what the potential in a finite line conductor is when it is aligned normal to the planes of plate charges generating a uniform field.

 

[Bill Roosa]

induced charge

If the top of the mast is at potential A and the bottom is at potential B then one of two things must happen. 1) There will be current flow. In this case the current has already flowed down from the sky and is now only trying to get back. Or 2 there will be a charge build up on the top and a charge deficiency on the bottom of the mast. This charge is a capacitive type and exactly cancels out the impressed electric field. The mast is a dipole. So if there is a 10 volt / ft impressed field and the mast is 60 ft long (and vertical) the top acquires electrons till there is a 600 volt charge on the tip. Or 300 volts + at top and 300 volts - at the bottom if the mast is not grounded, or 250 + and 350 - ....... depends on the exact conditions of the ground interaction. The common point is there is 600 volts potential between the ends and a linearly varying E-field that cancels out the impressed E-field. There is no net E-field along the mast. If there was some residual E-field then that would force electrons to move till charged built up (or dispersed) to cancel it.
I TOTALLY disagree that the mast will take on some mid-point potential with no charge concentration at the ends. There would be no counter E-field to cancel the impressed E-field so charge would flow….see above
In our case both things happen one after the other. The mast becomes a dipole till the charge at the top is enough to ionize the air. The resistance goes down and charge begins to flow till it is all equalized.

 

[T37 Phil]

If the top of the mast is at potential A and the bottom is at potential B then one of two things must happen. 1) There will be current flow. In this case the current has already flowed down from the sky and is now only trying to get back. Or 2 there will be a charge build up on the top and a charge deficiency on the bottom of the mast. This charge is a capacitive type and exactly cancels out the impressed electric field. The mast is a dipole.

Dipole of sorts. This is what I've been trying to remember. It's not capacitive, as there is no dielectric that separates the charge in the conductor. I haven't convinced myself still that your description here is correct. Also, see the edit on my last post.

I TOTALLY disagree that the mast will take on some mid-point potential with no charge concentration at the ends. There would be no counter E-field to cancel the impressed E-field so charge would flow….see above
In our case both things happen one after the other. The mast becomes a dipole till the charge at the top is enough to ionize the air. The resistance goes down and charge begins to flow till it is all equalized.

The problem is that walt is 100% correct that for a perfect conductor and a static analysis 1) the exterior surface is all one potential and 2) there is no field inside the conductor.

The second point explains why all the current in your electrical wiring flows on the outer surface, if you've ever heard that statement. On the other hand, it seems to fly in the face of the first statement. But still, a) home wiring does not use perfect conductors and b) I think I'm still missing some key point in all this stuff.

 

[Bill Roosa]

Lightning E-field and the earths static E-field

Two completely different things. The earth's static E-field comes from tha fact that the ionosphere is charged and the earth is a ground. It is static and has a very large gredient, like kV/ft.
A lightning (cloud) induced E-field is a completely different animal that comes from the cloud air currents physically (we think that is what is happening) moving charge around. The two add together but since the Earth's E-field essentially comprises the two plates (earth's surface and ionosphere) of a spherical capacitor we don't see/feel it at all.

The discussion got on to the "free space charge" (Earth’s static E-field) a while back and that is what I was addressing with the large capacitor experiment. Sorry for the confusion. I'll attempt to be more accurate when replying.

 

[Bill Roosa]

Don't need a dialectric

You do not need a dielectric to separate charge. All you need is an E-field (or H but lets not go there just yet)
If you get out some graph paper and make two pictures of the mast. One is the cloud impressed E-field and the other is the concentrated charge E-field from the mast you can then vector sum them to see that it actually does sum to 0 for every point inside the mast (no current flow after the mast gets charged up at its ends).

Basically the cloud impressed E-field causes the electrons to move up the mast and this would continue indefinably except the charge can’t leave the mast. So it gets concentrated. This continues till the E-field created by the charges on the mast exactly cancels out the cloud e-field. If the cloud E-field changes a little the mast would then have an excess (or deficiency) of charge at its tip and that excess E-field from the concentrated charge moves the electrons till there is no reason to move them any more aka no net E-field inside.

The charge is not maintained by an insulator that somehow let the charge up the mast but will not let it down it is maintained by the E-field.

I will grant that having a dielectric between your cap plates makes it a lot easier to store more charge with less voltage. But that is a completely different subject.

 

[walt]

f the top of the mast is at potential A and the bottom is at potential B then one of two things must happen. 1) There will be current flow. In this case the current has already flowed down from the sky and is now only trying to get back. Or 2 there will be a charge build up on the top and a charge deficiency on the bottom of the mast. This charge is a capacitive type and exactly cancels out the impressed electric field. The mast is a dipole. So if there is a 10 volt / ft impressed field and the mast is 60 ft long (and vertical) the top acquires electrons till there is a 600 volt charge on the tip. Or 300 volts + at top and 300 volts - at the bottom if the mast is not grounded, or 250 + and 350 - .......

Bill, if I understand correctly (some things you write, I don't understand such as what exactly is a "capacitive charge" when its on a conductor), you believe #2 is happening. Ie, that the mast becomes a diplole and that one end of this conductor (we don't need to worry about details such as "how" conductive for this discussion) is at a different potential than another.

I would invite you to google "conductor in an electric field" and read what other have to say about this - ie, electrostatic equilibrium. Ill list a few web sites here such as

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html

http://www.physicsclassroom.com/class/estatics/u8l4d.cfm

etc.. etc

The idea that the conductor in equilibrium will take on a single potential also happens to be the idea behind a Faraday cage. I don't believe you will find any reference saying that a Faraday cage in an electric field forms a "dipole" - in which case there would be a field inside the cage caused by the dipole.

 

[T37 Phil]

Bill, maybe it's just terminology again. A capacitor has a dielectric separating two charged surfaces, period. A capacitive effect may be experienced in less canonical situations. For example, the interface between a SS electrode and a salt solution is different than that between say a Pt, NaCl filled glass or AgCl coated Ag electrode and the same solution. The SS electrode passes current capacitively and therefore acts as a highpass filter. The dielectric in that case is really in the details of the surface chemistry of the interface and is beyond the scope of this forum. A conductor, on the other hand, has no such property. As such, I don't even see a capacitive effect in your description, only the dipole itself.

 

[Bill Roosa]

Yep that is how it works

From the second link you posted

"It was emphasized that when a conductor acquires an excess charge, the excess charge moves about and distributes itself about the conductor in such a manner as to reduce the total amount of repulsive forces within the conductor."

This is describing a "charged up conductor" (capacitive charging) NOT a conductor in an externally applied E-field.

The mast is not "at equilibrium" in the sense they are using it. When they say "at equilibrium" they mean no external E-field. They just magically pump some electrons into the conductor and then see what happens. The electrons distribute themselves (more or less) uniformly so as to have no net E-field.

I'll draw a picture as this is like diagnosing car noises over the phone. Gota go do some work now

 

[T37 Phil]

Bill, imagine that you take a 12V car battery, set it on a bench and connect a wire to the positive terminal. Assuming the wire is a perfect conductor, what is the potential at the other end of the wire? No dipole, but is it charged up? There is no capacitative effect in that picture. A capactive effect is simply the movement of charge at location A by placing some other charge nearby, say at B, that cannot move to A (as in a current), but that 'pushes' charge through a field effect [through the insulating material]. If you simply conduct charge someplace, that isn't automatically capacitive charging.

Edit: perhaps I've found an interpretation that might pan out. Let's say the mast is insulated from all other metal on the boat, no ground plate, nothing. If you now force current in through the bottom, the mast will be 'charged'. But how can you force charge in? There must be a capacitive effect through the 'dielectric' interface of the air. That means that whatever gadget you use to pump charge into the mast somehow involves the air in its circuit. In reality, you're suggesting that the gadget is everything that happens in a strike.

 

[T37 Phil]

From the second link you posted

Hey, by the way, that 2nd link you were talking about also mentions what I was saying about excess charge being on the outside surface of the conductor. Cool!

 

[walt]

Bill, certainly you can have charges "building up" at for example sharp points on a conductor. Notice that most of the reference don't point this out (because I don't think it affects the statement that the potential is uniform across the conductor).

But I would also challenge you to quantify how large of an electric field across the conductor these charges create. I think any fields caused by these charges across the conductor are completely in the noise - very close to zero.

 

[T37 Phil]

... I think any fields caused by these charges across the conductor are completely in the noise - very close to zero.

The field inside a conductor is zero. Not close, not noise, but exactly zero.

 

[walt]

Phil, I don't see any mechanism for applying charge to a mast in the case where we just have the electric field. Air may not be a perfect insulator but in the time frame we are considering here (ie, moving clouds creating a shadow charge on the ground), I think any actual "conductive" charging through the air could be considered negligible.

Here is another interesting thing to ponder. As you know, when one charge is seperated from another charge, the produced electric field is determined by dielectric constant of the material in between the charges. Air has a dielectric constant of one.

But water is "bipolar" - I'm not sure exactly what that means but one effect of this is that the dielectric constant of water is something like 80 (dependent on a bunch of variables such as frequency). But even at the frequencies we consider for a lightning strike, the dielectric constant of water is still very large. Water would make a very good capacitor dielectric except that it would also make a very lossy capacitor.

So what does the high dielectric constant of water mean.. One thing is that if I have a set of opposite charges and separate them by some distance, the voltage between the charges would be 80 times smaller in water than in air (at DC, some temp, etc).

I am guessing here - but think this probably has something to do with why a lightning strike will come down to the water surface and then follow the air just above the water surface for some time. The fields that the charge would create in water would be much smaller - ie, not enough to ionize water making the ionized air just above the surface a lower impedance path..

Just and idea.. could be wrong.

 

[T37 Phil]

Phil, I don't see any mechanism for applying charge to a mast in the case where we just have the electric field. Air may not be a perfect insulator but in the time frame we are considering here (ie, moving clouds creating a shadow charge on the ground), I think any actual "conductive" charging through the air could be considered negligible.

I agree. I was just spit balling to rescue the capacitive charging terminology. The large charge transfer happens only after a truly conductive path has been established and the bolt is in progress. But the conductive medium can influence the field shape pre-strike, I believe. That was your interest back there somewhere. What is the field and the potential in the mast like before conditions are ripe, just before a strike, during a strike, etc... If the ripe conditions were detectable by a potential change in the mast then you'd have a lightning warning system.

Here is another interesting thing to ponder. As you know, when one charge is seperated from another charge, the produced electric field is determined by dielectric constant of the material in between the charges. Air has a dielectric constant of one.

The field itself is determined by the [relative] permittivity of the medium. I guess dielectric constant is correct, but sounds odd in this context. H2O is a dipole because the bonds between the central oxygen and the hydrogens off at 120 degrees do not result in neutral sites in the molecule. The hydrogens are more 'giving' of their electron and the oxgen is more 'taking'. That makes the O end more negative and the H end (as defined by a line drawn between the two) is more positive. Lots of pictures if you do a web search. Very important and gives H2O many of its unusual properties including its ability to solvate ionic species. By the way, I'd say water molecules are dipoles rather than that water is bipolar (which makes it seem emotionally unstable or something).

So what does the high dielectric constant of water mean.. One thing is that if I have a set of opposite charges and separate them by some distance, the voltage between the charges would be 80 times smaller in water than in air (at DC, some temp, etc).

And smaller in salt water than in freshwater. I think that's on one of the site links you posted, particularly in the context of side flashes.

I am guessing here - but think this probably has something to do with why a lightning strike will come down to the water surface and then follow the air just above the water surface for some time. The fields that the charge would create in water would be much smaller - ie, not enough to ionize water making the ionized air just above the surface a lower impedance path..

Just and idea.. could be wrong.

Sounds plausible.

 

[Bill Roosa]

In less than a thousand words

Fig 1 is just a capacitor
Fig 2 is MS boat in salt water with a keel to mast ground and a cloud above it
Fig 3 is a mast isolated between the two plates of a capacitor. Off to the left side are three plots of the E-field a) of the cap, b) of the mast along the centerline due to the charges on it, and c) the net field along the centerline of the mast

As the capacitor charges in Fig 3 the mast will develop induced charges (just like the water surface under a charged cloud BTW). These develop to the extent that they cancel out the E-field. If there was an E-field inside the mast it would cause the charges to move till there was not one. Discharge the capacitor (not through the mast) and the charge concentration on the mast redistributes itself so there is (again) no E-field. You are not actually moving electrons into or out of the mast just rearranging them. The magnitude of the voltage across the mast (vertically) is just the magnitude of the induced E-field gradient times the height of the mast.

My hypothesis was that MS’s boat got inductively charged up to a few hundred or thousand volts. Not enough to cause a strike but enough to have a pretty good potential across the masthead-keel. The strike equalizes all the cloud and water surface charges, the induced E-field goes away very rapidly and MS is left with a charged up boat. This is electrically the same as a charged up capacitor. Now I’m not trying to say that it is a good capacitor but there is some charge there. Apparently more than enough (hundreds or thousands of volts) to zap his equipment as every wire that has some connection to ground discharges into the keel.

Thoughts?

 

[Bill Roosa]

Hey Main sail!

Looking back at your unfortunate event, is there a correlation between height above or below the waterline and damage? My suspicion is the further away vertically from the waterline the less likely the item was to be damaged.

 

[T37 Phil]

Bill, that at least answers one question. You're NOT saying that the boat is acting as a capacitor being charged. You're saying that the atmosphere is the capacitor being charged and the boat is "in the dielectric", as it were. What's the potential of the mast? That's the one question I have not answered for myself as yet. Being a conductor it should be the same potential everywhere (but which). In instances where you observe charge separation, the medium between the separated charge is generally something that conducts very poorly. For the atmosphere, that's air, right up until breakdown voltage. For the mast? There is no such region in the mast, so you need to explain how what you are describing is possible. In every theory I know (that is, remember), it's not possible.

 

[Bill Roosa]

In the diaelectric

Yes, that is one interpretation of it. The boat is "in the dielectric". Now take the boat and look at it in isolation. We don't care how it got polarized just that it is. So lots of electrons at the top and not so many at the keel. SURE does look like a capacitor to me. If I turn off the external field it will discharge JUST like a capacitor.

The voltage at the top of the mast is just the external E-field gradient times the mast height. Let’s not get into "free space potential" again as that is everywhere all the time and just acts like an overpressure on everything. Kinda like air pressure. My gage says 0 but we know it is really 14.7 psi compared to a vacuum. To determine an actual value for that voltage we can make some assumptions. The E-field is the change in voltage/distance between the two measurements.
Let’s make the following assumptions:

The earth is at 0 volts
Lightning starts on average 3000 ft above the ground
The dielectric strength of air is ? (Guessing) 12000 volts / ft at breakdown.

Just before the strike the air is on the verge of breaking down so I have the breakdown gradient applied. So that would mean that there is an equivalent voltage of 12000*3000 volts = 36,000,000 volts 3000 ft above our head or just 12000 volts/ft gradient. Should have seen this coming.
I can see that the maximum gradient is the dielectric strength for air. I’m such a dumb bunny sometimes.
So for a 60 ft mast you get a masthead potential; 60*12000 = 720,000 volts just as the strike begins. Probably a pretty fast rise time before that or I'd notice St. Elmos Fire

Somebody can supply the correct numbers but the math is sound. It does not really matter what height the lightning strikes from the voltage just before the current begins to flow is

V @ masthead = mast height * dielectric strength of air (moist, moving, at temp.....)