A Practical Maths Problem

[niall Y]

I'm making a series of ferrules in brass out of a piece of flat 1mm sheet. I am bending them around a 16mm ground, silver steel, rod. A simple question arises from this : what length should the brass be? Pi X 16mm gives me a circumference of, 50.272mm, which is initially what I cut the length to, but this isn't long enough.

If one is bending metal around a mandrel then some change has to be happening to the material to allow it to conform to the circular shape. The outer diameter of the ferrule is 18mm and it follows from this that the outer circumference will be 56.556mm. This is a difference between inner and outer of over 6mm. However the additional length needed, when measured, to close the gap is actually 3.1mm giving a true length of 53.372mm, which is closer to Pi X 17mm. Is this then the correct sum to establish the length of brass sheet needed?

 

[Yojevol]

Not really a maths problem, more of a materials behaviour characteristic. The inner 1/2mm will be compressed and the outer 1/2mm will be stretched giving a mean diameter of 17mm where the material is unchanged.
Brian

 

[Sachakins]

Pi x 17 is the centre line measurement of the brass length, 16 + 2 x half the thickness, hence 16+1= 17.
There is a bending stress between the inner and outer diameters, where inner one is compressed and outer surface is stretched, hence taking the mean diameter will a close approximation to the length needed, which at this thickness of brass should be accurate enough. If working with thicker material or and less malleable material then the calculation of the length for a set diameter is far more complicated and includes material deformation characteristics in the equation, here's an example

https://calculator.academy/roll-length-calculator/

 

[niemeyjt]

Pi x Diameter gives the circumference of a circle.

But if the rod is 16mm and the material is 1mm then the inside is 16mm but the outside is 18mm. So cut both ends at an angle to reflect that.

 

[Sachakins]

L = pi * (OD^2/4 – ID^2/4) / t

 

[Fergie 307]

Or maybe just buy a piece of suitably sized tube, then you will not have the weakness of a joint.

 

[Phil Pascoe]

I think the chance of bending that anything like accurately is extremely optimistic.

 

[niall Y]

Pi x 17 is the centre line measurement of the brass length, 16 + 2 x half the thickness, hence 16+1= 17.
There is a bending stress between the inner and outer diameters, where inner one is compressed and outer surface is stretched, hence taking the mean diameter will a close approximation to the length needed, which at this thickness of brass should be accurate enough. If working with thicker material or and less malleable material then the calculation of the length for a set diameter is far more complicated and includes material deformation characteristics in the equation, here's an example

https://calculator.academy/roll-length-calculator/

Thanks for the explanation and the link. I did feel that this was a well worn path others had trod and I was in danger of trying to "reinvent the wheel", so to speak.

Or maybe just buy a piece of suitably sized tube, then you will not have the weakness of a joint.

A sensible (and obvious) reply, so I feel a bit more information would help.
This is an experiment to test out a swage block jig I have made. This will be used for bending sterling silver ferrules, hence the need to iron- out all the problems before I get underway. One can buy sterling silver tube in very small diameters, but it does not seem to be available, in 18mm diameters.

 

[niall Y]

I think the chance of bending that anything like accurately is extremely optimistic.

Surprisingly, using the jig I have made, the bend is very neat and uniform. I certainly could not have done it quite so neatly with a pair of half-round pliers.

 

[niall Y]

Pi x Diameter gives the circumference of a circle.

But if the rod is 16mm and the material is 1mm then the inside is 16mm but the outside is 18mm. So cut both ends at an angle to reflect that.

This was one of my initial thoughts. But you would end up with very sharp bevels at both ends of the stock. that would only get more acute when the metal was bent In a sense the discrepancy between the inner and outer circumferences is lost at all stages around the bend and where the two ends meet the angle will to all intents and purposes , be 90 degrees.

 

[chaoticbob]

Hi Niall. Presumably these are for your whistles. Geoff Wooff (renowned pipemaker) describes his method here.
He says:
"The width of the ferrule, in the flat, is arrived at by taking the desired inside diameter,
adding to that the thickness of the sheet metal and multiplying this figure by Pi, 22/7 or
3.142.
Example:
Internal Dia. 15mm + thickness of sheet 0.6mm = 15.6 x Pi = 49mm.
or 0.59” + 0.0236” = 0.6136 X Pi = 1.928”
(Whether or not this is technically correct, it works)."
That actually amounts to taking the average of internal and external diameters as you have found.
He also gives info about annealing both brass and silver.
Bob.

 

[Fergie 307]

Thanks for the explanation and the link. I did feel that this was a well worn path others had trod and I was in danger of trying to "reinvent the wheel", so to speak.

A sensible (and obvious) reply, so I feel a bit more information would help.
This is an experiment to test out a swage block jig I have made. This will be used for bending sterling silver ferrules, hence the need to iron- out all the problems before I get underway. One can buy sterling silver tube in very small diameters, but it does not seem to be available, in 18mm diameters.

That makes perfect sense, although I might experiment with copper, as brass is tougher than silver, and you may find that what works in brass may not necessarily transfer to silver. If you wanted to do this a lot then probably worthwhile to buy a rolling machine, makes life much easier, and you can very easily make a range of different diameter tubes.

 

[Fergie 307]

Another option I suppose would be to make them in brass and then plate them. Would be easier and stronger, not to mention a great deal cheaper. Silver is one of the easiest metals to electro plate at home. Depends what they are for, and how important it is that they are solid silver. Sounds interesting

 

[J-G]

Apropos of nothing really - except that if the ferule were to be made of Stirling Silver at 1mm thick, it would be illegal to offer it for sale without being Hallmarked if it were longer than 14.06mm.

Incidentally, I would also do any prototype work in Copper rather than Brass exactly for the reasons cited by @Fergie 307

 

[Fergie 307]

You could also try nickel silver. Indistinguishable visually but much stronger and cheaper, very commonly used in watch cases in the days of pocket watches. Looks exactly like silver but much harder wearing, and doesn't tarnish. Also will behave very like brass so the results of your experiments should transfer well. It was also commonly used to make cigar tubes, so you might be able to repurpose if you could find some cheap ones the right size, depends how many you need.

 

[Phil Pascoe]

Apropos of nothing really - except that if the ferule were to be made of Stirling Silver at 1mm thick, it would be illegal to offer it for sale without being Hallmarked if it were longer than 14.06mm.

Incidentally, I would also do any prototype work in Copper rather than Brass exactly for the reasons cited by @Fergie 307

Being pedantic, it wouldn't be illegal to sell it without a hallmark, it would only be illegal to sell it as silver.

 

[Phil Pascoe]

Come on, Niall - let us into the secret - what are they for?

 

[niall Y]

Come on, Niall - let us into the secret - what are they for?

@chaoticbob , presumed correctly that these were for my "penny/ Irish" whistles. They have a ferrule at both end., The longer ones, that I am using my jig for, form part of the mouthpiece.

You could also try nickel silver.

I do have a small rod of this, that I manage to get cheaply. and will try turning some ferrules from it at a later date. I note that some whistle makers are offering this on their whistles under the euphemistic title of "German Silver" It would be really convenient if nickel silver was readily available in tube form in the size I need, but sadly I have not been able to find any.

Another option I suppose would be to make them in brass and then plate them.

An interesting idea, though I would have to modify my finishing method. At present I am holding the whistles by their ferrules in two chucks, at both the tailstock and headstock end. They are then turned to the finished size before sanding..

Also, I am not sure how the plating would perform at the mouthpiece, bearing in mind I'm not really a fan of that worn EPNS look.

 

[Phil Pascoe]

To protect the finish you could find some plastic tube with an inside diameter slightly smaller than the outside diameter of the ferrule, cut a piece the same length, slit it lengthways and slide it over the ferrule. You could use a larger diameter piece with a sliver removed, allowing it to hold tight.

 

[Fergie 307]

I think it was originally developed in China for decorative inlays in wood etc. I seem to recall reading somewhere that tbis wasnt just because it was cheaper, but also because it doesnt tarnish or stain the material it is inlaid into. Took a while for us western barbarians to work out what it was actually made of. Problem eventuall solved in Germany, hence one of its many names. My personal favourite is the American watch cases where it is referred to as Silveroid or coin silver. Often engraved " guaranteed genuine imitation silver" .