TriSeg2022

circadapt.components.cavity.TriSeg2022(model)

TriSeg2022 is designed to represent ventricles including interaction.

List of relevant tutorials

Documentation

TriSeg2022 is designed to represent ventricles including interaction.

Parameters

tau: float [-]

Constant to limit change in estimation of dV and dY

Signals

V: array

Volume of truncated septal wall

V0: array

Volume of truncated septal wall before finding mechanical equilibrium

VDot: array

Time-derivative of V, used to find the estimation V0

Y: array

Distance of wall-junction to center

Y0: array

Distance of wall-junction to center before finding mechanical equilibrium

YDot: array

Time-derivative of Y, used to find the estimation Y0

A TriSeg element consists of two cavities, encapsulated by three walls, one of them separating both cavities. Volumes of the left and right cavity are VcL and VcR, respectively. The left, septal and right wall volumes are indicated by VwL, VwS and VwR, respectively. For the derivation we use the assumption that mid-wall surfaces are spherical. The volumes enclosed by the left and right mid-wall volume skeleton, VL and VR, are the sum of cavity and enclosed wall volumes:

(1)\[ \begin{align}\begin{aligned} V_L = V_{c,L} + 0.5V_{w,L} + 0.5V_{w,S}\\ V_R = V_{c,R} + 0.5V_{w,R} + 0.5V_{w,S}\end{aligned}\end{align} \]

The mid-wall skeleton consists of 3 spherical mid-wall caps, connected by a common circular junction. The radius of this junction circle is indicated by symbol y. The volume enclosed by the septal mid-wall and the plane through the junction circle is indicated by symbol VmS. Note that bending of the septum towards the left ventricle implies a negative value for VmS. Given the cavity volumes VL and VR, the TriSeg geometry has two degrees of freedom, i.e., y and VmS. If y and VmS are known, geometry is known and mid-wall areas AwL, AwS and AwR, can be calculated. Assuming a linear relationship between mid-wall area Aw and wall tension T according to Eq. (.), deformation energy for all walls together can be determined as a function of y and VmS. The latter values are found by the solution with minimum deformation energy of the walls.

For each of the three mid-wall caps we determine wall area Aw as a function of cap junction radius y and cap volume V. Volume V and area Aw are written as a function of x and y, where x represents the maximum distance between cap wall and junction plane. It is derived:

(2)\[ \begin{align}\begin{aligned} V_m = \frac{\pi}{6} (x^3+3x y^2)\\ A_m = \pi (x^2+y^2)\end{aligned}\end{align} \]

Knowing y and V, from the 3rd order polynomial of Eq. (.), x is solved. Next, A_m is calculated with Eq. (.). Using Eq. (.), for deformation energy of the wall it is found:

(3)\[ E_m = \frac{1}{2} \frac{dt}{dA_m} (A_m - A_{m,0})^2\]

Parameters A_m0 and dt/dAm, determining the mechanical properties of the wall, are iteratively solved. Cap volumes VmL and VmR depend on septal cap volume VmS:

(4)\[ V_{m,L} = V_{m,S} - V_L V_{m,R} = V_{m,S} + V_R\]

Applying Eq. (12), deformation energy EwL, EwS and EwR for the three walls can be determined. Total elastic energy of the system equals the sum of energies of the three walls:

(5)\[ E_tot ( y ,V_{mS}) = E_{m,L}( y ,V_mS) +E_{mS} ( y ,V_{mS}) + E_mR ( y ,V_{mS})\]

By minimizing Etot, variables y and Vs are solved.

Solving strategy