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[Amps] RE : Another metalwork question

To: <craxd1@verizon.net>, <amps@contesting.com>
Subject: [Amps] RE : Another metalwork question
From: "hermans" <on4kj@skynet.be>
Date: Thu, 4 Aug 2005 00:10:42 +0100
List-post: <mailto:amps@contesting.com>
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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>> >
>> >
>> >
>> >
>> >--
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>> >
>> >
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>>
>>
>>_______________________________________________
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>
>Bill Aycock - W4BSG
>Woodville, Alabama 
>
>
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