[Amps] RE : Another metalwork question

Will Matney craxd1 at verizon.net
Thu Aug 4 03:46:15 EDT 2005 

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 at skynet.be>
>To: <craxd1 at verizon.net>; <amps at 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 at contesting.com [mailto:amps-bounces at contesting.com] De
>la part de Will Matney
>Envoyé : mercredi 3 août 2005 1:46
>À : amps at 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 at contesting.com
>[mailto:amps-bounces at contesting.com]
>>On
>>> >Behalf Of Will Matney
>>> >Sent: 02 August 2005 17:29
>>> >To: amps at 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
>>> >>
>>> >>
>>> >>
>>> >>--
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>>> >>28/07/2005
>>> >>
>>> >>
>>> >>
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>>> >
>>> >
>>> >
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>>>
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>>
>>Bill Aycock - W4BSG
>>Woodville, Alabama
>>
>>
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>
>
>
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