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Dual-Z0 Stubs
On May 9, 1:56*pm, wrote:
Tom, OK, I tried what you suggested. I put my loading coil midway up a 20ft vertical wire in the EZNEC model. I reduced the number of turns to lift the resonant frequency to 5.6MHz. EZNEC predicted that the magnitude of the current at the top of the coil would be 77% of the magnitude at the bottom. Then I removed the coil in the model, replaced it with a straight wire containing an EZNEC lumped load, and adjusted that load for antenna resonance at 5.6MHz again. I needed +j1630. Given the dimensions of the coil, the Corum calculator predicted a lumped circuit equivalent reactance of *+j1573, and it predicted a current fall-off across the coil of 78%. Hi Steve, OK, I'm wondering now exactly what "Corum calulator" you are using that predices "a current fall-off across the coil of 78%." The inductance calculator on the HamWaves website that I thought we were talking about doesn't seem to say anything about "current fall-off" in coils, though perhaps I'm missing it. Cheers, Tom |
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Tom Ring wrote:
And denigrating slide rules is silly. Most of the world that surrounds you was calculated with a slide rule's resolution. When used properly they give answers that are as accurate as is needed for engineering. I was one of the last classes in school to use a slide rule - they went to calculators the next year. I have to say that using a slide rule changed my outlook on math in all it's forms. Took a absolute idiot at math to the dilettante I am today. 8^) I use calculators all the time now, but I still have a slide rule that I use in the garage.... - 73 de Mike N3LI - |
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Hi Tom,
I should have been more explicit. I took the "Axial Propagation Factor" (4.372 rad/m) figure which was given by the HamWaves calculator and multiplied it by the coil length (155mm) to find the effective electrical length of the coil (38.83 degrees). Then I took cos(38.83)=0.779 as the fall-off in current across the coil. 73, Steve G3TXQ On May 11, 2:46*am, K7ITM wrote: Hi Steve, OK, I'm wondering now exactly what "Corum calulator" you are using that predices "a current fall-off across the coil of 78%." *The inductance calculator on the HamWaves website that I thought we were talking about doesn't seem to say anything about "current fall-off" in coils, though perhaps I'm missing it. Cheers, Tom |
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wrote:
I took the "Axial Propagation Factor" (4.372 rad/m) figure which was given by the HamWaves calculator and multiplied it by the coil length (155mm) to find the effective electrical length of the coil (38.83 degrees). Then I took cos(38.83)=0.779 as the fall-off in current across the coil. Interesting. W8JI's coil through which he measured a 3 nS delay was 100t, 2" dia, 10" long, #18 wire. http://www.w8ji.com/inductor_current_time_delay.htm Converting everything to metric and entering the data into the HamWaves calculator at 4 MHz, yields a calculated delay of 21.5 nS through the W8JI coil and a VF of ~0.04 at 4 MHz. So which are we to believe? W8JI's measurements or ON4AA's calculator. There's a 7x difference between 3 nS and 21 nS. -- 73, Cecil, IEEE, OOTC, http://www.w5dxp.com |
Dual-Z0 Stubs
On May 11, 2:26*am, wrote:
Hi Tom, I should have been more explicit. I took the "Axial Propagation Factor" (4.372 rad/m) figure which was given by the HamWaves calculator and multiplied it by the coil length (155mm) to find the effective electrical length of the coil (38.83 degrees). Then I took cos(38.83)=0.779 as the fall-off in current across the coil. 73, Steve G3TXQ Hi Steve, OK, so I suppose you are assuming that the current distribution will follow a cosine along electrical degrees of your antenna, with a maximum at the base/feedpoint. If that's the case, then would you not account for the bottom 10 feet of wire, about 20.5 electrical degrees? If I do that and assume 1 amp at the feedpoint, I should see about .9367 amps at 20.5 degrees and 0.5101 amps at (20.5+38.83) electrical degrees. 0.5101/.9367 would then be the ratio of currents between the ends of the coil, and that's 0.5446, only a 45.54 percent fall-off. In fact, it seems to me that the idea of cos(38.83 degrees) = .779 would imply a fall-off of 22.1%... and that tells me that perhaps I'm still not understanding your model very well. Maybe you are NOT assuming the current along the electrical degrees of the antenna, up from the feedpoint, will have a cosine distribution. At this point, I have to say that I'm just not at all sure what your model really is. Perhaps you are making different assumptions about the current distribution... Also, if you still have the model around, try adding a top hat to the upper wire. For simplicity, you can just use a simple "T" structure, where the top horizontal wire is, say, five feet long total. With such a configuration, what's the current distribution along the radiating element going to be? Of course, what I'm suggesting here is that one must be careful to test ones models at corner cases before putting too much faith in them, and even then, one must always be wary of cases where the model may go awry. Cheers, Tom |
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K7ITM wrote:
OK, so I suppose you are assuming that the current distribution will follow a cosine along electrical degrees of your antenna, with a maximum at the base/feedpoint. This is a good assumption for horizontal 1/2WL thin-wire dipoles as presented by Kraus. It doesn't seem to be valid for loaded vertical antennas where there is an instantaneous phase shift at the impedance discontinuities. There is a definite change in the slope of the current profile at such boundaries. And there is the nagging current bulge in the loading coil causing a rise in current in adjacent turns. Normally a current maximum would indicate a purely resistive impedance but that doesn't seem to be the case inside a loading coil. Years ago, I gave up on the current cosine argument for loaded mobile antenna current in favor of loading the coil with its characteristic impedance and using traveling wave current to measure the electrical length of the coil. -- 73, Cecil, IEEE, OOTC, http://www.w5dxp.com |
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On Sun, 10 May 2009 14:52:06 -0700, "Tom Donaly"
wrote: The presence of anything at all near the coil should lower its resonant frequency. Even the measuring apparatus should have an effect. I think it would require very careful planning and implementation to find an exact resonant frequency. You'd have to ask Richard Clark how to do it if you wanted high accuracy. I'm unwilling to find fault with either Corum or EZNEC at this point. Making accurate models can be just as hard as making valid experiments, and I wish you luck working with your models. I do urge you to make a coil to test your results against, though. It isn't difficult, and with a little help from some of your fellow experimenters, you should get results that are very close to being meaningful. 73, Tom Donaly, KA6RUH Hi Tom, Thanx for the flowers. In point of fact, inductor and capacitor standards are shielded. They are three terminal devices. To remove the effects of the shield you drive it at the same potential (which is to say the shield is floating with respect to everything/one around it). For the practicality of things, Reggie was never very far off the mark and you following him as an exemplar is suitable to other's inventions of proximities that have no defining moment in their references. I can well guess the remainder of your method as it was well defined in most Ham manuals (derived from conventional EE methods) when I read up on it 40 odd years ago. If Cecil every gets over this intellectual pebble in the path, please quote it so that I can see how much strain that ascent took. 73's Richard Clark, KB7QHC |
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Richard Clark wrote:
I note how little Corrum really has to offer when you had to take the same: effective electrical length of the coil (38.83 degrees) and change it (to the same effective electrical length? I think not.) to fit the same available wire, at the same specific frequency - only at a different height along the available wire. Richard, I explained that phenomenon in a posting last week which you obviously didn't read. Please go back and read my posting of 5-9-09 at 1:08pm to this thread. It is also explained on my web page at: http://www.w5dxp.com/shrtstub.htm -- 73, Cecil, IEEE, OOTC, http://www.w5dxp.com |
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Tom,
Firstly, I'm guilty of a "sloppy" choice of words. Whenever I've been using the phrase "drop off in current" I've meant the current at the top of the coil as a percentage of the current at the bottom. So when I've quoted 70% the current will have reduced by 30%. Apologies! Secondly, you're testing the limits of my understanding with the overall current distribution from base section, through the coil, to the top section. However I think the point is that you can't simply "add electrical degrees" through the various sections when the characteristic impedances of the sections are so disparate. That was Cecil's point in the very first posting. We also know that, as expected, summing the "degrees" for the three sections gets nowhere near a total of 90 degrees, so clearly you can't assume a cosine distribution that is contiguous across all three sections. I'll investigate what happens with a "top hat". 73, Steve G3TXQ On May 11, 6:56*pm, K7ITM wrote: Hi Steve, OK, so I suppose you are assuming that the current distribution will follow a cosine along electrical degrees of your antenna, with a maximum at the base/feedpoint. *If that's the case, then would you not account for the bottom 10 feet of wire, about 20.5 electrical degrees? *If I do that and assume 1 amp at the feedpoint, I should see about .9367 amps at 20.5 degrees and 0.5101 amps at (20.5+38.83) electrical degrees. *0.5101/.9367 would then be the ratio of currents between the ends of the coil, and that's 0.5446, only a 45.54 percent fall-off. In fact, it seems to me that the idea of cos(38.83 degrees) = .779 would imply a fall-off of 22.1%... and that tells me that perhaps I'm still not understanding your model very well. *Maybe you are NOT assuming the current along the electrical degrees of the antenna, up from the feedpoint, will have a cosine distribution. *At this point, I have to say that I'm just not at all sure what your model really is. Perhaps you are making different assumptions about the current distribution... Also, if you still have the model around, try adding a top hat to the upper wire. *For simplicity, you can just use a simple "T" structure, where the top horizontal wire is, say, five feet long total. *With such a configuration, what's the current distribution along the radiating element going to be? Of course, what I'm suggesting here is that one must be careful to test ones models at corner cases before putting too much faith in them, and even then, one must always be wary of cases where the model may go awry. Cheers, Tom |
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