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Curtate cycloids....What would Strad do?
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R Mac
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Joined: 05 Mar 2013
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Location: Near Phily USA

PostPosted: Tue Mar 12, 2013 1:39 pm    Post subject: Curtate cycloids....What would Strad do? Reply with quote

Just looking at the math for curtate cycloids makes my head hurt, and I have difficulty imagining a working luthier crunching numbers.

I am supposing that Strad (or anyone else) would have used physical, rather than mathematical models to develop their arching templates. In the case of curtates, I imagine the process to be something like this.

He would begin by cutting a pastboard disk whose circumference matched the (flat) width across the plate from nadir to nadir. This involves a compass, a scissors, and some trial and error.

He makes a hole in the disc whose distance from the center is 1/2 the height of the intended arching at that location.

He uses this disc like the old "spirograph" toy to scribe the curve on the template material, being sure to label it.

He then moves on the the next smaller width, and cuts the disc down to match.

He repeats the procedure with each successively smaller width, being careful to cross out the hole from the preceding width, so he does not get confused.

When this is all done, he should have a series of templates and a small useless disc full of holes. The templates get cut out carefully and the pasteboard disc goes in the fireplace.

Has anyone tried doing it this way?

Mac
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Michael Darnton
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PostPosted: Tue Mar 12, 2013 1:56 pm    Post subject: Reply with quote

Not in modern times that I know of. One of my workshop participants made a disc for me one year, though.

On Andrea Amati instruments the curve bottoms out on the purfling. The diameter of disk needed to make the templates for the c-bout and lower bouts are about 31.5mm and 63mm. Anyone who's been following what I do for a while knows of my fascination with the 31-32mm dimension, which is scribed on ALL of the extant Stradivari forms. The precise number in that range depends on the overall length of the instrument form.

For instance:


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actonern
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PostPosted: Tue Mar 12, 2013 2:09 pm    Post subject: Reply with quote

Rmac:

As I understand it, you need a different wheel diameter for every cross arch point... because each cross arch will have a different distance from minimus to minimus point.

For example if you used disk appropriate to the mid bout (around 32 mm in diameter) for the wide bottom bout, it would be too small to create a smooth arch edge to edge.

And I think the pencil hole would be the height distance between the minimus thickness of the plate and the top of the arch.

Sorry... I just noticed you say the same disk would be "cut down" for each smaller width point...
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R Mac
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PostPosted: Tue Mar 12, 2013 2:12 pm    Post subject: Reply with quote

Michael Darnton wrote:
Anyone who's been following what I do for a while knows of my fascination with the 31-32mm dimension, which is scribed on ALL of the extant Stradivari forms. The precise number in that range depends on the overall length of the instrument form.



Does that correspond to the historical size of a Cremonese "inch" or "dita" or whatever, or do you suppose that it is something more personal, like the length of the last joint of the luthier's thumb?

As an asside, when I am considering how the armor I build fits the wearer, I usually think in terms of "finger widths" or "palm widths" rather than inches. I started it as an affectation, but now it just seems normal.
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R Mac
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PostPosted: Tue Mar 12, 2013 2:17 pm    Post subject: Reply with quote

actonern wrote:
Rmac:

As I understand it, you need a different wheel diameter for every cross arch point... because each cross arch will have a different distance from minimus to minimus point.

For example if you used disk appropriate to the mid bout (around 32 mm in diameter) for the wide bottom bout, it would be too small to create a smooth arch edge to edge.

And I think the pencil hole would be the height distance between the minimus thickness of the plate and the top of the arch.

Sorry... I just noticed you say the same disk would be "cut down" for each smaller width point...


When I started writing that, I was thinking about a new disc for every point, but they it occurred to me that you just start with the biggest one and keep cutting it down. No sense wasting a lot of pasteboard.

Mac
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actonern
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PostPosted: Tue Mar 12, 2013 2:24 pm    Post subject: Reply with quote

My biggest problem when I did the actual wheel thing was to get the stupid little disks to roll vs. slide along an edge when tracing a line.

Once you get that worked out it is really quite precise what you end up with, and not hard to do. Put it this way... you could cut out wheels and draw arches for all cross points in the time it takes to watch one Simpsons episode...

perhaps the wheel edges were coated with something sticky? Lets see... Honey, egg white, gum arabic... Laughing
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R Mac
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PostPosted: Tue Mar 12, 2013 2:45 pm    Post subject: Reply with quote

actonern,

So, you actually did use discs! That's so cool!

I was worrying about how to get them to roll rather than slide, but did not get very far past the notion that putting gear teeth on them is out of the question.

I thought about putting sand paper on the base line. That would work, I think, but it don't seem vary "baroque". How about rosin? There should be some of that lying around the shop....

Mac
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Ken Pollard
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PostPosted: Tue Mar 12, 2013 2:45 pm    Post subject: Reply with quote

Been a while since I was on here -- don't know if this reply will 'take' or not.

One doesn't need to start with round things to draw round things. I don't know how they drew cc's in the old days, but they did know about them.

Here's a fun little video on drawing an ellipse.

http://www.youtube.com/watch?v=TfsfBT1xX1c
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Michael Darnton
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PostPosted: Tue Mar 12, 2013 2:48 pm    Post subject: Reply with quote

Ken, Mr. Math, is the one who made the disk in my workshop. Direct your math questions in his direction. Since my heartbeat starts going up when I have to type on the top row of the keyboard, he's a handy person to have around.
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actonern
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PostPosted: Tue Mar 12, 2013 4:34 pm    Post subject: Reply with quote

"...Anyone who's been following what I do for a while knows of my fascination with the 31-32mm dimension..."

It just struck me... the thickest part of the instrument top to back is about 64 mm... the ribs 32mm...

Da Vinci for sure! Laughing
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Michael Darnton
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PostPosted: Tue Mar 12, 2013 4:39 pm    Post subject: Reply with quote

Yup.
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actonern
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PostPosted: Tue Mar 12, 2013 5:41 pm    Post subject: Reply with quote

I'm going to go out and get me a 32 mm. varnish brush... I believe I'm on to something!
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John Masters
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PostPosted: Sun Mar 17, 2013 5:40 pm    Post subject: Re: Curtate cycloids....What would Strad do? Reply with quote

R Mac wrote:
Just looking at the math for curtate cycloids makes my head hurt, and I have difficulty imagining a working luthier crunching numbers.
Mac


To ease the pain and do away with disks, it is much more accurate and easier to calculate numbers you can measure at any point you wish. To calculate the T that corresponds to a given X is more elaborate, but not especially useful. You can choose a T close to what you in X and then use that X, it will give good results.

You can do these on an Excell spreadsheet. Just put the t variable in a colume and run it at intervals of .1 or whatever other increment you decide. I have made changes of variables in order to make the curtate cycloid centered on the Y axis and put the ends on Y=0. These formulas are for one cycle with ends wherever you want the deepest channel. This occurs at L. (Making L less than the width will make the rolling disk rotate more than one revolution because L is the circumference of the disk.)

The function is so-called parametric in t, but it is essentially Y=Cosine plotted on a distorted X-axis.


X = T+.5*h*sin(T*pi/L)
Y = .5*h*(1+cos(T*pi/L))

T runs from -L to +L where L is the half-width of the violin at the particular position on the long arch.h is the height of the long arch at that position.

choose t = T*Pi/L

X = t*L/Pi+.5*h*sin(t)
Y = .5*h*(1+cos(t))

Here, t runs from -Pi to +Pi

If you want an arching template, make a graph and cut it out. Many of you may have graphing programs, but make sure the scaling is full scale for any template you wish printed out.


Last edited by John Masters on Sun Mar 17, 2013 5:50 pm; edited 3 times in total
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actonern
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PostPosted: Sun Mar 17, 2013 5:44 pm    Post subject: Reply with quote

Hi John!

Good to see you here...

Apart from the "fun" of doing these calculations would you say there was any reason to avoid the software solution?

Best regards,

Ernie
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John Masters
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PostPosted: Sun Mar 17, 2013 5:57 pm    Post subject: Reply with quote

actonern wrote:
Hi John!

Good to see you here...

Apart from the "fun" of doing these calculations would you say there was any reason to avoid the software solution?

Best regards,

Ernie


I think that the software solution just hides what is being done. I have not seen the software solution for a long time. Can one choose more than one cycle of the CC?

Why not make your own software, the Excell spreadsheet is software, and you can see what is going on.

Also, I recall an adjustable variable. It is possible to make what I call "pseudocycloids." Here one modifies the X formula using a fraction of the height. Then the Y result is multiplied by the reciprocal of that number to bring the desired height back. Or just use the proper height on Y. It looks very close to a true CC, but the inflection (recurve) position has been repositioned.

I have made FEA illustrations of 2-D strips under various loads. There is no maximum stiffness at a perfect CC, but it may be a good middle ground from pseudocycloids that are close. I did not extend to 3-D, but I may do this in the future for domes that look like nipples. I did it in the past but was not secure in the method or results.
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