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On Mon, 23 Aug 2004 22:26:34 GMT, Richard Clark
wrote: Yes. No one is going to notice something less than half a dB (which by definition is unnoticeable). This may be true for upper HF, but for MF/low-HF dipoles the situation may be a bit different. The dipole length is inversely proportional to frequency and the skin depth (and hence AC resistivity) is directly proportional to the square root of frequency. Thus the net effect is that the total dipole resistance is inversely proportional to the square root of the frequency. Thus a dipole cut for the lower band has a larger resistance than an antenna cut for a higher bands, provided that the same wire type is used. The resistivity and skin depth depends on the material and apparently varies quite a lot depending of the type of stainless steel, since various sources give quite different values. The skin depth for copper at 1.8 MHz is about 50 um, while for some stainless steel, it appears to be around 200 um. Both values are well below the OP's 850 um conductor diameter. One source claimed that the stainless steel has a 52 times DC resistivity compared to copper, so factoring in the large skin depth, the AC resistivity at 1.8 MHz would be more than 12 times that of the copper wire of the same size. An other source specified the stainless steel resistivity as 43E-8 ohm/m and while the 0.85 um diameter conductor with 0.2 mm skin depth would have an effective cross section of 0.4 mm2 and with the 80 m length of the 1.8 MHz dipole, the total resistance would be 86 ohms. Compare this to the nominal 73 ohm radiation resistances for a half wave dipole and more than half of the transmitter power would be dissipated in the losses. However, since the current distribution is not uniform along the dipole, the effective losses are not that quite as bad. On the other hand, if the dipole is close to the ground, the radiation resistance is well below 73 ohms, so again, we are in the -3 dB efficiency ballpark value. With a 1500 W transmitter, the losses are about 1 W/m, this should help melting any frost accumulated on the wire during a cold night:-). Now the question is, is this 80 m stainless steel strong enough to support itself, even if supported in the middle at the feed point. For a copper wire with the same diameter, the resistance would have been below 10 ohms and the losses about 0.5 dB. Paul OH3LWR |
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On Tue, 24 Aug 2004 09:56:29 +0300, Paul Keinanen
wrote: Compare this to the nominal 73 ohm radiation resistances for a half wave dipole and more than half of the transmitter power would be dissipated in the losses. Hi Paul, And still, no one would notice. In fact, I gave the loss 100 Xs copper, lowered the frequency to 1MHz and saw roughly 6dB loss - the "standard" S-Unit, against which propagation variations would toss that around to wider variation. I suppose it wouldn't do for EME, but I don't know any work at 1MHz for EME. Still, and all, this determination is easily within the scope of the free distribution of EZNEC which allows any user to make an informed decision about a choice being OK. 73's Richard Clark, KB7QHC |
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On Tue, 24 Aug 2004 09:56:29 +0300, Paul Keinanen
wrote: Two corrections: An other source specified the stainless steel resistivity as 43E-8 ohm/m The unit of resistivity is of course ohm m (not ohm/m). With a 1500 W transmitter, the losses are about 1 W/m, this should help melting any frost accumulated on the wire during a cold night:-). With 50 % efficiency (-3 dBd gain), the losses are about 800 W and when divided evenly along 80 m will give 10 W/m (not 1 W/m), which in addition to frost will also melt ice and keep the birds away :-). Paul OH3LWR |
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