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[195.171.43.25]) by mx.google.com with ESMTP id d16si30132272wiv.38.2014.06.25.17.27.08 for ; Wed, 25 Jun 2014 17:27:09 -0700 (PDT) Received-SPF: none (google.com: owner-rsgb_lf_group@blacksheep.org does not designate permitted sender hosts) client-ip=195.171.43.25; Authentication-Results: mx.google.com; spf=neutral (google.com: owner-rsgb_lf_group@blacksheep.org does not designate permitted sender hosts) smtp.mail=owner-rsgb_lf_group@blacksheep.org; dkim=pass header.i=@mx.aol.com; dmarc=pass (p=REJECT dis=NONE) header.from=aol.com Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1WzxCo-0000Pw-Vr for rs_out_1@blacksheep.org; Thu, 26 Jun 2014 01:06:22 +0100 Received: from [195.171.43.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1WzxCo-0000Pn-Gc for rsgb_lf_group@blacksheep.org; Thu, 26 Jun 2014 01:06:22 +0100 Received: from omr-d10.mx.aol.com ([205.188.108.134]) by relay1.thorcom.net with esmtps (TLSv1:DHE-RSA-AES256-SHA:256) (Exim 4.82) (envelope-from ) id 1WzxCl-0001gu-Ie for rsgb_lf_group@blacksheep.org; Thu, 26 Jun 2014 01:06:21 +0100 Received: from mtaout-mbc02.mx.aol.com (mtaout-mbc02.mx.aol.com [172.26.221.142]) by omr-d10.mx.aol.com (Outbound Mail Relay) with ESMTP id D93B9700496C5 for ; Wed, 25 Jun 2014 20:06:16 -0400 (EDT) Received: from White (95-91-237-52-dynip.superkabel.de [95.91.237.52]) by mtaout-mbc02.mx.aol.com (MUA/Third Party Client Interface) with ESMTPA id 345343800008B for ; Wed, 25 Jun 2014 20:06:16 -0400 (EDT) Message-ID: <7A28FC8C2A664ABF9FA2291582E1D911@White> From: "Markus Vester" To: References: <53A97738.2070501@gmail.com> <8D15DE5531A8822-27BC-8EBC@webmail-m293.sysops.aol.com> <1799375869.9754147.1403629934148.JavaMail.root@comcast.net> <751419392.10617770.1403714825805.JavaMail.root@comcast.net> <53AB3EB8.5060705@gmail.com> Date: Thu, 26 Jun 2014 02:06:19 +0200 MIME-Version: 1.0 X-Priority: 3 X-MSMail-Priority: Normal Importance: Normal X-Mailer: Microsoft Windows Live Mail 12.0.1606 X-MimeOLE: Produced By Microsoft MimeOLE V12.0.1606 x-aol-global-disposition: G DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mx.aol.com; s=20121107; t=1403741176; bh=7BQ7+TSnm3Wz1EzZWpEV/GM/E8+0RWEW6+l+OzNWa7k=; h=From:To:Subject:Message-ID:Date:MIME-Version:Content-Type; b=fv56K/rqyQQiEmVFNWkIjP7gabiPS+Yn3sZ/SLgignvfi436ONP0hzpefD9BLM7z1 bghg6cvnTUin7MC/PswUWRbNzKFzeUJZRQ3n+sMwHWYx//c8olBB4xXAx8aPMzHi/6 9pf2TW2pdW0Q/SQuH5Ovjzdm9Vxhrv+/sR0fU2wk= x-aol-sid: 3039ac1add8e53ab63f87972 X-AOL-IP: 95.91.237.52 X-Spam-Score: 1.0 (+) X-Spam-Report: Spam detection software, running on the system "relay1.thorcom.net", has identified this incoming email as possible spam. The original message has been attached to this so you can view it (if it isn't spam) or label similar future email. If you have any questions, see the administrator of that system for details. Content preview: Mark, Tom, let me try to put some numbers to your question... Loop resistance: Wire diameter #8 AWG = 3.26 mm, skin depth at 8.3 kHz ~ 0.66 mm, AC cross section 6.7 mm^2, AC resistance 2.6 mohm/m, loop perimeter 244m, => wire resistance 0.62 ohm. Depending on ground conductivity, there will be additional eddy current losses in the ground, maybe on the order of an ohm. Plus some capacitor ESR and connection losses. Assume 2 ohms total resistance, giving i = 10 A from 200 watts. [...] Content analysis details: (1.0 points, 5.0 required) pts rule name description ---- ---------------------- -------------------------------------------------- -0.0 RCVD_IN_DNSWL_NONE RBL: Sender listed at http://www.dnswl.org/, no trust [205.188.108.134 listed in list.dnswl.org] 0.0 FREEMAIL_FROM Sender email is commonly abused enduser mail provider (markusvester[at]aol.com) -0.0 T_RP_MATCHES_RCVD Envelope sender domain matches handover relay domain -0.0 SPF_PASS SPF: sender matches SPF record 0.0 HTML_MESSAGE BODY: HTML included in message 0.0 T_DKIM_INVALID DKIM-Signature header exists but is not valid 1.0 FREEMAIL_REPLY From and body contain different freemails X-Scan-Signature: c6cbc5ce44df878784f4cb3a3fba4d0e Subject: Re: LF: VLF in Canada - earth loop Content-Type: multipart/alternative; boundary="----=_NextPart_000_0025_01CF90E3.38355930" X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.6 required=5.0 tests=HTML_40_50,HTML_MESSAGE, MISSING_OUTLOOK_NAME autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false Dies ist eine mehrteilige Nachricht im MIME-Format. ------=_NextPart_000_0025_01CF90E3.38355930 Content-Type: text/plain; charset="utf-8" Content-Transfer-Encoding: quoted-printable Mark, Tom, let me try to put some numbers to your question... Loop resistance: Wire diameter #8 AWG =3D 3.26 mm, skin depth at 8.3 kHz = ~ 0.66 mm, AC cross section 6.7 mm^2, AC resistance 2.6 mohm/m,=20 loop perimeter 244m, =3D> wire resistance 0.62 ohm. Depending on ground conductivity, there will be additional eddy current = losses in the ground, maybe on the order of an ohm. Plus some capacitor = ESR and connection losses. Assume 2 ohms total resistance, giving i =3D = 10 A from 200 watts. Loop area: a=3D3720 m^2, magnetic field at distance r towards the side = of the loop:=20 H =3D i*a / (4*pi*r^3) ~ 3000 Am^2 / r^3 Noise background at 9 kHz:=20 Approximately 3 to 10 uV/m/sqrt(Hz), but quite variable depending on = static background, time of day, noise blanking etc. Assuming 5 = uV/m/sqrtHz.=20 CW would be readable with 9 dB SNR in 50 Hz bandwidth =3D> required = signal Emin =3D 14 + 9 + 17 dB uV/m =3D 40 dBuV/m =3D 100 uV/m Assuming a magnetic receive antenna, this is equivalent to Hmin =3D Emin = / 377 ohm =3D 0.26 uA/m Solving for distance: rmax =3D (3000/0.26e-6)^0.33 m =3D 2.24 km=20 ... Stefan, nice estimate indeed! In reality, the receive loop would also need to be horizontal (ie = vertical axis), and may thus benefit from picking up much less of the = atmospheric noise. 20 dB lower noise would extend the range by a factor = of 2.15. Furthermore, G3XBM and others have demonstrated that in inhabited areas, = the field of the TX loop may be picked up by infrastructure like buried = cables or water conduits, inducing currents which may carry the signal = to much further distances in "utility aided earth mode". But... Though a horizontal loop (vertical axis) may have significant inductive = nearfields, it will not radiate much into the far field because it has = the wrong polarization. The associated horizontal E-fields are simply = being shunted by the ground. To "get out", you'd need to put the loop in = upright orientation (ie horizontal axis). Assuming the same loop area as = before, the radiation resistance (including the loop "image" below = conducting ground) would be Rrad =3D 2 * 31171 ohms * a^2 / lambda^4 =3D 0.5 microohms and radiated power EMRP =3D i^2 * Rrad =3D 50 uW But that would require a much larger mechanical construction, using = either very high supports or a much lengthier loop.=20 Employing the existing supports for an upright loop, the area would be = only eg. 610 m^2, giving Rrad =3D 0.013 microohms EMRP =3D 1.3 uW. A second option would be an earth antenna, grounded at the far end. You = might feed one corner of your rectangle against ground, and connect the = opposite corner to ground as well. This would be an earth loop of length = 61m*sqrt(2), and height 10 m above ground plus say 48 m below ground = (depending on conductivity) =3D> a=3D 5000 m^2. But here the ground = reflection is already "included" so there won't be the extra factor of = two for the image. Thus Rrad =3D 31171 ohms * a^2 / lambda^4 =3D 0.45 microohms However the total resistance of the two ground connections will be much = higher than the 2 ohms of the copper loop, say 200 ohms. So 200 watts = would radiate only=20 EMRP =3D 0.45 uW. You will be much better off using the existing horizontal loop as a top = load for a vertical "Marconi". Assuming there are only few and small = trees around it, effective height may be perhaps 80% of 10m, leading to Rrad =3D 1579 ohms * 8m^2/lambda^2 =3D 76.8 microohms Optimistically assuming around 200 ohms for coil and ground losses, you = coud get 1 A, giving EMRP =3D 77 uW, some two orders of magnitude stronger than the magnetic and earth loop = configurations. With around 1.4 nF capacitance and 30 kV voltage = withstand, you could then even QRO up to 800 Watts for 308 uW EMRP. = Which will easily put you in the ballpark of Dex' and Paul's successful = TA crossing. Bottom line: Winding a large coil will be well worth the effort.=20 And... if you want to go far, just forget about aural CW! This is only = for the biggies like SAQ, radiating about 30 kW. Best 73, Markus (DF6NM) From: DK7FC=20 Sent: Wednesday, June 25, 2014 11:27 PM To: rsgb_lf_group@blacksheep.org=20 Subject: Re: LF: VLF in Canada - earth loop Hello Mark,=20 200W in such a loop and 8.97 or 8.27 kHz, CW, my guess is that the = distance is less than 2 km. But it depends on the type of the RX antenna = (E or H field). 73, Stefan/DK7FC Am 25.06.2014 18:47, schrieb mbdittmar@comcast.net:=20 Hi -=20 I've been following the discussion on 9 khz transmit antennas with = some interest, as a friend of mine, Tom, KD0VBR, has done some = preliminary experimentation in this area ( using large loading coils w/ = vertical ). Tom asked me to post a query regarding some antenna ideas = he has, and it would be great to hear the group's thoughts on them and = whether they might work or not. 73, Mark AB0CW Below are his questions: -------------------------------------------------------------------------= ----- From: "Pouliot, Tom" To: "Dittmar, Mark" Sent: Wednesday, June 25, 2014 8:51:48 AM Subject: RE: LF: VLF in Canada - earth loop Hi Mark,=20 That would be great if you could post some questions about = transmitting loops for VLF. Thinking about a square loop 200' per side 10' off the ground #8 wire = with 200 watts into the antenna. How far away could one expect to be = able to detect a CW signal ? ( not QRSS) Another possibility would be a triangular loop 200' per side with a = 50' high apex. Tom -------------------------------------------------------------------------= ----- ------=_NextPart_000_0025_01CF90E3.38355930 Content-Type: text/html; charset="utf-8" Content-Transfer-Encoding: quoted-printable
 
Mark, Tom,
 
let me try to put some numbers to your=20 question...
 
Loop resistance: Wire diameter #8 AWG =3D 3.26 mm, skin=20 depth at 8.3 kHz ~ 0.66 mm, AC cross = section 6.7=20 mm^2, AC resistance 2.6 mohm/m,
loop perimeter 244m, =3D> wire = resistance 0.62=20 ohm.
Depending on ground=20 conductivity, there will be additional eddy current losses in the = ground,=20 maybe on the order of an ohm. Plus some capacitor ESR and = connection=20 losses. Assume 2 ohms total resistance, giving i =3D 10 A from = 200=20 watts.
 
Loop area: a=3D3720=20 m^2, magnetic field at distance r towards the side of the loop:=20
 H =3D i*a /=20 (4*pi*r^3) ~ 3000 Am^2 / r^3
 
Noise background at = 9 kHz:=20
Approximately 3 to = 10=20 uV/m/sqrt(Hz), but quite variable depending on static = background, time=20 of day, noise blanking etc. Assuming 5 = uV/m/sqrtHz.=20
CW would be readable = with 9=20 dB SNR in 50 Hz bandwidth =3D> required signal
 Emin =3D = 14 + 9 + 17 dB=20 uV/m =3D 40 dBuV/m =3D 100 uV/m
Assuming a magnetic = receive=20 antenna, this is equivalent to Hmin =3D Emin / 377 ohm =3D 0.26 = uA/m
 
Solving for=20 distance:
 rmax =3D = (3000/0.26e-6)^0.33=20 m =3D 2.24 km 
 
... Stefan, nice estimate indeed!
 
In reality, the = receive loop=20 would also need to be horizontal (ie vertical axis), and may thus = benefit from=20 picking up much less of the atmospheric noise. 20 = dB lower noise=20 would extend the range by a factor of 2.15.
 
Furthermore, G3XBM = and others=20 have demonstrated that in inhabited areas, the field of the TX = loop=20 may be picked up by infrastructure like buried cables = or water=20 conduits, inducing currents which may carry the signal to much = further=20 distances in "utility aided earth mode".
 
 
But...
 
Though a = horizontal loop=20 (vertical axis) may have significant inductive nearfields, it will not=20 radiate much into the far field because it has the wrong=20 polarization. The associated horizontal E-fields are simply being = shunted=20 by the ground. To "get out", you'd need = to=20 put the loop in upright orientation (ie horizontal axis). = Assuming the=20 same loop area as before, the radiation resistance (including the loop=20 "image" below conducting ground) would be
 Rrad =3D = 2 * 31171 ohms=20 * a^2 / lambda^4 =3D 0.5 microohms
and radiated = power
 EMRP =3D i^2 * = Rrad =3D 50=20 uW
But that would = require a much=20 larger mechanical construction, using either very high supports or a = much=20 lengthier loop.
 
Employing the = existing supports=20 for an upright loop, the area would be only eg. 610 m^2, = giving
 Rrad =3D 0.013 = microohms
 EMRP =3D 1.3=20 uW.
 
A second option = would be an=20 earth antenna, grounded at the far end. You might feed one corner of = your=20 rectangle against ground, and connect the opposite corner to ground = as=20 well. This would be an earth loop of length 61m*sqrt(2), and height 10 m = above=20 ground plus say 48 m below ground (depending on = conductivity) =3D> a=3D 5000 m^2. But here the ground reflection = is already=20 "included" so there won't be the extra factor of two for the image.=20 Thus
 Rrad =3D 31171 ohms * a^2 / lambda^4 =3D 0.45=20 microohms
However the total = resistance of=20 the two ground connections will be much higher than the 2 ohms of = the=20 copper loop, say 200 ohms. So 200 watts would radiate only
 EMRP =3D 0.45=20 uW.
 
You will be much = better off=20 using the existing horizontal loop as a top load for a vertical = "Marconi".=20 Assuming there are only few and small trees around it, effective height = may be=20 perhaps 80% of 10m, leading to
 Rrad =3D 1579 = ohms *=20 8m^2/lambda^2 =3D 76.8 microohms
Optimistically = assuming around=20 200 ohms for coil and ground losses, you coud get 1 A, = giving
 EMRP =3D 77 = uW,
some two orders of = magnitude=20 stronger than the magnetic and earth loop configurations. With around 1.4 nF capacitance and 30 kV = voltage=20 withstand, you could then even QRO up to = 800 Watts=20 for 308 uW EMRP. Which will easily put you in the ballpark of Dex' and = Paul's=20 successful TA crossing.
 
 
Bottom line: Winding = a large=20 coil will be well worth the effort.
And... if you want to go far, just forget about aural CW! This = is only=20 for the biggies like SAQ, radiating about 30 kW.
 
Best = 73,
Markus=20 (DF6NM)
 
 

From: DK7FC
Sent: Wednesday, June 25, 2014 11:27 PM
To: rsgb_lf_group@blacksheep.org= =20
Subject: Re: LF: VLF in Canada - earth = loop

Hello = Mark,=20

200W in such a loop and 8.97 or 8.27 kHz, CW, my guess is that = the=20 distance is less than 2 km. But it depends on the type of the RX antenna = (E or H=20 field).

73, Stefan/DK7FC

Am 25.06.2014 18:47, schrieb mbdittmar@comcast.net:=20
Hi -

I've been following the discussion on = 9 khz=20 transmit antennas with some interest, as a friend of mine, Tom, = KD0VBR, has=20 done some preliminary experimentation in this area ( using large = loading coils=20 w/ vertical ).  Tom asked me to post a query regarding some = antenna ideas=20 he has, and it would be great to hear the group's thoughts on them and = whether=20 they might work or not.  73, Mark AB0CW

Below are his questions:

From:=20 "Pouliot, Tom" <tomandmarti@hotmail.com>
To:=20 "Dittmar, Mark" <mbdittmar@comcast.net>Sent:=20 Wednesday, June 25, 2014 8:51:48 AM
Subject: RE: LF: VLF = in=20 Canada - earth loop

Hi Mark,=20
That=20 would be great if you could post some questions about transmitting = loops for=20 VLF.
Thinking about a square loop 200' = per side=20 10' off the ground  #8 wire with 200 watts into the antenna. How = far away=20 could one expect to be able to detect a CW signal ? ( not = QRSS)
Another=20 possibility would be a triangular loop 200' per side with a 50' high=20 apex.

Tom

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