## polycylindrical diffuser shape

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### polycylindrical diffuser shape

Hello All,

Scott and I happened to be flapping our jaws at each other recently, and the subject of polycylindrical diffuser shape came up. What is the natural shape of a bent piece of stiff material, like plywood, when you build a polycylindrical diffuser?

We agreed that it certainly wasn't of constant radius, so could not be a semicircle or chord of a circle. Scott suggested a catenary, which is the shape of a loose hanging chain supported at both ends. But this didn't seem quite right to me either, because I remembered some of the very specific math necessary to derive this curve.

So I clenched my teeth and looked into the mechanical engineering literature. Not being an ME myself, I have a natural aversion to the greater-than-2nd-order differential equations which stressed rods and beams seem to require for analysis. But I identified (hopefully correctly) the bent plywood case as corresponding to the ME's "axially compressed buckled rod with ends pinned."

"Axially compressed" means that the rod is subject to a compression force along its normally uncompressed length axis. "Ends pinned" means both ends are held in place, but are free to rotate and change angle about the end points. "Buckled" means that the compression force has caused a static deformation; that is, the force on the ends has exceded the critical, or maximum load (the formula for which Euler first derived).

The solution to this particular problem is surprisingly simple. It is given in both Engineering Solid Mechanics: Fundamentals and Applications, by Ragab and Bayoumi, pp. 398-400, and Buckling of Bars, Plates, and Shells, by Robert M. Jones, pp. 12-13 and 127. The displacement d as a function of position x along the total length L is:

d = C*sin(m*π*x/L)

where x is the position between 0 and L, C is some constant and m is an integer mode number [!!!]

The solution for mode 1, which the system will fall into if there are no constrains otherwise preventing this, and normalizing C=1 for readability, is:

d = sin(π*x/L)
where 0 <= x <= L

Which is half a sine wave.

Spooky, huh?
Terry Montlick Labs
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Terry Montlick

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### Re: polycylindrical diffuser shape

So, what does this mean to a layman trying to actuall accomplish this construction wise? Just yesterday, I posted a reply on another bbs to a thread author offering Scotts interpretation of building one of these poly devices in the simplist manner that I understood from a past post of Scotts. Maybe I'm missing something here, but does this mean that Scotts poly building technique doesn't work, or ALL polys with shapes other than a sine wave don't work...or what? Not being a math person, how would one achieve a "1/2 a sine wave"using the 48" width of a 1/4" plywood sheet if that is what makes this work properly? Thanks for any insight. I really want to understand this.
fitZ

homestudiobldr

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### Re: polycylindrical diffuser shape

homestudiobldr wrote:So, what does this mean to a layman trying to actuall accomplish this construction wise? Just yesterday, I posted a reply on another bbs to a thread author offering Scotts interpretation of building one of these poly devices in the simplist manner that I understood from a past post of Scotts. Maybe I'm missing something here, but does this mean that Scotts poly building technique doesn't work, or ALL polys with shapes other than a sine wave don't work...or what?

It means absolutely nothing of consequence to the layman.

Scott's poly building technique will work, because it simply applies pressure at the sides to bend out the middle. This will mathemagically form a half sine wave shape.

It is only of consequence to the intellectually curious and/or those of us who want to do fancy-schmantzy BEM or other modeling or simulation of poly diffusers.

- Terry
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### Re: polycylindrical diffuser shape

Terry Montlick wrote:d = sin(π*x/L)
where 0 <= x <= L

Obviously x needn't go as far as 0 or L, but would most often be a subset within those boundaries.

I had thought that sin() was an approximation for buckling, and that in reality it was either a less predictable parabola, or minor variations on a parabola depending upon the existance of weak spots.

In any event, it's neither sin, nor parabola, nor catenary -- because we all know you can take a piece of plywood and bend it into a circluar tube, or into a shape where the ends are further in than the extremities (i.e. an example where the ends are further in, not an example of the actual curve of bent plywood), neither of which can be represented with the math of a sin, parabola, or catenary:

Bob

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### Re: polycylindrical diffuser shape

Terry

I applaud your research on the ME maths - but I reckon the real shaped rendered does have some constraints arising from how bending forces propagate through a given substrate. For example the typical multi-ply panel most folks use to make a poly probably varies the resulting bent shape to some extent. For users this is distinction without a difference but those interested in measuring the resulting curve to try and match up reality with the maths be prepared for variance for this reason.

In trying to describe the curve, one of my early thoughts was that it was similar to a catenary - which is to say "the curve is the graph of the hyperbolic cosine function, which has a U-like shape, similar in appearance to a parabola...."

http://en.wikipedia.org/wiki/Catenary

Obviously there is a key difference - in that the the poly shape we are describing arises from buckling via compression, not just the tug of gravity. In commenting on that fact, Eric has suggested that the Euler - Bernoulli beam equation is the source for mathematical description of the resulting curve. Note how the bending mode on the left appears to match a typical poly precisely in both method and yield.

http://en.wikipedia.org/wiki/Buckling
SRF
Scott R. Foster

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### Re: polycylindrical diffuser shape

I would be more interested in how these came to be called polycylindrical diffusers. The prefix "poly-" = "many" as in "polygon" = "many-sided." Thus, "polyclindrical" means "many-cylindrical," which, in the contexts of diffusers, would be a device with a shape somewhat approximated by:

ooooooooo ...

I think "semi-" or "quasi-" might be more apropos.

"Quasicylindrical" = "partly cylindrical," but could be interpretted as "appearing to be, but not really, cylindrical." (Given the topic, that might not be all that bad...)

"Semicylindrical" could mean "partially cylindrical," but could also be interpretted as "half cylindrical." (Which might be appropriate in some cases?)

And, staying on topic, the sinusoidal shape introduces a whole new nomenclature problem!!!

Albeit, a problem I've just invented...

(Just don't call them polysinusoids! )
---Σοφός---

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Savant

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### Re: polycylindrical diffuser shape

Scott R. Foster wrote:I applaud your research on the ME maths - but I reckon the real shaped rendered does have some constraints arising from how bending forces propagate through a given substrate.
...
the poly shape we are describing arises from buckling via compression, not just the tug of gravity. In commenting on that fact, Eric has suggested that the Euler - Bernoulli beam equation is the source for mathematical description of the resulting curve. Note how the bending mode on the left appears to match a typical poly precisely in both method and yield.

http://en.wikipedia.org/wiki/Buckling

Exactly. The pictures are of the buckling of a thin column -- yet another synonym for a rod. I believe that these are all solutions to the Euler-Bernoulli equation, simply using different boundary conditions.

The curve on the left arises from pin-pin boundary conditions, and should be a half sine, like I said. The next one over has clamp-clamp boundaries. The next appears to be clamp at top and pin at the bottom. The last is clamp at the top and free at the bottom.
Terry Montlick Labs
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Terry Montlick

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### Re: polycylindrical diffuser shape

Bob wrote:
Terry Montlick wrote:d = sin(π*x/L)
where 0 <= x <= L

Obviously x needn't go as far as 0 or L, but would most often be a subset within those boundaries.

Actually, the equation says you do. The amount of bend at the max (L/2) corresponds to the amplitude of the sine.

To risk going deeper into this subject, suppose the left-most rod of the picture Scott posted was constrained to pass through a slot halfway down. That slot is at the same horizontal position as the top and bottom pins. The rod can no longer move freely at the center. It settles into the next possible equation solution, where the mode number m=2. This is a full period of a sine wave!

- Terry
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Terry Montlick

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### Re: polycylindrical diffuser shape

Terry:

If I find some time tonight, I'll draw you some pictures (graphs) to show you what I mean.
But in a nutshell, I wasn't talking about the algebraic limits of the equation, but rather the ranges of x for most physical poly's we have photos of, which would be from about 45 degrees through 135 degrees of a 360 degree (full period, one hump up, one hump down) sin wave.
Bob

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### Re: polycylindrical diffuser shape

It means absolutely nothing of consequence to the layman.

Scott's poly building technique will work, because it simply applies pressure at the sides to bend out the middle. This will mathemagically form a half sine wave shape.

Just to be clear to laymen... if you are doing it right you don't necessarily get an entire half of a waveform. Using a plywood panel you probably will just get the top bump of the wave shape - the math goes on to describe curves much more radical than would be sensible [or safe] to play around with.

Don't be like Bob!

And while this stuff gets interesting when discussing optimized shapes for causing diffusion it means nothing as a practical matter in making a real poly... just bend the panel an inch or so narrower than the at rest panel and mind you don't get you fingers caught in the cracks.
SRF
Scott R. Foster

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### Re: polycylindrical diffuser shape

In each case L is the same, but only a subset of x (a small range of x) is used, not the entire 0 <= x <= L :: at least for 'graph A' and 'graph B'. I can't imagine anyone doing 'graph C', which I think is 0 <= x <= 2L, but even the left up hump of 'graph C' doesn't look to me like the way plywood would bend.
Obviously I've played around with C a bit in this equasion: [ d = C*sin(m*π*x/L) ]

So when I wrote "Obviously x needn't go as far as 0 or L, but would most often be a subset within those boundaries", what I mean is
in 'graph A' the x goes from 0.44 to 0.56
in 'graph B' the x goes from 0.48 to 0.52
Bob

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### Re: polycylindrical diffuser shape

B will diffuse better.

SRF
Scott R. Foster

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### Re: polycylindrical diffuser shape

It means absolutely nothing of consequence to the layman.
Something in my belly was telling me the same thing. I just wanted to hear it from a non laymans belly.

Just to be clear to laymen... if you are doing it right you don't necessarily get an entire half of a waveform.

um........in the case of your construction strategy , at what point does one get it wrong?
homestudiobldr

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### Re: polycylindrical diffuser shape

B will diffuse better.

Why? Actully, and I may be wrong, but I was under the impression that polys don't diffuse at all. They scatter. How would a poly affect decay rates, as my perception of a perfectly diffuse sound field is one where all frequencies decay at the same rate, no? So how would a poly contribute to this?
homestudiobldr

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### Re: polycylindrical diffuser shape

There's a litle bit more about the Euler Bernoulli beam equation in the studiotips fac
viewtopic.php?f=8&t=48
Bob

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### Re: polycylindrical diffuser shape

B will diffuse better.

B is best because it looks the most like this one:

Bob

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### Re: polycylindrical diffuser shape

Bob wrote:
In each case L is the same, but only a subset of x (a small range of x) is used, not the entire 0 <= x <= L :: at least for 'graph A' and 'graph B'. I can't imagine anyone doing 'graph C', which I think is 0 <= x <= 2L, but even the left up hump of 'graph C' doesn't look to me like the way plywood would bend.
Obviously I've played around with C a bit in this equasion: [ d = C*sin(m*π*x/L) ]

So when I wrote "Obviously x needn't go as far as 0 or L, but would most often be a subset within those boundaries", what I mean is
in 'graph A' the x goes from 0.44 to 0.56
in 'graph B' the x goes from 0.48 to 0.52

Bob, the point is that the pin-pin boundary conditions give you a half a sine wave -- whether you want it or not. The angle which the diffuser makes at the edges is purely a matter of the amplitude C.

Watson got me up from his thunderstorm phobia, so before I get back to sleep ...

Renaming d to y and setting m=1 for the first buckling mode,
y = C*sin(π*x/L)

A little high school trig and freshman calculus gives us this angle from the depth and width of the diffuser. The tangent to the angle θ at any part of of the curve is:
dy/dx = π*C/L cos(π*x/L)

But the dy/dx tangent to θ is by definition simply tan(θ).

At the edge of the diffuser, x = 0, and thus cos(0) = 1, so
dy/dx = tan(θ) = π*C/L

This gives us a simple formula for the angle θx=0 at the edge of the diffuser:
θx=0 = atan(π*C/L)

Let's try a simple example where we choose C = L/π. Then,
θx=0 = atan(1) = 45 degrees

A good empirical test for the half sine wave buckling model is to simply measure the angle θx=0 that the edge makes for any given poly diffuser with total depth C and width L. This should be the same as from the equation above:
θx=0 = atan(π*C/L)

- Terry
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Terry Montlick

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### Re: polycylindrical diffuser shape

Here's my own graph to help:

Terry Montlick Labs
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Terry Montlick

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### Re: polycylindrical diffuser shape

homestudiobldr wrote:Actully, and I may be wrong, but I was under the impression that polys don't diffuse at all. They scatter.

That is a distinction without a difference... a poly is a better diffuser than anything else you can buy or build.

Its also a better "scatterer".
SRF
Scott R. Foster

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### Re: polycylindrical diffuser shape

Terry Montlick wrote:Exactly. The pictures are of the buckling of a thin column -- yet another synonym for a rod.

Here's another picture of a rod:

Terry Montlick Labs
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Terry Montlick

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