Radiation Resistance
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Radiation Resistance
Tom, W8JI wrote:
"You didn`t read something correctly." OK, here is the arithmetic. Radiation Resistance of a Short Electric Dipole: RR = 80 pi squared (L/lambda)squared Constant = 80 (8.97) = 790 But a short monopole has 1/2 the resistance of a short dipole. 790 / 2 = 395 All Reg asked for was the constant. Best regards, Richard Harrison, KB5WZI |
Radiation Resistance
Owen Duffy wrote:
Is that for uniform current as Reg asked? Reg asked for "uniformly distributed current". I took that to mean having a constant slope. Wonder what Reg really meant? -- 73, Cecil http://www.qsl.net/w5dxp |
Radiation Resistance
On Sun, 12 Mar 2006 20:46:21 GMT, Cecil Moore wrote:
Owen Duffy wrote: Is that for uniform current as Reg asked? Reg asked for "uniformly distributed current". I took that to mean having a constant slope. Wonder what Reg really meant? In the context of his use, I think the most probably reasonable interpretation of Reg's words is that the current is uniform at all points on the radiator. Yes, it does also have a constant slope (of zero), so you will ba able to argue a correct interpetation either way, even if the results are different. It was interesting how many different interpretations were made, and then how many different answers to such a simple questions, even a text book incorrectly quoted (yes, subject to your interpretation of Richard's interpretation of what was in Reg's mind. Reg will no doubt chuckle when he wakes in the morning. Owen -- |
Radiation Resistance
Richard,
Your calculation is OK as far as it goes. However, you overlooked the fact that "L" is different for the dipole and the monopole. The monopole has 1/2 the length of the dipole or 1/4 the length squared. The coefficient Reg asked for is therefore 4 times the number you quoted. 73, Gene W4SZ Richard Harrison wrote: Tom, W8JI wrote: "You didn`t read something correctly." OK, here is the arithmetic. Radiation Resistance of a Short Electric Dipole: RR = 80 pi squared (L/lambda)squared Constant = 80 (8.97) = 790 But a short monopole has 1/2 the resistance of a short dipole. 790 / 2 = 395 All Reg asked for was the constant. Best regards, Richard Harrison, KB5WZI |
Radiation Resistance
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Radiation Resistance
Owen Duffy wrote:
Yes, it does also have a constant slope (of zero), so you will ba able to argue a correct interpetation either way, even if the results are different. OK, I will change my statement to a "constant non-zero slope". I really think that what's Reg meant but obviously only Reg's opinion is important on that matter. :-) Reg will no doubt chuckle when he wakes in the morning. :-) When I chuckle with a hangover, it hurts. :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Radiation Resistance
Richard Harrison wrote:
Owen Duffy wrote: "Is that for uniform current as Reg asked?" Reg described a short vertical wire above a perfect ground. Without a capacitive hat or some such device, you have an open circuit at the tip of the antenna (zero current) and a finite current at the driven end of the wire. How would the current be uniform end to end? For the answer to that, open your Kraus again, and go to the beginning of the chapter (5) you quoted from. It's explained in the first paragraph. There's even a picture, Fig. 5-1. Roy Lewallen, W7EL |
Radiation Resistance
Richard Harrison wrote:
Tom, W8JI wrote: "You didn`t read something correctly." OK, here is the arithmetic. Radiation Resistance of a Short Electric Dipole: RR = 80 pi squared (L/lambda)squared Constant = 80 (8.97) = 790 But a short monopole has 1/2 the resistance of a short dipole. 790 / 2 = 395 All Reg asked for was the constant. If you'll read more in the chapter of Kraus you're quoting, you'll notice that L is the length of the dipole, not the length of a monopole. Do the proper substitution and you'll get the correct answer. Roy Lewallen, W7EL |
Radiation Resistance
Roy, W7EL wrote:
"There`s even a picture Fig 5-1" Yes, exactly as I speculated. Reg`s question that I tried to answer was: "What is the value of the constant C?" My answer is 395 and I`nm sticking with it until someone shows me the error in my ways. Best regards, Richard Harrison, KB5WZI |
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