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Re: [Amps] RE : Another metalwork question

To: amps@contesting.com
Subject: Re: [Amps] RE : Another metalwork question
From: Patrick Egloff <pegloff@gmail.com>
Reply-to: Patrick Egloff <pegloff@gmail.com>
Date: Thu, 4 Aug 2005 10:39:52 +0200
List-post: <mailto:amps@contesting.com>
Hi All,

Just my 2 cents...

Automobile means "auto mobile" = moves by itself.
No matter how many wheels it has 2, 3 or four !
So Cugnot is the inventor of the automobile...

Did C. Colombo discover America ?? It has been proved that not, but ...

73, Patrick TK5EP

2005/8/4, Will Matney <craxd1@verizon.net>:
> Aint that the honest truth!
> 
> Best,
> 
> Will
> 
> *********** REPLY SEPARATOR  ***********
> 
> On 8/4/05 at 7:17 AM PA3DUV wrote:
> 
> >That will never happen. They rather bring their production and engineering
> >over to China.
> >
> >Dick
> >
> >
> >----- Original Message -----
> >From: "hermans" <on4kj@skynet.be>
> >To: <craxd1@verizon.net>; <amps@contesting.com>
> >Sent: Thursday, August 04, 2005 1:10 AM
> >Subject: [Amps] RE : Another metalwork question
> >
> >
> >Why you guys dont adopt the metric sys.........? Its that more simple
> >for every body.
> >Seems European adopted GMT about a century ago in terms of an exchange
> >......but they'r still in the expectation.
> >
> >Jos
> >
> >-----Message d'origine-----
> >De : amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] De
> >la part de Will Matney
> >Envoyé : mercredi 3 août 2005 1:46
> >À : amps@contesting.com
> >Objet : Re: [Amps] Another metalwork question
> >
> >Bill,
> >
> >Sorry about that! I dont do metric conversions enough to remember
> >correctly sometimes. The 6.45 came from a conversion in a transformer
> >equation and it should have been divide by not multiply. The inches to
> >mm factor is 25.4 (1 inch = 25.4 mm). From the example below 0.045" =
> >1.143 mm. The same equation works, it's just you divide by, not multiply
> >by in the end to convert from one to another. However, for what you want
> >leave dividing 25.4 off at the end and it will be in mm. I know one
> >thing, I'll have to stop trying to think when I've been up so late.
> >
> >BA = (0.0078 * T + 0.0174 * R) * No. of deg. in bend
> >
> >14 Ga = 1.62814 mm
> >
> >For 0" radius, 90 Deg bend in aluminum;
> >
> >(0.0078 * 1.62814 + 0.0174 * 0) * 90
> >
> >(0.01269949 + 0) * 90 =
> >
> >0.01269949  * 90 = 1.14295428" or 1.14295428" / 25.4= 0.0449" (0.045")
> >or for ending pieces of flanges
> >
> >Best,
> >
> >Will
> >
> >*********** REPLY SEPARATOR  ***********
> >
> >On 8/2/05 at 11:29 AM Bill Aycock wrote:
> >
> >>Will-
> >>Where did the 6.45 come from? I think it is wrong.
> >>Bill
> >>
> >>At 07:14 AM 8/2/2005 -0400, Will Matney wrote:
> >>
> >>>Martin, I dont think much really changes as those factors are derived
> >>from
> >>>ratios. You should be able to add a multiplier of 6.45 to the formula
> >to
> >>>get metric sums. Try that to see what you get and substitute MM in
> >place
> >>>of the inch measurements then multiple the sum by 6.45. I think you'll
> >>get
> >>>the same just different measurment systems that way.
> >>>
> >>>BA = (0.0078 * T + 0.0174 * R) * No. of deg. in bend * 6.45
> >>>
> >>>Best,
> >>>
> >>>Will
> >>>
> >>>
> >>>*********** REPLY SEPARATOR  ***********
> >>>
> >>>On 8/2/05 at 5:57 PM Martin Sole wrote:
> >>>
> >>> >Will,
> >>> >
> >>> >Thanks for the info, very useful. Do you have the formula in a
> >suitable
> >>> >form
> >>> >for metric material? Haven't worked in imperial measurements for
> >over 30
> >>> >years and only see it now and again on odd bits of US made kit.
> >>Actually I
> >>> >am going to make a new plenum for my second Alpha and the original
> >is
> >>most
> >>> >certainly made to imperial measurements but nobody here would
> >>understand if
> >>> >I tried to replicate it that precisely so the new one will be made
> >to
> >>> >metric
> >>> >dimensions. Would definitely appreciate the drawing and picture,
> >mail
> >>away.
> >>> >
> >>> >Thanks
> >>> >
> >>> >Martin
> >>> >
> >>> >
> >>> >-----Original Message-----
> >>> >From: amps-bounces@contesting.com
> >[mailto:amps-bounces@contesting.com]
> >>On
> >>> >Behalf Of Will Matney
> >>> >Sent: 02 August 2005 17:29
> >>> >To: amps@contesting.com
> >>> >Subject: Re: [Amps] Another metalwork question
> >>> >
> >>> >Martin,
> >>> >
> >>> >On bending steel, you use 1/2 the material thickness to figure how
> >much
> >>to
> >>> >add in length. In other words you divide the material thickness in
> >half
> >>> >where there would be an imaginary center line (its neutral axis)
> >going
> >>> >trough it. Then when the steel is bent, a radius is formed on this
> >>> >imaginary
> >>> >center line even though the inside bend is a sharp 90 deg bend. So
> >>whatever
> >>> >the distance is around that small radius in the middle of the steel
> >is
> >>the
> >>> >material to be added. Now aluminum is a different story and there is
> >a
> >>> >formula for it too. What happens in aluminum, there is a shrinkage
> >on
> >>the
> >>> >inside radius and a stretching on the outside different than steel.
> >For
> >>> >aluminum see the formula and example below;
> >>> >
> >>> >BA = (0.0078 * T + 0.0174 * R) * No. of deg. in bend
> >>> >
> >>> >14 Ga = 0.0641"
> >>> >
> >>> >For 0" radius, 90 Deg bend in aluminum;
> >>> >
> >>> >(0.0078 * 0.0641 + 0.0174 * 0) * 90
> >>> >
> >>> >(0.00049998 + 0) * 90 =
> >>> >
> >>> >0.00049998 * 90 = 0.045" or about 3/64" or for ending pieces of
> >flanges
> >>> >make
> >>> >it 1/16" from bend line to end.
> >>> >
> >>> >BA = Bend Allowance
> >>> >R = Radius of bend on the inside, not the neutral axis.
> >>> >T = Material thickness
> >>> >
> >>> >I have a pic with this and a drawing if needed I can e-mail it to
> >you.
> >>Hope
> >>> >this helps.
> >>> >
> >>> >Best,
> >>> >
> >>> >Will
> >>> >
> >>> >
> >>> >*********** REPLY SEPARATOR  ***********
> >>> >
> >>> >On 8/2/05 at 2:46 PM Martin Sole wrote:
> >>> >
> >>> >>Well it is actually amp related, or will be at some point I hope
> >but
> >>> >>seeing how there is a great wealth of resource here it seems a good
> >>> >>place to start.
> >>> >>
> >>> >>Some time back I recall seeing an article, might have been in
> >Radcom,
> >>> >>might have been in QST. Think it had to be either one of those two
> >>> >>though. It addressed the process of marking out metalwork for
> >making
> >>> >>enclosures and explained how to allow the correct amount of
> >material
> >>> >>for bends etc. Was within the last year or two I think.
> >>> >>
> >>> >>Just hoping that somebody might recall where this was or maybe
> >point me
> >>> >>to another resource with similar information.
> >>> >>
> >>> >>Tks
> >>> >>Martin HS0ZED
> >>> >>
> >>> >>
> >>> >>
> >>> >>--
> >>> >>No virus found in this outgoing message.
> >>> >>Checked by AVG Anti-Virus.
> >>> >>Version: 7.0.338 / Virus Database: 267.9.7/60 - Release Date:
> >>> >>28/07/2005
> >>> >>
> >>> >>
> >>> >>
> >>> >>_______________________________________________
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> >>> >>Amps@contesting.com
> >>> >>http://lists.contesting.com/mailman/listinfo/amps
> >>> >
> >>> >
> >>> >
> >>> >
> >>> >--
> >>> >No virus found in this incoming message.
> >>> >Checked by AVG Anti-Virus.
> >>> >Version: 7.0.338 / Virus Database: 267.9.7/60 - Release Date:
> >28/07/2005
> >>> >
> >>> >
> >>> >
> >>> >--
> >>> >No virus found in this outgoing message.
> >>> >Checked by AVG Anti-Virus.
> >>> >Version: 7.0.338 / Virus Database: 267.9.7/60 - Release Date:
> >28/07/2005
> >>>
> >>>
> >>>
> >>>_______________________________________________
> >>>Amps mailing list
> >>>Amps@contesting.com
> >>>http://lists.contesting.com/mailman/listinfo/amps
> >>
> >>Bill Aycock - W4BSG
> >>Woodville, Alabama
> >>
> >>
> >>_______________________________________________
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> >
> >
> >
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> 
> 
> 
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