Return-Path: Received: (qmail 9482 invoked from network); 15 Jul 2001 09:24:03 -0000 Received: from unknown (HELO warrior-inbound.servers.plus.net) (212.159.14.227) by excalibur-qfe1-smtp-plusnet.harl.plus.net with SMTP; 15 Jul 2001 09:24:03 -0000 Received: (qmail 26618 invoked from network); 15 Jul 2001 09:23:06 -0000 Received: from unknown (HELO post.thorcom.com) (212.172.148.70) by warrior with SMTP; 15 Jul 2001 09:23:06 -0000 Received: from majordom by post.thorcom.com with local (Exim 3.16 #2) id 15Li3Z-0007k2-00 for rsgb_lf_group-outgoing@blacksheep.org; Sun, 15 Jul 2001 10:19:17 +0100 Received: from lepidachrosite.lion-access.net ([212.19.217.3]) by post.thorcom.com with esmtp (Exim 3.16 #2) id 15Li3Y-0007jx-00 for rsgb_lf_group@blacksheep.org; Sun, 15 Jul 2001 10:19:16 +0100 Received: from w8k3f0 (Rogue.Cartoons.dialin.freeler.nl [212.26.212.88]) by lepidachrosite.lion-access.net (I-Lab) with SMTP id 1B276CB0B7 for ; Sun, 15 Jul 2001 09:16:39 +0000 (GMT) Message-ID: <003501c10d0f$e8050c40$58d41ad4@w8k3f0> From: "Dick Rollema" To: rsgb_lf_group@blacksheep.org References: <004501c10cb2$a8fab840$828801d4@g4jnt> Subject: LF: Re: Re: Near field of 9kHz antenna IV Date: Sun, 15 Jul 2001 11:23:21 +0200 MIME-Version: 1.0 Content-Type: text/html; charset=windows-1252 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 5.00.2314.1300 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group Sender: Content-transfer-encoding: 8bit

To All from PAoSE

Andy, G4JNT wrote:

Wait a minute !

I = SQR(W / R) so current in the loop = 51.2A and voltage at the feed is this current in the 60 ohms reactance, ie. 3077V

Andy G4JNT

-----Original Message-----
From: Dick Rollema <d.w.rollema@freeler.nl>

To All from PA0SE

AO gives the feed point impedance as Z = 0.380 + j60.1 ohm and the efficiency as 0.03%.

So for 1 kW fed into the loop the current must be SQR(0.380 * 1000) = 19.5A.

The voltage at the feed point becomes 19.5 * 0.380 = 1172V.

Thanks Andy!

I made the same stupid mistake in my message nr. III on this subject (14-07-01 10.37)

The corrected figures are even more dramatic than for the loop.

To feed 1kW into the 180m high 15mm copper pipe the current must be 154A and the voltage on the antenne becomes 2.5MV! Perhaps the pipe would survive the current but the voltage makes the system completely unrealistic (corona).

And what would the required 286mH loading coil for that voltage and current look like ...

73, Dick, PAoSE