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#1
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dansawyeror wrote:
Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php Please note that the third wire to ground creates the unbalance that causes feedline radiation. You seem to be confusing cause and effect. The cause of the feedline radiation is the existence of that third wire, not SWR. All it proves is that feedline radiation is caused by that third wire path which unbalances the source currents. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#2
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If it is true it proves that a driving a "bad" load can cause the coax feedline
to radiate a significant portion of the feed energy. At this point I am not sure what word to use other then 'bad', but I due intend to perform experiments to measure and verify what is happening and the model. Dan Cecil Moore wrote: dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php Please note that the third wire to ground creates the unbalance that causes feedline radiation. You seem to be confusing cause and effect. The cause of the feedline radiation is the existence of that third wire, not SWR. All it proves is that feedline radiation is caused by that third wire path which unbalances the source currents. |
#3
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dansawyeror wrote:
If it is true it proves that a driving a "bad" load can cause the coax feedline to radiate a significant portion of the feed energy. You need to define "bad load". A "bad load" for unbalanced line is a balanced load, no matter what the impedance. A "bad load" for balanced line is an unbalanced load, no matter what the impedance. The third wire used in the aforementioned software is designed to unbalance the system, no matter what the impedance. To illustrate a balanced system, a fourth wire needs to be added in parallel with and about four inches away from the third wire. Then compare the currents in the third and fourth wires under conditions of changing loads. I due intend to perform experiments to measure and verify what is happening and the model. Please feel free to experiment but at least a dozen participants of this newsgroup already know what is happening and have been trying to tell you. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#4
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dansawyeror wrote:
If it is true it proves that a driving a "bad" load can cause the coax feedline to radiate a significant portion of the feed energy. At this point I am not sure what word to use other then 'bad', but I due intend to perform experiments to measure and verify what is happening and the model. dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php We are in a position now to say what is wrong with that program. Essentially, it violates the conservation of energy principle. Energy cannot exist as both reflected power and radiated power at the same time. If the SWR is 1400:1, the feedline cannot be radiating much because the reflected power is 99.7% of the forward power and both are based on differential currents which don't radiate. That leaves only 0.3% of the power available for radiating by the feedline. If 99.9% of the power is radiated by the coax braid, then the feedline SWR must be very close to 1.6:1 because the ratio of reflected power to forward power can be no more than 0.1/100. As with cake, you cannot have your reflected power and radiate it too. Reflected power plus radiated power cannot add up to more than the forward power. In fact, Pfor = Pref + Prad If 50% of the power is radiated by the antenna plus feedline, then 50% of the power is reflected and the SWR is 5.83:1. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#5
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Cecil Moore wrote:
If 99.9% of the power is radiated by the coax braid, then the feedline SWR must be very close to 1.6:1 because the ratio of Sorry, that should be 1.06:1 instead of 1.6:1. reflected power to forward power can be no more than 0.1/100. -- 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! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#6
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![]() Cecil Moore wrote: dansawyeror wrote: If it is true it proves that a driving a "bad" load can cause the coax feedline to radiate a significant portion of the feed energy. At this point I am not sure what word to use other then 'bad', but I due intend to perform experiments to measure and verify what is happening and the model. dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php We are in a position now to say what is wrong with that program. Essentially, it violates the conservation of energy principle. Energy cannot exist as both reflected power and radiated power at the same time. If the SWR is 1400:1, the feedline cannot be radiating much because the reflected power is 99.7% of the forward power and both are based on differential currents which don't radiate. That leaves only 0.3% of the power available for radiating by the feedline. If 99.9% of the power is radiated by the coax braid, then the feedline SWR must be very close to 1.6:1 because the ratio of reflected power to forward power can be no more than 0.1/100. As with cake, you cannot have your reflected power and radiate it too. Reflected power plus radiated power cannot add up to more than the forward power. In fact, Pfor = Pref + Prad If 50% of the power is radiated by the antenna plus feedline, then 50% of the power is reflected and the SWR is 5.83:1. Bear in mind that current flowing in a conductor will always generate a field. Alternating current flowing in a (straight) conductor will always generate a radiating electromagnetic field. That radiating field will convey energy if it is not nulled by another equal and opposite field. You need to demonstrate that all the fields resulting from all the currents are nulled in this scenario. Back into the woodwork. ac6xg |
#7
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Jim Kelley wrote:
Bear in mind that current flowing in a conductor will always generate a field. Alternating current flowing in a (straight) conductor will always generate a radiating electromagnetic field. That radiating field will convey energy if it is not nulled by another equal and opposite field. You need to demonstrate that all the fields resulting from all the currents are nulled in this scenario. It is true by definition for ideal coaxial feedlines in which the radiating fields are 100% nulled by the basic shielded design, i.e. solid copper outer tubing much thicker than the skin effect depth. The problem with the program is that the conclusions violate the conservation of energy principle. It is simply impossible to have a coax SWR of 1400:1 while radiating 99.9% of the power from the feedline. -- 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 =--- |
#8
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![]() "Cecil Moore" wrote in message ... dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php Please note that the third wire to ground creates the unbalance that causes feedline radiation. You seem to be confusing cause and effect. The cause of the feedline radiation is the existence of that third wire, not SWR. All it proves is that feedline radiation is caused by that third wire path which unbalances the source currents. -- 73, Cecil http://www.qsl.net/w5dxp Cecil How wrong would it be to say that the *coax* part of the line doesnt radiate at all? I see this as a situation where the *outer shield* of a transmission line is conducting current that radiates. It seems that a "balanced" antenna that is comprised of a single conductor and a L shaped conductor that includes the outer conductor of the coax, could be fed with a balanced line for modeling. Jerry |
#9
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Jerry Martes wrote:
"Cecil Moore" wrote in message ... dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php Please note that the third wire to ground creates the unbalance that causes feedline radiation. You seem to be confusing cause and effect. The cause of the feedline radiation is the existence of that third wire, not SWR. All it proves is that feedline radiation is caused by that third wire path which unbalances the source currents. -- 73, Cecil http://www.qsl.net/w5dxp Cecil How wrong would it be to say that the *coax* part of the line doesnt radiate at all? I see this as a situation where the *outer shield* of a transmission line is conducting current that radiates. It seems that a "balanced" antenna that is comprised of a single conductor and a L shaped conductor that includes the outer conductor of the coax, could be fed with a balanced line for modeling. Jerry What I was taught is that in a properly installed antenna system the coax will not radiate. If the antenna is not properly matched to the coax you get current flow along the outside of the coax shield. Dave WD9BDZ |
#10
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On Wed, 31 Aug 2005 16:17:42 -0500, "David G. Nagel"
wrote: What I was taught is that in a properly installed antenna system the coax will not radiate. If the antenna is not properly matched to the coax you get current flow along the outside of the coax shield. Dave WD9BDZ Please read: http://www.w2du.com/r2ch21.pdf |
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