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
> >>
> >>
> >>
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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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