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#1
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#2
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Hi Yuri,
And yet you are on Tom for just such a similar distinction of word choice. Does this illustrate the principles of reciprocity? ;-) 73's Richard Clark, KB7QHC Slight difference. Rauchians choose their words as weapons of smart ass attitude at "morons" who happen to be right, while Rs are wrong. They fail at the discussion of facts and arguments. I resort to (not too proud of it) responding and biting back in order to prove the point. There is reciprocity but not ideal, just like RF current behavior in real coil. And he did it for the fifth time, go figure. Fuggetaboutit! Yuri, K3BU |
#3
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#4
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Richard Clark wrote:
I will agree in this respect, but not all Toms are Rauchs. The term "Current Drop" is abhorrent to some (a pollution of technical language), an irritant to others, and inconsequential to many who simply enjoy the cat fight. Heh, heh, so you don't believe there is a current drop between the current maximum point and current minimum point on a transmission line with reflections? Seems to me going from 2 amps at a current maximum to 0.1 amps at a current minimum is a measurable drop in total current. Would you please provide a proof that going from 2 amps to 0.1 amps is NOT a drop in total current? Just one more example of trying to use lumped circuit analysis methods on distributed network problems. Are you guys ever going to learn? Hint: With distributed networks involving an appreciable percentage of a wavelength, there are definitely current drops in the series loop. This certainly applies to 75m Bugcatcher coils used on standing-wave mobile antennas. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
#5
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Cecil Moore wrote:
Richard Clark wrote: I will agree in this respect, but not all Toms are Rauchs. The term "Current Drop" is abhorrent to some (a pollution of technical language), an irritant to others, and inconsequential to many who simply enjoy the cat fight. Heh, heh, so you don't believe there is a current drop between the current maximum point and current minimum point on a transmission line with reflections? Seems to me going from 2 amps at a current maximum to 0.1 amps at a current minimum is a measurable drop in total current. Would you please provide a proof that going from 2 amps to 0.1 amps is NOT a drop in total current? Just one more example of trying to use lumped circuit analysis methods on distributed network problems. Are you guys ever going to learn? Hint: With distributed networks involving an appreciable percentage of a wavelength, there are definitely current drops in the series loop. This certainly applies to 75m Bugcatcher coils used on standing-wave mobile antennas. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- Next, Cecil, you're going to be talking about a "current gradient" and a "scalar current field." Here's a question for you, Cecil, and Richard Harrison, and Yuri, too: how do you take the gradient of the current at a point on a transmission line, and, if were possible to do so, what is the physical significance of the result? 73, Tom Donaly, KA6RUH |
#6
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Tom Donaly, KA6RUH wrote:
": how do you take the gradient of the current at a point on a transmission line?" Not sure I understand the question. Gradient is the rate of change and that`s the derivative of the current at a given point. Over a certain path it is the difference between the path ends and can be averaged for the path. For convenience, Kraus has collected transmission line formulas. I`m not a typist so I`ll just say they are near the end of the new edition, page 890. In the 1950 edition they can be fornd on pages 506 and 507, also near the end of the book. Work out your own example. Best regards, Richard Harrison, KB5WZI |
#7
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Richard Harrison wrote:
Tom Donaly, KA6RUH wrote: ": how do you take the gradient of the current at a point on a transmission line?" Not sure I understand the question. Gradient is the rate of change and that`s the derivative of the current at a given point. Over a certain path it is the difference between the path ends and can be averaged for the path. For convenience, Kraus has collected transmission line formulas. I`m not a typist so I`ll just say they are near the end of the new edition, page 890. In the 1950 edition they can be fornd on pages 506 and 507, also near the end of the book. Work out your own example. Best regards, Richard Harrison, KB5WZI Thank you, Richard, you just made my point. 73, Tom Donaly, KA6RUH |
#8
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Tom Donaly wrote:
Cecil Moore wrote: Heh, heh, so you don't believe there is a current drop between the current maximum point and current minimum point on a transmission line with reflections? Seems to me going from 2 amps at a current maximum to 0.1 amps at a current minimum is a measurable drop in total current. Next, Cecil, you're going to be talking about a "current gradient" and a "scalar current field." Here's a question for you, Cecil, and Richard Harrison, and Yuri, too: how do you take the gradient of the current at a point on a transmission line, and, if were possible to do so, what is the physical significance of the result? A total current gradient obviously exists on a transmission line with current minimums and maximums. You can locate those points with a simple pickup loop. The current gradient is caused by the superposition of forward and reflected current waves as described in any transmission line textbook. "Taking the gradient" seems to me to be unnecessary and just a logical diversion away from the qualitative conceptual discussion. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
#9
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Cecil Moore wrote:
Tom Donaly wrote: Cecil Moore wrote: Heh, heh, so you don't believe there is a current drop between the current maximum point and current minimum point on a transmission line with reflections? Seems to me going from 2 amps at a current maximum to 0.1 amps at a current minimum is a measurable drop in total current. Next, Cecil, you're going to be talking about a "current gradient" and a "scalar current field." Here's a question for you, Cecil, and Richard Harrison, and Yuri, too: how do you take the gradient of the current at a point on a transmission line, and, if were possible to do so, what is the physical significance of the result? A total current gradient obviously exists on a transmission line with current minimums and maximums. You can locate those points with a simple pickup loop. The current gradient is caused by the superposition of forward and reflected current waves as described in any transmission line textbook. "Taking the gradient" seems to me to be unnecessary and just a logical diversion away from the qualitative conceptual discussion. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- You and Richard need a refresher course in electromagnetics. I hope Yuri doesn't fall into the same trap. 73, Tom Donaly, KA6RUH |
#10
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Tom Donaly wrote:
Next, Cecil, you're going to be talking about a "current gradient" and a "scalar current field." Here's a question for you, Cecil, and Richard Harrison, and Yuri, too: how do you take the gradient of the current at a point on a transmission line, and, if were possible to do so, what is the physical significance of the result? 73, Tom Donaly, KA6RUH The standing wave current profile along, for example, a quarter wave radiator is a cosine function. The gradient then would be the derivative of the cosine function which is a -sine function. 73, ac6xg |
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