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#21
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![]() Richard Harrison wrote: 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 Is everyone from Texas like this? |
#22
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Roy, W7EL wrote:
"Do the proper substitutionn and you`ll get the correct answer." Yes. The warning also appears on page 137: "In developing the field expressions for the short dipole, which were used in obtaining (5-56), (5-56) is the value of radiation resistance, the restriction was made that lambda is much larger than the length of the dipole L." No problem there, Reg specified a short monopole. Kraus does a sample calculation for a short dipole. I used Kraus` data and got the same answer when duplicating his calculation. But Reg was not asking for an answer to a specific problem. Reg was asking for the value of the constant in a formula of the same form. Kraus gives it as 80 pi squared for a dipole.. This is 790. We know that a monopole has half the resistance of a dipole. Example: 73 ohms and 36.5 ohms. 790 / 2 = 395. That`s not a resistance, it is only the value of a constant which must be multiplied by (L/lambda) squared to give the radiation resistance of a very short monopole. Best regards, Richard Harrison, KB5WZI |
#23
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![]() Gene Fuller wrote: 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. 1.) He could have gotten length correct and assumed current was triangular. That would reduce radiation resistance by a factor of four. 2.) He could have assumed uniform current and gotten length wrong by a factor of two, and that would reduce radiation resistance by a factor of four. |
#25
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Gene Fuller wrote:
"However, you overlooked the fact that "L" is gifferent for the dipole and monopole." L is not a constant. L is a variable in another part of the formula. The difference in radiation resistance between a dipole and a monopole is a constant. It equals 2, not 4, not 8, or not 16. Best regards, Richard Harrison, KB5WZI |
#26
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Richard Harrison wrote:
We know that a monopole has half the resistance of a dipole. Example: 73 ohms and 36.5 ohms. 790 / 2 = 395. That`s not a resistance, it is only the value of a constant which must be multiplied by (L/lambda) squared to give the radiation resistance of a very short monopole. Does it matter that for a vertical that is 1/2 of the length of the dipole, (L/lamda)^2 is different by a factor of 4? -- 73, Cecil http://www.qsl.net/w5dxp |
#27
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Cecil, W5DXP wrote:
"Does it matter that for a vertical that the length of a dipole (L/lambda)squared is different by a factor of 4?" It doesn`t make a ratio different than two to one in the ratio of resistances of the 1/2-wave dipole to the 1/4-wave monopole. We are not comparing a monopole that is the the length of a dipole with the dipole. We are comparing a monopole that is 1//2 the length of a dipole to the dipole when we make the resistance ratio. The small dipole is working against a perfect ground in Reg`s specification. It would see its reflection in that perfect ground, so its equivalent length is doubled. Kraus` dipole is presumed to be in free space. Best regards, Richard Harrison, KB5WZI |
#28
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#29
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Richard Harrison wrote:
Cecil, W5DXP wrote: "Does it matter that for a vertical that the length of a dipole (L/lambda)squared is different by a factor of 4?" It doesn`t make a ratio different than two to one in the ratio of resistances of the 1/2-wave dipole to the 1/4-wave monopole. We are not comparing a monopole that is the the length of a dipole with the dipole. We are comparing a monopole that is 1//2 the length of a dipole to the dipole when we make the resistance ratio. Richard, Balanis doesn't say that the 'L' in the monopole formula is 1/2 the 'L' in the dipole formula. Does Kraus? -- 73, Cecil http://www.qsl.net/w5dxp |
#30
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Richard Harrison wrote:
Gene Fuller wrote: "However, you overlooked the fact that "L" is gifferent for the dipole and monopole." L is not a constant. L is a variable in another part of the formula. The difference in radiation resistance between a dipole and a monopole is a constant. It equals 2, not 4, not 8, or not 16. Best regards, Richard Harrison, KB5WZI Richard, I guess this must be the week for basic math explanations. Let's try it with numbers. The equation shown in Kraus "Antennas", 2nd edition, page 216, for the radiation resistance of a short dipole with constant current is: Rr = 80 pi^2 (L/lambda)^2 80 pi^2 is about 790, so the equation is rewritten as: Rr = 790 (L/lambda)^2 In the convention used by Kraus, "L" is the total length of the dipole. I presume the equivalent discussion is contained in the 1950 edition of "Antennas". As a test case, let's suppose that L is 8 meters, and lambda is 80 meters. We immediately see that Rr is 7.9 ohms. OK, now take the monopole over perfect ground that Reg mentioned. The monopole length that corresponds to one half of the test case dipole is 4 meters. The radiation resistance of the monopole is 3.95 ohms. So, the question becomes determining the correct coefficient for the Rr equation. (L/lambda) is now 0.05, not 0.10. Therefore, Rr = 3.95 ohms = X (0.05)^2 I believe you will discover that "X" must be 1580. This is set up using the definitions for L as stated in Kraus (dipole) and as stated by Reg in his monopole query. Of course you can set up your own rules, but that would be addressing a different problem. 73, Gene W4SZ |
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