Behavior of Clamped-Edge Circular Membrane

[Chrono G. Xay]

TL;DR Summary

In threads I had started several years ago I was asking questions where I had been attempting to derive a number of algebraic equations for predicting the behavior of a clamped-edge circular membrane. Though I’m still proud of the attempts, the approaches I presented back then modeled the behavior merely from a visual standpoint, so I was swiftly proven wrong.

This is a ‘Take Two’ of those attempts.
A fresh start, hopefully with more humility and less self-deprecation.

I am working on a project that would help guide aspiring drummers and percussionists in determining what thickness of drumhead to select for the tuning they want and/or the “feel” they want. (More on just what “feel” is later).

I’m trying to slowly work my way toward this starting with the tension a clamped-edge circular membrane. I understand that the unit for tension in this case is N/m, and is experienced by the membrane along its circumference. There’s a website called HyperPhysics where they actually have such an equation for the timpani: http://hyperphysics.phy-astr.gsu.edu/hbase/Music/cirmem.html

The equation they gave was:
$$f_1 = {0.766} \frac { \sqrt{T / σ}} {D}$$
I solved it for \text T to obtain
$$T = σ \left( \frac {D f_1} {0.766} \right) ^2$$
Based on some details they also provided… \text{mylar of thickness about 0.2 mm} I was able to define the area mass unit σ as $$σ = σ_M h$$ with σ_M as Mylar’s density of 1.390 kg/m³
and \text{“h”} as the membrane’s thickness. 7.5 mils is the closest to \text{0.2mm}, being a common drumhead thickness.

I emailed HyperPhysics to ask how they derived the multiplier “0.766” but they haven’t written back. The fact that such an equation has so little presence online is really confusing to me.[/sup]

 

[Baluncore]

If f is in hertz, not in radians per second, then I would expect the factor 0.766 to include a factor of 2*Pi.
See;
https://en.wikipedia.org/wiki/Vibra...embrane#Animations_of_several_vibration_modes
where the alpha coefficient for mode 01 = 2.40483
Notice that 2.40483 / Pi = 0.765481 ≈ 0.766
The coefficient 2.40483 comes from the roots of the Bessel function.

 

[Chestermiller]

Why do you want to know how it is derived?

Incidentally, $\sigma$ is called the "basis weight" of the membrane.

 

[Chrono G. Xay]

If f is in hertz, not in radians per second, then I would expect the factor 0.766 to include a factor of 2*Pi.
See;
https://en.wikipedia.org/wiki/Vibra...embrane#Animations_of_several_vibration_modes
where the alpha coefficient for mode 01 = 2.40483
Notice that 2.40483 / Pi = 0.765481 ≈ 0.766
The coefficient 2.40483 comes from the roots of the Bessel function.

I’ve seen the coefficients you’re referring to. That’s a good coincidence.

 

[Chrono G. Xay]

Why do you want to know how it is derived?

Incidentally, $\sigma$ is called the "basis weight" of the membrane.

I’m piecing together a Google Sheet which uses it. See attached. (Tension values are in lbs./ft)

Where I’m having trouble is when it comes to predicting the measurable change in overall diameter, not just the ‘speaking’ diameter.
I’m using the equation $$ΔL = \frac{F L_0} {A E}$$, where ΔL = ΔD, F = T, and L_0 = D_0, but is that the correct equation for a circular membrane? I understand that ‘E’ is the modulus of elasticity, but variable ‘A’ has me confused, which is why I’m thinking it’s not. (See attachment ‘IMG_6910’)

 

[Chrono G. Xay]

Why does “variable ‘A’ [have] me confused”?
I tried setting ‘A’ as the area of the membrane’s overall perimeter, $$A = \frac {π D_0 h} n,$$ where ‘n’ is the number of bolts applying tension to the clamped-edge circular membrane (so that I only get ΔD for one lug instead of two lugs opposite each other), and while it resulted in the correct units of measure the membrane’s height got cancelled out. I’m guessing the reason that formulation doesn’t work out is because, conceptually-speaking, the cross-section I defined probably intersects itself.

 

[Chestermiller]

If $\sigma$ is the magnitude of the isotropic in-plane stress in the membrane, then $$\sigma=\frac{\Delta D}{D_0}E$$

 

[Chrono G. Xay]

Wouldn’t the generalized Hooke’s Law $$σ = εE$$ with $$ε = \frac {ΔD} {D_0}$$ get modified when dealing with the specific case of cylindrical coordinates and axisymmetric stress instead of uniaxial, though?

I’m trying to better educate myself, but while I think I can follow along calculus is not my strong suit:
https://www.bu.edu/moss/mechanics-of-materials-strain/

http://ramanujan.math.trinity.edu/rdaileda/teach/s12/m3357/lectures/lecture_3_29.pdf

https://physics.stackexchange.com/questions/408987/can-hookes-law-be-derived/408992#408992

https://pkel015.connect.amazon.auck...ElasticityPolar/ElasticityPolars_Complete.pdf

 

[Chestermiller]

Wouldn’t the generalized Hooke’s Law $$σ = εE$$ with $$ε = \frac {ΔD} {D_0}$$ get modified when dealing with the specific case of cylindrical coordinates and axisymmetric stress instead of uniaxial, though?

Yes, sorry. You're right. it should be $$\sigma=\frac{E}{(1-\nu)}\epsilon$$

 

[Chrono G. Xay]

I’m trying to make sure I have my thoughts straight… solving for ΔD and substituting in the earlier formula for T I’m now getting $$ΔD = \frac {D_0 \left( {1 - v} \right) } E \left[ H d \left( \frac {D_{spk} f_1} {0.766} \right)^2 \right]$$

 

[Chrono G. Xay]

The way I’m currently understanding $T$ is that it’s like a tension unit itself (a… tension basis?), which if one wanted to know the total force on the membrane’s circumference they would multiply $T$ by the perimeter of the membrane’s “speaking” surface area, where the surface area’s speaking diameter does not include the extra radius past the rim of the drum’s shell’s bearing edge (perhaps one could refer to the extra radius as “skirting”), leading to $$D_{speaking} = D - \left( ΔD + 2R_{skirt} \right)$$
and assuming equal tension all around…
$$F_{total} = T \left( π D_{speaking} \right).$$
Considering this is now converted from tension units of $\frac N m$ to force units of $N$ parallel to the drum’s shell it would seem that if we wanted to we could now divide that force by the number of lugs around the drum’s rim for
$$F_{lug} = \frac {F_{total}} n.$$
I also currently understand that the tension a circular membrane experiences around its perimeter is similar to the polar graph of $$r \left( θ \right) = {10} - {1.5} \left| sin \left( \frac {nθ} 2 \right) \right|,$$ where ${10}$ and ${1.5}$ are numbers arbitrarily chosen to represent $T$ and the bending moment(?) of the drum’s rim. The peaks of the polar graph are where the lugs are located (See attached). For the sake of what I’m trying to model, though, I’m going to consider the flexing of the drum’s rim—in response to the force exerted by the rim on the clamped-edge of the membrane’s circumference—to be insignificant.

Am I making any big misunderstandings?

 

[Chestermiller]

I'm calling $D_0$ the diameter of the membrane before it is stretched (equal to the rim diameter), and D the diameter of the stretched membrane out to beyond the rim, and then fastened at the rim (without allowing it to contract)

There are 3 parts to the derivation. We have already done part 1 by establishing the relationship between the stress in the membrane and the amount that it has been stretched and then fastened to the rim.

Part 2 is recognizing what happens if a small region of the membrane is either pressed axially out of plane (say by our hand) or is oscillating vertically out of plane. As long as the axial displacement is small and not varying rapidly with radial location, the stress and strain in the membrane will not change significantly. However, the directionality of the internal stress within the membrane will change a little, and this will result in a restoring force component in the axial direction. This restoring force. would be trying to return the membrane to its original flat configuration.

Part 3 is to employ the understanding from part 2 to derive the force balance equation for an membrane that is oscillating axially out of plane. If we call $\sigma$ the in-plane stress within the membrane (determined in part 1) and $\delta(t,r)$ is the small axial displacement, and, if we perform a force balance on the region of the membrane between r and dr, and between $\theta$ and $\theta+d\theta$ (and assuming that the oscillation is axisymmetric), we obtain from Newton's 2nd law, $$\left[(\sigma h) rd\theta\frac{\partial \delta }{\partial r}\right]_{r+dr}-\left[(\sigma h) rd\theta\frac{\partial \delta }{\partial r}\right]_{r}=\rho h rdrd\theta \frac{\partial ^2 \delta}{\partial t^2}$$or, dividing by $rhdrd\theta$,$$\frac{\sigma }{r}\frac{\partial }{\partial r}\left(r\frac{\partial \delta}{\partial r}\right)=\rho \frac{\partial^2 \delta }{\partial t^2}$$

The left hand side of this equation represents the resultant of the axial components of the forces in the membrane and the right hand side represents mass times acceleration.

The previous equation is called a wave equation. The speed of propagation of a wave across the membrane is $c=\sqrt{\sigma/\rho}$

Are you OK with this so far?

 

[Chestermiller]

Solution to the wave equation for a drum: https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane

 

[Chestermiller]

TL;DR Summary: In threads I had started several years ago I was asking questions where I had been attempting to derive a number of algebraic equations for predicting the behavior of a clamped-edge circular membrane. Though I’m still proud of the attempts, the approaches I presented back then modeled the behavior merely from a visual standpoint, so I was swiftly proven wrong.

This is a ‘Take Two’ of those attempts.
A fresh start, hopefully with more humility and less self-deprecation.

I am working on a project that would help guide aspiring drummers and percussionists in determining what thickness of drumhead to select for the tuning they want and/or the “feel” they want. (More on just what “feel” is later).

I’m trying to slowly work my way toward this starting with the tension a clamped-edge circular membrane. I understand that the unit for tension in this case is N/m, and is experienced by the membrane along its circumference. There’s a website called HyperPhysics where they actually have such an equation for the timpani: http://hyperphysics.phy-astr.gsu.edu/hbase/Music/cirmem.html

The equation they gave was:
$$f_1 = {0.766} \frac { \sqrt{T / σ}} {D}$$
I solved it for \text T to obtain
$$T = σ \left( \frac {D f_1} {0.766} \right) ^2$$

The constant 0.766 is equal to the first zero of the zero order Bessel function ($J_0(x)$), 2.4048, divided by $\pi$

 

[Chestermiller]

The standing wave analytic solution to the wave equation in Post #12 is $$\Delta=J_0(\lambda r)(A \cos{c\lambda t}+B\sin{c\lambda t})$$where $J_0$ is the zero order Bessel function of the first kind, and $\lambda$ is a constant to be determined by the boundary condition at r = D/2:

$\delta=0$ at r = D/2. From this we have that $$J_0(\lambda D/2)=0$$, or $$\lambda=\frac{2}{D}\alpha _1$$ where $\alpha_1=2.4048$ is the smallest zero of $J_0(x)$. So we have: $$\delta = J_0(2.4048\frac{2r}{D})\left(A\cos{\frac{(2)(2.4048)c}{D}}+B\sin{\frac{(2)(2.4048)c}{D}}\right)$$So the fundamental frequency of oscillation is $$2\pi f=\frac{(2)(2.4048)c}{D}$$or $$f=\frac{(2.4048)c}{\pi D}=\frac{2.4048}{\pi}\frac{c}{D}=0.766\frac{c}{D}$$

 

[Chrono G. Xay]

I'm calling $D_0$ the diameter of the membrane before it is stretched (equal to the rim diameter), and D the diameter of the stretched membrane out to beyond the rim, and then fastened at the rim (without allowing it to contract)

There are 3 parts to the derivation. We have already done part 1 by establishing the relationship between the stress in the membrane and the amount that it has been stretched and then fastened to the rim.

Part 2 is recognizing what happens if a small region of the membrane is either pressed axially out of plane (say by our hand) or is oscillating vertically out of plane. As long as the axial displacement is small and not varying rapidly with radial location, the stress and strain in the membrane will not change significantly. However, the directionality of the internal stress within the membrane will change a little, and this will result in a restoring force component in the axial direction. This restoring force. would be trying to return the membrane to its original flat configuration.

Part 3 is to employ the understanding from part 2 to derive the force balance equation for an membrane that is oscillating axially out of plane. If we call $\sigma$ the in-plane stress within the membrane (determined in part 1) and $\delta(t,r)$ is the small axial displacement, and, if we perform a force balance on the region of the membrane between r and dr, and between $\theta$ and $\theta+d\theta$ (and assuming that the oscillation is axisymmetric), we obtain from Newton's 2nd law, $$\left[(\sigma h) rd\theta\frac{\partial \delta }{\partial r}\right]_{r+dr}-\left[(\sigma h) rd\theta\frac{\partial \delta }{\partial r}\right]_{r+dr}=\rho h rdrd\theta \frac{\partial ^2 \delta}{\partial t^2}$$or, dividing by $rhdrd\theta$,$$\frac{\sigma }{r}\frac{\partial }{\partial r}\left(r\frac{\partial \delta}{\partial r}\right)=\rho \frac{\partial^2 \delta }{\partial t^2}$$

The left hand side of this equation represents the resultant of the axial components of the forces in the membrane and the right hand side represents mass times acceleration.

The previous equation is called a wave equation. The speed of propagation of a wave across the membrane is $c=\sqrt{\sigma/\rho}$

Are you OK with this so far?

I’m following along so far, yes. I got confused only for a moment with the formula you presented just before you wrote

or, dividing by $rhdrd\theta$

because it seems like the form $A-A=B$, which I’m guessing was an accident. (I read through it a few times letter for letter to make sure I wasn’t just reading to fast)

I think I’m understanding most everything else, especially wave propagation speed. I’ve been trying to understand the two-dimension wave equation, and how it deviates from the one-dimensional wave equation, the former at least I understand is fundamental to the system I’m trying to model.

(For background, I only first got into trying to understand these two while pursuing curiosities about wound guitar strings; how their ‘clang tone’ differs from that of a plain string; deconstructing them; trying to model their behavior. I owe that initial mathematical curiosity to the transcript of a technical lecture by luthier Ralph Novak, that I read while taking guitar classes at my university: https://www.novaxguitars.com/info/technical.html
_ https://pubs.aip.org/asa/jasa/article/16/1_Supplement/102/683898/The-Clang-Tone-of-the-Pianoforte)

 

[Chrono G. Xay]

The standing wave analytic solution to the wave equation in Post #12 is $$\Delta=J_0(\lambda r)(A \cos{c\lambda t}+B\sin{c\lambda t})$$where $J_0$ is the zero order Bessel function of the first kind, and $\lambda$ is a constant to be determined by the boundary condition at r = D/2:

$\delta=0$ at r = D/2. From this we have that $$J_0(\lambda D/2)=0$$, or $$\lambda=\frac{2}{D}\alpha _1$$ where $\alpha_1=2.4048$ is the smallest zero of $J_0(x)$. So we have: $$\delta = J_0(2.4048\frac{2r}{D})\left(A\cos{\frac{(2)(2.4048)c}{D}}+B\sin{\frac{(2)(2.4048)c}{D}}\right)$$So the fundamental frequency of oscillation is $$2\pi f=\frac{(2)(2.4048)c}{D}$$or $$f=\frac{(2.4048)c}{\pi D}=\frac{2.4048}{\pi}\frac{c}{D}=0.766\frac{c}{D}$$

What I’m getting from this is, at least for this case, you want to carefully set your boundary conditions using one or more of the, let’s say, freeze-frame scenarios of the dynamic physical system you’re trying to model, then work your way outward while maintaining equalities until you get to the point where you can instantiate the variables you’re trying to solve for.

One place where I keep getting lost is how one gets from the mathematical analysis (e.g. the 2D wave equation) to the materials science portion of these kind of problems.

 

[Chestermiller]

I’m following along so far, yes. I got confused only for a moment with the formula you presented just before you wrotebecause it seems like the form $A-A=B$, which I’m guessing was an accident. (I read through it a few times letter for letter to make sure I wasn’t just reading to fast)

Yes. The 2nd subscript should have been r, which I've corrected.

I think I’m understanding most everything else, especially wave propagation speed. I’ve been trying to understand the two-dimension wave equation, and how it deviates from the one-dimensional wave equation, the former at least I understand is fundamental to the system I’m trying to model.

This equation is also one dimensional, r, except in cylindrical coordinates and axisymmetric.

 

[Chestermiller]

What I’m getting from this is, at least for this case, you want to carefully set your boundary conditions using one or more of the, let’s say, freeze-frame scenarios of the dynamic physical system you’re trying to model, then work your way outward while maintaining equalities until you get to the point where you can instantiate the variables you’re trying to solve for.

Where I keep getting lost is how one gets from the mathematical analysis (e.g. the 2D wave equation) to the materials science portion of these kind of problems.

There isn't much of a material science aspect to this, because the initial tensile stress is pre-established by mounting the skin on the drum. Then, because the displacement in the z - direction is small, the additional strain (and stress) caused by the displacement when oscillating is negligible, and the only thing that changes in the directionality of the skin surface, which has a small-but-significant axial force contribution. Imagine a tight string that slightly you press your finger on to produce a small component office normal to the original string.

 

[Chrono G. Xay]

[#12] I'm calling $D_0$ the diameter of the membrane before it is stretched (equal to the rim diameter), and D the diameter of the stretched membrane out to beyond the rim, and then fastened at the rim (without allowing it to contract)

[#19] There isn't much of a material science aspect to this, because the initial tensile stress is pre-established by mounting the skin on the drum.

I was reading back over these two posts of yours, and I wanted to double-check something with you, just because not everyone is familiar with acoustic drums:

This (https://en.m.wikipedia.org/wiki/File:Anatomy_of_a_Drumhead.jpg) is the physical system you were picturing when you wrote the above quoted posts, correct?

Here’s a quick four-minute video by the company Evans, a drumhead manufacturer:


For context, there are electric guitars with something called a ‘locking nut’ up where the tuners are. (See attached.) After all of the strings have been tuned to the desired musical pitches the locking nut’s clamping blocks are tightened down onto the strings so that the strings do not lose their tuning (this is especially common with electric guitars equipped with a ‘double-locking tremolo’, and has to do with ensuring tuning stability across all strings while playing using deep bends across the neck of the instrument or ‘divebombs’, which are notorious for throwing both that string and others out of tune). While clamped, however, the tuners themselves are rendered unusable.

Conventional acoustic drums do not have a locking mechanism other than, say, the bolts (AKA ‘tension rods’) that pull down on the rim, which pulls the drumhead’s collar (clamped to the edge of the drumhead’s skirt) down, which holds tension on the drumhead. The initial tension on the circular membrane is zero, with an initial—unstretched—net diameter of $D + 2L_{skirt,0}$. (This lacking tension-locking device does allow for other interesting devices to be used in its place, like this! https://www.drumtech.com/dts-drum-tuning-system)

 

[Chrono G. Xay]

In that case, what I’m thinking needs to be done is to seek a solution where one of the last steps look something like $$D = \int_0^σ u\left(σ\right)~dσ$$

 

[Chestermiller]

In that case, what I’m thinking needs to be done is to seek a solution where one of the last steps look something like $$D = \int_0^σ u\left(σ\right)~dσ$$

What is u, displacement?

 

[Chrono G. Xay]

What is u, displacement?

$u \left( σ \right)$ is a function of radial stretching in response to stress.

 

[Chestermiller]

Let r be the radial location in the stretched configuration of the skin of the material particles in the skin that were at radial location $r_0$ in the unstretched configuration. Then the strain in the skin in the radial direction is $$\epsilon_r=\frac{dr}{dr_0}-1$$ and the strain in the skin in the hoop direction is $$\epsilon_\theta=\frac{r-r_0}{r_0}$$. Since the skin deformation is axisymmetric, $$\epsilon_r=\epsilon_{\theta}=\frac{dr}{dr_0}-1=\frac{r-r_0}{r_0}$$The solution to this differential equation is $$r=(1+\epsilon)r_0$$with $$\epsilon_r=\epsilon_{\theta}=\epsilon$$So all radial lines stretch by the same factor $(1+\epsilon)$, with $$\epsilon=(1-\nu)\frac{\sigma}{E}$$where $\sigma$ is the stress in the skin. So $$\frac{r}{r_0}=\frac{d}{d_0}=1+(1-\nu)\frac{\sigma}{E}$$

 

[Chrono G. Xay]

Let r be the radial location in the stretched configuration of the skin of the material particles in the skin that were at radial location $r_0$ in the unstretched configuration. Then the strain in the skin in the radial direction is $$\epsilon_r=\frac{dr}{dr_0}-1$$ and the strain in the skin in the hoop direction is $$\epsilon_\theta=\frac{r-r_0}{r_0}$$. Since the skin deformation is axisymmetric, $$\epsilon_r=\epsilon_{\theta}=\frac{dr}{dr_0}-1=\frac{r-r_0}{r_0}$$The solution to this differential equation is $$r=(1+\epsilon)r_0$$with $$\epsilon_r=\epsilon_{\theta}=\epsilon$$So all radial lines stretch by the same factor $(1+\epsilon)$, with $$\epsilon=(1-\nu)\frac{\sigma}{E}$$where $\sigma$ is the stress in the skin. So $$\frac{r}{r_0}=\frac{d}{d_0}=1+(1-\nu)\frac{\sigma}{E}$$

Oh, of course! We were talking about that relation earlier.

 

[Chrono G. Xay]

This is what I think I’ve come to understand so far:

Starting with what I’m going to call an oscillation basis, $ω$, defined as: $$ω = \frac {2πf_0} {J_0 \left( \frac 2 {D_{speaking}} α_1 \right)}$$ We then get our stress basis, $σ$, defined as: $$σ = \frac {\left( ωD_{speaking} \right)^2} ρ$$ Now we find the amount of radial strain, $r$, using the formula $$r = r_0 \left[ 1 + \left( 1 + ν \right) \frac σ E \right]$$ Next we consider the geometry of the radial cross-section of a circular membrane, particularly where the membrane runs from the bearing edge of the drum’s cylindrical shell to the membrane’s hoop, clamped at the edge of the membrane’s skirt, the length of which is denoted as $L_{skirt,0}$ as the hypotenuse of a right triangle; the adjacent side of the triangle is the radial distance from the bearing edge to the hoop, denoted as $r_{hoop}$; and where $σ$ is still at least theoretically zero (i.e. where $σ_0 = 0$).

(Note: $r_0 = \frac {D_{speaking}} 2 + L_{skirt,0}$)

Solving for the opposing side, $δ$, using the Pythagorean Theorem yields $$δ = \sqrt {L_{skirt,0}^2 - r_{hoop}^2}$$ The last point of consideration is the number of revolutions, $n$, that each tension rod of the drum’s rim must make in order to pull the membrane’s hoop down by $δ$. A common thread for drum tension rods is #12-24 (24 threads per inch, $t$).

The number of revolutions is:$$n = \frac δ {t^{-1}}$$ For cases where the membrane is already under axisymmetric stress (i.e. where $σ_0 > 0$), $n$ simply becomes $$n = \frac {Δδ} {t^{-1}}$$ where $$Δδ = δ - δ_0$$ (Note: $t^{-1}$ could instead be thought of as $0.041 \bar 6 ~in/rev.$ or $1.058 ~mm/rev.$)

 

[Chestermiller]

This is what I think I’ve come to understand so far:

Starting with what I’m going to call an oscillation basis, $ω$, defined as: $$ω = \frac {2πf_0} {J_0 \left( \frac 2 {D_{speaking}} α_1 \right)}$$

If $\omega$ is the angular frequency in radians per second, then $$\omega=2\pi f_0=2\alpha_1\frac{\sqrt{\sigma/\rho}}{D}$$

We then get our stress basis, $σ$, defined as: $$σ = \frac {\left( ωD_{speaking} \right)^2} ρ$$

The stress is $$\sigma=\rho \left(\frac{\omega D}{2\alpha_1}\right)^2$$

Now we find the amount of radial strain, $r$, using the formula $$r = r_0 \left[ 1 + \left( 1 + ν \right) \frac σ E \right]$$

I don't understand the following part, and I don't understand why the skirt is being addressed at all. Can you please provide a sketch?

Next we consider the geometry of the radial cross-section of a circular membrane, particularly where the membrane runs from the bearing edge of the drum’s cylindrical shell to the membrane’s hoop, clamped at the edge of the membrane’s skirt, the length of which is denoted as $L_{skirt,0}$ as the hypotenuse of a right triangle; the adjacent side of the triangle is the radial distance from the bearing edge to the hoop, denoted as $r_{hoop}$; and where $σ$ is still at least theoretically zero (i.e. where $σ_0 = 0$).

(Note: $r_0 = \frac {D_{speaking}} 2 + L_{skirt,0}$)

Solving for the opposing side, $δ$, using the Pythagorean Theorem yields $$δ = \sqrt {L_{skirt,0}^2 - r_{hoop}^2}$$ The last point of consideration is the number of revolutions, $n$, that each tension rod of the drum’s rim must make in order to pull the membrane’s hoop down by $δ$. A common thread for drum tension rods is #12-24 (24 threads per inch, $t$).

The number of revolutions is:$$n = \frac δ {t^{-1}}$$ For cases where the membrane is already under axisymmetric stress (i.e. where $σ_0 > 0$), $n$ simply becomes $$n = \frac {Δδ} {t^{-1}}$$ where $$Δδ = δ - δ_0$$ (Note: $t^{-1}$ could instead be thought of as $0.041 \bar 6 ~in/rev.$ or $1.058 ~mm/rev.$)

 

[Chrono G. Xay]

Consider the picture shown at the top of this article, titled ‘Anatomy of A Conga’, particularly the part called “Drumhead Collar (Leg)”, noting the triangular shape of the collar outside of the “crown”, where the remaining length wraps over the “Drum Shell Bearing Edge”.

https://support.remo.com/hc/en-us/a...natomy-of-a-conga-Know-your-conga-drum-brand-

 

[Chestermiller]

Consider the picture shown at the top of this article, titled ‘Anatomy of A Conga’, particularly the part called “Drumhead Collar (Leg)”, noting the triangular shape of the collar outside of the “crown”, where the remaining length wraps over the “Drum Shell Bearing Edge”.

https://support.remo.com/hc/en-us/a...natomy-of-a-conga-Know-your-conga-drum-brand-

Why are you interested in the parts beyond the drum head?

 

[Chrono G. Xay]

Why are you interested in the parts beyond the drum head?

I’m interested in them because they’re a continuous portion of the membrane; as the rim pulls down on the hoop, which is clamped to the edge of the membrane’s collar, naturally the length of the collar affects the amount of elongation experienced until the desired musical pitch (and therefore tension) has been achieved.

(Up until now I thought there was no technical term for the portion of the membrane between the drumhead’s speaking diameter and the hoop, hence my use of “skirt”, but it’s actually called the ‘collar’.)

For this analysis I do not consider the added magnitude of elongation afforded by the collar to be insignificant.

 

[Chestermiller]

I’m interested in them because they’re a continuous portion of the membrane; as the rim pulls down on the hoop, which is clamped to the edge of the membrane’s collar, naturally the length of the collar affects the amount of elongation experienced until the desired musical pitch (and therefore tension) has been achieved.

(Up until now I thought there was no technical term for the portion of the membrane between the drumhead’s speaking diameter and the hoop, hence my use of “skirt”, but it’s actually called the ‘collar’.)

For this analysis I do not consider the added magnitude of elongation afforded by the collar to be insignificant.

It is not insignificant, but you also need to account for normal forces that result in frictional forces which affect the skin stresess and strains in membrane skin in this region. This is what allows the skin to be pulled down.

 

[Chrono G. Xay]

It is not insignificant, but you also need to account for normal forces that result in frictional forces which affect the skin stresess and strains in membrane skin in this region. This is what allows the skin to be pulled down.

It’s for that very reason that when lowering the pitch of a drum you intentionally take the pitch slightly lower then what you want it, then tune it *up* to where you actually want it. The same is done for guitar strings.
The rest of it will basically work itself out on its own once someone starts striking it with a stick.

I also just realized that I had mistyped a few of the earlier functions. I’m sorry. $$ω \left( f_0, D_s \right) = \frac {πf_0 D_s} {J_0 \left( \frac 2 {D_s} {α_1} \right)}$$ $$ρ \left( ρ_M, h_m \right) = ρ_M h_m$$ $$σ \left( ω, ρ \right) = \frac {ω^2} ρ$$ $$D_0 \left( D_s, L_{c,0} \right) = {D_s} + 2L_{c,0}$$ $$D \left( D_0, ν, E \right) = D_0 \left[ 1 + \left( 1 + ν \right) \frac σ E \right]$$ $$L_c \left( D, D_s \right) = \frac {D - {D_s}} 2$$ $$δ \left( L_{c}, r_h, h_c, h_r \right) = \sqrt {L_{c}^2 - r_h^2} - \left( h_c + h_r \right)$$ $$n \left( δ, t \right) = \frac δ {t^{-1}}$$ $$Δδ = δ - δ_0$$ In order of appearance…

$f_0$ = Desired fundamental frequency
$D_s$ = Speaking diameter of membrane
$α_1$ = 1st zero of First Order Bessel Function (2.4048)
$ρ$ = Basis weight
$ρ_M$ = 3D density of Mylar
$h_m$ = Height (thickness) of membrane
$σ$ = In-plane axisymmetric stress on membrane
$D_0$ = Unstretched membrane overall diameter
$L_{c,0}$ = Unstretched membrane collar length
$D$ = Stretched membrane overall diameter
$ν$ = Poisson’s ratio
$E$ = Modulus of elasticity
$L_c$ = Stretched membrane collar length
$δ$ = Required displacement of membrane rim’s tension rods after tension rods have been taken as low as they will go without increasing the in-plane stress $σ$ above theoretical zero.
$r_h$ = Extra radius of membrane hoop (from the center of the shell’s bearing edge to the inside diameter of the membrane’s hoop)
$h_c$ = Height (depth) of membrane collar
$h_r$ = Height (thickness) of membrane’s rim
$n$ = Decimal number of revolutions tension rods must make to achieve needed in-plane stress
$t$ = Number of threads per inch/mm on tension rods
$δ_0$ = Initial displacement of tension rods (assuming the membrane is already tuned to some initial fundamental frequency $f_{0,0}$; i.e. $σ_0 > 0$)

 

[Chestermiller]

It’s for that very reason that when lowering the pitch of a drum you intentionally take the pitch slightly lower then what you want it, then tune it *up* to where you actually want it. The same is done for guitar strings.
The rest of it will basically work itself out on its own once someone starts striking it with a stick.

I also just realized that I had mistyped a few of the earlier functions. I’m sorry. $$ω \left( f_0, D_s \right) = \frac {πf_0 D_s} {J_0 \left( \frac 2 {D_s} {α_1} \right)}$$ $$ρ \left( ρ_M, h_m \right) = ρ_M h_m$$ $$σ \left( ω, ρ \right) = \frac {ω^2} ρ$$ $$D_0 \left( D_s, L_{c,0} \right) = {D_s} + 2L_{c,0}$$ $$D \left( D_0, ν, E \right) = D_0 \left[ 1 + \left( 1 + ν \right) \frac σ E \right]$$ $$L_c \left( D, D_s \right) = \frac {D - {D_s}} 2$$ $$δ \left( L_{c}, r_h, h_c, h_r \right) = \sqrt {L_{c}^2 - r_h^2} - \left( h_c + h_r \right)$$ $$n \left( δ, t \right) = \frac δ {t^{-1}}$$ $$Δδ = δ - δ_0$$ In order of appearance…

$f_0$ = Desired fundamental frequency
$D_s$ = Speaking diameter of membrane
$α_1$ = 1st zero of First Order Bessel Function (2.4048)
$ρ$ = Basis weight
$ρ_M$ = 3D density of Mylar
$h_m$ = Height (thickness) of membrane
$σ$ = In-plane axisymmetric stress on membrane
$D_0$ = Unstretched membrane overall diameter
$L_{c,0}$ = Unstretched membrane collar length
$D$ = Stretched membrane overall diameter
$ν$ = Poisson’s ratio
$E$ = Modulus of elasticity
$L_c$ = Stretched membrane collar length
$δ$ = Required displacement of membrane rim’s tension rods after tension rods have been taken as low as they will go without increasing the in-plane stress $σ$ above theoretical zero.
$r_h$ = Extra radius of membrane hoop (from the center of the shell’s bearing edge to the inside diameter of the membrane’s hoop)
$h_c$ = Height (depth) of membrane collar
$h_r$ = Height (thickness) of membrane’s rim
$n$ = Decimal number of revolutions tension rods must make to achieve needed in-plane stress
$t$ = Number of threads per inch/mm on tension rods
$δ_0$ = Initial displacement of tension rods (assuming the membrane is already tuned to some initial fundamental frequency $f_{0,0}$; i.e. $σ_0 > 0$)

I really don't have anything to add to what I said in my previous posts.