Fitting top arch to 1724 Strad

[Johnmasters]

Bruce Carlson posted two side pictures of Strad longitudinal arching. I previously posted a class of functions in the form of a picture of some graphs.

The arch is nearly a cosine curve from 0 to 90 degrees. To fatten (not "flatten) it slightly in the more highly curved areas, I averaged this with cosine to the argument of 90degrees*(x/L)^1.3 .

The powers of x/L from 1.5 to 4 were shown previously in the previous long-arch thread. I did not spend a lot of time on this Strad arch. Various portions along a length can be fattened with higher-power expressions in small percentage. Given precise data, I think any arch could be represented. Also (ahem...) errors in the maker's scraping perhaps be discovered.

Notice that x/L to any power gives cosine=0 at x=L and 1 at x=0. That is why it is set up this way. All of the possible functions are set up to truncate at the inflection. That is, at the neck, the arch is to have no curvature. I also believe that any inflection is detrimental. If there IS one for cosmetic purposes, the inflection should be in the area where the wood is still thick in the block region.

The paper with the blue line was a graphic plot. I scaled it to fit the photo of the Strad.

The reason I say it was easier than I expected is that I did not know how many components of (x/L)^n would be needed. Also, these could have been added in various amounts, or multiplied.

It was originally thought that cos ( 90*( (x/L) + (x/L)^n1 + (x/L)^n2 ........)) might be needed.

Very likely, any of these forms could be used. What is surprising is that so few terms are needed to converge to a "good" arch. Only two in the case here.

In any case, all of these curves are infinitely differentiable. That was desired and why those functions were chosen. Another reason closely related to the curtate cycloid was also behind the idea. I won't belabor it unless someone is curious and wishes a short description.

[Anders Buen]

Bruce Carlson posted two side pictures of Strad longitudinal arching. I previously posted a class of functions in the form of a picture of some graphs.

The arch is nearly a cosine curve from 0 to 90 degrees. To fatten (not "flatten) it slightly in the more highly curved areas, I averaged this with cosine to the argument of 90degrees*(x/L)^1.3 .

The powers of x/L from 1.5 to 4 were shown previously in the previous long-arch thread. I did not spend a lot of time on this Strad arch. Various portions along a length can be fattened with higher-power expressions in small percentage. Given precise data, I think any arch could be represented. Also (ahem...) errors in the maker's scraping perhaps be discovered.

Notice that x/L to any power gives cosine=0 at x=L and 1 at x=0. That is why it is set up this way. All of the possible functions are set up to truncate at the inflection. That is, at the neck, the arch is to have no curvature. I also believe that any inflection is detrimental. If there IS one for cosmetic purposes, the inflection should be in the area where the wood is still thick in the block region.

The paper with the blue line was a graphic plot. I scaled it to fit the photo of the Strad.

The reason I say it was easier than I expected is that I did not know how many components of (x/L)^n would be needed. Also, these could have been added in various amounts, or multiplied.

It was originally thought that cos ( 90*( (x/L) + (x/L)^n1 + (x/L)^n2 ........)) might be needed.

Very likely, any of these forms could be used. What is surprising is that so few terms are needed to converge to a "good" arch. Only two in the case here.

In any case, all of these curves are infinitely differentiable. That was desired and why those functions were chosen. Another reason closely related to the curtate cycloid was also behind the idea. I won't belabor it unless someone is curious and wishes a short description.

Smart. Why do you need the function to be infinitely differentiable?

And expanding your function using a Taylor series would express exactly the same thing using a polynomial. Why does not the polynomial do it when it is expresses exactly the same function?

I do see the practical implications and benefits from using the simple function, eg in a spreadsheet etc for making a model.

[Anders Buen]

To follow up the idea of matching long arches to polynomials I took the arching pattern for the Ernst Neruda 1709 Strad in Otto Möckels book and graphed up the long arch of the top and back plate from manually read out arching heights. I measured up the distances from the upper end of the plate to where the iso height lines cross the central line using Adobe Acrobat on a pdf file of the arching maps. There is a tool for measuring distances in Acrobat and the Möckel plot does have a convenient scale to calibrate the tool against.

I enclose the Möckel plot and post the results in coming posts. I have used modified versions of these archings on hardagners, so I have an interest in possibly being able to digitalize them. I have also earlier done an attempt to use Hans Pluhars graduation program to map, and thus digitalize, these archings.

(I have been thinking about starting an own tread on this, but it is so close to what John does and it adds to an earlier discussion in using polynomials. They were doomed as not 'doing it'. And I think this example show that polynomial fits may work well, at least for this particular instrument)

[Anders Buen]

Here is the sequence on fitting the top long arch.

1. First comes the raw data as read out of Möckels chart both for the top and the back plate. In the next steps I have skipped the first and last point as these are guessworks and they make the lines less easy to fit. The neck end is to the left.

2. Then I show the best fitted lines using polynomials of order 2 up to the maximum I can get in my MS Excel version, 6. The closer the R^2 term is to 1 the better the line fits to the data points. You see that the two first best fitted lines, red and orange (polynomials of order 2 and 3), does not fit the data well in the central and bout regions. The second order polynomial consist of a parable and a linear term. (We shall see later that the back plate arch fit quite well to that parable with the linear added term). But the data is better described by higher order polynomials with the highest being the best, having the R^2 closest to 1.

3. The best fitted line through the raw data for the top plotted alone so you see how well the line fit the data points.

4. Then I show only the fitted line in a plot with an aspect ratio closer to the natural 1:1 for the height and length in mm (but not exact). Should give a better natural view of the top arch.

I will just add a comment up against what I saw on the the Ole Bull copy exhibition yesterday. Some of them had a longitudinal arch somewhat resembling the yellow and red line in plot #2.

[Anders Buen]

Here is the back plate fitting procedure and results.

1. First the raw data as read out of Möckels chart. From this stage on I skipped the first and last point (guessed by me) as they make the fitting somewhat less good.

2. Here we see the best fitted lines through the raw data. We see that already the first red polynomial and line does a pretty good job. Its a parable plus a linear term indicating that the highest point is south of the mid distance from the neck to the lower block.

3. Here we see the next best and first fitted lines in the same plot as the raw data. The difference between the 6th and 5th order fit is so small that I choose the simpler polynomial. Both the red and the blue line are close to the data, but we see that the blue is better then the red.

4. Here I have plotted the same figure as in 3. only using a more natural dimension between the x and y axes (but it is not exact). We barely see the red line, so the second order polynomial is describing the long arch pretty well.

5. The fitted line for the 5th order polynomial plotted close to as we would see it (but not exact)

6. The top and back plate best fitted lines to the long arch compared.

[Anders Buen]

For fun I measured the long arch of a Hardanger top I have here by Gunnar Røstad (Roestad) from the 1900-1910ds and used the same fitting procedure as the for the former Strad data. I just show the comparison of the best fitted line of the top arch of the Roestad with the Ernst Neruda for comparison.

The curves are pretty close, and closer than the top and the back of the Strad. One could be tempted to speculate if Roestad had used Moeckels data, but this top is made before Möckel published his book..

Reading out the data from the Roestad top I used a profile gauge with plastic pieces that are locked by a magnetic system that can fit any surface like this: http://www.hartvilletool.com/shared/images...arge/64252.JPEG.

[Anders Buen]

Finally I show a comparison of the long arch of a John H Tjønn hardanger top from the 1950ties to the Røstad and the 1709 Stradivari.

1. First the natural comparison of the arches, not easy to see the differences.

2. A plot where the arch height is stretched in comparison to the distance. You see the differences more clearly.

Usually the bridge is placed a little nonger north on a hardanger than on a violin and the f-holes are longer.

[Johnmasters]

Smart. Why do you need the function to be infinitely differentiable?

And expanding your function using a Taylor series would express exactly the same thing using a polynomial. Why does not the polynomial do it when it is expresses exactly the same function?

I do see the practical implications and benefits from using the simple function, eg in a spreadsheet etc for making a model.

I did not know much.............. but I knew the violins were virtually flat in the middle and that second derivative likely should go to zero at the neck. That is all. ............... Oh, and the thing I saw about CC which was the original hint. I will wait to see if anyone sees it or has an interest. Nice to have all the expanding functions of this sort.

The practical benefits and implications are likely different for the two of us. I will keep quiet about my own for the time being.

A Taylor series, of course must be truncated. The terms for Cos and Sin alternate in sign and converge slowly. There is also the need to use the chain rule because the argument of the cos is usually some power of x.... even a non-integer power.

By the time the sum of my two functions was fairly represented (over an entire pi/2 radians, there may be dozens or hundreds of terms in a polynomial. Worse for a product or a polynomial argument for the cosine. (And is the inclusion of a fractional power in a power series considered a polynomial? no, I guess not.) Fortunately, no need to prove anything about that.

I am sorry to criticize spline (and you) previously, but I did not think that a 10 or 20 component spline would be enough to see anything with physical implications.

That is basically why I eventually decided not to do a taylor seriies. Although, it would be nice to fit product, sum, and even the polynomial argument in a cosine. Also, too much work.

But they all have the same end boundary conditions. I think that was the key... Infinitely differentiable was perhaps not needed....... many weird combinations (sum, product, add terms in cos argument) likely will work because of these end boundary conditions of the cos(x^n).

[Johnmasters]

Anders,

I did not really answer your question. (why infinitely differentiable? Maybe not needed, just turned out to be inherent in the approach.) But it may be good insurance.

Consider the curtate cycloid and rolling wheels..... shift t by pi so that maximum height is at x=0. Then pitch a and b, and the additive constant to y = C + ..... Finally, forget about rolling wheels.

One winds up with cos(t) plotted on a distorted x-axis with x >>> t(1 + A*sin(t))... (sign change because t >>> t-pi)

I like cosine........ and it is close. I just decided to change the distortion of the x-axis. I know that cosine(t) on x=(a+b*cos(t)) is a bit backwards, but at least in some sense, cos(x) on x = a power series in 't' could be made to do what I wanted. A lot of things can be made to converge quickly to an actual arch.. There must be a physical reason for this, and I aim to find it out.

The problem with curtate cycloid is that it curves too quickly where I want the function to go flat toward x=0. The next easiest thing was to make the argument of cosine a power series in x. Then I played with the graphing tool ** and found that fractional powers could be used.

There are a lot of things that all have the same end boundary conditions, so I used various arguments in cosine functions.

Now, I have committed the cardinal sin of the true mathematician. I told you I experimented with the graphing tool, and even justified my approach.

A REAL mathematician strolls into the room, writes something on the chalkboard and quietly grasps his hands behind his back. And says nothing. Great for dramatic effect and very impressive to many women students.

** Mathgv, user-friendly and free if some readers here don't have it......

[Johnmasters]

(I have been thinking about starting an own tread on this, but it is so close to what John does and it adds to an earlier discussion in using polynomials. They were doomed as not 'doing it'. And I think this example show that polynomial fits may work well, at least for this particular instrument)

Fitting the curve was only a demonstration. Polynomials still have a problem for me. I still have in mind what end boundary conditions are needed. Ultimately, I want a physical model.

By the way, could you tell me if a spline is more than a best-fit polynomial? What is "best fit"? There must be some consideration of what the polynomial does outside of the ends, otherwise it may do something weird.

How is a spline made? My main interest is in the end boundary conditions.

[Anders Buen]

By the way, could you tell me if a spline is more than a best-fit polynomial? What is "best fit"? There must be some consideration of what the polynomial does outside of the ends, otherwise it may do something weird.

How is a spline made? My main interest is in the end boundary conditions.

I do not know how a spline is made. But I saw Ted White draw up the longitudinal arch of a violin top using only two or three points and a spline function in Rhino, a drawing program for designing 3D objects. It looked pretty efficient. I think the line can be handled with different kind of "handles" to make it fit what you want dragging around with the mouse pointer.

[Johnmasters]

I do not know how a spline is made. But I saw Ted White draw up the longitudinal arch of a violin top using only two or three points and a spline function in Rhino, a drawing program for designing 3D objects. It looked pretty efficient. I think the line can be handled with different kind of "handles" to make it fit what you want dragging around with the mouse pointer.

I used to have a pirate copy of Rhino. Now I have BobCad v.23.

Splines don't suggest a physical way to actually make a violin arch. Suppose an end compression on a special arch caused the entire elevation to be multiplied by a single constant.... Now THAT would be special. No change of shape, only a change of scale. Maybe indefinite differentiation would be useful here. I want to ask Woodhouse, but he is either on vacation or is ignoring my letters.

I argued before for a ventral peg to keep the arch from flattening while one bends the top down to tapered ribs... Can you figure it out? Hint: no reasonable craftsman needs locating pins to glue on a top. I never used them, and I am just a self-taught maker.

[Anders Buen]

Splines don't suggest a physical way to actually make a violin arch. Suppose an end compression on a special arch caused the entire elevation to be multiplied by a single constant.... Now THAT would be special. No change of shape, only a change of scale.

You do not seem know what splines are yet, John. But my main intrest is to describe natural arches which polynomials seem to do very well. Much better than I anticipated before trying it. Although calculations of the one dimensional effect on the arches may teach us something basic. I believe the method is too far from the natural one in a plate where in plane effects in other direction than only the longitudinal is going to have a large effect.

There are also wood moisture effects that are taking palce that have larger practical implications than the tensions alone. Wood creep and shrink at different rates in the different directions and it affect the arch in such a way tat it matter to to the setup of the violin. These effect work together of course.

I do not think we will see the change of scale thing wil happen in a violin top. That is a fancy idea from playing with curves and idea and not good wood intuition.

I argued before for a ventral peg to keep the arch from flattening while one bends the top down to tapered ribs... Can you figure it out? Hint: no reasonable craftsman needs locating pins to glue on a top. I never used them, and I am just a self-taught maker.

I have always thought that the rib taper was linear from the lower to the upper block. I have seen tapered ribs at the sides of the upper bout that was intentionally made so to stiffen the cross arch at the upper bout. I think it works. I think my hardangers has been made without the taper of the ribs. You do not need to reach over the top as there is no playing in position.

[Anders Buen]

I argued before for a ventral peg to keep the arch from flattening while one bends the top down to tapered ribs... Can you figure it out? Hint: no reasonable craftsman needs locating pins to glue on a top. I never used them, and I am just a self-taught maker.

I think you are prestressing your top before mounting the bass bar, fitting the bar and then glueing it in while the top is under tension. Then realeasing your top it has a shape that will fit your tapered ribs.

[Johnmasters]

You do not seem know what splines are yet, John.

I asked YOU just now, and you said that you did not know either ...

There are also wood moisture effects that are taking palce that have larger practical implications than the tensions alone. Wood creep and shrink at different rates in the different directions and it affect the arch in such a way tat it matter to to the setup of the violin. These effect work together of course.

I do not think we will see the change of scale thing wil happen in a violin top. That is a fancy idea from playing with curves and idea and not good wood intuition.

No, it usually will not. All you need to do is dig in your heels, and we will see who gets where, and why.

Nothing wrong with fancy ideas and playing with curves. Nothing wrong with my wood intuition, I think.

I think there must be a variational exercise which could state the maximal increase in potential energy as the top is compressed a given small amount. (holding the edges at z=0) In other words, a stiffest possible arch in compression. I don't give a damn about wood. Do it for an isotropic metal, and you will have done something.

I have always thought that the rib taper was linear from the lower to the upper block. I have seen tapered ribs at the sides of the upper bout that was intentionally made so to stiffen the cross arch at the upper bout. I think it works. I think my hardangers has been made without the taper of the ribs. You do not need to reach over the top as there is no playing in position.

There are visible scoring marks on Strads (or perhaps their moulds) in the upper bouts, as I understand it. Stradofear ?? got your ears on?

The continuous taper top to bottom is utterly and exactly what is not seen in Cremona violins.

[Anders Buen]

I asked YOU just now, and you said that you did not know either ...

Yes, and I did not claim that the curve needed to explain physics. It will fit the shape. Splines is easy to google if we really want to know.

Now I do think that one could speculate that the polynomial arches can explain some physics. Maybe the orders over two does explain deformations? Maybe the second order polynomials is where many would start off? It is a loose speculation. I am not quite sure that such an arch is good for the sound.

[Anders Buen]

"In mathematics, a spline is a special function defined piecewise by polynomials. In interpolating problems, spline interpolation is often preferred to polynomial interpolation because it yields similar results, even when using low-degree polynomials, while avoiding Runge's phenomenon for higher degrees."

Read more: http://en.wikipedia.org/wiki/Spline_(mathematics)

[Johnmasters]

Yes, and I did not claim that the curve needed to explain physics. It will fit the shape. Splines is easy to google if we really want to know.

Now I do think that one could speculate that the polynomial arches can explain some physics. Maybe the orders over two does explain deformations? Maybe the second order polynomials is where many would start off? It is a loose speculation. I am not quite sure that such an arch is good for the sound.

Why in God's name would I want to simply find an algebraic expression for an arch? Whatever splines are, they are just a best fit and also used in CAD to make smooth curves.

You will need order 3 to find non-constant second derivative... I don't need to tell you that. I know that second derivative is not exactly curvature in the physical sense. But I want continuity of both curvature and changes in curvature. Maybe more.

Maybe polynomials is where some would start off. Maybe most. I thought of it briefly and discarded it. My fitting functions need specified end boundary conditions. (excuse the arrogance: I am not "most".)

My assumption is that the second derivative is zero at the neck end, and goes flat toward x=0. Maybe that is false, but I find that an inflection near the neck end causes a downward buckling which I don't like.

I ignore the roll of the bar because early bars did not seem to extend beyond the flat part. The arch must be optimal even before the bar.

I am forgetting sound for now for two reasons. 1.) nobody can say why internal stresses affect sound, nor how. 2.) I am thinking purely of static stability. I don't believe the arches as seen are deformed back-like arches as I have said. They are too perfectly shaped for that. ("Perfect" means very nice and no irregularities in the best examples in spite of the odd outlines.)

The shapes are related to CC. I am willing to accept that the cross arches are near to CC even if Strad did not know what they were. They remain so after many years, humidity or not. I suggest that the long arch has properties that are related....... and for a reason.

[Johnmasters]

"In mathematics, a spline is a special function defined piecewise by polynomials. In interpolating problems, spline interpolation is often preferred to polynomial interpolation because it yields similar results, even when using low-degree polynomials, while avoiding Runge's phenomenon for higher degrees."

Read more: http://en.wikipedia.org/wiki/Spline_(mathematics)

I looked it over. It seems that spline components are chosen for fast convergence, presumably with some special conditions. My expansion functions would then be a special sort of spline component. But they have a physical interpretation which seems appropriate.

[Johnmasters]

I think you are prestressing your top before mounting the bass bar, fitting the bar and then glueing it in while the top is under tension. Then realeasing your top it has a shape that will fit your tapered ribs.

That is right, and it seems to work well.. But the present excercise leaves out the bar, and for a good reason which I have already stated. I also ignore the eventual bend of the top at the upper ribs.

"Then realeasing your top it has a shape that will fit your tapered ribs." This part is wrong. There is nothing to account for the tapered ribs......... that is what the ventral post would be for.

[Anders Buen]

My assumption is that the second derivative is zero at the neck end, and goes flat toward x=0. Maybe that is false, but I find that an inflection near the neck end causes a downward buckling which I don't like.

The neck extend a bit into the top in hte neck end, then there is a block. Is there still a downward buckling when the upper block is included in the model?

The shapes are related to CC. I am willing to accept that the cross arches are near to CC even if Strad did not know what they were. They remain so after many years, humidity or not. I suggest that the long arch has properties that are related....... and for a reason.

I understand that the spruce wood creep over time. Eg the lower holes in the f-holes look like droplets and are not perfectly round as they were new. I wonder if that is related only to humidity changes or if this is a slow process that goes on over the centuries. If the plate becomes narrower the arch should go flatter over time.

Sam Z say that his fiddles arch in the top rises. David B indicate an opposite direction.

Old wood is not more resistant to humidity changes than new wood, according to Rex Thompsons experiments. So does spruce still shrink over time irrespective of humidity changes?

As the width is larger at the widest at the bouts, the effects should be largest there.

[David Burgess]

Sam Z say that his fiddles arch in the top rises. David B indicate an opposite direction.

Looks like I'll need to clarify.

Total arching height, at the center, measured from the rib gluing surface can go up, even as the center section flattens. The flattening in this case is due to the bulging of the upper and lower bouts due to compression. Whether there is an increase or decrease in total height at the bridge position can depend on a number of things, such as the original arching height and shape, graduations, whether the bar is sprung, string angle, and tightness of the soundpost. Sam came from the Morel shop, so it's likely that he fits posts tighter than I do.

Lest anyone should think that a sprung bar, or a stronger bar, or a tighter post is a strategy for eliminating arching distortion, it doesn't exactly work that way. For example, if a post is tight enough to reduce overall flattening near the bridge, it will tend to raise the inner part of the treble ff so it no longer aligns with the outer part, perhaps create a larger bulge on the lower treble side, and distort the back more severely. Sprung bars or stronger bars can help hold the center up, but contribute to greater distortion where the load is transferred. If one shines reflected light under the fingerboard and tailpiece in the upper and lower bouts, one can see a depression, and sometimes a concave kink where the bar is attached on most old fiddles. The bar has held things in shape at the attachment point, but the surrounding top will still want to do its own thing.

Lest anyone think these are isolated opinions, none of this is news to restorers. While some of these arching threads were running, I received an email from a restorer who reads here, who mentioned that one expensive old Italian instrument he takes care of undergoes aching restoration every 25 to 30 years.

What was the original shape of a Strad long arch? In post #124 of THIS THREAD , Fjodor posted a picture which shows the original bar from the Messiah. What do you think? Does it show much of a "plateau"?

[Anders Buen]

What was the original shape of a Strad long arch? In post #124 of THIS THREAD , Fjodor posted a picture which shows the original bar from the Messiah. What do you think? Does it show much of a "plateau"?

I guess you mean this picture? I have added a straight line on the bar. Difficult to say clearly that it is not flat in the centre.

In sacconis book there are pictures of two small baroque bars by Niccolo Amati and Andrea Amati. The first one is not flat while the second is flat, if not saddle shaped. But that is different makers than Strad.

[David Burgess]

I guess you mean this picture? I have added a straight line on the bar. Difficult to say clearly that it is not flat in the centre.

Actually, I found it a little easier in the original photo, comparing the curvature of bar with whatever that brown thing is directly in front of it.

Also, keep in mind that this is a much shorter bar, so an awfully long portion of it would need to be flat in order to fit the plateaus we see today.

Of course, bars will distort too, which can be noticed by making a template of a bar, the then matching it with the bar many years later. I was thinking that maybe the Messiah bar had a kinder environment, and maybe a shorter service life than some other bars.

But as I keep saying, it's easy enough for anyone to do their own accelerated experiments.

[Anders Buen]

Actually, I found it a little easier in the original photo, comparing the curvature of bar with whatever that brown thing is directly in front of it.

Also, keep in mind that this is a much shorter bar, so an awfully long portion of it would need to be flat in order to fit the plateaus we see today.

Yes the picture here is not very sharp. Thanks for sharing your experience on the matter. There seem to be many effects taking pace with the arching. I remember that you also say that its easier to increase the arch a little than reducing it in arching corrections. Weisshaar does not recommend sprung bars. I assume what you describe above may be the reason.

The old bass bars are some 3-5cm shorter or so?

I enclose two bigger pics of the bar with the line in different positions.