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Re: Topband: COAX RX loop question.

To: "Jose M. Valdes R. YV5LIX" <yv5lix@cantv.net>,<topband@contesting.com>
Subject: Re: Topband: COAX RX loop question.
From: "Tom Rauch" <w8ji@contesting.com>
Reply-to: Tom Rauch <w8ji@contesting.com>
Date: Thu, 5 Feb 2004 13:29:48 -0500
List-post: <mailto:topband@contesting.com>
> resonance", but at 1.810 MHz 20 foot is ~0.037 lambda in the air, and
since
> the velocity factor for RG-58U is .66 shouldn't the actual overall length
of
> the coax be 13.2 feet? (20 foot X .066 = 13.2) electrically speaking for
> radio waves traveling through a RG-58U 13.2 foot will be 0.037 lambda at
> 1.810.

Jose,

The velocity factor of the cable has little to do with the electrical length
of the antenna. The coax doesn't not behave like coax in a loop.

The outside layer of the coax shield is the actual antenna. The jacket of
the coax would only affect wave velocity.

The conductor inside just is a transformer winding working against the
inside layer of the shield. The velocity factor length may affect the size
of the tuning capacitor, but not how the loop works (unless it will not
tune). The key to balance is all in the shield.

I have something at: http://www.w8ji.com/magnetic_receiving_loops.htm
but no construction details.

If you do a web search you will find many loop articles. Some good and some
bad. Good articles have the gap in the shield at the top, and the center
conductor exiting for tuning at the bottom where the shield is shorted
across from side to side. Bad articles show the shield gap near the bottom.

73 Tom


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