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Curtate Cycloids. Again


violins88

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4 hours ago, Roger Hill said:

If you follow the arch shape line from any interior point along the line, you will at some point come to the edge, i.e. the point at which the top plate is glued to the lining.  This is the end of the arch.  as you approach the end of the arch,the angle between the rib and the arch shape line (extended) is very close to a right angle.

 

Now I am confused.

What do you mean by "from any interior point along the line"?  I agree that the interior line appears the intersect with the rib at right angles, but the curtate cycloid shape is on the outside surface.  What has curtate cycloid shape on the outside to do with right angle on the inside?

What am I missing?

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2 minutes ago, Janito said:

Now I am confused.

What do you mean by "from any interior point along the line"?  I agree that the interior line appears the intersect with the rib at right angles, but the curtate cycloid shape is on the outside surface.  What has curtate cycloid shape on the outside to do with right angle on the inside?

What am I missing?

My interpretation of that is that at any point on the arch, the tangent to the line points to the edge. As you get closer to the edge the arch becomes steeper.

But then, I know nothing, and that is probably wrong.

 

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"What do you mean by "from any interior point along the line"?" 

A point on the arch shape curve that is not at either end.  If you carve a perfect CC on the outside and use a uniform thickness for the top, both the arch shape curve and the parallel line along the inside surface of the plate will also be curtate cycloids to a close approximation.  You will find a lot of clarification if you simply print out a transverse ct scan picture and draw the middle line of the top and look at what I am saying.  If you are a windows user, open Windows accessories and find "snipping tool" to make this easy.

"But then, I know nothing, and that is probably wrong."

yup, its wrong.  As you get closer to the end, the tangent becomes horizontal and not steeper.

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Don:

I know what a cycloid looks like.   You know what a cycloid looks like.  We both know what a cycloid approximation to a violin arch looks like, with the low point at the minimum of the trough.  I guess no nit is too fine to be left unpicked.  You are not contributing to the understanding of the point I made for those who do not have the knowledge of mathematics that you and I have.  You know very well from what I said that I meant the end of the arch shape curve as I defined it.  You don't have to prove to me that you are smart, I know that. I've made my point, no good deed goes unpunished.  I'll bow out now and move on to other things where I don't have to be harangued with pedantic bullshit.

Roger Hill

 

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If I do understand  what Roger is saying, the feature so common  in these old violins, often missed, is this. 

The inside  of the plate is hollowed out, and the hollowing approaches the rib gluing  surfaces  in a smooth curve, not with a ridge or an angle. The recurve that we see on the outside of the plate is fully reflected on the inside, at least to its lowest point. Obviously, rather than rising again to form the edge, it flattens out to form the rib gluing surfaces. Even in strongly arched plates whose  arching rises quickly from the edge, this is true. 

I found it hard to get my head around this, because carving the plates as I learned to do at school, unless I hollowed the inside right to the edge, I had a really  thick margin that the edge fluting  (as I knew it) couldn't fix. 

In order to achieve this, one must start from a much  thinner edge, or carve the fluting  very deep, or move the fluting inward. In fact, some combination  of  all three, depending on the arching. I don't use templates, but 'draw' the arching  shape on the wood with my gouge, always  with  this in mind.

I'm sure  many of you spotted this , or were told about it from day one, but it took me  years to get it, and to pluck up the courage to carve boldly enough  to make my violins that way.

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