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Apparently the "CUP HOLDER" has a long history. The 1944 Spitfire cupholder above has me a tad confused about the manufacturing process. I am looking at the encased wired that is running the perimeter of the sloped top side edges.

I think this is simply a way of terminating the sheet metal so as not to leave an overly sharp edge or snag points. In order to create it, I originally thought it was just an edge tab rolled over to form a tube then the wire was inserted. Then I thought they might of used the wire are part of the forming process and rolled the edge around it. My final thought was that its actually a circular tube with 26 Gauge wall with a 14 Gauge wire insert as described at the top of the page item "C".

So my question is how would one make this "tube" of a different gauge than the adjacent sheet metal which it would be attached to? I noted no welding symbols so I am curious as to the process.

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  • $\begingroup$ youtu.be/CKflVjeCUu4 ... you can extrapolate on possible automated process $\endgroup$
    – jsotola
    Jun 24 at 4:52
  • $\begingroup$ 24Ga is thin enough to wrap around the wire. $\endgroup$
    – Abel
    Jun 24 at 10:38
  • $\begingroup$ @jsotola Thanks for the video. After watching it I also found this linked video which showed how to do it with a beading machine? $\endgroup$
    – Forward Ed
    Jun 24 at 11:21
  • $\begingroup$ @Abel thank you for the insight. I am pretty clueless when it comes to the working with sheet metal side of the house. I basically just know that it can be bent, folded, rolled, embossed/stamped, punched and sheared pretty easy. It was interesting to watch the machining process of the rollers to form edges. Because the rolled edge is adjacent to the angle and they are specified as 26 Ga and 20 Ga respectively, does that mean the 20 gauge edge is flattened further to 26 Ga thickness before the edge wire bead is made? $\endgroup$
    – Forward Ed
    Jun 24 at 11:27
  • $\begingroup$ 20Ga in C is just a L bracket butted up against the edge and riveted to 24Ga A. Unless the wire has a rectangular cross section, I haven't found meaning in the 26 yet. A is wrapped around the wire. First Angle Projection views... $\endgroup$
    – Abel
    Jun 24 at 13:05

1 Answer 1

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As comments mentioned, and your second guess surmised, the 24 Gauge sheet metal A is wrapped around the (solid) 14 Gauge wire.

The confusion appears to be the result of a mistake on the print. The callouts of A B and C refer to materials each part is made from, with designations as to the quantity required to make this part.

  • A is cut and bent from 11.5"x13.5" of 24 Gauge sheet metal

  • B is 2 pieces, 4" each of 0.6"x0.5" 20 Gauge L angle

  • C is 6.1" of 0.8"x0.55" 20 Gauge L angle

  • 26" of 14 Gauge wire is likely what was meant by the error.

Material is Alclad so aluminum chart likely applies for Gauge to thickness.

The print is not without other errors/omissions - the Tenax fastener appears to be shown on only two of the three views.

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  • $\begingroup$ I noted that the fastener is on in 2 of the three views as well. The other one that got me was the bottle hole is 4" diameter with the center being located 3" away from the a reference line. This puts the hole tangent to the face of the wire bead limit. Then in the side view the opposite edge of the same wire bead is dimensioned as being 1" away from nearly the same reference line. In my opinion either the 1", 3" or 4" diameter is in error. Fun times spotting this stuff then trying to figure out what to do. $\endgroup$
    – Forward Ed
    Jun 24 at 16:33
  • $\begingroup$ Thanks for also for confirming that using the Aluminum Gauge chart was the right option. I did not have any other reference material for that one so it was the assumption I had made. $\endgroup$
    – Forward Ed
    Jun 24 at 16:36
  • $\begingroup$ Hole does not appear to be in error. Look at what face it is on. There's a 1" deep box if you look at the rightmost view. Wire is on right wall of box while hole is on left. $\endgroup$
    – Abel
    Jun 24 at 17:24
  • $\begingroup$ 2" Radius of Hole + 1" box = 3" to center of hole, but since the 1 inch references one side of the wire bead, and the hole is tangent to the other side of the wire bead, the positioning of something is out by the width of a wire bead. If 1" referenced the other side of the bead that would solve the situation, or if the hole was tangent to the other side of the wire bead would also solve it, or reduce the radius of the hole by the diameter of the wire bead. At least that is what my summation of the numbers. I could be wrong. $\endgroup$
    – Forward Ed
    Jun 24 at 17:54

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