DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws;
  s=s1024; d=snet.net;
  h=Received:X-Yahoo-SMTP:X-YMail-OSG:X-Yahoo-Newman-Property:Reply-To:From:To:Subject:Date:Message-ID:MIME-Version:Content-Type:Content-Transfer-Encoding:X-Priority:X-MSMail-Priority:X-Mailer:In-Reply-To:Importance:X-MimeOLE;
  b=Axj6T2WxElq6A68Hu6Adm6kSVaza95W1sE1dE+7rAIMap1P/1qsEF77CKyMNej94ygUsP8mI7LUoSZkXuf42qCCPBVEVLWeX4FEw99ivkSPXMo18OVihkxk0CHlG9QdF+7I2thiebLFuqoQBDklL2qHX+614j8IoZDh/BKsFCzY=  ;
From: "Paul A. Cianciolo" <paulc@snet.net>
To: <rsgb_lf_group@blacksheep.org>
Date: Mon, 29 Mar 2010 18:36:27 -0400
Message-ID: <MFEJIKCLCMPDFHOKMANAMEAFOFAA.paulc@snet.net>
MIME-Version: 1.0
In-Reply-To: <0F2B77385FC04A07A080811EE2E61079@JimPC>
Importance: Normal
DomainKey-Status: good (testing) 
Subject: LF: RE: Re: PF per Meter  dependant on wire size?
Content-Type: text/plain;
	charset="iso-8859-1"
Content-Transfer-Encoding: 7bit
Sender: owner-rsgb_lf_group@blacksheep.org
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org


Hi Jim

Thank you for the formula for calculating "C" of a wire.
Also the information about adding additional wires is valuable.

There are 2 methods I could try with my 55 meters of wire.
With the use of fiberglass spreaders, multiple wires could be spread apart
at the top of the tower forming a fan of the wires.
With a single point on the shack end where the wires would all would
combine.

Probably better would be to support 4 wires equally spaced on a fiber glass
poles 9 meters long on both the tower end and the shack end.

This seems too easy Jim, there must be a point of diminishing returns.

Knowing that the wire is 55 meters long, and 1 wire measures at 340 pf would
you hazard a guess as to 4 wires 1 meter apart each the value of "C"

Do you think 700 pf is achievable?

Thank you for reading

PaulC
W1VLF




-----Original Message-----
From: owner-rsgb_lf_group@blacksheep.org
[mailto:owner-rsgb_lf_group@blacksheep.org]On Behalf Of James Moritz
Sent: Sunday, March 28, 2010 4:09 PM
To: rsgb_lf_group@blacksheep.org
Subject: LF: Re: PF per Meter dependant on wire size?


Dear Paul, LF Group,

One formula for C per unit length:

C = 24/Log(4H/d) picofarads/m , H= height, m, d = diameter, m

So the capacitance depends on the logarithm of the ratio of height over wire
diameter, which only varies a little for a large change in diameter, e.g for
1mm wire at 10m high = 5.2pF/m, 10mm diameter wire at 10m high 6.7pF/m.
Actually this formula only applies to an infinitely long, uniform,
horizontal, straight wire. In practice, the presence of ends, downleads,
things on the ground, insulation on the wire, etc. etc. will all have an
effect, and are difficult to calculate, so 6pF/m is usually as good an
estimate as you are likely to get.

Adding multiple wires will increase capacitance. If the wires are many
metres apart (spacing large compared to height), you can multiply the
capacitance by the number of wires. But usually, the wires are more closely
spaced, and there is less increase in capacitance. The figures I have to
hand are for two 1mm wires 100mm apart, C is higher by 39% compared to a
single wire, 1m apart and C is 68% higher.

Cheers, Jim Moritz
73 de M0BMU

----- Original Message -----
From: "Paul A. Cianciolo" <paulc@snet.net>
To: <rsgb_lf_group@blacksheep.org>
Sent: Sunday, March 28, 2010 8:24 PM
Subject: LF: PF per Meter dependant on wire size?


> Hello,
>
> I see from previous posts that a number of  aprox 6 PF what size wire is
> this for?
>
> Increasing wire size should in "C"  as should several conductors in
> parallel
> spaced a few feet apart.
>
> PauLC
>
> W1VLF
>
>
>
>