Natural resonant frequency of uncoupled top

[GregHolmberg]

In the 4DOF model, on page AII-7. Equ. AII 2-21, we see the natural resonant frequency of uncoupled top with small damping:

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ω^2 = (K + κA^2) / m

This is quite different than the relationship used for frequency, stiffness, and mass used when calculating the monopole mobility:

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ω^2 = K / m

I'm a little confused about what the effective area of the top is doing in the first one. Is this the "small damping"?

In one calculation I did, the first one gave me 185 Hz, while the second gave 159, so the difference is not small.

Alternatively, we could say that the frequency is 185, but that the masses are different, at 31.6 g and 42.9 g.

The reason I ask is, if I have a 4DOF model that says I want build the top to an uncoupled frequency of 185 Hz, and a stiffness of K = 42,700 N/m, what's my target effective mass of the area of the piston (inside the Chladni lines)? Is it 43 or 32g?

Would I expect the resulting monopole mobility to be 1/SQRT(42700 * 0.0429) = 23E-3 or would it be 1/SQRT(42700 * 0.0316) = 27E-3 ?

Basically, my question is: what is the meaning of the κA^2 term? Why is it included some times but not others?

Greg

[GregHolmberg]

Or, asked another way, what is K-bar on page AII-7?

He says the equations have the form:

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ω^2 = K-bar / m

In which case, for the top,

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K-bar = K + κA^2

But what which should be used to calculate MM or frequency, K or K-bar?

Greg

[Trevor Gore]

Basically, my question is: what is the meaning of the κA^2 term? Why is it included some times but not others?

The simple answer is that MM is meant as a simple and useful way (i.e. not involving a huge amount of maths, or a detailed understanding thereof) of gaining a measure of the responsiveness of a guitar. By its very nature it is fairly "quick and dirty" (though based on reasonable principles) and that is much of the point. If it wasn't "quick and dirty", it wouldn't get used. The method of measuring MM plugs the sound hole and so uncouples the top from the Helmholtz oscillator, but it doesn't uncouple it from the stiffness of the air as a spring inside the box.

That extra stiffness of the air is what the κA^2 term is about. Including the κA^2 term in MM is fine if you want to do it, but it then places MM back in the complexity of the 4-DOF model, rather than the usability and measureability of the standard definition.

Greg, it may be worth you checking out the history of usage of the 3-DOF model, the most accessible being Howard Wright's PhD thesis. What I learnt from doing that (and what precipitated the 4-DOF model) was that the 3-DOF model effective masses are just "fiddle factors" that are used to match the model to a certain measured frequency response and they bear very little relationship to any real component masses of the instrument in question. The 4-DOF model suffers similarly (because at the end of the day it is just a simple analytical model) but it has the benefit of the masses at least approaching some sort of reality and so is useful for examining the relationships and sensitivities between parameters. To get answers that are truly meaningful at the wood-cutting level a much more sophisticated model is required. These are usually very complex finite element models. When I last surveyed the literature (which was a while ago) someone had made a very detailed FE model but hadn't included the air loading (I can't recall the reason why - maybe it was just too much complexity) which, of course renders the model pretty much unusable for design/construct purposes (unless the guitar is intended to be played in a vacuum!). So you may be expecting too much of these simple models.

Anecdote: I was teaching the modal tuning course in Colorado, local altitude 6000 feet. All the main resonances of the guitar I had brought to do demonstrations on were different from home in Sydney (elevation 0 feet in my workshop at high tide). Changing the air density in the 4-DOF model to match 6000 feet explained the differences I was measuring quite accurately. Moral of the story, always have in the back of your mind that everything is an approximation, but some things are more approximate than others.

[GregHolmberg]

Thanks, Trevor. That makes sense.

So, if I designed with a target stiffness K, and I could estimate the mass m of the top within the piston area (I guess, within the Chladni line of the monopole mode) including the bridge and bridge plate, I should not expect an uncoupled frequency f = SQRT(K/m)/2π, right?

So, for example (using values from your model on page 2-37), if I set a target K=42,700 N/m, and estimated the mass of my top panel, braces, and bridge in the center 30% of the top m=43 g, then I would not get a top with a frequency of SQRT(42700/0.043)/2π = 158.6 Hz.

To calculate the expected frequency, I would have to also estimate the internal volume of the box (V=0.0141 m^3, to get kappa, κ) and the area of the top piston, A=0.039 m^2. And then I could use f = SQRT((K + κA^2)/m)/2π to calculate my expected frequency of 184.7 Hz.

I think I can do that. Having the expected top frequency would help me validate my brace dimensions.

And then, the coupled frequency of the top would be the second peak in the SPL graph from the 4DOF model, 189.3 Hz.

Do I have that right?

Greg

[Trevor Gore]

Any frequency you calculate, by whatever method, is only as valid as the assumptions under which the method was developed.

Remember that the published 4-DOF model and input parameters was for a single specific guitar with the K and m values sized so that the model output matched that guitar's specific frequency response curve. It is not necessarily valid for any other specific guitar, nor is it a generalised guitar model when using those inputs. It is really only "valid" once you calibrate it for the type of guitar you want to build - which means building the guitar first, calibrating the model and then the model is likely only "valid" for small perturbations in the input parameters. The model was not intended to be used in "reverse".

Now, that doesn't mean you shouldn't try to use it in reverse, but you need to be aware of its quite severe limitations when attempting that. So whilst your approach above may be arithmetically correct (I think, I haven't checked it in detail), it doesn't mean you will necessarily get a meaningful result that is useful. But there is only one way to find out...try it and see.

Once you have a prototype guitar built and have calibrated the model (not an exercise for the faint of heart!) you should be able to fairly accurately predict what will happen if (for example) you change the sound hole diameter by a couple of millimeters or the back stiffness by 20% or so (or take it to 6000 feet).

[GregHolmberg]

Thanks, Trevor. I get it. I will have to fit the model to the prototype if I want any accuracy.

My hope is that when coming up with a completely new design, where I know what frequencies I want for the first three resonances, a model will at least give me a target top stiffness (brace dimensions/mass), back stiffness, sound-hole area, and cavity volume to use in the prototype. Then it's a matter of re-fitting the model and iterating the prototype. But I hope the model puts me in the ballpark and gets me to the final iteration faster. If I can get there in three builds rather than ten, the model will have saved me a lot of time.

I just wanted to make sure I was using the equations correctly. It sounds like I am.

I guess one could say that, in the example above, the monopole mobility using 1/SQRT(Km) is 23.3, but the true MM = 1/SQRT((k + κA^2)m) is 20.0. Because the volume of the box and the area of the piston matter. Two guitars of different body dimensions but the same K and m, don't really have the same MM.

Greg

[Trevor Gore]

I guess one could say that, in the example above, the monopole mobility using 1/SQRT(Km) is 23.3, but the true MM = 1/SQRT((k + κA^2)m) is 20.0. Because the volume of the box and the area of the piston matter. Two guitars of different body dimensions but the same K and m, don't really have the same MM.

Yes, but the true MM is always 1/SQRT(Km), because that's how I defined it!

I recognise that there are other ways of estimating both K and m, which may be more "accurate" in specific circumstances, but all the ones I've tried are more involved and and detract from the simplicity of the current method, which I would argue is adequate for its intended purpose.

As I mentioned previously, everything is an approximation, but some things are more approximate than others.