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Effects of Central Airway Shunting on the Mechanical Impedance of the Mouse Lung

Abstract

The mechanical properties of the lung are embodied in its mechanical input impedance, which it is interpreted in physiological terms by being fit with a mathematical model. The normal lung is extremely well described by a model consisting of a single uniformly ventilated compartment comprised of tissue having a constant-phase impedance, but to describe the abnormal lung it frequently becomes necessary to invoke additional compartments. To date, all evidence of regional mechanical heterogeneity in the mouse lung has been assumed to be of the parallel variety. We therefore investigated the use of a serial heterogeneity model, relative to parallel heterogeneity and homogeneous models, for describing impedance spectra in mice subjected to a variety of interventions designed to make their lungs heterogeneous. We found that functional evidence of the finite stiffness of the airway wall in mice with airways obstruction can sometimes be apparent in lung impedance below 20 Hz. The model estimates of airway stiffness were smaller than direct estimates obtained from micro-CT images of the lung in vivo, suggesting that the conducting airways alone are likely not the precise anatomical correlate of proximal functional stiffness in the lung. Nevertheless, we conclude that central airway shunting in mice can sometimes be an important physiological phenomenon.

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Acknowledgments

The authors acknowledge Michael Kavouksorian for assistance with the micro-CT image analysis and the support of NIH grants HL087788 and NCRR COBRE PP15557.

Author information Authors and Affiliations
  1. Vermont Lung Center, Department of Medicine, University of Vermont, 149 Beaumont Avenue, HSRF 228, Burlington, VT, 05405, USA

    Benjamin L. Schwartz, Ron C. Anafi, Minara Aliyeva, John A. Thompson-Figueroa, Gilman B. Allen, Lennart K. A. Lundblad & Jason H. T. Bates

Authors
  1. Benjamin L. Schwartz
  2. Ron C. Anafi
  3. Minara Aliyeva
  4. John A. Thompson-Figueroa
  5. Gilman B. Allen
  6. Lennart K. A. Lundblad
  7. Jason H. T. Bates
Corresponding author

Correspondence to Jason H. T. Bates.

Additional information

Associate Editor Kenneth R. Lutchen oversaw the review of this article.

Appendix Appendix

The length and orientation of an airway is identified by a line drawn axially along it using the GE software tools. This line is oriented with respect to the x, y, and z directions of the three-dimensional image by the angles θ x , θ y , and θ z , respectively (Fig. A1, a). Cuts through the airway for the purposes of determining its diameter were made in the xy plane, which intersects the (assumed circular) airway walls in an ellipse with eccentricity ε and a major axis that makes an angle θ with the x axis. Given θ x , θ y , and θ z , we need to calculate ε and θ.

Figure A1

(a) The geometry of a circular airway located at angles θ x , θ y , and θ z with respect to the x, y, and z axes of a three-dimensional lung image. (b) The ellipse made by the intersection of the xy plane with the airway

The line \( \overline{OA} \) runs from the origin of the axes through the center of the airway, as shown in Fig. A1. At the point A, we drop a line down so that it is normal to the xy plane and intersects the plane at point B. The project of \( \overline{OA} \) on the xy plane is the line \( \overline{OB} , \) and θ is the angle that \( \overline{OB} \) makes with the x axis. The projection of the line \( \overline{OB} \) on the x axis is the line \( \overline{OD} , \) and the projection on the y axis is \( \overline{OC} . \) OAD is then a right triangle with hypotenuse \( \overline{OA} . \) Therefore,

$$ \overline{AD} = \overline{OA} \sin \theta_{x} $$

(A.1)

\( \overline{AB} \) is the projection of \( \overline{OA} \) onto the z axis, so

$$ \overline{AB} = \overline{OA} \cos \theta_{z} $$

(A.2)

ADB is a right triangle with hypotenuse \( \overline{AD} , \) so

$$ \overline{DB}^{2} = \overline{AD}^{2} - \overline{AB}^{2} = \overline{OA}^{2} \left( {\sin^{2} \theta_{x} - \cos^{2} \theta_{z} } \right). $$

(A.3)

Similarly,

$$ \overline{CB}^{2} = \overline{AC}^{2} - \overline{AB}^{2} = \overline{OA}^{2} \left( {\sin^{2} \theta_{y} - \cos^{2} \theta_{z} } \right) $$

(A.4)

The angle θ that the major axis of the ellipse makes with the y-axis is thus

$$ \theta = \tan^{ - 1} \sqrt {{\frac{{\sin^{2} \theta_{y} - \cos^{2} \theta_{z} }}{{\sin^{2} \theta_{x} - \cos^{2} \theta_{z} }}}} $$

(A.5)

The eccentricity of the ellipse arises from the elongation of the major axis by the factor 1/cos(θ z ) due to the tilt of the airway away from the z direction, and is equal to the ratio of the major to the minor axis lengths. Thus,

$$ \varepsilon = {\frac{1}{{\cos \theta_{z} }}} $$

(A.6)

The line profiles collected from an airway lie in the xy plane and make angles θ i with the y-axis. The angles they make with the major axis of the ellipse are thus

$$ \phi_{i} = \theta_{i} - \theta $$

(A.7)

for i = 1,2,…,6.

Let the major axis of the ellipse be parallel to the α-axis and the minor axis with the β-axis, with the origin of the axes at the center of the ellipse. The equation of the ellipse in these axes is

$$ {\frac{{\alpha^{2} }}{{r^{2} /\cos^{2} \theta_{z} }}} + {\frac{{\beta^{2} }}{{r^{2} }}} = 1 $$

(A.8)

Consider a line segment of length q that makes an angle ϕ with the major axis of the ellipse (Fig. A1, b), passes through its center, and intersects opposite walls at the points (α, β) and (−α, −β). Then we have

$$ \alpha = \frac{q}{2}\cos \phi ,\quad \beta = \frac{q}{2}\sin \phi $$

(A.9)

so

$$ r = \frac{q}{2}\sqrt {\cos^{2} \phi \cos^{2} \theta_{z} + \sin^{2} \phi } $$

(A.10)

Thus, if the ith line drawn through the center of an airway in the xy plane shows a distance between opposite walls of q, and the line makes an angle θ i with the x-axis when the major axis of the airway ellipse makes an angle θ with the x-axis, the corrected distance d (=2r) between opposite walls is

$$ d = q\sqrt {\cos^{2} \left( {\phi_{i} - \phi } \right)\cos^{2} \theta_{z} + \sin^{2} \left( {\phi_{i} - \phi } \right)} $$

(A.11)

We tested this approach on a length of plastic tubing angled at 145.3, 107.5, and 118.9° with respect to the x, y, and z axes, respectively. Measuring multiple cuts through the tubing as described in “Methods,” we determined the internal diameter to be 0.295 mm (SD 0.038). According to the manufacturer, the diameter was 0.290 mm.

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Schwartz, B.L., Anafi, R.C., Aliyeva, M. et al. Effects of Central Airway Shunting on the Mechanical Impedance of the Mouse Lung. Ann Biomed Eng 39, 497–507 (2011). https://doi.org/10.1007/s10439-010-0123-2

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