Article: 221624 of rec.radio.amateur.antenna From: "Steve Nosko" Subject: Re: db Question Date: Mon, 13 Feb 2006 14:37:41 -0600 Message-ID: References: <11uqbj78re1ad11@corp.supernews.com> If Roy's answer works for you, skip mine. I think an answer of mine on this must be in the archives. Rule number one: The Decibel is a measure of the ratio of two power levels ONLY, ALWAYS and NEVER anything else. Rule Number two. We like to violate rules and drive newbies crazy. However, I'll go back to Rule Number One first. The so called "Voltage dB", "Current dB" and "Power dB' are all figments of your (well some people's)imagination. I used to have fun with new Engineers tripping them up on this. "Why are Voltage dBs twice as big as power dB's?". This is a false-reasoned question. While, it is a fact that there is only one "kind" of dB, there are, however, three ways to _calculate_ them depending on what quantities you have at hand. Because Power is Volts times Amps, there is something special happening when we do the math. Lets call this V x A for this discussion. For a simple case of doubling the power - or a power ratio of two to one: In order to get twice the power, we need to change the Voltage AND Current enough so that VxA is twice what we started. 1 Volt and 1 Amp gives 1 Watt For 2 Watts we'll need: 1.414 Volts and 1.414 Amps 1.414 x 1.414 = 2 Looking, we see that the voltage did NOT double. It went up less that twice. However, we can calculate the decibel difference from the voltage difference OR the power difference. Using the "Power dB" formula for dB we get the correct 3.whatever dB and using the "Voltage dB" formula we get the VERY SAME amount of 3.whatever dB. So the dB are not different, only the starting quantities and, therefore the formulas. Rule Number 2: We can, and, we do use the voltage formula ( if it's there, use it... just like the mountain climber's answer to why does he climb the mountain...because it's there) in cases where we are not actually doing power ratio comparisons. Why? Because it is easy to compare large differences this way..that is, in the log format. Frequently, voltage gain in OP amp circuits is shown in dB even though there is practically an INFINITE power gain. This is because the input resistance is very high and the output resistance is very low and ignoring the resistance and ONLY looking at the voltage gives erroneous dB numbers. HOWEVER, those in the field know and understand this and all is well in the universe, so to speak. By measuring the voltage gain, and using the dB formula, it is easier to show 50 dB "Voltage" gain than showing a gain of 100,000. Silly Engineers! 73, Steve, K9DCI "dansawyeror" wrote in message news:kNGdnXLtYPCDp3DeRVn-jg@comcast.com... > There are three terms, idb, pdb, and vdb. They are not the same. idb and pdb > have the same value: 10*log10 (i/i0), while vdb is 20*log10(v/v0). > > Dan > > > Roy Lewallen wrote: > > Robert11 wrote: > > > >> Hello: > >> > >> When a table gives the attenuation (at some freq.) for a length of a > >> particular coax in db, > >> are they referring to db in voltage or db in power ? > > > > > > Both. When voltage and power ratios are measured across the same single > > impedance, the ratio of voltages in dB is the same as the ratio of power > > in dB. That is, dB in voltage equals dB in power if only one impedance > > is involved. That's why there are different formulas for dB voltage and > > power ratios, to make this happen. And it's the case if a coax cable is > > terminated in its characteristic impedance so the input and output > > impedances are the same. It's under this condition that cable loss is > > specified, so the dB loss represents the loss of both voltage and current. > > > > But when you compare the power at the output of a coax cable with the > > power at the input and the cable isn't terminated in its characteristic > > impedance, output and input impedances can be very different. The power > > loss will be somewhat higher than the specification (probably not much, > > when the matched loss is only 2 dB), but you can have much less or more > > voltage at the output than the input. So the power and voltage loss in > > dB can be very different, and you can actually have voltage gain -- > > although it can be argued that defining dB for a ratio of voltages > > across two different impedances is a bit shaky and perhaps not too > > meaningful. > > > >> > >> Is is correct to ask a question as simply as the following: > >> > >> If the attenuation is given as, e.g., 2 db, what Percentage therefore > >> of a received signal is "lost" > >> going thru the coax length ? > > > > > > 100 * (1 - 10^(-dB/10)) ~ 37% is the fraction of power lost. > > > > 100 * (1 - 10^(-dB/20)) ~ 21% is the fraction of voltage lost. > > > > These assume that the coax is terminated with its characteristic impedance. > > > > And you don't need to put "lost" in quotation marks. It is truly lost as > > a signal, having been turned into heat. > > > > Roy Lewallen, W7EL Article: 221625 of rec.radio.amateur.antenna From: "Steve Nosko" Subject: Re: demodulator & detector Date: Mon, 13 Feb 2006 14:49:10 -0600 Message-ID: References: <1139513435.028624.39500@g43g2000cwa.googlegroups.com> None. Different terms for the same thing. Perhaps time is the difference... In the early days, "Detector" was used. Why? Who knows! It worked. You listen for a weak signal and if you were good, you could "detect" one. Therefore that part of the receiver was the "Detector". As we grew and became more sophisticated, we "modulated a signal on a carrier". Since we used the same "detector" to capture it, we "detected" the signal. If getting on a plane is boarding and getting off is "de-planeing" then the opposite of modulating must be "de-modulating". It's like why is "Random Access Memory" called RAM when ROM "Read Only Memory" is also "Random Access". We're just a bunch of silly techno-geeks. 73, Steve, K9DCI "Ron J" wrote in message news:1139513435.028624.39500@g43g2000cwa.googlegroups.com... > Good morn', > > What is the difference between a demodulator and a detector? I was > reading the datasheet and I noticed a block diagram that says FM > detector and another diagram that says FM demodulator. > > Thanks! > Article: 221626 of rec.radio.amateur.antenna From: chuck Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: Message-ID: Date: Mon, 13 Feb 2006 21:33:03 GMT Technically, is it not energy that leaves the transmitter and is received by the receiver? Actual integration being performed by later receiver stages and/or the human ear/brain? Chuck, NT3G Reg Edwards wrote: > It is power which leaves the transmitter. > > Power is what is received by the receiver. > > So the S-meter is a power meter. > > It is a great pity that the usual S-meter is scaled partly in S-units > and partly in decibels relative to S9. > > But given that 1 S-unit = 6 dB, the modern meter scale fits very > nicely between receiver internal noise level and the receiver overload > point. > > And, for example, it's so much easier to report signal strength as S5 > rather than 0.047 micro-microwatts. > > In the same vein, we could, of course, report a signal strength of 20 > dB over S9 as S12 and be done with decibels. > ---- > Reg, G4FGQ. > > Article: 221627 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Mon, 13 Feb 2006 15:44:17 -0800 Message-ID: <11v26ek3oi94fab@corp.supernews.com> References: Steve Nosko wrote: >> . . . > Power, on the other hand, cannot be directly measured it has to be > calculated. Power, we know, is Volts times Amps or E x I. There are other > ways to figure out power which I won't go into here, but this is how we must > be satisfied doing it. > . . . Power meters are in wide use and directly measure power by detecting the amount of heat it produces. Roy Lewallen, W7EL Article: 221628 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: Message-ID: Date: Mon, 13 Feb 2006 23:52:59 GMT chuck wrote: > Technically, is it not energy that leaves the transmitter and is > received by the receiver? Technically, RF energy passing a point/plane during a unit of time is RF power (joules/sec). We can't have one without the other. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221629 of rec.radio.amateur.antenna From: chuck Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: Message-ID: Date: Tue, 14 Feb 2006 01:42:02 GMT OK, I'm a little confused. Well, maybe more than a little. Starting with energy as the "ability to do work" and power as the rate at which energy is "transformed" into work, things quickly get muddy. Energy passing through an imaginary surface (or point or plane) would not actually do any work in passing through, and in fact would retain its full potential to do work after having passed through. What then is power density? Is it the amount of work that the energy passing through a unit area of the surface "could have done" had it been actually and fully "captured" at that surface? There is no real power at that surface, is there? While power (and work) absolutely require energy, it strikes me as metaphysical whether all energy ultimately does do work and produce power. I don't think physics requires that, and it seems that lot of radiated energy is not obviously being transformed into work. So is energy without power really impossible, Cecil? Been away from this for a longer bit than I'm comfortable mentioning. Chuck. NT3G Cecil Moore wrote: > chuck wrote: > >> Technically, is it not energy that leaves the transmitter and is >> received by the receiver? > > > Technically, RF energy passing a point/plane during a unit > of time is RF power (joules/sec). We can't have one without > the other. Article: 221630 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Mon, 13 Feb 2006 17:58:59 -0800 Message-ID: <11v2eb78uukil1a@corp.supernews.com> References: chuck wrote: > OK, I'm a little confused. Well, maybe more than a little. > > Starting with energy as the "ability to do work" and power as the rate > at which energy is "transformed" into work, things quickly get muddy. Power is the rate at which energy is transferred or used. Period. > Energy passing through an imaginary surface (or point or plane) would > not actually do any work in passing through, and in fact would retain > its full potential to do work after having passed through. Yes. > What then is power density? Is it the amount of work that the energy > passing through a unit area of the surface "could have done" had it been > actually and fully "captured" at that surface? No. Power is not an "amount of work", nor is power density. Power is the rate at which power is being transferred, so it can tell you only the rate at which work can be done, not the amount. The rate at which energy is being passed through a given cross sectional area of a surface is the power density. If you integrate the power going across the boundary for some period of time, you then know the amount of energy which has passed, and therefore the amount of work which can be done. There is no real power at > that surface, is there? There is real energy passing that surface, and the rate at which it's passing is the power at that surface. So yes, there is. > While power (and work) absolutely require energy, it strikes me as > metaphysical whether all energy ultimately does do work and produce > power. I don't think physics requires that, and it seems that lot of > radiated energy is not obviously being transformed into work. So is > energy without power really impossible, Cecil? Cecil loves metaphysical arguments, so this is an ideal question for him. > Been away from this for a longer bit than I'm comfortable mentioning. Try finding a basic physics textbook at your local library. Most high school level texts should cover the topic adequately. For a more mathematical and quantitative treatment, a freshman level college text would be fine. A couple of the more popular ones are Resnick & Halliday, and Weidner & Sells. In the older editions of both, at least, electrical phenomena are covered in the second volume. However, power, work, and energy aren't restricted to electricity so are covered in general terms in the first volume. Roy Lewallen, W7EL Article: 221631 of rec.radio.amateur.antenna From: Steve Silverwood Subject: Re: About Benjamin Franklin and Lightning Message-ID: References: <1sKdnXYDRdG9rE_eRVn-hw@comcast.com> Date: Tue, 14 Feb 2006 02:10:47 GMT In article <1sKdnXYDRdG9rE_eRVn-hw@comcast.com>, A.K01@comcast.net says... > > While researching lightning protection and grounding > > in general I stumbled across the following: > > > > < http://www.physicstoday.org/vol-59/iss-1/p42.html > > > > > Benjamin Franklin's groundbreaking (pun fully intended) > > work on the subject. > > > > john, N5DWI > > The USA Mint is releasing two silver dollar collectables on Franklin; > one, as scientist, the other as statesman. The coins are available in > either Silver Proof or uncirculated condition. Both of which will conduct electricity just fine, I bet! ;-) -- -- //Steve// Steve Silverwood, KB6OJS Fountain Valley, CA Email: kb6ojs@arrl.net Article: 221632 of rec.radio.amateur.antenna From: Steve Silverwood Subject: Re: comet vert?no radals Message-ID: References: <49xAf.15283$F_3.8561@newssvr29.news.prodigy.net> Date: Tue, 14 Feb 2006 02:10:48 GMT In article <49xAf.15283$F_3.8561@newssvr29.news.prodigy.net>, mycall@arrl.net says... > Reg Edwards wrote: > > Yes, whatever the explanation, it's just a load of old wives' tales > > and bafflegab. > > Quoting from the article: "If Comet had claimed that the CHA-250B > was a world-beating miracle antenna, we would have blasted it with > both barrels. Comet doesn't make such claims, however. Comet's > literature merely states that the antenna will radiate a signal > and provide a low SWR on all bands without the use of radials. > In this respect, the CHA-250B performs as advertised. It is neither > a miracle nor a fraud." (The article says it's about equal to a > mobile whip antenna.) At least Comet is being honest in its claims. -- -- //Steve// Steve Silverwood, KB6OJS Fountain Valley, CA Email: kb6ojs@arrl.net Article: 221633 of rec.radio.amateur.antenna From: elgoog Subject: Re: demodulator & detector Date: 14 Feb 2006 02:44:18 GMT Message-ID: References: <1139513435.028624.39500@g43g2000cwa.googlegroups.com> > "Ron J" wrote in message > news:1139513435.028624.39500@g43g2000cwa.googlegroups.com... >> >> What is the difference between a demodulator and a detector? It is an Off Topic question in an usenet ng focused on _antennas_ by yet another drive-by Google poster Article: 221634 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Mon, 13 Feb 2006 19:49:43 -0800 Message-ID: <11v2kqqkfvqt7ca@corp.supernews.com> References: <11v2eb78uukil1a@corp.supernews.com> Correction: Roy Lewallen wrote: > . . . > No. Power is not an "amount of work", nor is power density. Power is the > rate at which power is being transferred, so it can tell you only the > rate at which work can be done, not the amount. . . I of course meant ". . .Power is the rate at which *energy* is being transferred. . ." Thanks, Owen! Roy Lewallen, W7EL Article: 221635 of rec.radio.amateur.antenna From: "Mr Fed UP" References: <11up7id72jhuv5b@corp.supernews.com> <5YnHf.1139$S03.729@bignews1.bellsouth.net> <8u1vu19cv2q7f90j0ods38805gsjev2c5n@4ax.com> <5sPHf.27484$Jd.10088@newssvr25.news.prodigy.net> Subject: Re: Anti-oxidant grease question Message-ID: Date: Mon, 13 Feb 2006 22:51:38 -0600 Well for those of us who don't get around so well, it was nice to have a product name to ask for. Home Depot had an even smaller tube 1/2 oz and I got some steel wool to buff things up first. both less than $6. I probably would have been an hour finding that tiny tube hanging in the store. With more days behind me now than ahead.... and hoping my next antennas will be new ones. It should last me till my end. Kinda like buying a tube of water pipe compound once you use it once the rest hangs around forever. LOL Thanks for brand name to ask for. Saved lots of steps I can't spare these days. 73 K4TWO Gary "Jim - NN7K" wrote in message news:5sPHf.27484$Jd.10088@newssvr25.news.prodigy.net... > Not to mention, that Home Depot (and Lowes), have even smaller bottles , 4 > oz,$6.00 , and think even 2 oz tubes, for even less! And a little goes a > long way! Jim NN7K > > John Ferrell wrote: >> A little more info: >> http://lists.contesting.com/archives//cgi-bin/mesg.cgi?a=Towertalk&i=v01530500aeaf6ee61189%40%5B207.65.48.146%5D >> >> On Sat, 11 Feb 2006 10:22:01 -0700, Larry Benko >> wrote: >> >> >>>Come on folks. This stuff is cheap compared to maintenance. A 1 minute >>>web search finds Noalox at Home Depot stores for $8.95 for an 8oz. >>>bottle. Penetrox is about the same price but not sold by Home Depot. >>> >>>Larry, W0QE >> >> John Ferrell W8CCW Article: 221636 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: Message-ID: Date: Tue, 14 Feb 2006 05:04:09 GMT chuck wrote: > So is energy without power really impossible, Cecil? Roy has already answered the basic question from an engineering viewpoint. Physicists often consider "power" to have a different definition than the engineering definition. From the IEEE Dictionary: "power - The rate of generating, transferring, or using energy." (agrees with Roy) From "University Physics" by Young and Freedman: "power is the time rate at which work is done." i.e. only the "using energy" portion of the engineering definition. A certain physicist I know will argue that no work is being done in a lossless transmission line so there is no power there. He will say the existence of 100 watts at 1000 points along the line means there must be 100,000 watts in the line. He will say that a Bird wattmeter doesn't measure watts. He will say that a power generating plant doesn't generate power and a transmission line doesn't transfer power. He will also say that reflected power doesn't exist because it is not doing any work. He will say that for an EM wave in free space, ExH has the dimensions of watts but it isn't power because no work is being done. As Roy indicated, engineers have a wider definition of "power". Energy without power is certainly possible, e.g. a DC battery with zero current. However, for a constant steady-state power level associated with an EM wave, energy and power are inseparable. I like to use a one-second long lossless transmission line in some of my examples because it is impossible to hide the joules. A one-second long line with 200 watts forward power and 100 watts reflected power contains 300 joules that have been generated but have not reached the load. Since EM wave energy cannot stand still (or slosh around side to side) it is only logical to assume that 200 of those joules are in the forward wave and 100 of those joules are in the reflected wave both traveling at the speed of light. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221637 of rec.radio.amateur.antenna From: "Sal M. Onella" References: <11ucnvqhck36j37@corp.supernews.com> Subject: Re: Thru-Wall Coax Feedthrus Message-ID: Date: Mon, 13 Feb 2006 21:46:52 -0800 "J. Mc Laughlin" wrote in message news:11ucnvqhck36j37@corp.supernews.com... > Foam is not a good idea. It can become very difficult to remove. > > Best is to use cooper wool. It will not rust nor flake and beasties do > not eat it. The stuff has become hard to find, however. Marine supply stores sell bronze wool. I think the boaters use it for cleaning brass fittings, presumably without scratching them. (I used it as EMI packing for some leaky old pull-boxes.) It may be close enough to copper wool to function the same. Article: 221638 of rec.radio.amateur.antenna From: "Joel Kolstad" Subject: Re: db Question Date: Mon, 13 Feb 2006 21:55:34 -0800 Message-ID: <11v2s9fsga187b@corp.supernews.com> References: <22778-43EE18F7-598@storefull-3258.bay.webtv.net> "Richard Harrison" wrote in message news:22778-43EE18F7-598@storefull-3258.bay.webtv.net... > In the telephone industry, the most common reference level is one > milliwatt. Ditto the RF industry. > Signal loss is common. Pretty ubiquitous... something about the three laws of thermodynamics... ;-) > Other dB units are easily contrived. There are dBw (dB referenced to one > watt), dBk (dB referenced to one kilowatt), dBRAP (dB above reference > acoustical power which is defined at 10 ro the minus 16 watts), etc, The cable TV industry likes to use dBuV, for whatever odd reason. Article: 221639 of rec.radio.amateur.antenna From: "Michael" Subject: Receiving VMS (Vessel Monitoring Signals) from Inmarsat-C Message-ID: Date: Tue, 14 Feb 2006 13:12:28 +0100 I have been playing around with receiving AIS (Automatic Identification System) signals. AIS is used on ships, it is used to send information about your position, etc via VHF to the surrounding ships. But know this started to get a bit boring. So know I wan to try to se if it is possibly to receive the VMS signals from Inmarsat-C. The VMS system is manly used on fishing vessels. On the ship there is a box with a gps and satellite tranciver. The position information is send each hour, to a land earth station, were it is distributed to the different authorities. The first thing is to receive the signals. But also to decrypt the data can be complicated. Any help or advices are welcome. Also if you have some good links to people ho have been dealing with projects like this, could be useful. Cheers Michael Article: 221640 of rec.radio.amateur.antenna From: John Ferrell Subject: Re: Anti-oxidant grease question Message-ID: References: <11up7id72jhuv5b@corp.supernews.com> <5YnHf.1139$S03.729@bignews1.bellsouth.net> <8u1vu19cv2q7f90j0ods38805gsjev2c5n@4ax.com> <5sPHf.27484$Jd.10088@newssvr25.news.prodigy.net> Date: Tue, 14 Feb 2006 12:26:45 GMT Re: With more days behind me now than ahead.... There are a lot of us with that condition! Time and physical demands have become important considerations on projects. And the increasing memory problems that make it harder than ever to find where I put things... On Mon, 13 Feb 2006 22:51:38 -0600, "Mr Fed UP" wrote: >Well for those of us who don't get around so well, it was nice to have a >product name to ask for. >Home Depot had an even smaller tube 1/2 oz and I got some steel wool to >buff things up first. >both less than $6. I probably would have been an hour finding that tiny >tube hanging in the store. >With more days behind me now than ahead.... and hoping my next antennas >will be new ones. It should last me till my end. Kinda like buying a tube >of water pipe compound >once you use it once the rest hangs around forever. LOL >Thanks for brand name to ask for. Saved lots of steps I can't spare these >days. > >73 K4TWO Gary > > John Ferrell W8CCW Article: 221641 of rec.radio.amateur.antenna From: chuck Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: <+WPYJQGG1Z8DFAQ5@ifwtech.co.uk> Message-ID: <2wkIf.18412$vU2.10689@newsread3.news.atl.earthlink.net> Date: Tue, 14 Feb 2006 13:00:46 GMT Thank you all for your assistance. It will take a little time (and a little work) for the difference in physicist/engineer definitions of power to sink in. Chuck, NT3G Ian White GM3SEK wrote: > chuck wrote: > >> >> Energy passing through an imaginary surface (or point or plane) would >> not actually do any work in passing through, and in fact would retain >> its full potential to do work after having passed through. >> >> What then is power density? > > > The full name is power *flux* density, implying the rate at which energy > *flows through* unit area of a defined reference plane. SI units are > watts per square metre. > >> Is it the amount of work that the energy passing through a unit area >> of the surface "could have done" had it been actually and fully >> "captured" at that surface? > > > Yes, that is the implication - except that it's the *rate* of energy > capture, ie the amount that could be captured from unit area in unit time. > > This is only a concept, because it isn't physically possible to > intercept the power flux through unit area of an EM wavefront - your > wave-catcher would disturb the wavefront around its edges, and the > shadow behind it would be filled in by diffraction. However, very > similar concepts apply to power flux in a transmission line - and in > that case you really *could* capture the exact steady-state power flux > at any point, by cutting the line and substituting a dummy load of the > correct impedance. > > >> There is no real power at that surface, is there? >> > That rather depends on your personal definitions of the words "real", > "at" and possibly "is" :-) I think you'd caught it correctly in the > previous paragraph... but if you squeeze too hard, the waves will slip > through your fingers. > > > > Article: 221642 of rec.radio.amateur.antenna Subject: Re: More ATAS-120 problems From: "Doc" Date: Tue, 14 Feb 2006 16:22:15 GMT Message-ID: <1139934061_8985@sp6iad.superfeed.net> References: <5JGdneVCIKrJu3jenZ2dnUVZ_sCdnZ2d@comcast.com> I would think that the radio is not recognizing the ATAS, so, try turning on that ability. I know nothing about the FT-857D, but would think there should be a 'menu' thingy to turn the ATAS feature on?? - 'Doc Article: 221643 of rec.radio.amateur.antenna From: "Bill Ogden" References: <+WPYJQGG1Z8DFAQ5@ifwtech.co.uk> <2wkIf.18412$vU2.10689@newsread3.news.atl.earthlink.net> Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: Date: Tue, 14 Feb 2006 11:39:54 -0500 I think the differences in this discussion go back to very fundamental definitions of power and energy (and power flux, which might not be the same as power). One set of terminology came from classical Thermodynamics and the other set from more general principles. More fun may be had dealing with the Second Law (which arrived from two very different viewpoints) and, for example, quantum effects in the same discussion. The discussion can quickly hinge on precise meanings of common words, such as "power." Mixing terminology produces such wonderful concepts as measuring 100 watts flowing at 100 points in a transmission line and concluding that we have found 10,000 watts. Bill W2WO Article: 221644 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: What's the best way to measure radiated power from an antenna? Date: Tue, 14 Feb 2006 17:01:24 +0000 (UTC) Message-ID: References: <1139930218.609008.164890@g43g2000cwa.googlegroups.com> The only way to determine power radiated from an antenna at HF is to calculate it from its dimensions. To simplify matters, one can neglect ground and other losses. In which case radiated power will be about 99.99 percent of power input from the transmitter. About 90 to 95 percent is what you would get - PROVIDED a way can be found to measure it. Otherwise just take the calculated figures as being near enough. The directions and elevations in which power is radiated are, of course, the answers to yet more important questions. ---- Reg. Article: 221645 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 17:10:40 +0000 (UTC) Message-ID: References: <+WPYJQGG1Z8DFAQ5@ifwtech.co.uk> <2wkIf.18412$vU2.10689@newsread3.news.atl.earthlink.net> Why unnecessarily complicate matters with words. Watts = Amps times Volts. . . . . and that's all there is to it. ---- Reg. Article: 221646 of rec.radio.amateur.antenna From: "Steve Nosko" Subject: Re: Coax Losses ? Date: Tue, 14 Feb 2006 11:37:00 -0600 Message-ID: References: <11usilchl6j771e@corp.supernews.com> Bob, If you are comfortable with Lew's explanation, this shouldn't confuse things, but one minor additional concept. Regarding your comment: "Is it correct for me to say that the actual losses really aren't all that significant or meaningful, aqnd (sic) that a good receiver, which I have, can easily make up for them ?" While on HF atmospheric noise usually dominates, loss before the receiver can not always be "made up" by a receiver. If we go up to an otherwise dead frequency, say, up around 20 or 30 MHz where/when there is no significant atmospheric noise, we have a slightly different situation. S/N is still the thing to consider, but the noise of the receiver front end may now be the dominant noise. Now, consider that the loss of the feedline causes a reduction of the signal, but, of course, has no affect on the receiver noise. This feed line loss lowers the signal and degrades the S/N ratio. The only way to get better is to either reduce the feedline loss or reduce the receiver noise (noise figure). 73, Steve, K9DCI "Roy Lewallen" wrote in message news:11usilchl6j771e@corp.supernews.com... > You have the right idea. Signal/noise ratio is all that counts. > > When receiving HF, feedline loss almost never matters. The reason is > that atmospheric noise is strong, and unless a receiver is exceptionally > noisy and/or an antenna is exceptionally lossy, the atmospheric noise > will be much greater than the receiver noise. > > The input to the receiver consists of the signal and atmospheric noise, > to which the receiver adds its own noise. As long as the atmospheric > noise is much greater than the receiver noise, you won't hear the > receiver noise. Any attenuation in the antenna system will attenuate > both the signal and the atmospheric noise equally, so the signal/noise > ratio, which determines what you can hear, doesn't change. Any gain > ahead of or within the receiver has the same effect. > > Atmospheric noise declines as the frequency increases, so it might be > possible to start hearing some receiver noise over it as you approach 30 > MHz, particularly if your receiver is unusually noisy and/or the antenna > system unusually lossy. As soon as receiver noise becomes audible over > the atmospheric noise, the rules change. Then, attenuation ahead of the > receiver will reduce the signal but not the receiver noise -- which is > now the "noise" part of the signal/noise ratio --, so it *will* decrease > the signal/noise ratio. > > There's a very simple test to determine whether reducing the loss will > improve your ability to hear signals. Tune your receiver to a spot with > no signals in the frequency band of interest. Turn up the volume so you > can clearly hear the background noise. Then disconnect your antenna(*). > If the noise decreases, it means that atmospheric noise is dominating, > so reducing loss won't help the signal/noise ratio. If the noise doesn't > decrease, you're hearing receiver noise with the antenna connected and > you'd benefit by reducing losses ahead of the receiver. You'll probably > find that the noise will decrease by quite a few dB when you disconnect > the antenna, and this represents the amount of loss you can add before > your ability to hear signals suffers. > > When receiving HF, about the only way you can improve your signal/noise > ratio is by using a directional antenna. This will reduce the > atmospheric noise coming from unneeded directions. Particular kinds of > antennas can also improve the signal/noise ratio if the dominant noise > is man made and coming from a nearby source. > > (*) In a really marginal case, you might need to replace the antenna > with a dummy load for this test -- an ordinary small 47 or 51 ohm > resistor will suffice -- but it usually isn't necessary. > > Roy Lewallen, W7EL > > Robert11 wrote: > > Hi, > > > > Just getting started with all of this, and want to say a quick thanks to > > everyone for all the help. > > > > Another question: > > > > Am I interpreting this more or less correctly: > > > > Looking at all the different types of coax available. > > Will be for a receive-only HF antenna. > > Antenna will have to be in the backyard about 150 feet or so from house. > > > > The db losses are beginning to add up; at the upper limit of my interest of > > 30 MHz > > we are getting close to around 3.5 db or so. > > > > Is it correct for me to say that the actual losses really aren't all that > > significant or meaningful, > > and that a good receiver, which I have, can easily make up for them ? > > > > That the only thing of real concern would be the S/N ? > > > > What are the caveats to my statement above ? > > > > Thanks, > > Bob > > > > > > Article: 221647 of rec.radio.amateur.antenna From: fmmck@aol.com (Fred McKenzie) Subject: Re: help appreciated with unusual j pole feed Date: Tue, 14 Feb 2006 13:52:20 -0500 Message-ID: References: In article , "kevin" wrote: > hi there, i'm looking for some help with a j pole antenna, it's actually a > vhf marine antenna so i hope it will be ok to ask here. Kevin- One version of the J-Pole antenna has the cable connected an inch or two above the connection between the two vertical rods. The following article >from my archives, shows how it could be connected, except it is based on Television Twin Lead instead of the two rods yours has. If you use a mono-spaced font like Courier, it should be clear. The conductor above the 1/4" gap is not essential, it just comes with the twinlead! Fred --------------- 2M Jpole with 300 ohm twin lead From: kinsman@ektools.UUCP (Andrew A. Kinsman) Newsgroups: rec.ham-radio Organization: Eastman Kodak, Dept. 47, Rochester NY The following is a description of a J-Pole antenna made from 300 ohm TV twin-lead. Quite a few of the hams in the area have successfully built one or more of these antennas. They have quite a few advantages which include improved performance for 2-meter HTs, portability, and last, but never least, they are inexpensive. | | do not short this end. | | | | | | | | | | 54" | | | | | | | | | 1/4" gap | | | | | | 16 3/4" coax ctr conductor=>* *<= coax shield | | 1 1/4" | | -*- solder the twin leads together at bottom 1. Start with 54 1/2" of TV twin lead (foam core does not work as well) 2. Strip 1/2" of insulation at bottom and solder wires together. 3. Measure 1 1/4" from soldered wires and strip insulation on both sides. This is the solder point for a coax feedline. 4. Measure 16 3/4" from bottom and cut out 1/4" notch on one side. 5. Feed with a length of RG58U coax. Tape coax at feedpoint to the twin lead for strength and weather protection. I have also attached an alligator clip to the plastic on the top of the antenna in order to conveniently hang it. I am successfully using one in my attic to avoid any CCNR problems. I have another that I can take with me camping, sailing, etc. This design has appeared on many BBSs and in club newsletters; the earliest reference that I know of is a Jan. 1984 D.A.R.C. antenna article by James Burks, KA5QYV. Enjoy, N6UBM ------------------ Article: 221648 of rec.radio.amateur.antenna From: Mike Coslo Subject: Re: demodulator & detector Date: Tue, 14 Feb 2006 14:16:21 -0500 Message-ID: <43F22C85.7080402@psu.edu> References: <1139513435.028624.39500@g43g2000cwa.googlegroups.com> <1139890757.855341.190720@g44g2000cwa.googlegroups.com> Ron J wrote: > So sorry, I figured that the antenna folks would no more about it > because after all, there are modulators and demodulators in our > transmitter and receiver circuits. What else is interesting in the > antenna world? Did the DLM and EH antenna combine into one? ;-) > Not to worry, Ron, cuz his post was just another drive-by flame from a poster who prefers to remain anonymous. Most who do try to remain anonymous have a good reason - and it's not the one that they give ;^) - 73 de Mike KB3EIA - Article: 221649 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: <+WPYJQGG1Z8DFAQ5@ifwtech.co.uk> <2wkIf.18412$vU2.10689@newsread3.news.atl.earthlink.net> Message-ID: Date: Tue, 14 Feb 2006 19:40:08 GMT Reg Edwards wrote: > Watts = Amps times Volts. Only for DC or in-phase AC/RF. Volt-Amps = Amps times Volts = SQRT(watts^2 + vars^2) Watts = Amps times Volts times cos(A) = real power Vars = Amps times Volts times sin(A) = reactive power Reference: "Alternating Current Circuits", Kerchner/Corcoran, 3rd edition, (C) 1938, 1943, 1951. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221650 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: What's the best way to measure radiated power from an antenna? Date: Tue, 14 Feb 2006 14:09:21 -0600 Message-ID: <7288-43F238F1-1133@storefull-3251.bay.webtv.net> References: Reg, G4FGQ wrote: "Otherwise just take the calculated figures as being near enough." Standard AM broadcast stations usually do that once their performance has been proved by field measurements. A point in the system just before the power from the transmitter branches out to the towers in an antenna array is designated as the "common point" of the system. Here the impedance and current are measured to determine the power radiated by the system. Losses are considered insignificant, An r-f ammeter permanently installed here constantly monitors the station`s power. Best regards, Richard Harrison, KB5WZI Article: 221651 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 12:30:58 -0800 Message-ID: <11v4fg75rhomo98@corp.supernews.com> References: <+WPYJQGG1Z8DFAQ5@ifwtech.co.uk> <2wkIf.18412$vU2.10689@newsread3.news.atl.earthlink.net> Reg Edwards wrote: > Why unnecessarily complicate matters with words. > > Watts = Amps times Volts. > > . . . . and that's all there is to it. > ---- Only if you're talking about instantaneous values, which people seldom do (but should). At any instant of time, the power at a point (that is, the rate of energy flow past the point) is the volts time the amps at that point at that instant. But if you look at, say, the energy moving between the inductor and capacitor in a high Q parallel resonant LC circuit, you'll find that nearly all the energy moving in one direction moves back on the alternate half cycle. That is, the instantaneous value of the power is positive for one half cycle and negative for the other. The average power is therefore relatively small, much smaller than the product of RMS volts times RMS amps. In the extreme theoretical case of a lossless circuit, the average power is zero, so there's no net long term energy flow even though both V and I can be very large. Roy Lewallen, W7EL Article: 221652 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: What's the best way to measure radiated power from an antenna? Message-ID: <48g4v116grjnbr4p6ducdptdvmi7cv67rk@4ax.com> References: <1139930218.609008.164890@g43g2000cwa.googlegroups.com> Date: Tue, 14 Feb 2006 21:03:39 GMT On Tue, 14 Feb 2006 17:01:24 +0000 (UTC), "Reg Edwards" wrote: >The only way to determine power radiated from an antenna at HF is to >calculate it from its dimensions. > >To simplify matters, one can neglect ground and other losses. In which >case radiated power will be about 99.99 percent of power input from >the transmitter. Reg, I may be misuderstanding what you mean by "power input from the transmitter", is it the same as the transmitter power output as measured at the transmitter? As you know, ground losses can be a major loss component in the case of some types of antennas, for instance short Marconi type antennas. If "other losses" as used above includes transmission line losses, it is not infrequent that we see people describe configurations where transmission line losses are in the range 10dB to 20dB (ie radiated power less than 10% to 1% of the transmitter output respectively). Owen -- Article: 221653 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 15:03:05 -0600 Message-ID: <6236-43F24589-857@storefull-3255.bay.webtv.net> References: <11v4fg75rhomo98@corp.supernews.com> Roy, W7EL wrote: "The average power is therefore relatively small, much smaller than the ptoduct of RMS volts times RMS amps." I have not read the thread, but I recall from some old memory store that rms volts times rms amps is one of the definitions of "average power". Best regards, Richard Harrison, KB5WZI Article: 221654 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> Date: Tue, 14 Feb 2006 21:25:32 GMT On Tue, 14 Feb 2006 15:03:05 -0600, richardharrison@webtv.net (Richard Harrison) wrote: >I have not read the thread, but I recall from some old memory store that >rms volts times rms amps is one of the definitions of "average power". Only in a DC circuit, or a purely resistive load in an AC circuit. Owen > >Best regards, Richard Harrison, KB5WZI -- Article: 221655 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> Date: Tue, 14 Feb 2006 21:29:09 GMT On Tue, 14 Feb 2006 21:25:32 GMT, Owen Duffy wrote: >On Tue, 14 Feb 2006 15:03:05 -0600, richardharrison@webtv.net (Richard >Harrison) wrote: > > >>I have not read the thread, but I recall from some old memory store that >>rms volts times rms amps is one of the definitions of "average power". > >Only in a DC circuit, or a purely resistive load in an AC circuit. I shouldn't use that work ONLY!!! Only in a DC circuit, or a in an AC circuit (loop) where the current and voltage measured are in phase. In an AC circuit where the voltage and current are not in phase you must multiply the product of the RMS voltage and RMS current by the cosine of the phase difference to get real power (which is what I think you mean by "average power"). Owen -- Article: 221656 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 14:10:27 -0800 Message-ID: <11v4lao6jo4i89b@corp.supernews.com> References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> Richard Harrison wrote: > Roy, W7EL wrote: > "The average power is therefore relatively small, much smaller than the > ptoduct of RMS volts times RMS amps." > > I have not read the thread, but I recall from some old memory store that > rms volts times rms amps is one of the definitions of "average power". Time to dust off your old circuit analysis text, then. Pay special attention to the discussion of "imaginary power" or "vars". Roy Lewallen, W7EL Article: 221657 of rec.radio.amateur.antenna From: "J Tabor" Subject: WinCAP Wizard - VOACAP - freeware interface and tools Date: Tue, 14 Feb 2006 16:10:48 -0600 Message-ID: <11v4lbodor2aoab@corp.supernews.com> Greetings, You are invited to download and use the new version of WinCAP Wizard. The default download is known as QSL-ware. Which basically means freeware. This QSL-ware version provides unlimited point-to point and NCDXF beacon predictions, prefix search, numerous prediction views and more. www.taborsoft.com The registered version adds coverage analysis and user batch prediction modes. A 60-day trial key is available upon request. If you use a competing tool, not to worry, they can surely add a few of the new features. Cheers, KU5S -- email sent to: denews@taborsoft.com is discarded without being seen. Sorry for any inconvenience. Article: 221658 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 14:15:20 -0800 Message-ID: <11v4lk1dlthe02f@corp.supernews.com> References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> Owen Duffy wrote: > > I shouldn't use that work ONLY!!! > > Only in a DC circuit, or a in an AC circuit (loop) where the current > and voltage measured are in phase. In an AC circuit where the voltage > and current are not in phase you must multiply the product of the RMS > voltage and RMS current by the cosine of the phase difference to get > real power (which is what I think you mean by "average power"). > Of course, that only works when the voltage and current are sinusoidal and of the same frequency. More generally, the average power is 1/T times the integral over T of v(t) * i(t) dt, where T is the interval over which it's being averaged. If the waveforms are periodic, an interval of one cycle can be used for T. Roy Lewallen, W7EL Article: 221659 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> <11v4lk1dlthe02f@corp.supernews.com> Date: Tue, 14 Feb 2006 23:11:55 GMT On Tue, 14 Feb 2006 14:15:20 -0800, Roy Lewallen wrote: >Owen Duffy wrote: >> >> I shouldn't use that word ONLY!!! >> >> Only in a DC circuit, or a in an AC circuit (loop) where the current >> and voltage measured are in phase. In an AC circuit where the voltage >> and current are not in phase you must multiply the product of the RMS >> voltage and RMS current by the cosine of the phase difference to get >> real power (which is what I think you mean by "average power"). >> > >Of course, that only works when the voltage and current are sinusoidal >and of the same frequency. Yes, implied by the "in phase" condition. Thinking that through further brings a third case to the "ONLY" conditions, and that is if the circuit is entirely resistive (eg real power is the product of Vrms and Irms if the waveform is square and the circuit contains only resistances). > More generally, the average power is 1/T >times the integral over T of v(t) * i(t) dt, where T is the interval >over which it's being averaged. If the waveforms are periodic, an >interval of one cycle can be used for T. > >Roy Lewallen, W7EL -- Article: 221660 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 15:39:04 -0800 Message-ID: <11v4qgt4om34v85@corp.supernews.com> References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> <11v4lk1dlthe02f@corp.supernews.com> Owen Duffy wrote: > . . . > Thinking that through further brings a third case to the "ONLY" > conditions, and that is if the circuit is entirely resistive (eg real > power is the product of Vrms and Irms if the waveform is square and > the circuit contains only resistances). If you look at the definition of average (as in my previous posting), you'll see that when the load is purely resistive, average power = 1/T * the integral over T of v^2(t) / R dt or 1/T * the integral over T of i^2(t) * R dt, for any waveform. And using the definition of RMS(*), you can see that this is exactly Vrms^2 / R or Irms^2 * R respectively, again for any waveform. So Pavg = Vrms * Irms for any waveform, as long as (and only as long as) the load is purely resistive. Again, the average and RMS values can be calculated for any interval (as long as they're the same), but a single cycle is adequate to determine the long-term average and RMS values of periodic waveforms. (*) frms = Sqrt(avg(f^2(t))) = Sqrt(1/T * integral over T of f^2(t) dt) Roy Lewallen, W7EL Article: 221661 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Tue, 14 Feb 2006 15:51:00 -0800 Message-ID: <11v4r79bmunnb4@corp.supernews.com> References: <11v4fg75rhomo98@corp.supernews.com> <6236-43F24589-857@storefull-3255.bay.webtv.net> <11v4lk1dlthe02f@corp.supernews.com> This is probably a good place to mention that people interested in the relationship between RMS voltage and current and average power (and the uselessness of the RMS value of power) can find an explanation at http://eznec.com/Amateur/RMS_Power.pdf. It doesn't use any mathematics more advanced than a square and square root, so any amateur should be able to understand it. Some time ago I was surprised to find this to be one of the most frequently visited pages at my web site, apparently due to a link in the Wikipedia entry for RMS. Roy Lewallen, W7EL Article: 221662 of rec.radio.amateur.antenna From: "David G. Nagel" Subject: Re: Stealth antenna question Date: Tue, 14 Feb 2006 19:02:38 -0600 Message-ID: <11v4vdvs7jeerab@corp.supernews.com> References: <5eudnRTUMuRG9m_eRVn-qA@comcast.com> Joe S. wrote: > I am considering installing a stealth antenna on the porch of my first-floor > apartment and would like some suggestions. Here's the deal. > > -- First floor apartment. > -- Porch is a concrete slab, 10 X 10. > -- The ceiling of the porch is the underside of the balcony of the apartment > above me. It's all wooden. > -- Thus, the ceiling of my porch is 10 X 10, wood. > > How about I build a 40-meter dipole, 67 feet long, feed with coax and a > balun in the center. Attach the coax and the balun to one of the corner > posts that supports the balcony above my porch. Run the two sides of the > dipole in opposite directions, but, instead of stretching the wire straight > as I normally would if hanging the dipole from trees, towers, etc., what if > I run the wires around the underside of the balcony deck? This would give > me a horizontal loop made up of two pieces of wire, each 33.5 feet long, > with the loop being 10 feet on a side, thus, each end of the dipole would > run around underneath the balcony deck and come back almost to the center of > the dipole. > > That is -- the balun would be secured to one of the corner posts. One piece > of wire, 33.5 feet long would run straight out from the balun along the 2 X > 8 rim joist for 10 feet, make a 90-deg turn (20 ft), run another 10 feet and > make another 90-deg turn (30 ft), and end up 3.5 feet after the third turn. > The other end of the dipole would do the same thing, only run in the > opposite direction. At the balun, the two wires making up the dipole would > run at a 90-deg angle to each other. And, the two wires would overlap (or > be only a few inches apart) for 27 feet of their length. > > I probably didn't explain this idea too well but I'd like to hear some > suggestions. Thanks. > > What ever you do make sure that your neighbors upstairs aren't exposed to excessive RF Radiation. Dave WD9BDZ Article: 221663 of rec.radio.amateur.antenna From: "Sal M. Onella" References: <1139930218.609008.164890@g43g2000cwa.googlegroups.com> Subject: Re: What's the best way to measure radiated power from an antenna? Message-ID: <40xIf.58600$V.39941@fed1read04> Date: Tue, 14 Feb 2006 19:14:40 -0800 "MRW" wrote in message news:1139930218.609008.164890@g43g2000cwa.googlegroups.com... > Happy Valentine's Day! > > In terms of high powered transmitter (at least 500W), what is the best > way to measure the output power from an antenna? I had pictured an > in-line power sensor between the matching network and the antenna, but > then at 500W I don't think I'll get close to the transmitter. > > My friend actually proposed of finding out how long it will cook an egg > if exposed to a certain output power and used that as a reference to > approximate whether the power output. But then again I think you'd > initially need to cook a couple of eggs in a controlled environment > before a nice curve can be derived. > > Also, during our engineering expo, a power company showcased their > infrared tools to detect whether or not a power line is overloaded, > shorted, or working normally. I was wondering if there is a similar > tool used on antennas. > > Thanks! > The other answers aren't wrong, but measuring power within the antenna/transmission line system can be misleading because not all the power measured there is actually radiated. Some is dissipated as heat in the matching unit (if used) or reflected back to the transmitter. Radiated power doesn't go in all directions, either. The output power from an antenna is often described as such-and-such "field strength" at some standard distance and it is determined only with expensive test equipment. Because of directional effects, making just one measurement often isn't enough. You set up equipment at a fixed distance from the antenna (perhaps a mile) with a special receiver, one with a meter calibrated for input signal strength at the antenna terminals. A spectrum analyzer may also be used, but you have to be very careful about overloading its wide-open front end, not with your transmitter, necessarily, but with every other darn thing in town. You feed it with a particular test antenna, one which will be supplied with a set of calibration charts. Read the microvolts from the receiver, correct for cable losses and use the calibration charts for the antenna to yield the signal strength in standard units, like microvolts/meter. I do not remember the antenna names, but I do remember one receiver was the Singer NM-25. That mama weighed about sixty pounds. If you go to a vendor site, like www.tucker.com, you can appreciate the variety of equipment available for quantitative measurements. I worked as a certified EMI Engineer from 1988 until 2002 and I hope I have provided some useful information. Article: 221664 of rec.radio.amateur.antenna Subject: Re: Stealth antenna question From: Dave Oldridge References: <5eudnRTUMuRG9m_eRVn-qA@comcast.com> Message-ID: Date: Wed, 15 Feb 2006 05:22:59 GMT "Joe S." wrote in news:5eudnRTUMuRG9m_eRVn-qA@comcast.com: > I am considering installing a stealth antenna on the porch of my > first-floor apartment and would like some suggestions. Here's the > deal. > > -- First floor apartment. > -- Porch is a concrete slab, 10 X 10. > -- The ceiling of the porch is the underside of the balcony of the > apartment above me. It's all wooden. > -- Thus, the ceiling of my porch is 10 X 10, wood. > > How about I build a 40-meter dipole, 67 feet long, feed with coax and > a balun in the center. Attach the coax and the balun to one of the > corner posts that supports the balcony above my porch. Run the two > sides of the dipole in opposite directions, but, instead of stretching > the wire straight as I normally would if hanging the dipole from > trees, towers, etc., what if I run the wires around the underside of > the balcony deck? This would give me a horizontal loop made up of two > pieces of wire, each 33.5 feet long, with the loop being 10 feet on a > side, thus, each end of the dipole would run around underneath the > balcony deck and come back almost to the center of the dipole. > > That is -- the balun would be secured to one of the corner posts. One > piece of wire, 33.5 feet long would run straight out from the balun > along the 2 X 8 rim joist for 10 feet, make a 90-deg turn (20 ft), run > another 10 feet and make another 90-deg turn (30 ft), and end up 3.5 > feet after the third turn. The other end of the dipole would do the > same thing, only run in the opposite direction. At the balun, the two > wires making up the dipole would run at a 90-deg angle to each other. > And, the two wires would overlap (or be only a few inches apart) for > 27 feet of their length. > > I probably didn't explain this idea too well but I'd like to hear some > suggestions. Thanks. The real problem is going to be that winding the antenna back on itself like that has the effect of lowering its radiation resistance. I ran it up in Mininec and it comes out to less than 1 ohm at resonance. At that rate, you'd be better off making a dipole of two screwdriver antennas back to back. -- Dave Oldridge+ ICQ 1800667 Article: 221665 of rec.radio.amateur.antenna From: gsm@mendelson.com (Geoffrey S. Mendelson) Subject: 300 ohm folded dipole from ARRL Handbook, early 1990's Date: Wed, 15 Feb 2006 08:11:03 +0000 (UTC) Message-ID: Hi, the early '90s ARRL handbooks had construction information for a simple folded dipole made of 300 oHm twinlead. The article was not in the '96 edition or later. The antenna was made of 300 ohm twinlead. It was fed in the center by a 300 ohm twinlead feeder to a matching capacitor which made it aproximate 50 ohms. The capacitor could be either a regular capacitor (I used silver mica WWII surplus) or a "stub" of more twinlead. At the end of the feed line, which was part of the antenna, you could attach it directly to a 50 omh unbalanced transciever or a 50 ohm coax of any length. The length of the antenna was one half wavelength without the velocity factor and then shorted at the wavelength points compensating for the velocity factor. For example, if it was 10 meters long for a 20 meter dipole, the shorts were about 4 meters from the center for twinlead with a velocity factor of 0.8. If anyone has a copy of the article and could scan it for me, I'd appreicate it, or even better yet, a pointer to an online version of it. Thanks in advance and 73, BTW, if you are also in Israel and asking yourself "where does he get the twinlead"? I don't. I brought it with me when I moved here in '96. Geoff. -- Geoffrey S. Mendelson, Jerusalem, Israel gsm@mendelson.com N3OWJ/4X1GM IL Voice: (07)-7424-1667 IL Fax: 972-2-648-1443 U.S. Voice: 1-215-821-1838 Visit my 'blog at http://geoffstechno.livejournal.com/ Article: 221666 of rec.radio.amateur.antenna From: Iain Kelly Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Date: Wed, 15 Feb 2006 10:54:17 +0000 Message-ID: References: Geoffrey S. Mendelson wrote: > Hi, the early '90s ARRL handbooks had construction information for a > simple folded dipole made of 300 oHm twinlead. The article was not in > the '96 edition or later. > > The antenna was made of 300 ohm twinlead. It was fed in the center by > a 300 ohm twinlead feeder to a matching capacitor which made it aproximate > 50 ohms. The capacitor could be either a regular capacitor (I used silver > mica WWII surplus) or a "stub" of more twinlead. > > At the end of the feed line, which was part of the antenna, you could > attach it directly to a 50 omh unbalanced transciever or a 50 ohm coax > of any length. > > The length of the antenna was one half wavelength without the velocity > factor and then shorted at the wavelength points compensating for > the velocity factor. > > For example, if it was 10 meters long for a 20 meter dipole, the shorts > were about 4 meters from the center for twinlead with a velocity factor > of 0.8. > > If anyone has a copy of the article and could scan it for me, I'd > appreicate it, or even better yet, a pointer to an online version > of it. > > Thanks in advance and 73, > > BTW, if you are also in Israel and asking yourself "where does he get > the twinlead"? I don't. I brought it with me when I moved here in '96. > > Geoff. > Sounds like a scaled up version of an antennas I built from a design called 'The Mighty Wide 6M Dipole'... The design came from the internet I think, I seem to recall G0IER published it on his website. I have also seen the scaling information for other bands... Hope that's of some help... -- 73, Iain M0PCB/P Article: 221667 of rec.radio.amateur.antenna From: gsm@mendelson.com (Geoffrey S. Mendelson) Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Date: Wed, 15 Feb 2006 12:51:05 +0000 (UTC) Message-ID: References: Iain Kelly wrote: > Sounds like a scaled up version of an antennas I built from a design > called 'The Mighty Wide 6M Dipole'... The design came from the internet > I think, I seem to recall G0IER published it on his website. I have also > seen the scaling information for other bands... Thanks, but it led to some questions. Here is the web page I found for it: http://www.qsl.net/g3pto/6m_dipole.html >From my calculations, assuming his center frequency was 51 mHz, he used a size factor of 142.494 for the total length of the antenna. This seems odd to me, as it would translate to 467.5 feet which is number I've never seen before. Either something is amiss in my calculations, or he used a different center frequency or just an unusual number. The shorted length makes perfect sense, the ratio of total to shorted is .8425 which is a reasonable velocity factor for 300 ohm twinlead. The length of the stub makes no sense to me at all, but I'm sure he had his reasons. Using the proportions he used, I wrote a perl program to calculate the size for any frequency. For six meters, it matches his dimensions with a center frequency of 50.1 mHz: for frequency 50.100 mHz. total length 9 feet 3 inches or 2.844 meters. shorted length length 8 feet 10 inches or 2.396 meters. difference length length 0 feet 8 inches or 0.224 meters. stub length length 0 feet 11 inches or 0.298 meters. For 20 meters (14.175 mHz center frequency) I got: for frequency 14.175 mHz. total length 32 feet 11 inches or 10.052 meters. shorted length length 32 feet 9 inches or 8.469 meters. difference length length 0 feet 7 inches or 0.792 meters. stub length length 0 feet 5 inches or 1.052 meters. Anyone have any ideas? Will this work? How do I calculate the stub length? The ARRL Handbook antenna (not to be confused with the ARRL Antenna Handbook, which he used) was slightly different, although this would work too if my calculations are correct. Geoff. -- Geoffrey S. Mendelson, Jerusalem, Israel gsm@mendelson.com N3OWJ/4X1GM IL Voice: (07)-7424-1667 IL Fax: 972-2-648-1443 U.S. Voice: 1-215-821-1838 Visit my 'blog at http://geoffstechno.livejournal.com/ Article: 221668 of rec.radio.amateur.antenna From: "Dale Parfitt" References: Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Message-ID: Date: Wed, 15 Feb 2006 14:17:58 GMT > > For 20 meters (14.175 mHz center frequency) I got: > for frequency 14.175 mHz. > total length 32 feet 11 inches or 10.052 meters. > shorted length length 32 feet 9 inches or 8.469 meters. > difference length length 0 feet 7 inches or 0.792 meters. > stub length length 0 feet 5 inches or 1.052 meters. > > > Anyone have any ideas? Will this work? How do I calculate the stub > length? > Think you have some errant decimal points in the last three English/metric conversions W4OP Article: 221669 of rec.radio.amateur.antenna From: gsm@mendelson.com (Geoffrey S. Mendelson) Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Date: Wed, 15 Feb 2006 15:11:03 +0000 (UTC) Message-ID: References: Dale Parfitt wrote: > Think you have some errant decimal points in the last three English/metric > conversions Thanks for noticing that. I had the feet part of them being zero, when they should have been greater. It was stupid testing, it worked right at 51 mHz, where they were zero. I didn't look at the 14.1 mHz numbers closely enough. Here is the correct numbers: for frequency 14.100 mHz. total length 33 feet 1 inches or 10.106 meters. shorted length length 32 feet 11 inches or 8.514 meters. difference length length 2 feet 7 inches or 0.796 meters. stub length length 3 feet 5 inches or 1.057 meters. Just another proof that programmers should NOT do quality assurance on their own work. 73, Geoff. -- Geoffrey S. Mendelson, Jerusalem, Israel gsm@mendelson.com N3OWJ/4X1GM IL Voice: (07)-7424-1667 IL Fax: 972-2-648-1443 U.S. Voice: 1-215-821-1838 Visit my 'blog at http://geoffstechno.livejournal.com/ Article: 221670 of rec.radio.amateur.antenna From: "PA3HGT" Subject: ZM30 antenna-analyzer Date: Wed, 15 Feb 2006 16:27:42 +0100 Message-ID: <45gvjdF6j1m6U1@individual.net> Hello, There's a Yahoo group for those who is interested in the Palstar ZM30 antenna-analyzer. http://groups.yahoo.com/group/palstarzm30/ Kind regards, Hans Article: 221671 of rec.radio.amateur.antenna From: Gene Fuller Subject: Re: Orthogonality relation between modes in Dielectric-Lined Circular References: <1139991837.392371.176510@g14g2000cwa.googlegroups.com> Message-ID: Date: Wed, 15 Feb 2006 15:41:16 GMT Did you actually read the paper you referenced? Two points: The authors note on the third page that they could not find any prior derivation of the orthonormality relationships for this configuration. The paper was published in 2000, so it should be an easy library research project to find any subsequent papers that reference this work. Appendix A goes through the derivation of the orthonormality relationships in excruciating detail. Every step is outlined, including the integration tricks. It is unlikely that you would find any paper, or even a textbook, that is more detailed. 73, Gene W4SZ skatsis@ee.duth.gr wrote: > I am working on dielectric-lined circular waveguides and I am searching > for a reference on the orthogonality of modes. The only thing that I > found is in the link below (at the bottom of the message) > but I am looking for something with more details. > > Assume that we have a uniform cylindrical waveguide with N concentric > dielectric layers [my case of interest is when N=2, and we have the > waveguide with a thin inner dielectric layer (second layer N=2) and the > rest is vacum (air-> N=1-> first "layer")]. > I need an orthogonality relation that applies between any two modes. > The fact is that I already have found an orthogonality relation (see > the link below) but I need a more detailed analysis. > > http://www.bnl.gov/ATF/experiments/Dielectric/2-Theory_of_Wakes.pdf > > IF ANYONE KNOWS SOMETHING PLEASE ANSWER THIS MESSAGE. > Article: 221672 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's References: Message-ID: Date: Wed, 15 Feb 2006 17:31:12 GMT Geoffrey S. Mendelson wrote: > Hi, the early '90s ARRL handbooks had construction information for a > simple folded dipole made of 300 oHm twinlead. The article was not in > the '96 edition or later. I have the '93 edition and cannot find it but I remember how to do it. > The antenna was made of 300 ohm twinlead. It was fed in the center by > a 300 ohm twinlead feeder to a matching capacitor which made it aproximate > 50 ohms. The capacitor could be either a regular capacitor (I used silver > mica WWII surplus) or a "stub" of more twinlead. Assuming the SWR on the 300 ohm twinlead is 1:1, you want to install a capacitor that causes an SWR of 6:1 on the twinlead between the capacitor and the 1:1 choke/balun. 300+j0 in parallel with 0-jXc needs to result in 60-j120 ohms. The reactance of the capacitor should be -j150 ohms and it should be positioned 0.062 wavelength from the 1:1 choke/balun. For instance, for 7.2 MHz, the capacitor should be 147pf and be positioned 7.7 ft. from the 1:1 choke/balun assuming a VF of 0.9. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221673 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's References: Message-ID: Date: Wed, 15 Feb 2006 17:40:43 GMT Geoffrey S. Mendelson wrote: > From my calculations, assuming his center frequency was 51 mHz, he used > a size factor of 142.494 for the total length of the antenna. This seems > odd to me, as it would translate to 467.5 feet which is number I've > never seen before. 468/f is the length of a dipole adjusted for end effects. > Anyone have any ideas? Will this work? How do I calculate the stub > length? Why not use a Xc = -j150 cap? I get 0.176 wavelength for a capacitive stub which may be inconvenient at HF. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221674 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: New Program: OP_ANGLE Date: Wed, 15 Feb 2006 17:45:24 +0000 (UTC) Message-ID: New Program - OP_ANGLE.exe This program assists with understanding, choosing and setting the operating angles of Class-B and Class-C RF power amplifiers. The operating angle underlies all performance characteristics. Although RF operating conditions for tubes can be set up using manufacturers' data sheets, the program refers to use of a tube's Operating Characteristic Curves. Download program OP_ANGLE in a few seconds from website below. Not zipped up. Run immediately. File size = 39 kilobytes. ---- ........................................................... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp ........................................................... Article: 221675 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Date: Wed, 15 Feb 2006 10:38:57 -0800 Message-ID: <11v6ta6f01hv26@corp.supernews.com> References: Geoffrey S. Mendelson wrote: > . . . > The shorted length makes perfect sense, the ratio of total to shorted is > .8425 which is a reasonable velocity factor for 300 ohm twinlead. > . . . Actually, shorting the antenna at an intermediate point between the center and ends rather than just at the ends doesn't make much sense. The feedpoint impedance of a folded dipole consists of four times (or other ratio if the conductors are different in size or there are more than two folded conductors) the impedance of a standard dipole, in parallel with two series connected shorted stubs. The dipole consists of the two conductors in parallel. This behaves as a single fat wire which has the effective velocity factor of ordinary insulated wire, around 0.97 or 0.98 that of bare wire. That's why a folded dipole is about the same overall length as a standard dipole. The stubs have the physical length of half the dipole, or a bit shy of a quarter free space wavelength. Unlike the dipole part, they operate as transmission lines, so their velocity factor is around 0.8 -- a value which varies somewhat with cable construction. Folded dipoles are sometimes shorted about 0.8 of the way from the center to the ends in an apparent attempt to make the stubs an electrical quarter wavelength, resulting in their impedance being very high as seen at the feedpoint. But if the intermediate short circuit isn't done, the effect of the somewhat longer stubs is only to add a bit of capacitive reactance across the feedpoint. This lowers the antenna resonant frequency roughly 50 kHz at 14 MHz, and has very little effect on the feedpoint resistance. Antenna resonance can be restored by simply shortening the antenna a little. So why bother with the intermediate short? Roy Lewallen, W7EL Article: 221676 of rec.radio.amateur.antenna From: "John, N9JG" References: <11v6ta6f01hv26@corp.supernews.com> Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Message-ID: Date: Wed, 15 Feb 2006 20:12:11 GMT The 1986 handbook has the details for this type of antenna on page 33-14, under the title of "Simple Antennas for HF Portable Operation." For 7.15 MHz, the value of the capacitor is 152 pF, and the length of the matching stub is 6'11-1/2". An open-end stub, made from twin lead, of length 20'-1/2" can be substituted for the capacitor. "Roy Lewallen" wrote in message news:11v6ta6f01hv26@corp.supernews.com... > Geoffrey S. Mendelson wrote: >> . . . >> The shorted length makes perfect sense, the ratio of total to shorted is >> .8425 which is a reasonable velocity factor for 300 ohm twinlead. >> . . . > > Actually, shorting the antenna at an intermediate point between the center > and ends rather than just at the ends doesn't make much sense. > > The feedpoint impedance of a folded dipole consists of four times (or > other ratio if the conductors are different in size or there are more than > two folded conductors) the impedance of a standard dipole, in parallel > with two series connected shorted stubs. The dipole consists of the two > conductors in parallel. This behaves as a single fat wire which has the > effective velocity factor of ordinary insulated wire, around 0.97 or 0.98 > that of bare wire. That's why a folded dipole is about the same overall > length as a standard dipole. The stubs have the physical length of half > the dipole, or a bit shy of a quarter free space wavelength. Unlike the > dipole part, they operate as transmission lines, so their velocity factor > is around 0.8 -- a value which varies somewhat with cable construction. > Folded dipoles are sometimes shorted about 0.8 of the way from the center > to the ends in an apparent attempt to make the stubs an electrical quarter > wavelength, resulting in their impedance being very high as seen at the > feedpoint. > > But if the intermediate short circuit isn't done, the effect of the > somewhat longer stubs is only to add a bit of capacitive reactance across > the feedpoint. This lowers the antenna resonant frequency roughly 50 kHz > at 14 MHz, and has very little effect on the feedpoint resistance. Antenna > resonance can be restored by simply shortening the antenna a little. So > why bother with the intermediate short? > > Roy Lewallen, W7EL Article: 221677 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's References: <11v6ta6f01hv26@corp.supernews.com> Message-ID: Date: Wed, 15 Feb 2006 22:44:00 GMT John, N9JG wrote: > The 1986 handbook has the details for this type of antenna on page 33-14, > under the title of "Simple Antennas for HF Portable Operation." For 7.15 > MHz, the value of the capacitor is 152 pF, and the length of the matching > stub is 6'11-1/2". An open-end stub, made from twin lead, of length 20'-1/2" > can be substituted for the capacitor. It might be interesting to some to explain how/why this works. On a feedline with reflections, there are purely resistive current maximum points existing every half-wavelength up and down the feedline. Since the resistance at the current maximum point is often in the ballpark of 50 ohms, this is often a logical point at which to connect the transmitter/tuner. That's why I feed my 130 foot dipole at a current maximum point on each HF band. However, for a folded dipole with a 300 ohm resonant impedance fed with 300 ohm feedline, there are no reflections and therefore no current maximum points because there are no standing waves. So the trick is to cause a 50 ohm current maximum point to occur by causing reflections and standing waves on a short piece of series matching section. A parallel capacitor can often accomplish this function. Obviously, a capacitive stub can do the same thing. Note: the following calculations were done with a paper Smith Chart and therefore suffer from some inaccuracies. A parallel capacitive reactance of -j150 ohms will shift the 300+j0 ohm impedance to 60-j120 ohms. Some may want to refresh their memories on the parallel/series and series/parallel impedance equations. 60-j120 ohms will cause an SWR of ~6:1 from the point where the capacitor is installed on a flat 300 ohm feedline to the necessary 1:1 choke/balun. If we divide the 300 ohm Z0 by the SWR, we will obtain the resistance at the 6:1 SWR current maximum point. That value is 300/6 = 50 ohms and it exists ~0.062 wavelengths farther along than the 60-j120 point where the cap is located. The cap needs to cause an SWR of 9:1 if flat 450 ohm line is used and 12:1 if flat 600 ohm line in used. Needless to say, the Z0 of the line determines the value of capacitive reactance that is necessary to accomplish that function. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221678 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Message-ID: References: <11v6ta6f01hv26@corp.supernews.com> Date: Wed, 15 Feb 2006 23:04:48 GMT On Wed, 15 Feb 2006 22:44:00 GMT, Cecil Moore wrote: >It might be interesting to some to explain how/why this works. ... Some thoughts on why it might not work to plan, though it should be close to plan and capable of fine adjustment if needed. Worth noting that the characteristic impedance of some lines is quite different to their market labelling, especially 300 ohm and 450 ohm lines. In this part of the world, nominal 300 ohm TV ribbon is more like 360. Wes measured a range of nominal 450 ohm lines to be from 360 to 400 ohms. Although the calculated answer might be a shunt capacitor of 147pF, in practice a close value would be used and both the dipole length and feedline length could be juggled for better match if needed. Alternatively, a shunt stub could be used and tuned along with the series line length. The feed point impedance of the dipole will depend on its environment to some extent, and may be reactive if formulas are used to cut it to length. Of course, it is a one-band matching arrangment. Owen -- Article: 221679 of rec.radio.amateur.antenna From: "kd5sak" References: <5eudnRTUMuRG9m_eRVn-qA@comcast.com> <8LGdnXuY8oVfNG7eRVn-gg@comcast.com> Subject: Re: Stealth antenna question Message-ID: <2wOIf.33304$H71.29885@newssvr13.news.prodigy.com> Date: Wed, 15 Feb 2006 23:08:46 GMT "Joe S." wrote in message news:8LGdnXuY8oVfNG7eRVn-gg@comcast.com... > > "jimbo" wrote in message > news:C-SdnR6M3OKqpG7eRVn-sw@comcast.com... > XYL and I are refugees from Katrina -- we were living in an apartment in > Bay Saint Louis, MS, and building a house nearby -- it was all washed away > in Katrina and we took a big uninsured loss. We are now looking for a > home here in NE Tennessee -- 1,500 feet or more above sea level. Now there is one of the better decisions I've heard from recipents of Katrinas free demolition services (G) God grant you and your family a long and safe life at this more fortuitous elevation. Oh, and good luck with the antenna farm acquisition, also. Harold KD5SAK Article: 221680 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Wed, 15 Feb 2006 17:13:45 -0600 Message-ID: <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> References: <6236-43F24589-857@storefull-3255.bay.webtv.net> Roy Lewallen, W7EL wrote: "The average power is therefore relatively small, much smaller than the product of RMS volts times RMS amps." RMS is short for root-mean-square. RMS is synonymous with the "effective value" of a sinusoidal waveform. Therefore, the average power for the time period of one complete cycle or any number of complete cycles is the product of the effective volts times the effective amperes. See page 19 of "Alternating Current Fundamentals" for derivations of the proof. Average power is exactly the product of rms volts times rms amps in usual circumstances. Best regards, Richard Harrison, KB5WZI Article: 221681 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: References: <6236-43F24589-857@storefull-3255.bay.webtv.net> <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> Date: Thu, 16 Feb 2006 00:08:53 GMT On Wed, 15 Feb 2006 17:13:45 -0600, richardharrison@webtv.net (Richard Harrison) wrote: >Roy Lewallen, W7EL wrote: >"The average power is therefore relatively small, much smaller than the >product of RMS volts times RMS amps." > >RMS is short for root-mean-square. RMS is synonymous with the "effective >value" of a sinusoidal waveform. > >Therefore, the average power for the time period of one complete cycle >or any number of complete cycles is the product of the effective volts >times the effective amperes. Leaving aside your new confusing term "effective value", if you multiply Vrms by Irms in an AC circuit you get Apparent Power (units are Volt Amps or VA). Apparent Power is the vector sum of two quadrature components Real Power (Watts) and Reactive Power (VAR). Real Power is the thing you describe when you talk about average power. It is Real Power that is the rate of flow of energy averaged over a complete AC cycle. > >See page 19 of "Alternating Current Fundamentals" for derivations of the >proof. > >Average power is exactly the product of rms volts times rms amps in >usual circumstances. Depends on what you mean by "usual circumstances". Your rule does not apply if there is a phase difference between V and I, which is commonly the case in power distribution, and is commonly the case in RF where loads circuit impedances may have a reactive component. To an electrician, (Real) Power = Vrms * Irms * PF where PF (the power factor) is the cosine of the phase angle between V and I. This isn't engineering stuff, sparkies know and apply it every day. Owen -- Article: 221682 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's References: <11v6ta6f01hv26@corp.supernews.com> Message-ID: Date: Thu, 16 Feb 2006 00:25:57 GMT Owen Duffy wrote: > Cecil Moore wrote: >>It might be interesting to some to explain how/why this works. > > Some thoughts on why it might not work to plan, though it should be > close to plan and capable of fine adjustment if needed. Close is all most of us need. Once the 50 ohm SWR is below about 10:1, the fine adjustments are relatively easy. If the SWR decreases with adjustments, we are going the right direction. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221683 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Wed, 15 Feb 2006 18:21:29 -0600 Message-ID: <23080-43F3C589-1341@storefull-3256.bay.webtv.net> References: <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> The book I referred to in a previous posting is: "Alternating Current Fundamentals" by John R. Duff. It is a 1963 edition that was copyrighted in 1961. Library of Congress Catalog Card Number is 61-15728. Publisher in the U.S,A, is Litton Educational Publishing, Inc. In Canada the simultaneous publisher is Delmar Publishers. Best regards, Richard Harrison, KB5WZI Article: 221684 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Wed, 15 Feb 2006 19:00:16 -0800 Message-ID: <11v7qm1bn0lh745@corp.supernews.com> References: <6236-43F24589-857@storefull-3255.bay.webtv.net> <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> Richard Harrison wrote: > Roy Lewallen, W7EL wrote: > "The average power is therefore relatively small, much smaller than the > product of RMS volts times RMS amps." > > RMS is short for root-mean-square. RMS is synonymous with the "effective > value" of a sinusoidal waveform. > > Therefore, the average power for the time period of one complete cycle > or any number of complete cycles is the product of the effective volts > times the effective amperes. No, I'm sorry, that isn't true. The average power isn't the product of the product of the RMS voltage times the RMS current, except in the single circumstance of their being in phase. > See page 19 of "Alternating Current Fundamentals" for derivations of the > proof. I don't have this book, but I know that in the past you've quoted from books without having fully understood the context of the quote. I'm sure that's the case here. Any electrician or technician should know that for sinusoidal waveforms, Pavg = Vrms * Irms * cos(theta) where theta is the phase angle between V and I. And hopefully you can see with a few moments and a calculator that if theta = 90 degrees, Pavg = zero regardless of V and I. > Average power is exactly the product of rms volts times rms amps in > usual circumstances. Perhaps your "usual circumstances" are that the load is purely resistive. But that's not "usual circumstances" for a host of applications. Roy Lewallen, W7EL Article: 221685 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Thu, 16 Feb 2006 04:26:34 +0000 (UTC) Message-ID: References: <6236-43F24589-857@storefull-3255.bay.webtv.net> <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> How about P = Square(V) / R watts or P = Square(I) * R watts for no phase angles. Article: 221686 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Wed, 15 Feb 2006 21:19:01 -0800 Message-ID: <11v82q6i7mglmc3@corp.supernews.com> References: <6236-43F24589-857@storefull-3255.bay.webtv.net> <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> Reg Edwards wrote: > How about P = Square(V) / R watts > > or P = Square(I) * R watts > > for no phase angles. Those are fine by me. Just for the record, though: V is the voltage across R, I is the current through R. This is important when there are other components, hence possibly other values of V and I, in the circuit. P can be average and V and I RMS; or P, V, and I can all be functions of time -- the formulas are ok in either case. And finally, just for the fuss-budgets, we're assuming R isn't varying with time. Picky as this might seem, it's awfully important to make perfectly clear just what a formula applies to and what it doesn't. Otherwise it's sure to be misapplied in situations where it isn't valid. (Clarification reduces the chance of misapplication from certain to only highly likely.) Roy Lewallen, W7EL Article: 221687 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Wed, 15 Feb 2006 23:40:49 -0600 Message-ID: <17265-43F41061-356@storefull-3253.bay.webtv.net> References: <11v82q6i7mglmc3@corp.supernews.com> Roy Lewallen, W7EL wrote: "V is the voltage across R, I is the current through R." Yes. The voltage drop across a resistor is always in-phase with the current through the resistor as there is no energy storage in a resistor. Best regards, Richard Harrison, KB5WZI Article: 221688 of rec.radio.amateur.antenna From: David Harmon Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: <447d1c37.338298984@news.west.earthlink.net> References: <6236-43F24589-857@storefull-3255.bay.webtv.net> <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> Date: Thu, 16 Feb 2006 06:36:59 GMT On Wed, 15 Feb 2006 17:13:45 -0600 in rec.radio.amateur.antenna, richardharrison@webtv.net (Richard Harrison) wrote, >RMS is short for root-mean-square. RMS is synonymous with the "effective >value" of a sinusoidal waveform. > >Therefore, the average power for the time period of one complete cycle >or any number of complete cycles is the product of the effective volts >times the effective amperes. That is true for the Special Case where your load is a pure resistance. In general, no. Article: 221689 of rec.radio.amateur.antenna From: "Jeff Dieterle" References: <10cd5$43eb4ef3$424980b2$32629@COMTECK.COM> <11un62s35hpkk9c@corp.supernews.com> Subject: Re: Another Car Radio static question Date: Thu, 16 Feb 2006 07:28:28 -0500 Message-ID: <515c3$43f46fec$4249809e$13977@COMTECK.COM> I've noticed something else on this problem, below the 3-phase electrical conductors, there is either a telephone or catv wire. The intense periods occur when I drive by some type of device that loops this cable in to a pear shaped configuration. I guess it's not a periodic sound but appears that way because they are evenly spaced and I drive the same speed. Maybe if anyone is still following this thread and that sheds anymore light they'll respond. "Dave Platt" wrote in message news:11un62s35hpkk9c@corp.supernews.com... > In article <10cd5$43eb4ef3$424980b2$32629@COMTECK.COM>, > Jeff Dieterle wrote: > >>Is the person or company responsible for creating this type of >>interference >>obligated to correct the cause? > > It depends. > > If they're radiating RF in excess of FCC part 15 limits, and do not > have a license to transmit in whatever frequency band they are using, > then they're very probably obligated to fix the problem or to shut > down whatever equipment is radiating. > > If they're transmitting in a frequency band / allocation for which > they have a proper license, and if their transmission isn't leaking > spurious frequencies in excess of the FCC limits, then they're > probably quite legal. In this situation, what you'd hearing would be > considered to be "undesired reception" - that is, your car radio is > being overwhelmed by a strong signal outside of its normal passband - > and the FCC would consider this the fault of your radio. > > Without doing some spectrum analysis and direction finding it's hard > to distinguish the two from the symptoms you report. > > -- > Dave Platt AE6EO > Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior > I do _not_ wish to receive unsolicited commercial email, and I will > boycott any company which has the gall to send me such ads! Article: 221690 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question References: <6236-43F24589-857@storefull-3255.bay.webtv.net> <23081-43F3B5A9-311@storefull-3256.bay.webtv.net> Message-ID: Date: Thu, 16 Feb 2006 13:25:03 GMT Reg Edwards wrote: > How about P = Square(V) / R watts > > or P = Square(I) * R watts > > for no phase angles. Actually, the "R" implies an impedance of R+j0, i.e. a phase angle of zero. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221691 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: New Program: OP_ANGLE Date: Thu, 16 Feb 2006 15:28:47 +0000 (UTC) Message-ID: References: To anyone who may be interested. In the notes to this new program I omitted to say that the computed components of plate current constitute the first few significant terms of a Fourier Series. There's the DC component, the fundamental frequency component, and the 2nd, 3rd, 4th and 5th harmonic components. The amplitude of the harmonics decrease as the harmonic order increases. Amplitudes are the peak sinewave values (not rms). If the components, including the DC component, are added together taking into account their signs, then their sum is very nearly equal to the computed peak instantaneous plate current. Which has been adjusted to equal the actual value which flows in the tube. If the Fourier Series had been extended to even higher harmonics then the sum of the series would have converged to the peak instantaneous value of plate current exactly. This is a check on the accuracy of the Fourier Analysis performed by the program. There's ample oportunity for bugs. If an amplifier is used as a frequency multiplier then power can be extracted from the plate by an LC circuit tuned to the required harmonic. The correct plate load impedance is the required peak RF volts across it divided by the peak amplitude of the plate current component. There is NOT a conjugate match. The internal resistance of PA's is NOT 50 ohms as is commonly believed. With pentodes, beam tetrodes and transistors (transistors have operating angles) it is often much higher. There, I've said it again! ---- Reg, G4FGQ. Article: 221692 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Thu, 16 Feb 2006 09:20:34 -0600 Message-ID: <7288-43F49842-1369@storefull-3251.bay.webtv.net> References: Owen Duffy wrote: "Leaving aside your new confusing term "effective value", if you multiply Vrms by Irms in an AC circuit you get Apparent Power (units are volt amps or VA." Exactly, except the term "effective value is as old as a-c power calculations. The value of the a-c volt was chosen to produce the same effect, lamps as bright, heat as warm, as d-c does. My electronic dictionary says: "rms amplitude - Root-mean-square amplitude, also called effective amplitude. The value assigned to an alternating current or voltage that results in the same power dissipation in a given resistance as dc current or voltage of the same numerical value.' Best regards, Richard Harrison, KB5WZI Article: 221693 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: New Program: OP_ANGLE Date: Thu, 16 Feb 2006 18:24:53 +0000 (UTC) Message-ID: References: Program OP_ANGLE The Fourier Analysis of plate current has now been extended from the 5th up to the 7th harmonic. Otherwise the program has not been changed except for the issue date. If you are still interested, download the new version and use it to overwrite the first. It seems that for almost exact agreement between peak plate current and the sum of the harmonic components it is necessary to extend the analysis up to around the 20th harmonic. There's not enough room on the screen to display results. ---- ........................................................... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp ........................................................... Article: 221694 of rec.radio.amateur.antenna From: "Phil" References: Subject: Re: 300 ohm folded dipole from ARRL Handbook, early 1990's Message-ID: Date: Thu, 16 Feb 2006 21:11:03 GMT "Geoffrey S. Mendelson" wrote in message news:slrndv5o0l.jp4.gsm@cable.mendelson.com... > Hi, the early '90s ARRL handbooks had construction information for a > simple folded dipole made of 300 oHm twinlead. The article was not in > the '96 edition or later. > > The antenna was made of 300 ohm twinlead. It was fed in the center by > a 300 ohm twinlead feeder to a matching capacitor which made it aproximate > 50 ohms. The capacitor could be either a regular capacitor (I used silver > mica WWII surplus) or a "stub" of more twinlead. > > At the end of the feed line, which was part of the antenna, you could > attach it directly to a 50 omh unbalanced transciever or a 50 ohm coax > of any length. > > The length of the antenna was one half wavelength without the velocity > factor and then shorted at the wavelength points compensating for > the velocity factor. > > For example, if it was 10 meters long for a 20 meter dipole, the shorts > were about 4 meters from the center for twinlead with a velocity factor > of 0.8. > > If anyone has a copy of the article and could scan it for me, I'd > appreicate it, or even better yet, a pointer to an online version > of it. > > Thanks in advance and 73, > > BTW, if you are also in Israel and asking yourself "where does he get > the twinlead"? I don't. I brought it with me when I moved here in '96. > > Geoff. > > -- > Geoffrey S. Mendelson, Jerusalem, Israel gsm@mendelson.com N3OWJ/4X1GM > IL Voice: (07)-7424-1667 IL Fax: 972-2-648-1443 U.S. Voice: > 1-215-821-1838 > Visit my 'blog at http://geoffstechno.livejournal.com/ I played a lot with a folded dipole for 14 MHz, both with twin lead and with two parallel wires about 20 cm apart. To feed it, I used a simple 75 ohms to 300 ohms Neosid transformer currently/formerly intended to connect a 75 ohms VHF television set to an antenna with twin lead. Depending on the country, such a bal/un is called a "symetriser", a "bazooka", a "pig's nose", etc... This small pig's nose can cope with 100 watts output on 14 MHz and higher. I never tried it on lower frequencies. I expect it would get rather warm on 80 m. Article: 221695 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: References: <7288-43F49842-1369@storefull-3251.bay.webtv.net> Date: Thu, 16 Feb 2006 22:00:56 GMT On Thu, 16 Feb 2006 09:20:34 -0600, richardharrison@webtv.net (Richard Harrison) wrote: >Owen Duffy wrote: >"Leaving aside your new confusing term "effective value", if you >multiply Vrms by Irms in an AC circuit you get Apparent Power (units >are volt amps or VA." > >Exactly, except the term "effective value is as old as a-c power >calculations. The value of the a-c volt was chosen to produce the same >effect, lamps as bright, heat as warm, as d-c does. > >My electronic dictionary says: >"rms amplitude - Root-mean-square amplitude, also called effective >amplitude. The value assigned to an alternating current or voltage that >results in the same power dissipation in a given resistance as dc >current or voltage of the same numerical value.' Richard, your discussion here is limited to DC circuits. Your proposition in another post that Vrms * Irms gives the Real Power (ie indicates net energy flow over time) does apply to DC circuits, but it does not apply generally. You have cited a text book to support your position, however it is likely that you have misinterpreted the text book. Work this example through with your textbook: We have 120Vrms AC 60Hz (sinusoidal) impressed across a load of 85 ohms of resistance and an ideal inductor of 85 ohms reactance in series. The load impedance is 85+j85. The Circuit current is 120/(|85+j85|) or 1Arms. The power dissipated in the resistance is 85W, and since it is the only resistance dissipating power, the Real Power for the entire circuit is 85W. The circuit Apparent Power is 120 * 1 or 120VA, the Reactive Power is 85VAR. The circuit Vrms * Irms does not give the Real Power for this circuit. Owen -- Article: 221696 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Thu, 16 Feb 2006 17:03:29 -0600 Message-ID: <981-43F504C1-1526@storefull-3254.bay.webtv.net> References: Owen Duffy wrote: "Work this example through with your textbook." Don`t need the textbook. I`ve been working these for over 60 years. Pythagoras gave us the solution in ancient times when electricity was produced by rubbing an amber rod with an animal pelt. The impedance is close enough to 120 ohms. I=E/Z E= 1 ampere Power=Isquared x R Power = 85 watts Best regards, Richard Harrison, KB5WZI Article: 221697 of rec.radio.amateur.antenna From: Ken C Subject: Aux Antenna for Wlakman FM radio? Message-ID: Date: Thu, 16 Feb 2006 18:28:43 -0500 I am tryng to use a Sansa M240 MP3 player and FM radio on my motorcycle, with the output going into a helmet sound system (Starcom1). This means that the headphones -- which usually act as the FM antenna -- will not be used. I would like to try using a splitter on the headphone jack, with one leg to the helmet sound system and the other leg to an antenna. My first thought was a simple 1/4-wave dipole to run around the fairing perimeter. Would that work? Would I need to add some passive components or some matching because the device expects the headphones to be the antenna? And, Yes, the Starcom1 also interfaces to a TH-F6A tri-bander. Ken KC2JDY Article: 221698 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Message-ID: References: <981-43F504C1-1526@storefull-3254.bay.webtv.net> Date: Thu, 16 Feb 2006 23:38:16 GMT On Thu, 16 Feb 2006 17:03:29 -0600, richardharrison@webtv.net (Richard Harrison) wrote: >Owen Duffy wrote: >"Work this example through with your textbook." > >Don`t need the textbook. I`ve been working these for over 60 years. > >Pythagoras gave us the solution in ancient times when electricity was >produced by rubbing an amber rod with an animal pelt. The impedance is >close enough to 120 ohms. > >I=E/Z >E= 1 ampere > >Power=Isquared x R >Power = 85 watts Nothing like Vrms * Irms though, is it? > >Best regards, Richard Harrison, KB5WZI -- Article: 221699 of rec.radio.amateur.antenna From: richardharrison@webtv.net (Richard Harrison) Subject: Re: For Roy Lewallen et al: Re Older Post On My db Question Date: Thu, 16 Feb 2006 18:31:51 -0600 Message-ID: <7814-43F51977-1546@storefull-3257.bay.webtv.net> References: Owen Duffy wrote: "Nothing like Vrms times Irms through it is it?" Yes it is. Sorry I fat fingered (E=1ampere). It is I=1 ampere. 1 ampere times the 85 volts dropped across an 85-ohm resistor by 1 ampere = 85 watts, same as 1x1x85=85, or 85x85/85=85. All methods of calculation must give the same answer, and they do if you don`t make a mistake. Best regards, Richard Harrison, KB5WZI Article: 221701 of rec.radio.amateur.antenna From: "SignalFerret" References: Subject: Re: Aux Antenna for Wlakman FM radio? Message-ID: Date: Fri, 17 Feb 2006 04:27:04 GMT The shield on the headphone cable is used as the antenna for such devices. As for matching, how good of a match can a headphone cable be? Inside most radios the connection is a single cap, in the pF range, between the jack and the RF front end. Oh, and some kind of choke for DC continuity for the audio part. Strip away some of the plastic jacketing around cable at the FM radio end, and solder on an extra wire to the shield. Route the wire some place convenient. You'll have to experiment for best reception. Just keep it away from any noise and power sources, like spark plugs, transmitters, Illudium P-36 Explosive Space Modulators. Life's a highway -- get on and ride, Robert N3LGC "Ken C" wrote in message news:a72av11gq4grf1btela8httgq2ukejuoc2@4ax.com... >I am tryng to use a Sansa M240 MP3 player and FM radio on my > motorcycle, with the output going into a helmet sound system > (Starcom1). This means that the headphones -- which usually act as > the FM antenna -- will not be used. > > I would like to try using a splitter on the headphone jack, with one > leg to the helmet sound system and the other leg to an antenna. My > first thought was a simple 1/4-wave dipole to run around the fairing > perimeter. Would that work? Would I need to add some passive > components or some matching because the device expects the headphones > to be the antenna? > > And, Yes, the Starcom1 also interfaces to a TH-F6A tri-bander. > > Ken KC2JDY Article: 221702 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: using coax shield to create a loading coil ? Date: Fri, 17 Feb 2006 10:16:27 -0800 Message-ID: <11vc4ntcub1k91@corp.supernews.com> References: dansawyeror wrote: > Good morning. I would like to experiment with making a high Q coil for > creating a tuned radial counterpoise. Reg's program predicts a coil of > about 70 uH will create a match. One way to create such a coil would be > to wind coax and use the shield as the conductor. Besides the obvious > loss of physical stability due to lack of a form what are the > limitations or drawbacks from using the shield? The braided shield will be substantially lossier than a solid wire or tube of the same diameter. And the relatively poor quality dielectric used for the outside of the cable will also reduce the Q somewhat. So coax isn't a good choice for your stated objective of making a high Q coil. Roy Lewallen, W7EL Article: 221703 of rec.radio.amateur.antenna From: "Hal Rosser" References: Subject: Re: using coax shield to create a loading coil ? Message-ID: Date: Fri, 17 Feb 2006 21:25:06 -0500 You may have better luck by using sone soft copper tubing. Dimensional stability would be one advantage. You could use ScotchKote or some insulating paint if you needed it to be insulated. "dansawyeror" wrote in message news:YqudnU1MnL5ddmjeRVn-qQ@comcast.com... > Good morning. I would like to experiment with making a high Q coil for creating > a tuned radial counterpoise. Reg's program predicts a coil of about 70 uH will > create a match. One way to create such a coil would be to wind coax and use the > shield as the conductor. Besides the obvious loss of physical stability due to > lack of a form what are the limitations or drawbacks from using the shield? > > Thanks - Dan - kb0qil Article: 221704 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: using coax shield to create a loading coil ? Date: Fri, 17 Feb 2006 18:38:07 -0800 Message-ID: <11vd24h21qefea@corp.supernews.com> References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> Mike Speed wrote: >> Two factors. One is surface roughness. The other is caused by the >> current having to continually move from one group of wires to another as >> it travels. > > Interesting, but how do you know the current is moving as you say? Skin effect is well known. On a good conductor at high frequencies, current is concentrated very near the surface. When a bundle of wires ducks under another in the direction of current flow, the current has to migrate to the outside again, which means it has to move from one conductor to another. There's no question that it happens -- what's a bit harder to pin down is just how much loss typically results. >> I'm just now doing some research on how significant these >> effects are, but so far I've found out they're very noticeable. > > Ok. I'm curious: What equipment are you using for the research? Books, and to a lesser extent the web. Information about this is scattered among a number of sources. Quite a few discuss surface roughness in a general way, but there's a particularly good explanation, analysis, and something of a quantitative treatment in Johnson & Graham's _High-Speed Signal Propagation: Advanced Black Magic_. The effect of weaving is harder to track down -- most authors simply assume coax shield conductivity loss to be negligible, and don't deal with woven conductors in any other context. But it really isn't, if you're interested in good accuracy. And of course when the braided conductor is the primary conductor, it becomes much more important. I know Tom, W8JI has done some measurements on braided vs solid strap, and I'll be asking him for more information before long. I do know that he found a very significant difference, and I have a great deal of respect for his experience, measurements, and opinions. >> But hey, you don't have to believe me. Make up some coils and measure >> their Q -- it's not hard at all. Then stick them outside for a while and >> measure them again. > > What would be a good way to measure Q? The way I do it is by resonating the inductor with a parallel air variable capacitor. It's important to keep it away from just about everything. I couple in and out with a very small (typically 1 pF at HF) capacitor, and make sure that the impedances of the source and detector are either very high or quite low (say 50 ohms) to minimize loss. I use a signal generator for the source and a scope for the detector. Using a frequency counter connected to the signal generator, I measure the resonant frequency and -3dB points. The Q is the ratio of the center frequency to the 3dB bandwidth. For convenience, I made a 3dB pad I can switch in an out of the signal generator. With this, you don't even need a linear detector, and a diode and meter would do. My measurements have been within about 5 - 10% of readings with a good HP Q meter on the few occasions when I've compared them. That's close enough for my purposes. >> Or do like most amateurs do -- make the coils, >> discover that you can talk to far away places "barefoot", and declare >> that they "work". >> > > Uugghh, don't I know. Many people have worked the world with 1 watt, knowing that's what they were running. A lot more have worked the world with 1 watt, thinking they were running 100. Ignorance is bliss. Roy Lewallen, W7EL Article: 221705 of rec.radio.amateur.antenna From: "Sal M. Onella" References: Subject: Re: using coax shield to create a loading coil ? Message-ID: Date: Fri, 17 Feb 2006 20:48:02 -0800 "dansawyeror" wrote in message news:YqudnU1MnL5ddmjeRVn-qQ@comcast.com... > Good morning. I would like to experiment with making a high Q coil for creating > a tuned radial counterpoise. Reg's program predicts a coil of about 70 uH will > create a match. One way to create such a coil would be to wind coax and use the > shield as the conductor. Besides the obvious loss of physical stability due to > lack of a form what are the limitations or drawbacks from using the shield? > > Thanks - Dan - kb0qil You have cited the major limitation: physical form. matching coils for many broadband transmit antennas use wound coils of roughly that dimension, although they are made of copper tubing that exhibits much more stability, once shaped. May I suggest the use of PVC pipe as a form for winding the coax? Tape it in place while you're "monkeying" and then fiberglass it when you like what you have. (Auto supply stores sell the fiberglass for doing auto body work -- it's durable & light-weight.) KD6VKW ET USN (ret) Article: 221706 of rec.radio.amateur.antenna From: Owen Duffy Subject: Re: using coax shield to create a loading coil ? Message-ID: <84hdv1lsug1m323jk3hjiss5otn1udq06l@4ax.com> References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> Date: Sat, 18 Feb 2006 07:22:15 GMT On Fri, 17 Feb 2006 17:25:28 -0800, Roy Lewallen wrote: >... Or do like most amateurs do -- make the coils, >discover that you can talk to far away places "barefoot", and declare >that they "work". Roy, As we dumb amateur radio down to make it attractive to the disinterested masses in a desperate and mistaken persuit of increasing the number of licenced hams, this is becoming the new standard of understanding in the redefined amateur radio. I wrote comment on an a recent article in Australia's ham radio magazine "Amateur Radio" that was an example of the declaration of not just something that works, but "something that really works" though it looks to be quite inefficient on at least one band. The comment is at http://www.vk1od.net/blog/index.php?op=ViewArticle&articleId=21&blogId=1 . Supporters argue "amateur radio is about having QSOs, so if you have QSOs then the antenna works... QED". Though antenna systems remains one of the few areas of amateur radio where hams can cost effectively design solutions specific to their location and needs, the lower competency standard of the new "communicator" style amateur does not support a soundly based understanding of antenna systems. We frequently hear the argument that there is no need to understand electronics for modern amateur radio where commercial radios are the norm, but forget electronics for a moment, how many hams understand a common three component passive network that is so often employed with variable results, the ATU. Get used to it! Amateur radio is being transformed to "I just wanna talk on the radio". Owen -- Article: 221707 of rec.radio.amateur.antenna From: Ken C Subject: Re: Aux Antenna for Wlakman FM radio? Message-ID: References: Date: Sat, 18 Feb 2006 08:51:50 -0500 On Fri, 17 Feb 2006 04:27:04 GMT, "SignalFerret" wrote: >The shield on the headphone cable is used as the antenna for such devices. >As for matching, how good of a match can a headphone cable be? Inside most >radios the connection is a single cap, in the pF range, between the jack and >the RF front end. Hmm. Well, the shield on the shielded stereo cable between the player and the helmet sound module might pick up some interference from the motorcycle's electrics. What components would you suggest that I install on that cable to filter that out? Basically, I would want to filter out everything that is not audio frequency (20 Hz - 20 kHz) or between 88-108 MHz. The player headphones have a main cable that is 29" long before splitting to each headphobne. This is 1/4 wave for FM BCB. I could, as you suggest, splice in a 29" wire at the player end of the cable. Ken KC2JDY Article: 221708 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: using coax shield to create a loading coil ? Date: Sat, 18 Feb 2006 06:26:14 -0800 Message-ID: <11vebk92jcjnlbf@corp.supernews.com> References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <43F6A025.3020707@comcast.net> dansawyeror wrote: > Roy, > > Thank you. It is a quick experiment to build a test coax coil and > measure the Q. That should produce enough evidence to test a counterpoise. > > In the mean time the research to build a coil out of copper tubing > continues. So far the only alternative I can conceive is to make a > wooden form and wrap the coil on the outside. > > Thanks again - Dan A copper tube will definitely produce an improved Q. If you do make comparative measurements of ones made from tubing and from coax, please post the results. A real problem in maintaining the Q of coils outside in the weather is keeping water from getting between the turns. Water is very lossy stuff at HF, and it has a very high dielectric constant. The two combine to make it a real Q killer if it gets into any region of high electric field strength. A bit of accumulated dust mixed with the water makes it worse yet. So if you anticipate leaving the coil on a form and exposed to the weather, also check the Q when the coil is wet. See http://www.eznec.com/Amateur/Articles/Portable_Feed_Lines.pdf for results of measurements of wet and dry 300 ohm twin lead. It's not quite the same situation, but the loss mechanism is essentially the same. I recommend that you do some modeling or just simple calculating, if you haven't done so already, to see just how high the Q has to be in order to keep overall loss acceptable. One final thing to keep in mind -- I've heard reports of poor performance of elevated verticals being tracked down to badly imbalanced currents in the radials. Apparently even small physical differences among the radials can cause one or two to hog all the current. If this is so, it seems to me that making them more sharply tuned by inductive loading might make this effect even worse. So when you get the thing up, I suggest measuring the current in each radial. This is easily done with a toroid core with a few turns for the secondary and a fairly low R across the secondary. It's been discussed a number of times on this newsgroup, the last time quite recently. Good luck! Roy Lewallen, W7EL Article: 221709 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: What the? Date: Sat, 18 Feb 2006 06:29:53 -0800 Message-ID: <11vebr41b1une5d@corp.supernews.com> References: <1140240942.959492.40540@g47g2000cwa.googlegroups.com> KG0WX wrote: > OK - I just bought a new in the box 1955 Heath GD-1B grid dipper > with the extra low band coil set. $25. The dipper works FB and I > used an old SO-239 chassis jack plus a bit of 22g wire and a cut > down pill bottle to make a 4t coupling loop big enough to "dip" - > that is I inserted the dipper's coil inside the pill bottle coil > attached > to my feedline. > > Now here is where I make myself look stupid - My antenna's best > SWR is at 7185 with a 1.1:1 but the dipper showed max dip at > 6990 (I tuned into the dipper with my K2). What's going on here? > > Someone educate me.....lol > > 73's de Ken KG0WX 1. Although it's usually close for simple antennas, the point of minimum SWR isn't necessarily the resonant frequency. The dipper measures resonance. 2. The coupling loop adds inductance to the circuit, lowering its resonant frequency. Roy Lewallen, W7EL Article: 221710 of rec.radio.amateur.antenna From: Roy Lewallen Subject: Re: using coax shield to create a loading coil ? Date: Sat, 18 Feb 2006 06:38:31 -0800 Message-ID: <11vecbaciouhcda@corp.supernews.com> References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <1140266886.722102.279930@z14g2000cwz.googlegroups.com> Mike Speed wrote: > Again, interesting, but what's been outlined so far is not scientific. > For something of this nature to be of any utility, it must be grounded > in science. I can assure the readers that all the effects I've discussed are soundly based on very well known principles. Anyone truly interested in the topic can find ample confirmation of what I've said, although it might take a bit of digging. The Johnson and Graham text is an excellent place to start. What's lacking is good measured data for typical shields, and even that's going to have limitations because of the wide variations among cables and manufacturers. But even some rules of thumb will be useful. But you've shown an interest in the topic. Why don't you make some measurements of coils made from tubing and from coax shields and report back? Roy Lewallen, W7EL Article: 221711 of rec.radio.amateur.antenna From: Jim Higgins Subject: Re: using coax shield to create a loading coil ? Message-ID: References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <1140266886.722102.279930@z14g2000cwz.googlegroups.com> Date: Sat, 18 Feb 2006 16:00:16 GMT On 18 Feb 2006 04:48:06 -0800, "Mike Speed" wrote: >> When a bundle of wires ducks under another in the direction of current flow, the current > has to migrate to the outside again, > >> There's no question that it happens > > >> Books, and to a lesser extent the web. Information about this is > >> I have a great deal of respect for his experience, measurements, and opinions > >Again, interesting, but what's been outlined so far is not scientific. >For something of this nature to be of any utility, it must be grounded >in science. In order to understand that something is grounded in science, one must first know science. Article: 221712 of rec.radio.amateur.antenna From: "RST Engineering" References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <1140266886.722102.279930@z14g2000cwz.googlegroups.com> Subject: Re: using coax shield to create a loading coil ? Date: Sat, 18 Feb 2006 09:46:51 -0800 Message-ID: Hm. Just searched the FCC database for a SPEED, MI... and nothing pops. QRZ.COM gives quite a few (39) hits for SPEED, but none with the first name or middle initial that you can make a "mike" from. Hm. Jim "Mike Speed" wrote in message news:1140266886.722102.279930@z14g2000cwz.googlegroups.com... Article: 221713 of rec.radio.amateur.antenna From: "kevin" Subject: Re: help appreciated with unusual j pole feed Date: Sat, 18 Feb 2006 18:20:59 -0000 Message-ID: References: hi fred, thanks for your input, if i read you correctly this is cut for 2mtr and the one i'm working on is actually a marine vhf, (usually I think cut for 156.8mHz), i realise this is a ham group but hoped it would be ok to ask here. i have found some online calculators, maybe i can use the information from them to recut this design. thanks fred kevin "Fred McKenzie" wrote in message news:fmmck-1402061352200001@aca8cd90.ipt.aol.com... > In article , "kevin" > wrote: > > > hi there, i'm looking for some help with a j pole antenna, it's actually a > > vhf marine antenna so i hope it will be ok to ask here. > > Kevin- > > One version of the J-Pole antenna has the cable connected an inch or two > above the connection between the two vertical rods. The following article > from my archives, shows how it could be connected, except it is based on > Television Twin Lead instead of the two rods yours has. If you use a > mono-spaced font like Courier, it should be clear. The conductor above > the 1/4" gap is not essential, it just comes with the twinlead! > > Fred > > --------------- > 2M Jpole with 300 ohm twin lead > From: kinsman@ektools.UUCP (Andrew A. Kinsman) > Newsgroups: rec.ham-radio > Organization: Eastman Kodak, Dept. 47, Rochester NY > > The following is a description of a J-Pole antenna made from 300 ohm > TV twin-lead. Quite a few of the hams in the area have successfully > built one or more of these antennas. They have quite a few advantages > which include improved performance for 2-meter HTs, portability, and > last, but never least, they are inexpensive. > > | | do not short this end. > | | > | | > | | > | | > | | > 54" | | > | | > | | > | | > | 1/4" gap > | | > | | > | | 16 3/4" > coax ctr conductor=>* *<= coax shield > | | > 1 1/4" | | > -*- > solder the twin leads together at bottom > > 1. Start with 54 1/2" of TV twin lead (foam core does not work as well) > 2. Strip 1/2" of insulation at bottom and solder wires together. > 3. Measure 1 1/4" from soldered wires and strip insulation on both > sides. This is the solder point for a coax feedline. > 4. Measure 16 3/4" from bottom and cut out 1/4" notch on one side. > 5. Feed with a length of RG58U coax. Tape coax at feedpoint to the > twin lead for strength and weather protection. > > I have also attached an alligator clip to the plastic on the top of > the antenna in order to conveniently hang it. I am successfully using > one in my attic to avoid any CCNR problems. I have another that I can > take with me camping, sailing, etc. > > This design has appeared on many BBSs and in club newsletters; the > earliest reference that I know of is a Jan. 1984 D.A.R.C. antenna > article by James Burks, KA5QYV. > > Enjoy, > N6UBM > ------------------ Article: 221714 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Accuracy of Q meters Date: Sat, 18 Feb 2006 18:31:22 +0000 (UTC) Message-ID: There are inumerable uses for solenoidal wound coils. Over the years there have been been hundreds of discussions and contributions to newsgroups about the Q of single-layer solenoid coils. But what is ENTIRELY missing is the measured data for particular coil dimensions and frequency. Does nobody have a Q meter? It would appear nobody has any confidence in Q meters in the HF range. QUESTION: What is the measuring accuracy of the best commercial Q meters in the ranges of 1 to 100, 100 to 500, 500 to 2000 and above? Or do not manufacturers state measuring accuracy? Are they themselves uncertain of what its all about? A subsidiary question is what use is made of Q values after a measurement has been made? Does an inacurate Q value matter very much anyway? Please give numbers in your reply. For once I confess to a minor troll. But I hope I get a few sensible answers to sensible questions. Nil answers will be considered to be of equal consequence to the others. ---- Reg, G4FGQ Article: 221715 of rec.radio.amateur.antenna From: stananger Subject: Under Eave Antenna Message-ID: Date: Sat, 18 Feb 2006 19:06:19 GMT I would like to install an under eave antenna at my house. Mostly for listening but maybe some low power transmitting. What I need is some input from all the experts here as to what supports would work best. Also I dont want it to upset the other half too much so it would have to be "pleasing" to the eye or damn near invisible. As I dont have any plans or thoughts as to what would work best I am open to all serious suggestions and ideas. It will be a loop fed with open wire feeders is all I have in mind at this time. Stan AH6JR Article: 221716 of rec.radio.amateur.antenna From: John Ferrell Subject: Re: using coax shield to create a loading coil ? Message-ID: <52kev19u0mrk27424soa8q7m2co00uojta@4ax.com> References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <43F6A025.3020707@comcast.net> <11vebk92jcjnlbf@corp.supernews.com> Date: Sat, 18 Feb 2006 19:45:55 GMT A visit to the kitchen supplies of your local department store will reveal a large assortment of coil forms and covers. Until recently I have limited my antenna work to vhf-uhf. In that circumstance I have frequently punched a snug hole in a plastic container bottom and placed it in a bell shaped fashion over lumped LC components. It has worked best for me to leave the bottom open for ventilation. I have not done it yet but preliminary plans for a base loaded 160 meter vertical have me considering an inverted plastic garbage can as a weather shield. The EZNEC models may spare me the effort on that one! On Sat, 18 Feb 2006 06:26:14 -0800, Roy Lewallen wrote: >dansawyeror wrote: >> Roy, >> >> Thank you. It is a quick experiment to build a test coax coil and >> measure the Q. That should produce enough evidence to test a counterpoise. >> >> In the mean time the research to build a coil out of copper tubing >> continues. So far the only alternative I can conceive is to make a >> wooden form and wrap the coil on the outside. John Ferrell W8CCW Article: 221717 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: using coax shield to create a loading coil ? Date: Sat, 18 Feb 2006 19:53:53 +0000 (UTC) Message-ID: References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <1140266886.722102.279930@z14g2000cwz.googlegroups.com> <11vecbaciouhcda@corp.supernews.com> "Roy Lewallen" wrote > I can assure the readers that all the effects I've discussed are soundly > based on very well known principles. ==================================== Roy, you seem to have forgotten proximity effect. If one calculates the Q of a coil from HF skin resistance of the wire and from coil inductance, one gets ridiculously high values of Q. Other producers of coil calculators appear to have forgotten this too. That's if they were ever aware of it. I have a coil, about 4 inches long, about 1.7 inches in diameter, with about 90 close-wound turns of 1mm diameter wire, which has an inductance of about 100 micro-henrys. The measured value of Q at 1.9 MHz is about 240. This makes the proximity effect about 3.5 or 4 times the effect of simple HF wire skin resistance. This is a large amount. This is the first time such information has been appeared on a newsgroup or published in bibles anywhere else. They didn't have Q meters 120 years ago, in Heaviside's time, when such factors were first considered. My findings are incorporated in program SOLNOID3 which estimates Q (and other characteristics) for coils of various dimensions. There are, of course, other factors which influence Q which is a relatively unimportant coil characteristic. What do you do with Q once you have taken the trouble to find it? The other more important things will already be apparent. ---- ........................................................... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp ........................................................... Article: 221718 of rec.radio.amateur.antenna From: dplatt@radagast.org (Dave Platt) Subject: Re: Under Eave Antenna Date: Sat, 18 Feb 2006 20:09:30 -0000 Message-ID: <11vevnqj56grv22@corp.supernews.com> References: In article , stananger wrote: >I would like to install an under eave antenna at my house. >Mostly for listening but maybe some low power transmitting. >What I need is some input from all the experts here as to what supports >would work best. Also I dont want it to upset the other half >too much so it would have to be "pleasing" to the eye or damn near >invisible. As I dont have any plans or thoughts as to what would work best >I am open to all serious suggestions and ideas. It will be a loop fed with >open wire feeders is all I have in mind at this time. One possibility would be the sort of standoffs which have traditionally been used to support runs of 300-ohm twinlead. They're something like a long-shafted eye bolt, with a plastic insert in the center of the eye to hold the twinlead away from the metal. Some of these are intended for mounting on a metal strap on an antenna mast (these have machine threads on their shaft), while others have a wood-screw-like threading and a sharp point and can be screwed into a wood beam. You'd probably want the latter. Radio Shack still sells 'em (15-853, $2.99 for a set of four) and they're probably available elsewhere as well. Another possibility would be to use short pieces of PVC or fiberglass tubing, mounted to the eves using one- or two-screw conduit brackets. 4"-6" pieces of 3/4" or 1" PVC tubing, with 1/4" holes cross-drilled in one end for the wire, then primed and painted to match your house's color scheme, might be cosmetically-acceptable. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! Article: 221719 of rec.radio.amateur.antenna From: Cecil Moore Subject: Re: Under Eave Antenna References: Message-ID: Date: Sat, 18 Feb 2006 20:44:16 GMT stananger wrote: > I would like to install an under eave antenna at my house. > Mostly for listening but maybe some low power transmitting. > What I need is some input from all the experts here as to what supports > would work best. Also I dont want it to upset the other half > too much so it would have to be "pleasing" to the eye or damn near > invisible. As I dont have any plans or thoughts as to what would work best > I am open to all serious suggestions and ideas. It will be a loop fed with > open wire feeders is all I have in mind at this time. I have used insulated cup hooks screwed into the wood through which I ran a loop of solid insulated wire. Both cup hooks and insulation are available in a number of colors. -- 73, Cecil http://www.qsl.net/w5dxp Article: 221720 of rec.radio.amateur.antenna From: "J. Mc Laughlin" Subject: Re: using coax shield to create a loading coil ? Date: Sat, 18 Feb 2006 16:24:43 -0500 Message-ID: <11vf40oo8o04l19@corp.supernews.com> References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <1140266886.722102.279930@z14g2000cwz.googlegroups.com> To extend from Ian's remarks: In some critical applications, the use of coax cable with braided outer Cu conductor can cause problems. RF charge flow (current) in the braid experiences a non-linear circuit resulting in harmonic distortion or IM or both. Just made coax can have a very low level of non-linearity with the effect increasing with age (and probable corrosion). Ag plated Cu braid seems to have less of the non-linear effect - perhaps because of a poorer mechanism for current to move from one wire to another. The effects are small, but can be important in certain applications. Solid Cu outer conductors have advantages beyond mechanical and power-handling. 73 Mac N8TT -- J. Mc Laughlin; Michigan U.S.A. Home: JCM@Power-Net.Net "Ian White GM3SEK" wrote in message > > Then it gets worse. Even the thinnest film of corrosion can disrupt the > contact between copper strands in a braid. Unless the current density is > large enough to break down this film, it means the RF current is forced > to flow into the interior of the braid. Again the exact geometry is hard > to visualize, but again the physics dictate that if an isolated > 'filament' of current is forced to flow beneath a conducting surface, > the voltage drop per unit length must increase - in other words, the RF > resistance must increase. > > Scientific deduction has told us that all these effects must exist. > Whatit cannot tell us is how big they are in real braid, or how > important they are in practice. For that we'll need some measured > numbers. > > You have two choices here: either look for existing measurements from > people who have demonstrated their competence and scientific approach; > or do it yourself. > > > > -- > 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) > http://www.ifwtech.co.uk/g3sek Article: 221721 of rec.radio.amateur.antenna Message-ID: <43F79B1F.86148B6A@sympatico.ca> From: Dave Holford Subject: Re: Under Eave Antenna References: Date: Sat, 18 Feb 2006 17:09:35 -0500 stananger wrote: > I would like to install an under eave antenna at my house. > Mostly for listening but maybe some low power transmitting. > What I need is some input from all the experts here as to what supports > would work best. Also I dont want it to upset the other half > too much so it would have to be "pleasing" to the eye or damn near > invisible. As I dont have any plans or thoughts as to what would work best > I am open to all serious suggestions and ideas. It will be a loop fed with > open wire feeders is all I have in mind at this time. > > Stan > AH6JR I have considered the same idea, however my soffit, fascia and eavestrough are all metal, and not electrically connected to each other. Does anyone have any suggestions on how far below these I would have to string my wire? I have also considered bonding both the individual sections of each, and bonding all three together; but I'm not sure how successful such an attempt would be. Dave VE3HLU Article: 221722 of rec.radio.amateur.antenna Message-ID: <43F7ACBD.2A293466@shaw.ca> From: Irv Finkleman Subject: Re: Under Eave Antenna References: Date: Sat, 18 Feb 2006 23:24:18 GMT stananger wrote: > > I would like to install an under eave antenna at my house. > Mostly for listening but maybe some low power transmitting. > What I need is some input from all the experts here as to what supports > would work best. Also I dont want it to upset the other half > too much so it would have to be "pleasing" to the eye or damn near > invisible. As I dont have any plans or thoughts as to what would work best > I am open to all serious suggestions and ideas. It will be a loop fed with > open wire feeders is all I have in mind at this time. > > Stan > AH6JR I ran a loop on the roof using plastic 35mm film containers as standoffs. I see no reason why you couldn't hang them under the eaves. A small dab of silicone seal was sufficient to glue the bottles in place. I used a soldering iron tip to put holes into and out of the bottles and strung the wire through them. My downlead was light 300 ohm twinlead. The antenna worked well and was up for a few years. A simple firm pull removed the bottles when I moved and didn't leave any damage on the shingles. I didn't worry about things like radiation angles, radiation resistance, UV deterioration and all those things that can drive you nuts if you think about them long enough -- I just worked loads of stations and had a pile of fun. It's cheap and simple. Irv VE6BP -- -------------------------------------- Diagnosed Type II Diabetes March 5 2001 Beating it with diet and exercise! 297/215/210 (to be revised lower) 58"/43"(!)/44" (already lower too!) -------------------------------------- Visit my HomePage at http://members.shaw.ca/finkirv/index.html Visit my Baby Sofia website at http://members.shaw.ca/finkirv4/index.htm Visit my OLDTIMERS website at http://members.shaw.ca/finkirv5/index.htm -------------------- Irv Finkleman, Grampa/Ex-Navy/Old Fart/Ham Radio VE6BP Calgary, Alberta, Canada Article: 221723 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: using coax shield to create a loading coil ? Date: Sat, 18 Feb 2006 23:50:11 +0000 (UTC) Message-ID: References: <11vc4ntcub1k91@corp.supernews.com> <1140207382.882131.11500@z14g2000cwz.googlegroups.com> <11vctsaacbdfa57@corp.supernews.com> <1140228740.616862.294460@g14g2000cwa.googlegroups.com> <11vd24h21qefea@corp.supernews.com> <1140266886.722102.279930@z14g2000cwz.googlegroups.com> <11vecbaciouhcda@corp.supernews.com> Q meters, above the range of 150, fall into the same category as so called SWR meters above the range of 1.5 Neither are of much use. Just an opinion! ---- Reg. Article: 221724 of rec.radio.amateur.antenna From: "Reg Edwards" Subject: Re: Under Eave Antenna Date: Sat, 18 Feb 2006 23:50:11 +0000 (UTC) Message-ID: References: <43F79B1F.86148B6A@sympatico.ca> Just hang up some wire and see what happens. I make no predictions but you will probably be pleased with results. ---- Reg. Article: 221725 of rec.radio.amateur.antenna From: John Ferrell Subject: Smith Chart vs JAVA Message-ID: Date: Sun, 19 Feb 2006 01:15:07 GMT In my quest to understand Smith Charts better I ran across a very interesting site at Agilent http://www.educatorscorner.com/index.cgi?CONTENT_ID=2482 Unfortunately, it would not work since I did not have Java on my machine. A little Google time later, I located the necessary free download (after reading of legal maneuvers) and installed it. Then the link worked fine. Nice site, very helpful. Since they offer a free download I thought it best to capture my own copy of this for the future. After I download & unzip it, it does not want to function from my machine. I would rather spend the computer time with antennas rather than trying to figure out this computer puzzle. Anyone have a SIMPLE solution? TIA, John Ferrell W8CCW Article: 221726 of rec.radio.amateur.antenna From: Fabian Kurz Subject: Re: Smith Chart vs JAVA Date: 19 Feb 2006 01:24:37 GMT Message-ID: <45pvmlF7tr9iU1@news.dfncis.de> References: John Ferrell wrote: > Since they offer a free download I thought it best to capture my own > copy of this for the future. After I download & unzip it, it does not > want to function from my machine. I would rather spend the computer > time with antennas rather than trying to figure out this computer > puzzle. Anyone have a SIMPLE solution? Just downloaded it and it works without any problems. Do you receive any error message? What do you see at the bottom of the "SChart.html" file (which is part of the ZIP-File), where the Java-Aplett should appear? 73, -- Fabian Kurz, DJ1YFK/AD5UR * Dresden, Germany * http://fkurz.net/ Online Log: http://dl0tud.tu-dresden.de/~dj1yfk/log.html Article: 221727 of rec.radio.amateur.antenna From: "Hal Rosser" References: Subject: Re: using coax shield to create a loading coil ? Message-ID: Date: Sat, 18 Feb 2006 20:31:18 -0500 > May I suggest the use of PVC pipe as a form for winding the > coax? Tape it in place while you're "monkeying" and then fiberglass it when > you like what you have. (Auto supply stores sell the fiberglass for doing > auto body work -- it's durable & light-weight.) > > KD6VKW > ET USN (ret) > > That sounds like a good idea - and I have used pvc forms myself. I wonder if ecasing the coil in that 'Great Stuff' spray foam would be ample protection ? my 2¢ W4PMJ Article: 221728 of rec.radio.amateur.antenna From: John Ferrell Subject: Re: Smith Chart vs JAVA Message-ID: References: <45pvmlF7tr9iU1@news.dfncis.de> Date: Sun, 19 Feb 2006 01:40:24 GMT Wow! That was quick. What I see is like what happens on a HTML download when a picture is missing. If I load the SChart.html file into FrontPage and split the view with the source, the part that is associated with the missing component is This experiment requires a Java-enabled Web Browser. I don't know if the system is unable to find the .jar or the .class files or cannot resolve them. My html skills are minimal and Java skills = nil. On 19 Feb 2006 01:24:37 GMT, Fabian Kurz wrote: >John Ferrell wrote: >> Since they offer a free download I thought it best to capture my own >> copy of this for the future. After I download & unzip it, it does not >> want to function from my machine. I would rather spend the computer >> time with antennas rather than trying to figure out this computer >> puzzle. Anyone have a SIMPLE solution? > >Just downloaded it and it works without any problems. Do you >receive any error message? What do you see at the bottom of the >"SChart.html" file (which is part of the ZIP-File), where the >Java-Aplett should appear? > >73, John Ferrell W8CCW Article: 221729 of rec.radio.amateur.antenna From: John Ferrell Subject: Re: Smith Chart vs JAVA Message-ID: References: <1pifv1l0gc9ji32ipa3os3gfma0en7nvn5@4ax.com> <9kjfv19mg4aikb97jtkbhe68vph8jb68mj@4ax.com> Date: Sun, 19 Feb 2006 02:34:51 GMT I am using Internet Explorer 6. That is probably the key. I will give it try on my Linux machine. I will look into another for windows as well. >>Hi John, >> >>I unzipped all the contents into a dedicated folder. >> >>I double-clicked the file SChart.html located inside the dedicated >>folder. Works fine for Firefox. > >Same here. Maybe he's using a non-standard browser like IE [g]. > >Another option: http://www.qsl.net/ac6la/xlzizl.html John Ferrell W8CCW Article: 221730 of rec.radio.amateur.antenna From: Allodoxaphobia Subject: Re: Smith Chart vs JAVA Date: 19 Feb 2006 03:09:57 GMT Message-ID: References: <1pifv1l0gc9ji32ipa3os3gfma0en7nvn5@4ax.com> <9kjfv19mg4aikb97jtkbhe68vph8jb68mj@4ax.com> On Sun, 19 Feb 2006 02:34:51 GMT, John Ferrell wrote: > >>>I unzipped all the contents into a dedicated folder. >>> >>>I double-clicked the file SChart.html located inside the dedicated >>>folder. Works fine for Firefox. >> >>Same here. Maybe he's using a non-standard browser like IE [g]. >> >>Another option: http://www.qsl.net/ac6la/xlzizl.html > I am using Internet Explorer 6. > That is probably the key. I will give it try on my Linux machine. > I will look into another for windows as well. Works Just Fine here in konqueror and Firefox on Mandrake 10.2. Jonesy -- Marvin L Jones | jonz | W3DHJ | linux 38.24N 104.55W | @ config.com | Jonesy | OS/2 *** Killfiling google posts: Article: 221731 of rec.radio.amateur.antenna From: "J. Mc Laughlin" Subject: Re: Thru-Wall Coax Feedthrus Date: Sun, 19 Feb 2006 00:32:11 -0500 Message-ID: <11vg0iqfnrkpkb1@corp.supernews.com> References: <11ucnvqhck36j37@corp.supernews.com> Dear Mr. Onella: Thanks for the suggestion. I shall ask for it in the local marine supply store. 73 Mac N8TT -- J. Mc Laughlin; Michigan U.S.A. Home: JCM@Power-Net.Net "Sal M. Onella" wrote in message news:N8eIf.57771$V.14814@fed1read04... > > "J. Mc Laughlin" wrote in message > news:11ucnvqhck36j37@corp.supernews.com... > > Foam is not a good idea. It can become very difficult to remove. > > > > Best is to use cooper wool. It will not rust nor flake and beasties > do > > not eat it. The stuff has become hard to find, however. > > Marine supply stores sell bronze wool. I think the boaters use it for > cleaning brass fittings, presumably without scratching them. (I used it as > EMI packing for some leaky old pull-boxes.) > > It may be close enough to copper wool to function the same. > > Article: 221732 of rec.radio.amateur.antenna From: "J. Mc Laughlin" Subject: Re: New 4nec2 version Date: Sun, 19 Feb 2006 00:42:35 -0500 Message-ID: <11vg168klbelde8@corp.supernews.com> References: <1138119779.443881.171030@g49g2000cwa.googlegroups.com> <43D82AEB.1020009@comcast.net> <1138275019.217196.21700@g14g2000cwa.googlegroups.com> <277Cf.110589$AP5.60182@edtnps84> <11usiuk6fbnjr3b@corp.supernews.com> Dear Frank: You are welcome. I too date to the day when one punched cards. However, for most tasks, I find Roy's front-end greatly to speed setting up the wires. For some really large models, I use card images. Computations are performed on a dedicated computer that has stone-simple hardware. 73 Mac N8TT -- J. Mc Laughlin; Michigan U.S.A. Home: JCM@Power-Net.Net "Frank" wrote in message news:tItHf.2202$Be4.1294@clgrps13... > Thanks for the info Mac. I see EZNEC/4 is $600.00, compared to $795.00 for > GNEC. I have no trouble with the FORTRAN compiled program on my PC, > although the graphics leave a little to be desired. Not sure what I will > do to improve it yet. Not that interested in GUI inputs, as I find it far > easier to edit NEC code directly. > > 73, > > Frank > VE6CB > > > "J. Mc Laughlin" wrote in message > news:11usiuk6fbnjr3b@corp.supernews.com... > > Dear Frank: > > I run licensed NEC4 with an appropriate version of EZNEC on a PC. The > > OS is W2000. > > 73 Mac N8TT > > > > -- > > J. Mc Laughlin; Michigan U.S.A. > > Home: JCM@Power-Net.Net > > "Frank" wrote in message > > news:277Cf.110589$AP5.60182@edtnps84... > > > >> > >> You mention "Nec 4 cards". Do you actually support NEC 4? I plan to > > obtain > >> a license for NEC 4 in the near future, and your interface looks > >> interesting; The cost of GNEC from Nittany Scientific is a bit high, but > >> thought it is probably the only way to run NEC 4 on a PC. > >> > >> Frank > >> > >> > > > > > >