Ping John Masters and all the techies..........

[Roger Hill]

Continuing from the Marilyn Wallin thread:

QUOTE (Roger Hill @ Mar 27 2010, 07:28 PM)

I am not convinced of this, John. Each chain section hangs as it would if alone and suspended at the end-points, i.e. the tension is continuous. I don't think you will find a kink there. If so, it would be visible to the naked eye and it is not. Try it and see.

John Masters

There is no kink because the first derivative or slope is continuous. But is the second derivative continuous? I find that hard to believe if the sections suddenly change density.

Well, John, let me expose my thinking regarding catenaries and buckling and second derivatives. If there is an error in my thinking I am sure you and a dozen others will point it out. I am a big boy, I promise to neither stamp my feet nor cry if I am wrong.

First, lets consider a chain of length S, attached at the origin at one end and hanging from a second attachment point some distance away at height h. (We are assuming, of course, that s is greater than the distance from the origin.) The shape of the chain is given by

y = H/w(cosh wx/H – 1)

y’ = sinh w/Hx

where H is the horizontal component of the tension in the chain and w is the weight per unit length. If you follow the derivation given here:

http://mysite.du.edu/~jcalvert/math/catenary.htm

you find that

H = (w/8h)(S^2 - 4h^2)

And

y’ = sinh(wx/H)

Substituting for H gives

y’ = sinh(8hx/(S^2 – 4h^2)

independent of the weight per unit length of the chain, from which I conclude that y” is also independent of the weight per unit length of chain. (But I think we already knew that, all catenaries are created equal and their shape depends only on their length and vertical separation of the ends.)

Now, suppose we make an inhomogeneous catenary of kevlar thread of length Sk and and logging chain of length Sl. Attach the kevlar thread to the origin and attach the logging chain to an attachment point at some height h, less than Sl. The logging chain and the kevlar thread each assume catenary shapes which have their slopes continuous, essentially zero at the point of connection. Neither slope has any dependence on its weight/length, only on the vertical separation of it’s end points and length and the same is true of its second derivative. The catenary of the thread thinks that there is a near infinite amount of kevlar thread running outward in order to get the tension it experiences at the attachment point, while the logging chain thinks that it is attached to the origin directly. If we can find some Unobtainium thread, having zero weight/length and infinite E, the inhomogeneous catenary becomes simply the catenary of the chain, displaced to the right by the length of the Unobtainium thread. I just do not see how a discontinuity in the second derivative arises due to the property of all catenaries that the weight/length cancels out of the equations. Another way of saying this is to note that all the forces are proportional to the mass/length. Of greatest importance is the property of the catenary that it attempts to make all loading forces co-linear with the curve of the arch, thereby inhibiting the buckling. If the second derivative were discontinuous, this property would not obtain for the inhomogeneous catenary, but I just don’t see how that happens. Thus, I am still a proponent of the inhomogenous catenary for the long arch shape.

If I have a "kink" in my thinking, please explain it to me.

[stradofear]

If one is thinking precisely, as I know John always is, the ONLY lines that have a consistent curvature from one end to the other are circles and their sections, or straight lines. Nothing else. If you maintain your "catenary" has a consistent curvature, then it's a circle in disguise.

[Wm. Johnston]

independent of the weight per unit length of the chain, from which I conclude that y” is also independent of the weight per unit length of chain. (But I think we already knew that, all catenaries are created equal and their shape depends only on their length and vertical separation of the ends.)

Now, suppose we make an inhomogeneous catenary of kevlar thread of length Sk and and logging chain of length Sl. Attach the kevlar thread to the origin and attach the logging chain to an attachment point at some height h, less than Sl. The logging chain and the kevlar thread each assume catenary shapes which have their slopes continuous, essentially zero at the point of connection. Neither slope has any dependence on its weight/length, only on the vertical separation of it’s end points and length and the same is true of its second derivative. The catenary of the thread thinks that there is a near infinite amount of kevlar thread running outward in order to get the tension it experiences at the attachment point, while the logging chain thinks that it is attached to the origin directly. If we can find some Unobtainium thread, having zero weight/length and infinite E, the inhomogeneous catenary becomes simply the catenary of the chain, displaced to the right by the length of the Unobtainium thread. I just do not see how a discontinuity in the second derivative arises due to the property of all catenaries that the weight/length cancels out of the equations. Another way of saying this is to note that all the forces are proportional to the mass/length. Of greatest importance is the property of the catenary that it attempts to make all loading forces co-linear with the curve of the arch, thereby inhibiting the buckling. If the second derivative were discontinuous, this property would not obtain for the inhomogeneous catenary, but I just don’t see how that happens. Thus, I am still a proponent of the inhomogenous catenary for the long arch shape.

If I have a "kink" in my thinking, please explain it to me.

I had written something else here but then started looking into this a little. I haven't found a good answer yet but found these links which may be of interest for arches.

The shapes produced by viscous threads, you can get catenaries and a non-catenary shape that looks potentially interesting for long arches. http://www.aps.org/units/dfd/pressroom/papers/seiwert.cfm

A generalization of a curtate cycloid, http://mathworld.wolfram.com/Roulette.html

Wikipedia has some information on the curves produced by non-uniform hanging chains but the change in density appears to be continuous in these cases so they don't really answer your question. http://en.wikipedia.org/wiki/Catenary#Variations

[Roger Hill]

Thanks, William. This one: http://www.ams.org/notices/201002/rtx100200220p.pdf

seems to be on point, although I have yet to digest it.

Michael, I am only quibbling over mathematical esoterica. I think John is looking for the smoothest analytic function that looks strad-like. I think the inhomogeneous catenary accomplishes what he wants, but then again, it may not. It certainly is composed of more than one analytic function, or perhaps none at all if I am wrong about each of the chain sections hanging as individual catenaries, with only the join position as determining which two specific catenaries would result from the hanging of a two section chain. I will continue playing with such minutiae in preference to corners, scrolls, beestings, and the things that more normal violin hobbyists obsess over

[Michael_Molnar]

I would like someone to produce a file of data points that define a "Strad" curve. Once we have that we can argue over which curve(s) fit it.

Until that happens this thread is pointless. We must agree on something specific to argue about.

[Craig Tucker]

I would like someone to produce a file of data points that define a "Strad" curve. Once we have that we can argue over which curve(s) fit it.

Until that happens this thread is pointless. We must agree on something specific to argue about.

I wish I was brief and to the point, like this.

Alas, I fear it is not to be.

[Wm. Johnston]

I would like someone to produce a file of data points that define a "Strad" curve. Once we have that we can argue over which curve(s) fit it.

Until that happens this thread is pointless. We must agree on something specific to argue about.

Is the only point for violinmaking or study to arrive at something Strad-like? I don't think so. If that is the case a lot of very highly rated old violinmakers have failed.

Based on the original post and the posts that lead to this one, I thought this was just about ways to arrive at a design for a smooth pleasing arch.

[Johnmasters]

independent of the weight per unit length of the chain, from which I conclude that y” is also independent of the weight per unit length of chain. (But I think we already knew that, all catenaries are created equal and their shape depends only on their length and vertical separation of the ends.)

If I have a "kink" in my thinking, please explain it to me.

I am reading this with a minute to go before I leave for orchestra. I will read this and William's letter tonight.

for sure, the center section will always be a catenary. What about the ends. Take 10 feet each end to be anchor chains from the Forestal. Hang them a 10 feet apart and tie a 10.5-foot length of jeweler's chain between the ends.

Is the entire three-componet chain a catenary? I don't think so. The light chain has to pull in the heavy ones a bit. Try it with a 9-foot jewelers chain. The tension in the center chain is not from only its own gravity. And the jewele's chain does not help the anchor chain in either case. I will admit that there is only one shape of catenary, but different sizes and a short section of one will look different at a different scaling.. It is like a parabola in that sense.

I will look at your derivation, but it seems something is missing,

Thanks, William. This one: http://www.ams.org/notices/201002/rtx100200220p.pdf

seems to be on point, although I have yet to digest it.

Michael, I am only quibbling over mathematical esoterica. I think John is looking for the smoothest analytic function that looks strad-like. I think the inhomogeneous catenary accomplishes what he wants, but then again, it may not. It certainly is composed of more than one analytic function, or perhaps none at all if I am wrong about each of the chain sections hanging as individual catenaries, with only the join position as determining which two specific catenaries would result from the hanging of a two section chain. I will continue playing with such minutiae in preference to corners, scrolls, beestings, and the things that more normal violin hobbyists obsess over

I need analytic functions to find elevations of mesh points for an FEA model. If it looks Strad that is OK, but I am not trying to go there. I want the most smoothly-changing curvature possible. Buckling seems directly concerned with curvature, and I want NO discontinuities in curvature. That is why I want to get some nice curve that is not made of grafted sections.

I decided to terminate the longitudinal arch with no recurve, end forces seem to buckle the edge downward (and I ignore the extra wood of the neckblock except for a simple doubling of the shell elements there.) I want the arch to come down nice and convex, but approaching flat just as one gets to the end. This looks good, also familiar actually.

I find that the second derivative of a curtate cycloid goes to zero at cos(t) = -b/a. I divided dy/dt by dx/dt and then differentiated again by t. No need to further divide by dx/dt as I was setting 2nd derivatie to zero where dx/dt was not zero. I am using the section of CC up to the place where cos(t)=-a/b (about 2.1 radians for a flat CC). I can flatten a central area in a way to not hurt gradual change of curvature, but it will deviate from CC a little. The extra requirement overly constrains the CC.

I will look over the postings now.

If one is thinking precisely, as I know John always is, the ONLY lines that have a consistent curvature from one end to the other are circles and their sections, or straight lines. Nothing else. If you maintain your "catenary" has a consistent curvature, then it's a circle in disguise.

I did not want constant curvature (which you call consistent curvature), I want "smoothly and gradually changing curvature." It is a bit of detail point. But I always hand-rubbed my plates for lumps and tried to eliminate "discontinuous curvatures" ever since I started in high school. It seemed a natural thing to desire a smooth blend of curvatures. I am sure you do it also. The math thing here is to see if there is a way to find a top least likely to buckle under compression by actually going after math curves that have that property in the first place.

If there is no real finding, at least they look nice and graceful.

[Johnmasters]

Roger, I have not read carefully yet, but your H is the horizontal component of force which my ridiculous exagerated example was meant to look at. I think that this H is discontinuous over the join. But I will keep looking at it. Too tired tonight to make much of it.

[Roland]

I would like someone to produce a file of data points that define a "Strad" curve. Once we have that we can argue over which curve(s) fit it.

Until that happens this thread is pointless. We must agree on something specific to argue about.

Yes. We talk about arching theories all the time but rarely do we compare our theories with the archings of actual instruments. William Johnston did this here - have a look how many replies he got.

I tried something similar in this topic.

[Torbjörn Zethelius]

IMHO

The long arch is the function of the crossing of inside diagonal catenaries with catenaries across the upper and lower bouts. It's the same for both the back and belly plates. It was never conceived as a curve per se.

The photo illustrates my point.

Torbjörn

[Roger Hill]

Roger, I have not read carefully yet, but your H is the horizontal component of force which my ridiculous exagerated example was meant to look at. I think that this H is discontinuous over the join. But I will keep looking at it. Too tired tonight to make much of it.

Hi John:

H cannot be discontinuous at the join, otherwise and we do not obtain static equilibrium. Thinking about extreme examples with kevlar thread and heavy chain, we could obtain a curve that is for practical people a square function. It will clearly tend to buckle at the bends, which are in reality corners. I suspect that your intuition is in fact, correct and I am wrong.

Looking at the article in the link, the author does an integration to obtain y'. If we divide the curve into two sections demarked by Xo, to get the curve for points x <= Xo, we must do the integral up to Xo and

y' = sinh(wx/H)

the ordinary catenary shape.

for x> Xo,

y' = sinh (wXo/H + w'(x-Xo))

and you quickly discover that the second derivative is discontinuous at Xo.

I Bow to your superior intuition

[Johnmasters]

Hi John:

H cannot be discontinuous at the join, otherwise and we do not obtain static equilibrium. Thinking about extreme examples with kevlar thread and heavy chain, we could obtain a curve that is for practical people a square function. It will clearly tend to buckle at the bends, which are in reality corners. I suspect that your intuition is in fact, correct and I am wrong.

Looking at the article in the link, the author does an integration to obtain y'. If we divide the curve into two sections demarked by Xo, to get the curve for points x <= Xo, we must do the integral up to Xo and

y' = sinh(wx/H)

the ordinary catenary shape.

for x> Xo,

y' = sinh (wXo/H + w'(x-Xo))

and you quickly discover that the second derivative is discontinuous at Xo.

I Bow to your superior intuition

I hate bowing .. Just got up at 1:00 pm Saturday. As I went to bed, I realized that rescaling of different portions of the cycloid and taking portions is likely important. Also, no torques at any point along chain... that is why a chain is postulated in the first place.

I don't know the answer yet. But consider a heavy chain in a deep catenary. Decide where to cut out a central section. Fix those points in space and cut out the section. Tie any length of thin chain to those points. Now, all three sections are of a catenary shape. That is certain.

Now, unfasten those points... the shape will change. Will it change to a different section and scaling of a catenary? I don't know... but I am suspicious. But I am ready to read the proofs.

[Wm. Johnston]

.

[Roger Hill]

I looked up one derivation of the catenary in a mechanics book last night. In this book they first solve for the arclength S of the curve, do some manipulating, integrate again, and then arrive at the catenary. I think the important part for this discussion is this equation,

dy/dx = w*S/T

dy/dx is the slope of the chain, w is mass per length, S is arclength, and T is tension. So this actually gets to the answer for your question without actually solving for the curve. In your inhomogeneous catenary w is discountinuous at the junction but T is continuous and I'm pretty sure S is as well (unless you have to define it differently on the two sides of the junction). So the result is that because w has a sudden change then so does the slope of the curve.

When I have time later I might solve for the exact solution just to play with it some.

I don't believe the slope is discontinuous, only the second derivative. Substitute x = Xo in the equations above to see this. I will have to study this some more before I am filled with the great certainty I normally exhibit

[STRAD~STYLE]

Fight!

[Wm. Johnston]

.

[Johnmasters]

I don't believe the slope is discontinuous, only the second derivative. Substitute x = Xo in the equations above to see this. I will have to study this some more before I am filled with the great certainty I normally exhibit

y = H/w(cosh wx/H – 1)

y’ = sinh w/Hx

I assume C is some constant.

H = (w/8h)(S^2 - 4h^2)

And

y’ = sinh(wx/H)

Substituting for H gives

y’ = sinh(8hx/(S^2 – 4h^2)

independent of the weight per unit length of the chain, from which I conclude that y” is also independent of the weight per unit length of chain. (But I think we already knew that, all catenaries are created equal and their shape depends only on their length and vertical separation of the ends.)

I agree that they are all the same shape, fastened at y= plus infinity. Different scalings, however. When you select the separation of the ends, at finite y, and a sagita you are introducting a scaling.

The shape comes from minimizing the gravitational potential energy without any torques along the line. You integrate along the entire curve and minimize the pootential energy by variational methods on the shape.

Lets make the ends at the same y. If you splice in a light chain in the middle of a heavy one whose ends are at y=o, then its projected fastening points,y, will not be known unless you do the variational trick all over again.

[Wm. Johnston]

Looking at the article in the link, the author does an integration to obtain y'. If we divide the curve into two sections demarked by Xo, to get the curve for points x <= Xo, we must do the integral up to Xo and

I can find the solution that you posted believable and they seem to have the correct behavior for situations where one section is very dense and the other is very low density. But I think there could be something tricky going on.

[Johnmasters]

I can find the solution that you posted believable and they seem to have the correct behavior for situations where one section is very dense and the other is very low density. But I think there could be something tricky going on.

The problem that I am having is with the integration part. The author of the article left out the upper limit of integration on the y' integral which is what I had to think about a while. Since the integration limits on x are 0 and x (bad notation!) I think the integration limits on y' would need to be y'(x=0) and y'(x). To take the discontinuous mass density into account you have to integrate one density up to the junction and then start integrating the other side, which you did. The potential problem that I see is with the integration limits on y' at that junction. For the central section you have to integrate on y'(0) to y'(x0) in the limit that you approach x0 from the left. Then on the other side of the junction you have to integrate from y'(x0) to y'(x) while approaching x0 from the right. If you skip over this step then I think you might be forcing y'(x0) to be the same from both directions which may or may not be correct (that's the original question which started this), or you should end up with some arbitrary constants. I'll have to look at an alternate derivation again later, maybe it will be clearer then.

I think you will find that there are two integration paths to be minimalized. One for each segment. For the middle segment, you do not even know the length of the chain. You cannot assume that it is attached at the same points as the outer lengths. In fact, it won't be if the length densities are different.

[Wm. Johnston]

.

[Don Noon]

I don't believe the slope is discontinuous, only the second derivative. Substitute x = Xo in the equations above to see this. I will have to study this some more before I am filled with the great certainty I normally exhibit

I'm not interested in the math, but thinking about the limiting case of heavy chain attached to weightless fishing line:

The slope has to be continuous, as there is only tension at any point, including the joint... i.e. the slope of the chain and the slope of the fishing line have to be the same at the joint, otherwise there would be an unbalanced sideforce.

The next derivative (curvature) will not be continuous. The weightless line will always have zero curvature, and the chain will have some curvature.

[Wm. Johnston]

I'm satisfied with this argument. Seems reasonable that if you had two heavy chains connected with a bit of fishing line at the bottom that the chains would take their typical shape except spread farther appart and that the fishing line would just be in a straight line connecting the bottom pieces.

I think I'll remove some of my earlier comments which were centered around this this point since the it was easily dealt with in words.

[Ken Pollard]

The next derivative (curvature) will not be continuous. The weightless line will always have zero curvature, and the chain will have some curvature.

Kinda hard to form a catenary with a weightless line, isn't it?

I think you're right on a practical level.

[Johnmasters]

Kinda hard to form a catenary with a weightless line, isn't it?

I think you're right on a practical level.

Yes, Don is confirming the views established earlier. That is why he is right.