c75e8d6Draft of cymbal-bow-v1: the flat disk bowed into a shallow paraboloidJeremy Magland 1# cymbal-bow-v1: bowed bronze disk struck with a drum stick, geometrically nonlinear
d9f5f11cymbal-bow-v1 is no longer a draftJeremy Magland 3
5This is the second challenge of the cymbal tier. It is [cymbal-flat-disk-v1](../cymbal-flat-disk-v1/README.md) with one change: the disk is no longer flat. Its mid-surface is a shallow paraboloid rising 25 mm from the rim to the center, the bow of a cymbal without its bell, and the thickness is uniform along the normal. Material, damping, strike, listener, output format and error measures are unchanged, so the difference between the two challenges is the effect of curvature alone. Curvature couples stretching and bending already in the linear regime, which raises the low modes and changes their radiation, and it adds quadratic terms to the nonlinearity, so the pitch glide and the cascade of the flat disk take a different form. The deliverable is again six seconds of pressure at the listener from the surface acceleration.
7The machine-readable version of this specification is [challenge.json](challenge.json).
9## Model
11### Reference geometry
13Let $R$ be the disk radius and $H$ the bow height. The mid-surface is the graph
15$$
16S = \{ (x, y, Z(r)) : r = \sqrt{x^2 + y^2} \le R \}, \qquad Z(r) = H \left( 1 - \frac{r^2}{R^2} \right),
17$$
19a paraboloid with its apex at height $H$ on the axis and its rim in the plane $z = 0$. Its unit normal, chosen with positive $z$ component, is
21$$
22\mathbf{n}(x, y) = \frac{(2 H x / R^2,\; 2 H y / R^2,\; 1)}{\sqrt{1 + 4 H^2 r^2 / R^4}}.
23$$
25The body is the set of points within half a thickness of the mid-surface along its normal,
27$$
28\Omega = \{ \mathbf{X} + s\, \mathbf{n}(\mathbf{X}) : \mathbf{X} \in S,\ |s| < h/2 \},
29$$
31so the thickness is $h$ everywhere when measured normal to the surface. Its boundary consists of the top face $S^+ = \{ \mathbf{X} + (h/2) \mathbf{n} \}$ (the convex side, facing $+z$), the bottom face $S^- = \{ \mathbf{X} - (h/2) \mathbf{n} \}$, and the rim, the ruled surface swept by the normal segment at $r = R$. The maximum slope of the mid-surface, at the rim, is $2H/R \approx 0.246$, and its radius of curvature at the apex is $R^2 / (2H) \approx 0.826$ m. The greatest departure from the plane $z = 0$ is $H + h/2$ at the apex.
33### Material and equations of motion
35The material, the strain measure, the Saint Venant-Kirchhoff stress with Kelvin-Voigt viscosity, the mass-proportional loss, and the equation of motion in the reference configuration are exactly those of cymbal-flat-disk-v1:
37$$
38\rho\, \frac{\partial^2 \mathbf{u}}{\partial t^2} = \nabla \cdot (\mathbf{F} \mathbf{S}) - 2 \sigma_0 \rho\, \frac{\partial \mathbf{u}}{\partial t}, \qquad
39\mathbf{S} = \lambda\, \mathrm{tr}(\mathbf{E} + \eta\, \dot{\mathbf{E}})\, \mathbf{I} + 2 \mu\, (\mathbf{E} + \eta\, \dot{\mathbf{E}}),
40$$
29fdfcaDefine the output acceleration with the rigid motion linearizedJeremy Magland 42with $\mathbf{E}$ the Green-Lagrange strain and the same constants. All of $\partial\Omega$ is traction-free except for the strike on $S^+$, the body starts at rest, and it is free, so the strike imparts momentum and angular momentum; the output is defined with the rigid motion linearized, as for the flat disk.
44### The strike
46The strike acts on the top face and pushes along $-z$, as a stick coming down vertically:
48$$
49\mathbf{F} \mathbf{S}\, \mathbf{N} = -\, p(\mathbf{y}, t)\, \mathbf{e}_z \quad \text{on } S^+, \qquad
50p(\mathbf{y}, t) = \frac{F_{\max}}{C}\; \phi\!\left( \frac{r'(\mathbf{y})}{R_s} \right) \psi\!\left( \frac{t}{T_s} \right),
51$$
53where for a point $\mathbf{y} = (y_1, y_2, y_3)$ of the top face $r'(\mathbf{y}) = \sqrt{(y_1 - x_s)^2 + (y_2 - y_s)^2}$ is its horizontal distance from the strike center, $\phi$ and $\psi$ are the raised-cosine bump and the Hann pulse of the flat-disk challenge, and
55$$
56C = \int_{S^+} \phi\!\left( \frac{r'(\mathbf{y})}{R_s} \right) dS(\mathbf{y})
57$$
59normalizes the pressure so that the total downward force is $F_{\max}\, \psi(t / T_s)$ and the impulse is $F_{\max} T_s / 2$, exactly as before. The footprint is the part of the top face whose horizontal projection lies within $R_s$ of $(x_s, y_s)$; since $dS = dA / (\mathbf{n} \cdot \mathbf{e}_z)$, $C$ is $c_\phi \pi R_s^2$ divided by a local cosine close to $0.98$, and its value to six digits is given in the parameter table for convenience. Solvers may compute it themselves.
61## Parameters
63All parameters of cymbal-flat-disk-v1 are unchanged; the bow height is new and the listener point is restated relative to the rim plane.
65| Symbol | Value | Unit | Meaning |
66| --- | --- | --- | --- |
67| $R$ | 0.2032 | m | disk radius (16 in diameter) |
68| $H$ | 0.025 | m | bow height, apex above the rim plane |
69| $h$ | 0.001 | m | thickness, normal to the mid-surface |
70| $E$, $\nu$, $\rho$ | 110e9, 0.34, 8800 | Pa, , kg/m^3 | B20 bronze |
71| $\sigma_0$, $\eta$ | 1.1, 1.2e-8 | 1/s, s | damping constants |
72| $(x_s, y_s)$ | (0.17272, 0) | m | strike center (horizontal), $0.85 R$ |
73| $R_s$, $T_s$, $F_{\max}$ | 0.005, 0.001, 1000 | m, s, N | strike footprint radius, duration, peak force |
74| $C$ | see challenge.json | m^2 | strike normalization integral |
75| $\rho_a$, $c_a$ | 1.204, 343 | kg/m^3, m/s | air |
76| $\mathbf{x}_L$ | (0.06, 0.08, 0.5) | m | listener point, above the rim plane |
78The listener is 0.5 m above the rim plane, and therefore about 0.475 m above the apex; it is the same point as in the flat-disk challenge up to the half thickness, which was measured from the mid-plane there. The strike point on the mid-surface is at height $Z(0.85 R) \approx 6.9$ mm and the local slope is about 0.21.
80With these constants the shell's membrane (ring) frequency is $c_L / (2 \pi R_c) \approx 725$ Hz with $c_L = \sqrt{E / (\rho (1 - \nu^2))} \approx 3760$ m/s and $R_c$ the apex radius of curvature. Below it, modes that stretch the mid-surface are stiffened by the membrane strain and their frequencies rise above those of the flat disk, while the lowest modes of a free shallow shell, which are nearly inextensional, rise only modestly; the appendix gives the shallow-shell equations from which solvers can estimate the linear spectrum before computing it.
82## Output signal
84The output is the pressure at the listener point defined by the Rayleigh-type integral over the top face of the acceleration component along the reference normal,
86$$
87s(t) = \frac{\rho_a}{2\pi} \int_{S^+} \frac{1}{|\mathbf{x}_L - \mathbf{y}|}\; \mathbf{n}(\mathbf{y}) \cdot \frac{\partial^2 \mathbf{u}}{\partial t^2}\!\left( \mathbf{y},\; t - \frac{|\mathbf{x}_L - \mathbf{y}|}{c_a} \right) dS(\mathbf{y}), \qquad \text{in Pa},
88$$
90with $\mathbf{n}$ the unit normal of the reference top face and the integrand zero for negative retarded times. For the flat disk this is the Rayleigh integral; for the curved surface it is no longer the exact field of a baffled radiator, and it is adopted as a defined output with the same form so that the two challenges are directly comparable. A later challenge will replace it with the coupled acoustic problem.
29fdfcaDefine the output acceleration with the rigid motion linearizedJeremy Magland 92The acceleration in this integral is the total acceleration with the rigid motion linearized. At each instant let $\mathbf{c}(t) \in \mathbb{R}^3$ and $\mathbf{Q}(t) \in SO(3)$ be the mass-weighted best-fit rigid placement of the body, minimizing $\int_\Omega \rho\, |\mathbf{x} + \mathbf{u} - \mathbf{c} - \mathbf{Q} \mathbf{x}|^2\, dV$; let $\boldsymbol{\omega}$ be the angular velocity of $\mathbf{Q}$ and $\boldsymbol{\alpha} = \dot{\boldsymbol{\omega}}$; and let $\tilde{\mathbf{u}} = \mathbf{Q}^{\mathsf{T}} (\mathbf{x} + \mathbf{u} - \mathbf{c}) - \mathbf{x}$ be the displacement relative to that frame. The acceleration used is $\mathbf{a} = \ddot{\mathbf{c}} + \boldsymbol{\alpha} \times \mathbf{x} + \partial^2 \tilde{\mathbf{u}} / \partial t^2$, and $\mathbf{n} \cdot \mathbf{a}$ replaces $\mathbf{n} \cdot \partial^2 \mathbf{u} / \partial t^2$ above. For small motions $\mathbf{a}$ is $\partial^2 \mathbf{u} / \partial t^2$; the two differ only by the centripetal and Coriolis terms of the rigid motion. The body is free and the strike imparts momentum and angular momentum, so in the exact model it keeps turning slowly after the strike, and without this linearization the output would contain the accelerations of that turning, which no thin-plate model represents and which is not part of the sound. The convention keeps the translational and rotational kick of the strike in the output, since $\ddot{\mathbf{c}}$ and $\boldsymbol{\alpha}$ are nonzero while the strike acts, and it coincides with what modal thin-plate solvers compute when they include their rigid modes. The optional field export uses the same linearization: the displacement $\tilde{\mathbf{u}} + \mathbf{c} - \mathbf{x} + \boldsymbol{\theta} \times \mathbf{x}$ with $\boldsymbol{\theta}$ the rotation vector of $\mathbf{Q}$, which for a thin-plate solver is its transverse displacement with the rigid modes included.
c75e8d6Draft of cymbal-bow-v1: the flat disk bowed into a shallow paraboloidJeremy Magland 94The wav has the same format as before: 48 kHz, mono, 32-bit float, 288000 samples, sample $k$ equal to $s(k / f_s) / s_\text{ref}$, no normalization. The scale $s_\text{ref}$ (`output.signal_scale`) will be fixed from the first solutions so that the peak sample is near 0.5, before any solution is registered.
96## Error measures and reference
98The same two error measures as for cymbal-flat-disk-v1 (relative waveform and spectrogram $\ell^2$ errors, the waveform error primary) and the same reference-selection rule. With a strike of this strength the response is strongly nonlinear and long waveform agreement between different methods is not expected; the spectrogram error is the more informative measure.
100## Optional field export
102As for the flat disk, but the quantity is the normal displacement of the mid-surface, $\mathbf{u} \cdot \mathbf{n}$, evaluated at the mid-surface point $(x_i, y_j, Z(r_{ij}))$ above each grid point: datasets `x`, `y` (61 points from $-R$ to $R$), `t` (3001 frames at 1 kHz) and `u` of shape (3001, 61, 61) in meters, NaN outside the disk.
104## Appendix: shallow-shell reduction
106For a shallow shell with initial mid-surface height $w_0(x, y) = Z(r)$ and uniform thickness, the Marguerre (Donnell-Mushtari-Vlasov) equations extend the von Kármán system of the flat disk by the initial shape:
108$$
109\rho h\, \frac{\partial^2 w}{\partial t^2} + 2 \sigma_0 \rho h\, \frac{\partial w}{\partial t} + D\, \nabla^4 \left( w + \eta\, \frac{\partial w}{\partial t} \right) = \mathcal{L}(w + w_0, \Phi) + p, \qquad
110\nabla^4 \Phi = -\frac{E h}{2} \left[ \mathcal{L}(w, w) + 2\, \mathcal{L}(w_0, w) \right],
111$$
113with $w$ the additional normal (here vertical) displacement, $\Phi$ the Airy stress function, $\mathcal{L}(f, g) = f_{xx} g_{yy} + f_{yy} g_{xx} - 2 f_{xy} g_{xy}$, and $D = E h^3 / (12 (1 - \nu^2))$. For the paraboloid $\nabla^2 w_0 = -4H/R^2$ is constant and $\mathcal{L}(w_0, g) = -(2H/R^2) \nabla^2 g$. The terms linear in $w$ couple bending and stretching and are what raise the low modes; the quadratic terms give the shell a nonlinearity with a preferred direction, so the pitch glide need not be downward. These equations assume small slopes and are an approximation of the three-dimensional model above; at a maximum slope of 0.25 the approximation is reasonable but not exact, and the difference is part of what a converged three-dimensional solution will measure. Free edges require vanishing bending moment and effective shear for $w$ and vanishing in-plane tractions, which for the Airy function on a simply connected domain means $\Phi = \partial \Phi / \partial n = 0$ on the rim.