THE HALF-HOLE MODEL

Figure 1 shows a cross-sectional view of a woodwind tonehole. In the low-frequency limit, the hole dimensions are usually small in comparison with the acoustic wavelength, thus the acoustic behaviour may be characterised by a lumped acoustic element. For an open hole, the behaviour is approximately that of a pure inertance, while for a closed one it approximately corresponds to a pure compliance [10].

Figure 1: Cross-sections of a woodwind tonehole.

For intermediate tonehole states (partially open holes or ``half-holes''), the tonehole volume $V_b$ can be divided into an ``open part'' that behaves as an inertance, and a ``closed part'' that behaves as a compliance. These volumes operate in parallel, thus the half-hole load impedance is:

$$Z_{s}(\omega) = \frac{j \omega L}{1 - \omega^{2} L C},$$ (2)

Figure 2 shows the network equivalent of this model.

Figure 2: Electrical network representation of the half-hole model.

The half-hole compliance ($C$) and inertance ($L$) are given by:

$$C =(1-g) \cdot \frac{S_{b} t}{\rho c^2} \quad,\quad L = \frac{\rho t_{e}}{g S_{b}},$$ (3)

where the parameter $g$ expresses the tonehole state, defined as the ratio between open and total tonehole volume. The tonehole height $t$ is defined such that its product with the tonehole surface $S_{b}$ equals the geometric volume $V_{b}$ [3]:

$$t = t_{w} + 0.25 b (b/a) \left( 1 + 0.172 (b/a)^{2} \right),$$ (4)

The tonehole effective length $t_e$ is similar to $t$, though it includes inner and outer length-correction terms. The value for $t_e$ given in [3] is frequency-dependent, though at low frequencies the following approximation is sufficiently accurate:

$$t_{e} = t + b \left( 1.4 - 0.58 (b/a)^2 \right)$$ (5)

An additional effect of inserting a hole in a woodwind bore is that the effective acoustic length of the bore is slightly reduced on both sides of the hole [3, 10]. This length-correction depends on the tonehole series equivalent length, for which we found a simplified expression that applies to both open and closed tonehole state:

$$t_{a} = \frac{0.47 (b/a)^4}{1 + 0.62(b/a)^2 + 0.64(b/a)}$$ (6)

The total main bore negative length correction for a tonehole with series equivalent length $t_a$ is $l_{a} = -(a/b)^2 t_{a}$ [3]. Thus if the lengths of the main bore sections on each side of the tonehole are $l_1$ and $l_2$, they should be corrected to $l_1 + l_{a}/2$ and $l_2 + l_{a}/2$, respectively. Because the length-correction is very small, this formulation differs only slightly from the series impedance formulation in [3].