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#1
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Jim Kelley wrote in message ...
Tom Bruhns wrote: There is no such thing as "the voltage" between the ends of your excited dipole at an instant in time. Perhaps they meant the voltage 'across' the ends of the dipole. The ends should always be an electrical half-wave out of phase, right? There should only be two instants of time during a period when the difference in potential from end to end is zero. What are you saying exactly, Tom? I'm saying that if you measure the voltage between two points on a good conductor, in a path along that conductor, it will be very small. The electric field is always perpendicular to a perfect conductor at the surface of that conductor. For a conductor with resistance, the drop along it is I*R, and therefore the nearby electric field is in general not quite perpendicular, but unless it's a darned inefficient antenna, it's very nearly so. I'm also saying that the voltage (potential) between two points depends, in general, on the path you take between the two points. You should be _especially_ aware of that fact when you're in the presence of time-varying magnetic fields, such as you have around a powered antenna. As I said, if you measure the potential along a line perpendicular to the antenna, it will be large (when the antenna is excited with some power). I fully expect the electric field to be high near the wire, but perpendicular to the wire, NOT parallel to it. Cheers, Tom |
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#2
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Tom Bruhns wrote:
I'm also saying that the voltage (potential) between two points depends, in general, on the path you take between the two points. You should be _especially_ aware of that fact when you're in the presence of time-varying magnetic fields, such as you have around a powered antenna. In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
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#3
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Ian White, G3SEK wrote:
In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. How about using an artificial ground at the measurement point? -- 73, Cecil, W5DXP |
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#4
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Cecil Moore wrote:
Ian White, G3SEK wrote: In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. How about using an artificial ground at the measurement point? Define one, if you can! What are its properties, and how would you achieve them? It's rather like the early wireless users who "earthed" their receivers to the aspidistra pot in the corner of the room - after all, it contained earth, so why didn't it work? -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
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#5
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How 'bout you model it with your concept of an "artificial ground", and
let us know the result? You can measure the voltage with EZNEC by connecting the two points to be measured with a wire and inserting a zero amplitude current source in the wire. The source will act like an open circuit, and the voltage will be reported in the Source Data output. After you've determined the voltage relative to your "artificial ground", modify the "artificial ground" and note the effect on the voltage. Then see if you can figure out what the voltage is between the "artificial ground" and the Earth. Or, give us your justification for assuming that it's zero. If it is zero, via what path? As Tom has been saying, the voltage between two points depends on the path you take between them. Roy Lewallen, W7EL Cecil Moore wrote: Ian White, G3SEK wrote: In practice that will means that the voltage you measure between say the end of a whip and ground will depend on how you choose to route the connecting leads to the voltmeter, and how you connect to ground... and above (below?) all on what you define "ground" to be. How about using an artificial ground at the measurement point? -- 73, Cecil, W5DXP |
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#6
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Roy Lewallen wrote:
After you've determined the voltage relative to your "artificial ground", modify the "artificial ground" and note the effect on the voltage. Then see if you can figure out what the voltage is between the "artificial ground" and the Earth. Or, give us your justification for assuming that it's zero. If it is zero, via what path? As Tom has been saying, the voltage between two points depends on the path you take between them. Wow, you sure ASSume a lot from a simple question. Let's turn it around and you guys prove that the voltage at the ends of a dipole is less than or equal to the feedpoint voltage even though a florescent light bulb is brighter at the ends. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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#7
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Cecil wrote,
Wow, you sure ASSume a lot from a simple question. Let's turn it around and you guys prove that the voltage at the ends of a dipole is less than or equal to the feedpoint voltage even though a florescent light bulb is brighter at the ends. -- 73, Cecil http://www.qsl.net/w5dxp What causes the light to light up, Cecil, an E field, or a V field? You can't seem to get over the fact that the voltage between two points in a time varying E field may not be unique and thus not measurable. Go read the book I told you to read, Cecil, and you'll understand why. 73, Tom Donaly, KA6RUH |
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#8
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Tdonaly wrote:
What causes the light to light up, Cecil, an E field, or a V field? You can't seem to get over the fact that the voltage between two points in a time varying E field may not be unique and thus not measurable. Go read the book I told you to read, Cecil, and you'll understand why. Why not just tell me to read the Bible where God is the cause of everything - same difference. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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#9
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Tdonaly wrote:
What causes the light to light up, Cecil, an E field, or a V field? According to quantum electrodynamics, fields don't exist. So are you talking about photons or virtual photons above? In either case, voltage can still be measured by a voltmeter. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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#10
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Tom Bruhns wrote:
I'm saying that if you measure the voltage between two points on a good conductor, in a path along that conductor, it will be very small. True for DC and RF traveling waves. Not true for standing waves. A 1/2WL dipole is a *standing-wave* antenna. What do you get when you measure the voltage between the voltage maximum and voltage minimum on a feedline with a 10:1 SWR? Exactly the same principle applies to *standing-wave* antennas. -- 73, Cecil, W5DXP |
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