Let's clear up the lightning/car safety issue

[Oblivionaire]

Originally posted by: JohnCU If you drew a circuit, the resistance of the tires is so high it would appear as an open circuit, a floating node above ground.

The resistance of the tires isn't high enough to impede the flow of current in the presence of a voltage as high as lightning now is it? As a matter of fact neither is the resistance of the air between the clouds and ground.

 

[JohnCU]

Originally posted by: Oblivionaire

Originally posted by: JohnCU If you drew a circuit, the resistance of the tires is so high it would appear as an open circuit, a floating node above ground.

The resistance of the tires isn't high enough to impede the flow of current in the presence of a voltage as high as lightning now is it? As a matter of fact neither is the resistance of the air between the clouds and ground.

my point is that the car is not grounded.

 

[BrownTown]

The resistance of the air is not what the lightning eventually flows threw. The lightning discharge is actually flowing threw a very conductive stream of plasma between the cloud and the sky. Obviously same thing woudl happen to your tires, they would be destroyed by the amount of energy released. Just trying to point out that this is not a linear relationship between voltage and current that would follow ohms law.

 

[bobsmith1492]

First, though, there has to be enough current flow to start the plasma stream, and hence, an extremely high potential.

 

[Oblivionaire]

Originally posted by: JohnCU

Originally posted by: Oblivionaire

Originally posted by: JohnCU If you drew a circuit, the resistance of the tires is so high it would appear as an open circuit, a floating node above ground.

The resistance of the tires isn't high enough to impede the flow of current in the presence of a voltage as high as lightning now is it? As a matter of fact neither is the resistance of the air between the clouds and ground.

my point is that the car is not grounded.

And my point is that at voltages as high as lightning that will strike the ground, it may as well be, because those tires will not insulate your car from lightning trying to reach earth ground.

We are after all discussing lightning level voltages and currents in this thread, and not small voltages or currents as in your DMM, vehicle electrical system or house current.

 

[BrownTown]

Not sure what the whole tire discussion is about, but lightning will have absolutely zero problem jumping from your cars frame to ground, even a medium voltage distribution line has enough voltage to jump that far, high voltage lines and lightning are sucessive orders of magnitude higher. Anyways, im sure people ahve died in a car, people ahve even died in the middle of their own homes, but certinaly a car is better protection than nothing.

 

[bobsmith1492]

Most likely the only people who have died in a car in a thunderstorm would have to have been in a convertible or fiberglass/carbon/plastic frame, although even fiberglass might conduct better than air and direct the flow around the occupants. If you're in a metal vehicle, you'd have to be safe. I suppose there could be some potential between the top and bottom of the car during a strike due to the massive current flow, but I don't see how that could kill you since the flow of current won't go straight through the car when it has an extremely low impedance path along the skin (sheet metal) around the car. Now, I COULD see it messing with your car's engine and computer and whatnot, causing it to fail and leading to an accident, but that's a sort of secondary effect.

 

[BrownTown]

Well, I took the 5 seconds needed to look this up on google, and it is infact the skin effect that is protecting you in a car. The charges will repel each other and flow along the surface of the car and then jump to ground. If your car is fiberglass then your screwed, but a metal car *should* be safe, even [theoretically] if you are touching the metal frame.

 

[JohnCU]

Originally posted by: BrownTown
Well, I took the 5 seconds needed to look this up on google, and it is infact the skin effect that is protecting you in a car. The charges will repel each other and flow along the surface of the car and then jump to ground. If your car is fiberglass then your screwed, but a metal car *should* be safe, even [theoretically] if you are touching the metal frame.

I did the same, but did you see my argument against the skin effect? You can also find things on google that talk about the car being a faraday cage.

 

[duragezic]

So now the question should be: what cars have a mostly metal skin and which do not?

 

[Born2bwire]

Originally posted by: JohnCU

Originally posted by: BrownTown
Well, I took the 5 seconds needed to look this up on google, and it is infact the skin effect that is protecting you in a car. The charges will repel each other and flow along the surface of the car and then jump to ground. If your car is fiberglass then your screwed, but a metal car *should* be safe, even [theoretically] if you are touching the metal frame.

I did the same, but did you see my argument against the skin effect? You can also find things on google that talk about the car being a faraday cage.

The skin effect is a bit of a misnomer here since it is used in reference to electromagnetic waves. But the skin effect and a Faraday cage are the same thing. When an EM wave strikes a PEC, currents are induced in such a way as to cancel out the EM wave inside the PEC. In the same manner, a Faraday cage in regards to a static field arranges its surface charges to cancel out the static Electric field. But since an ideal Faraday cage is an enclosed PEC shell, it will work on EM waves too. The skin effect deals with the fact that we do not have perfectly electrical conductors. Despite the enormous conductivity of most metals, since they do not behave as PEC's, then an EM wave can penetrate as an evanescent wave into the metal. Skin depth is just a number of merit that describes how far a wave can penetrate into the medium and decrease its amplitude by 1/e.

I do not know if I would model the lightening as an delta function. The currents, yes, but in my mind the electric field would locally be more like a step function. A large potential is built up between the cloud and the ground/car, so a large static E-field arises. Then the charges are dissipated between the cloud and ground through the conducting plasma. So maybe it would be better of thinking like you have a huge voltage potential built up across a cap with its ground lead shorted to a large sink, and then shorting the unconnected lead to the sink with a wire.

Either way, the Fourier transform of a delta function is 1/(2\pi) \int dy e^{ixy}. So we can see that the frequency domain of the delta function is a constant across all frequencies. Which simply means that a delta function is the superposition of all frequencies with equal weights. So if we were to have an EM wave of delta function in the time domain, and it strikes a Faraday cage constructed of reasonable conductivity, then my intuition is that the transmitted signal into the cage would look something like an exponential in the frequency domain, with the maximum at DC and an exponential drop-off to zero as frequency goes up. The drop-off should be very drastic. And since we will only transmit into the cage very very low frequencies, it would be safe to assume that we will not achieve any resonant waves in the cage since the wavelengths of the signals will be far larger than the physical dimensions of our cage.

As for the tires, I do not think that they are really a significant factor. Your car is going to be grounded on a regular basis, when you touch the car to open the door, touch the gas pump to the car to fill it up, etc. In addition, the potential difference for the discharge is very large and really there probably is not going to be a very large potential difference between the ground and the car in comparison to that of the clouds and ground. And I would expect that the resistive properties of the rubber tires would only mean that the currents running through the car will sink to the ground through the air as opposed to the tires. When the lightning strikes the car, the charge distribution will start to equalize between the cloud and the car. And since the car is probably not going to be able to sink much charge and the charge is going to build up on the surface of the metal, then it will quickly gain a very large potential in respect to the ground and allow either the currents to overcome the resisitivity of the tires or the dielectric of the air.

 

[JohnCU]

Originally posted by: Born2bwire

Originally posted by: JohnCU

Originally posted by: BrownTown
Well, I took the 5 seconds needed to look this up on google, and it is infact the skin effect that is protecting you in a car. The charges will repel each other and flow along the surface of the car and then jump to ground. If your car is fiberglass then your screwed, but a metal car *should* be safe, even [theoretically] if you are touching the metal frame.

I did the same, but did you see my argument against the skin effect? You can also find things on google that talk about the car being a faraday cage.

The skin effect is a bit of a misnomer here since it is used in reference to electromagnetic waves. But the skin effect and a Faraday cage are the same thing. When an EM wave strikes a PEC, currents are induced in such a way as to cancel out the EM wave inside the PEC. In the same manner, a Faraday cage in regards to a static field arranges its surface charges to cancel out the static Electric field. But since an ideal Faraday cage is an enclosed PEC shell, it will work on EM waves too. The skin effect deals with the fact that we do not have perfectly electrical conductors. Despite the enormous conductivity of most metals, since they do not behave as PEC's, then an EM wave can penetrate as an evanescent wave into the metal. Skin depth is just a number of merit that describes how far a wave can penetrate into the medium and decrease its amplitude by 1/e.

I do not know if I would model the lightening as an delta function. The currents, yes, but in my mind the electric field would locally be more like a step function. A large potential is built up between the cloud and the ground/car, so a large static E-field arises. Then the charges are dissipated between the cloud and ground through the conducting plasma. So maybe it would be better of thinking like you have a huge voltage potential built up across a cap with its ground lead shorted to a large sink, and then shorting the unconnected lead to the sink with a wire.

Either way, the Fourier transform of a delta function is 1/(2\pi) \int dy e^{ixy}. So we can see that the frequency domain of the delta function is a constant across all frequencies. Which simply means that a delta function is the superposition of all frequencies with equal weights. So if we were to have an EM wave of delta function in the time domain, and it strikes a Faraday cage constructed of reasonable conductivity, then my intuition is that the transmitted signal into the cage would look something like an exponential in the frequency domain, with the maximum at DC and an exponential drop-off to zero as frequency goes up. The drop-off should be very drastic. And since we will only transmit into the cage very very low frequencies, it would be safe to assume that we will not achieve any resonant waves in the cage since the wavelengths of the signals will be far larger than the physical dimensions of our cage.

As for the tires, I do not think that they are really a significant factor. Your car is going to be grounded on a regular basis, when you touch the car to open the door, touch the gas pump to the car to fill it up, etc. In addition, the potential difference for the discharge is very large and really there probably is not going to be a very large potential difference between the ground and the car in comparison to that of the clouds and ground. And I would expect that the resistive properties of the rubber tires would only mean that the currents running through the car will sink to the ground through the air as opposed to the tires. When the lightning strikes the car, the charge distribution will start to equalize between the cloud and the car. And since the car is probably not going to be able to sink much charge and the charge is going to build up on the surface of the metal, then it will quickly gain a very large potential in respect to the ground and allow either the currents to overcome the resisitivity of the tires or the dielectric of the air.

:thumbsup: damn good response. i guess i feel a little dumber after reading that and not being able to throw out that kind of response myself, but atleast i understood what you wrote.

 

[silverpig]

Originally posted by: duragezic
So now the question should be: what cars have a mostly metal skin and which do not?

SUVs ftw!

 

[0verload]

I have heard that lightning strike due to separation of charge. Nothing unusual there. But these charges attract each other. When the force is great enough they travel through the air from multiple points. When one trail from the cloud and one from the ground connect, bam, lightning strikes. If it hits your car it travels through the more conductive parts that free electrons to allow current flow. When it is finish there is excess charge left over that was insufficient to breakdown the dielectric between the car and ground. If you immediately get out of your car you complete the connection with the lower resistance of your body and zap, you get fried! I believe also that the opposite charge accumulates above the cloud, which is why you see flashes in the clouds during a storm.

Another question I have heard often that I will bring up here is: which direction does current flow in a lightning bolt?

 

[LeoDioxide]

Both ways, don't let anyone tell you otherwise.

I can't really agree with the "hop out of your car and get zapped" idea you put forth. With the massive forces of induction at play, let us suppose the cloud is positive and the ground + car negative. For the lightning to strike, your car and the ground around it become very negative, air becomes ionized, electricity flows (in a sense).

Suppose the lightning strikes from the cloud to your car. Your car builds up a charge and imparts it towards the ground, same as if the cloud weren't there, you would end up with a neutral charge. The process can be thought of as another lightning strike from the car to the ground if need be, and at such high voltages the charge would flow through the tires, completely overriding the resistivity of hard rubber.

I'm going to assume that the resistivity of the rubber is less than that of air, after checking wikipedia.

 

[wwswimming]

Originally posted by: BigT383
The people inside the car are safe because the energy travels through the metal around them. Even though it is a longer route, there is less resistance since metal is a much better conductor.

what happens if the current surge from the lightning tosses a few sparks in the direction of the gas tank ?

i'm wondering what (if anything) the car companies do to design for this possibility (besides a fire extinguisher).

 

[Born2bwire]

Originally posted by: wwswimming

Originally posted by: BigT383
The people inside the car are safe because the energy travels through the metal around them. Even though it is a longer route, there is less resistance since metal is a much better conductor.

what happens if the current surge from the lightning tosses a few sparks in the direction of the gas tank ?

i'm wondering what (if anything) the car companies do to design for this possibility (besides a fire extinguisher).

Probably nothing. The gas in your tank generally does not have enough oxygen or vapors to combust. You would probably need to be bone dry to be able to have combustion in your tank and then there would not be much to burn.

 

[Biftheunderstudy]

 

[Biftheunderstudy]

Tires are most definitely not good insulators, rubber yes, the metal wires running through the tires no.
I live on a farm and we used tires to hold down the bottom of a fence mistakenly thinking they would not ground out the electric cattle fence running above them.
Turns out the metal wires conduct VERY well, the fence then is grounded and the cows get out
The fence is at about 5-6 kV and about 1E-6 amps.

 

[kenji4life]

To go back ... to the future!
1.21 GIGAWATTS? GREAT SCOTT!!
http://www.seeing-stars.com/Locations/B...outhouseStruckByLightning(smaller).JPG

 

[BrownTown]

lol, and the makers of the movie knew so little about electricity that they though it was pronnounced "Jiggawatts", how dumb can you get? Also, 1.21 Gigawatts is about the size of the largest nuclear or fossil units, so that amount of power is not exactly imppossible to obtain.

 

[JohnCU]

Originally posted by: BrownTown
lol, and the makers of the movie knew so little about electricity that they though it was pronnounced "Jiggawatts", how dumb can you get? Also, 1.21 Gigawatts is about the size of the largest nuclear or fossil units, so that amount of power is not exactly imppossible to obtain.

IIRC jiggawatts is the correct pronounciation, but no one says it lik ethat.

 

[BrownTown]

yeah, and my computer has a 2 jiggahertz processor and 80 jiggabytes of memmory...

seriously, nobody I have ever heard pronounced them that way

 

[bobsmith1492]

Originally posted by: JohnCU

Originally posted by: BrownTown
lol, and the makers of the movie knew so little about electricity that they though it was pronnounced "Jiggawatts", how dumb can you get? Also, 1.21 Gigawatts is about the size of the largest nuclear or fossil units, so that amount of power is not exactly imppossible to obtain.

IIRC jiggawatts is the correct pronounciation, but no one says it lik ethat.

I don't think so; how do you pronounce giddyup? Giggle? Giddy? That is a hard G sound, not a soft j sound in all of those cases; giga is the same setup, hence, the same pronunciation.

However, the dictionary agrees with you.... http://www.bartleby.com/61/24/G0122400.html

YearrrrggghhH!!! Well, I dunno. English is a strange language, if I do say so myself.

 

[BrownTown]

let me just point out that languages are constantly evolving, and given the fact that EVERYONE uses the hard G in this case I would have to say that this has now become the correct pronunciation.