To: | [email protected] |
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Subject: | RE: LF: Re: "T" versus "L"aerial |
From: | "Dick Rollema" <[email protected]> |
Date: | Sat, 03 Jan 2004 12:30:09 +0100 |
In-reply-to: | <000001c3d009$acf7a620$c7e47f50@Smisan> |
References: | <[email protected]> <000001c3d009$acf7a620$c7e47f50@Smisan> |
Reply-to: | [email protected] |
Sender: | <[email protected]> |
To All from PA0SE, Mike, PC4M, wrote: At 02:50 1-1-04, you wrote: Dear Dick / Bob and Lofers, In the computer simulation no resistances were included. That means that the 1 kW fed to the aerial is completely radiated. Even an extremelly short vertical with no top load would do so and produce the calculated 29.9mV/m at 10km Bu the point raised by Bob, ZL2CA, was that the current in the single wire topload of the "L" would generate a horizontally polarised field. In the "T" the currents in the two topload wires flow in opposite directions so the horizontally polarised fields caused by these currents would at least partially cancel each other. The horizontally polarised field is radiated as a sky wave and the power in it detracts from that in the vertically polarised field of the ground wave. If the above reasoning were correct it could be expected that the "T" would produce a stronger ground wave than the "L" because less power disappears in the horizontally polarised sky wave. The simulation has shown that this is not the case. The subject of losses in the earth and surrounding objects has been treated very well by Jim, M0BMU, in his e-mail. 73, Dick, PA0SE -----Oorspronkelijk bericht----- |
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