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Reductio ad absurdum - short antennae do not radiate well
Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Now there's no difficulty in feeding all that power into that 1/4 inch because it is so short compared to a wavelength that there is a uniformity of voltage and current along it, and it will be essentially the same as that existing in the last gnat's cock of the coax. Attach a hi-impedance scope probe to the end of that 1/4 inch and all the power being delivered through the coax will be detectable right at the tip of that 1/4 inch. Now, will that 1/4 inch antenna radiate all the power that is being successfully fed to it at 3.6MHz, or will the configuration behave merely as an open-circuit with all the power being reflected back down the coax? A number of contributors to this NG claim that the 1/4 inch stub antenna will radiate the full 100W at 3.6MHz, but I fear that they are sadly mistaken and that their associated infantile outbursts are because they are in denial either about their error, or cannot face up to a challenge to their seemingly-religious faith as to what is happening. |
Reductio ad absurdum - short antennae do not radiate well
gareth wrote:
Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. A piece of coax is not an antenna. Now there's no difficulty in feeding all that power into that 1/4 inch because it is so short compared to a wavelength that there is a uniformity of voltage and current along it, and it will be essentially the same as that existing in the last gnat's cock of the coax. Wrong and shows an utter lack of understanding how antennas work. Attach a hi-impedance scope probe to the end of that 1/4 inch and all the power being delivered through the coax will be detectable right at the tip of that 1/4 inch. But not the current in that 1/4 inch piece, which is the important part. Now, will that 1/4 inch antenna radiate all the power that is being successfully fed to it at 3.6MHz, or will the configuration behave merely as an open-circuit with all the power being reflected back down the coax? The amount of power the will be successfully fed to it at 3.6MHz will be miniscule. A number of contributors to this NG claim that the 1/4 inch stub antenna will radiate the full 100W at 3.6MHz, If and only if the impedance of the 1/4 inch stub is matched to 50 Ohms without losses in the matching device and the resistance of the 1/4 inch stub is much less than the resistive input impedance of the 1/4 inch stub. This is something you just can not seem to understand no matter how many times it is repeated or shown to be true. but I fear that they are sadly mistaken and that their associated infantile outbursts are because they are in denial either about their error, or cannot face up to a challenge to their seemingly-religious faith as to what is happening. I fear you are incapable of understanding how any antenna works. -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
gareth wrote:
Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Now there's no difficulty in feeding all that power into that 1/4 inch because it is so short compared to a wavelength that there is a uniformity of voltage and current along it, and it will be essentially the same as that existing in the last gnat's cock of the coax. Attach a hi-impedance scope probe to the end of that 1/4 inch and all the power being delivered through the coax will be detectable right at the tip of that 1/4 inch. Now, will that 1/4 inch antenna radiate all the power that is being successfully fed to it at 3.6MHz, or will the configuration behave merely as an open-circuit with all the power being reflected back down the coax? A number of contributors to this NG claim that the 1/4 inch stub antenna will radiate the full 100W at 3.6MHz, but I fear that they are sadly mistaken and that their associated infantile outbursts are because they are in denial either about their error, or cannot face up to a challenge to their seemingly-religious faith as to what is happening. The resistive impedance of your 1/4 inch stub will be a small fraction of an Ohm. Let us assume 0.001 Ohms though it would likely be much smaller than that and I do not want to spend the effort on getting the actual value for your nonsense. Since P=I^2R and power in your scenario is 100 W and the total resistance is 50.001 Ohms, the current is 1.414 A. Again since P=I^R and the antenna has a resistive impedance of 0.001 Ohms, the power in the antenna is 0.002 W. Do look up impedance matching and why it is used. -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
"gareth" wrote in message
... Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? |
Reductio ad absurdum - short antennae do not radiate well
On 10/27/2014 7:38 PM, gareth wrote:
"gareth" wrote in message ... Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? You are right. Very little of the power reaching the end of that cable will be radiated. Most will be reflected back down the cable toward the source. When you add a 50 ohm termination you will still have very little radiated. Nearly all of the power will be dissipated in the resistor. I think you have a very clear picture of what is happening. -- Rick |
Reductio ad absurdum - short antennae do not radiate well
gareth wrote:
"gareth" wrote in message ... Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? Assuming an antenna impedance of 0.001 Ohms, about 0.002 W. 2. How much of that power is radiated by that short antenna? About 0.002 W. 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? Yes. 4. How much of the power is dissipated in the 50 ohm resistor? About 99.998 W. 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? Nil. Still no clue how antennas work, I see. -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
rickman wrote:
On 10/27/2014 7:38 PM, gareth wrote: "gareth" wrote in message ... Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? You are right. Very little of the power reaching the end of that cable will be radiated. Most will be reflected back down the cable toward the source. When you add a 50 ohm termination you will still have very little radiated. Nearly all of the power will be dissipated in the resistor. I think you have a very clear picture of what is happening. Nope, it just shows he does not understand the concept of impedance matching. -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
"Brian Reay" wrote in message
... He doesn't want to understand, he wants to stir up a long thread he can pepper with abuse Untrue. and claim he 'won', Untrue. Once again you project your own failings. Also untrue, because I am always willing to accept that I may be wrong, unlike yourself who regularly changes the subject in order to avoid being challenged; a strategy for which you are notorious. while others who disagreed with him didn't understand the subject. Untrue, although they might have misconstrued and shot off at a tangent, as you are wont to do. As strange as it sounds, this seems to give him some kind of 'buzz', perhaps his self esteem is so low that he is driven to such measures. Untrue. Perhaps he is just a nasty individual. Physician, heal thyself. Once again, Brian, you have jumped into a technical thread only to originate abusive remarks but without making any technical contribution. Why do you behave like that? |
Reductio ad absurdum - short antennae do not radiate well
"gareth" wrote in message ... Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? 6. Of those who claim that a short antenna will radiate all the power fed to it, how many will realise that for any power to be dissipated in the resistor, it must have been successfully fed to that short antenna in the first place? 7. Of those who suggest that impednace matching is a serious consideration, how many will realise that at 3.6MHz, that the 1/4" short antenna is the standard practice to connect the end of the coax to the dummy load with a bit of wire? |
Reductio ad absurdum - short antennae do not radiate well
Jeff wrote in :
However, if you do succeed in feeing it, either practically or mathematically by simulation, then all of the power that ends up in the short antenna will be radiated apart from that lost as heat in the resistance of the element. My imagination by now is going hyperbolic, I'm beginning to have visions of this Mythical Tiny Beast of an antenna, as a short fat gold cross, perhaps operating on the Hail Mary passband. Surreal, no? Miraculous certainly, if it could exist. But I suspect even the resistance of pure gold is too low for a hotline to God. :) |
Reductio ad absurdum - short antennae do not radiate well
"gareth" wrote in message
... Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? 6. Of those who claim that a short antenna will radiate all the power fed to it, how many will realise that for any power to be dissipated in the resistor, it must have been successfully fed to that short antenna in the first place? 7. Of those who suggest that impednace matching is a serious consideration, how many will realise that at 3.6MHz, that the 1/4" short antenna is the standard practice to connect the end of the coax to the dummy load with a bit of wire? Setting aside dielectric and slot antennae, an antenna is a conductor into which power has been passed. Once the power has entered that antenna, how it got there by feed, and how the antenna is terminated are irrelevant. In the example above, the short piece of wire coupling the coax to a dummy load is a short antenna, and yet it clearly is not radiating all the power being fed to it, as most of it will be dissipated in the dummy load. Nevertheless, it is a valid example of power being fed to a short antenna which is not all being radiated, because a short antenna does not radiate efficiently, which is where we came in. |
Reductio ad absurdum - short antennae do not radiate well
"gareth" wrote in message ... Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Now there's no difficulty in feeding all that power into that 1/4 inch because it is so short compared to a wavelength that there is a uniformity of voltage and current along it, and it will be essentially the same as that existing in the last gnat's cock of the coax. Attach a hi-impedance scope probe to the end of that 1/4 inch and all the power being delivered through the coax will be detectable right at the tip of that 1/4 inch. Now, will that 1/4 inch antenna radiate all the power that is being successfully fed to it at 3.6MHz, or will the configuration behave merely as an open-circuit with all the power being reflected back down the coax? A number of contributors to this NG claim that the 1/4 inch stub antenna will radiate the full 100W at 3.6MHz, snip In this example, the transmitter delivers 100w to the coax, but only a small part of that is delivered to the 1/4 inch "antenna". The 1/4 inch will radiate the power delivered to it just as well as a to a full sized antenna. Matching and radiation are two different subjects. |
Reductio ad absurdum - short antennae do not radiate well
gareth wrote:
"gareth" wrote in message ... Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? 6. Of those who claim that a short antenna will radiate all the power fed to it, how many will realise that for any power to be dissipated in the resistor, it must have been successfully fed to that short antenna in the first place? Nope. Consider a 100V source, a 1 Ohm resistor and a 100 Ohm resistor in series. The power dissipated by the 1 Ohm resistor is 0.99 W and the power dissipated in the 100 Ohm resistor is 99 W. Phrases like "power fed to it" are meaningless word salad. You can apply voltage and you can apply current, but you can NOT feed power. 7. Of those who suggest that impednace matching is a serious consideration, how many will realise that at 3.6MHz, that the 1/4" short antenna is the standard practice to connect the end of the coax to the dummy load with a bit of wire? Non sequitur. -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
gareth wrote:
"gareth" wrote in message ... Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? 6. Of those who claim that a short antenna will radiate all the power fed to it, how many will realise that for any power to be dissipated in the resistor, it must have been successfully fed to that short antenna in the first place? 7. Of those who suggest that impednace matching is a serious consideration, how many will realise that at 3.6MHz, that the 1/4" short antenna is the standard practice to connect the end of the coax to the dummy load with a bit of wire? Setting aside dielectric and slot antennae, an antenna is a conductor into which power has been passed. Once the power has entered that antenna, how it got there by feed, and how the Power does not enter an antenna, voltage and/or current enter an antenna. antenna is terminated are irrelevant. In the example above, the short piece of wire coupling the coax to a dummy load is a short antenna, and yet it clearly is not radiating all the power being fed to it, as most of it will be dissipated in the dummy load. As Ohms law says it should because of the impedance of the antenna compared to the impedance of the resistive load. Nevertheless, it is a valid example of power being fed to a short antenna There is no such things as "power being fed to" anything; you can only apply voltage or current. The only place were power is fed is in Star Trek. which is not all being radiated, because a short antenna does not radiate efficiently, which is You have been shown repeatedly that this is false. -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
Jeff wrote:
Actually, here is a better example, because it represents the situation found in many shacks. Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding, and thereby forming a short antenna. The short antenna, only 1/4 inch long is immediately terminated by a 50 ohm resistance. 1. How much of the power from the coax is fed into that short antenna despite the claimed (by others) impedance mismatch? 2. How much of that power is radiated by that short antenna? 3. If all the power that is fed to the short antenna is radiated, does the 50 ohm resistor dissipate any of it? 4. How much of the power is dissipated in the 50 ohm resistor? 5. How much of the power is reflected back down the coax because of the impedance mismatch of that (very) short antenna? It depends on how you connect the 50ohm load. The ground side of the 50 ohm load must go somewhere, so logic would dictate that it was connected to the outer of the coax. If that case the 1/4 inch is so small compared to a wavelength that it will not present much of a mismatch to the coax at 3.6MHz, the coax will see 50ohms in series with a very small inductance and even smaller stray capacitance, virtually all of the power will be dissipated in the load, a small ammopunt will be reflected. ie you have just added a small lumped inductor in series with the 50ohms. To be pendatic, the short wire has a small resistance and a small impedance, both of which are in milliohms. Since P=I^2R and the R is very, very small, the P dissipated both by radiation and heat are very small. This example has nothing to do with how short whips radiate. You keep harping back to feed lines which are not what is under discussion. We all agree that practically matching short whips is where the problem lies. However, there is unequivocal proof that IF power is got into a short whip by some means or other it is ALL radiated or lost as heat in the resistance of the element. Read Kraus, Jasik, or do some NEC simulations yourself and you will see and hopefully stop this ridiculous quest for an answer that is just not there. Jeff -- Jim Pennino |
Reductio ad absurdum - short antennae do not radiate well
Wayne wrote:
"gareth" wrote in message ... Consider 100W at 3.6MHz propagating along some 50 ohm coax, which terminates suddenly but with 1/4 inch of the central conductor protruding. Now there's no difficulty in feeding all that power into that 1/4 inch because it is so short compared to a wavelength that there is a uniformity of voltage and current along it, and it will be essentially the same as that existing in the last gnat's cock of the coax. Attach a hi-impedance scope probe to the end of that 1/4 inch and all the power being delivered through the coax will be detectable right at the tip of that 1/4 inch. Now, will that 1/4 inch antenna radiate all the power that is being successfully fed to it at 3.6MHz, or will the configuration behave merely as an open-circuit with all the power being reflected back down the coax? A number of contributors to this NG claim that the 1/4 inch stub antenna will radiate the full 100W at 3.6MHz, snip In this example, the transmitter delivers 100w to the coax, but only a small part of that is delivered to the 1/4 inch "antenna". The 1/4 inch will radiate the power delivered to it just as well as a to a full sized antenna. Matching and radiation are two different subjects. Correct but it is much easier to see if one looks at it in terms of voltage and current and calculates the power at any given point. The equivelant circuit to a 100 W, 50 Ohms transmitter is a 141.4 V voltage source with a 50 Ohm resistor. The impedance of this antenna is a very small value, assume 0.001 Ohms as a reasonable value for analysis. There is 141.4 V applied to 50.001 Ohms, so the current is 141.4/50.001. The power in the antenna is I^2R, which is (141.4/50.001)^2*0.001, which my calculator says is 0.008 W. Where is the other 99.992 W? It is dissipated in the internal resistance of the transmitter and given real world coax, some of it is dissipated as heat in the coax because of the huge VSWR. -- Jim Pennino |
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