Constant-Strength Vortex Sheet

(Adapted from David Pate's 2017 doctoral dissertation A Surface Vorticity Method for Wake-Body Interactions, and Pate and German (2018), A Surface Vorticity Panel Method)

The velocity induced by a vortex sheet at a target point $\mathbf{x}$ is the integral of the Biot-Savart law over the surface of the sheet:

\[\begin{align*} \mathbf{u}(\mathbf{x}) = \frac{1}{4\pi} \int\limits_S \frac{\mathbf{r} \times \boldsymbol\gamma}{\Vert \mathbf{r} \Vert^3} \,\mathrm{d} S' ,\end{align*}\]

where $\mathrm{r} = \mathbf{x}' - \mathbf{x}$ is the vector directed from the target to the differential $\mathrm{d}S'$.

In order to evaluate this integral, we follow the procedure put forth by Pate 2017. First, we define a local coordinate system at the projection of point $\mathbf{x}$ unto the plane of the triangle as shown below:

(figure retrieved from Pate 2017)

The projection point $\mathbf{p}_x$ (or $P_B$ in the figure above) is calculated as $\mathbf{p}_x = \mathbf{x} - \left[ (\mathbf{x} - \mathbf{p}_1)\cdot \hat{\mathbf{n}} \right] \hat{\mathbf{n}}$, and the orthonormal basis $(\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3)$ is completely arbitrary except for requiring $\mathbf{b}_3 = \hat{\mathbf{n}}$. We also define the vector directed from $\mathbf{p}_x$ to the differential as $\boldsymbol\rho \equiv \mathbf{x}' - \mathbf{p}_x$. We then define the following three triangles $S_1, S_2, S_3$ that will be used for the integration:

(figure retrieved from Pate 2017)

where, $\mathbf{q}_i = \mathbf{p}_i - \mathbf{p}_x$. $S_1$ is the triangle of vertices $(\mathbf{p}_x, \mathbf{p}_1, \mathbf{p}_2)$, $S_2$ is $(\mathbf{p}_x, \mathbf{p}_2, \mathbf{p}_3)$, and $S_3$ is $(\mathbf{p}_x, \mathbf{p}_3, \mathbf{p}_1)$. More generally, the vertices of triangle $S_i$ in the global coordinate system are $(\mathbf{p}_x, \mathbf{p}_i, \mathbf{p}_{i+1})$, with $\mathbf{p}_{4} = \mathbf{p}_{1}$. In the coordinate system with origin at $\mathbf{p}_x$ and basis $(\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3)$, the vertices of triangle $S_i$ are

\[\begin{align*} \boxed{ (\mathbf{0}, \mathbf{q}_1, \mathbf{q}_{2})_i = (\mathbf{p}_x, \mathbf{p}_i - \mathbf{p}_x, \mathbf{p}_{i+1} - \mathbf{p}_x) } \end{align*}\]

For each triangle $S_i$, Pate then proceeds to define the following orthonormal basis $(\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3)$:

(figure retrieved from Pate 2017)

\[\begin{align*} \mathbf{c}_3 & \equiv \mathbf{b}_3 = \hat{\mathbf{n}} \\ \mathbf{c}_2 & \equiv \frac{\mathbf{q}_2 - \mathbf{q}_1}{\Vert \mathbf{q}_2 - \mathbf{q}_1 \Vert} \\ \mathbf{c}_1 & \equiv \mathbf{c}_2 \times \mathbf{c}_3 \end{align*}\]

The vector $\boldsymbol\rho$ is then decomposed in this basis as

\[\begin{align*} \boldsymbol\rho & = u \mathbf{c}_1 + v \mathbf{c}_2 \end{align*}\]

where

\[\begin{align*} u & \equiv \boldsymbol\rho \cdot \mathbf{c}_1 \\ v & \equiv \boldsymbol\rho \cdot \mathbf{c}_2 .\end{align*}\]

The full-distance vector $\mathbf{r}$ (pointing from $\mathbf{x}$ to the differential) is decomposed as

\[\begin{align*} \mathbf{r} & = u \mathbf{c}_1 + v \mathbf{c}_2 - z \mathbf{c}_3 \end{align*}\]

where

\[\begin{align*} z & \equiv \left( \mathbf{x} - \mathbf{p}_1 \right) \cdot \hat{\mathbf{n}} ,\end{align*}\]

while the constant strength of the vortex sheet is decomposed as

\[\begin{align*} \boldsymbol\gamma & = a_{00} \mathbf{c}_1 + b_{00} \mathbf{c}_2 ,\end{align*}\]

where

\[\begin{align*} a_{00} & \equiv \boldsymbol\gamma \cdot \mathbf{c}_1 \\ b_{00} & \equiv \boldsymbol\gamma \cdot \mathbf{c}_2 .\end{align*}\]

The numerator in the integral of the Biot-Savart law then becomes

\[\begin{align*} \mathbf{r} \times \boldsymbol\gamma & = z b_{00} \mathbf{c}_1 - z a_{00} \mathbf{c}_2 + (u b_{00} - v a_{00}) \mathbf{c}_3 .\end{align*}\]

In order to avoid the singularity when the denominator of the integral becomes zero, we add a small offset $\delta$ to the distance of the denominator defining $\boxed{r \equiv \sqrt{u^2 + v^2 + z^2 + \delta^2}}$. We also add the same offset to the height of the target as $\boxed{h \equiv \sqrt{z^2 + \delta^2}}$, effectively defining a thickness for the sheet.

The integral over the original triange $S$ can then be expressed in terms of $S_1$, $S_2$, and $S_3$ as

\[\begin{align*} \int\limits_S \frac{\mathbf{r} \times \boldsymbol\gamma}{r^3} \,\mathrm{d} S' = \sum\limits_i \int\limits_{S_i} \frac{\mathbf{r} \times \boldsymbol\gamma}{r^3} \,\mathrm{d} S_i' ,\end{align*}\]

The integral terms $H_{00}$, $H_{10}$, and $H_{01}$ are given in Pate 2017, Appendix A.4:

\[\begin{align*} & H_{00} = \frac{1}{h} \tan^{-1}\left.\left( \frac{al}{ a^2 + h^2 + h\sqrt{l^2 + a^2 + h^2}} \right)\right\rvert_{l_1}^{l_2} \\ & H_{10} = \left. \left[ \frac{l}{\sqrt{l^2 + a^2}} \ln\left( \sqrt{l^2 + a^2 + h^2} + \sqrt{l^2 + a^2} \right) - \ln\left( l + \sqrt{l^2 + a^2 + h^2} \right) - \frac{l \ln h }{\sqrt{l^2 + a^2}} \right] \right\rvert_{l_1}^{l_2} \\ & H_{01} = \left. \left\{ \frac{a}{\sqrt{l^2 + a^2}} \left[ \ln h - \ln\left( \sqrt{l^2 + a^2 + h^2} + \sqrt{l^2 + a^2} \right) \right] \right\} \right\rvert_{l_1}^{l_2} \end{align*}\]

and the integral over $S_i$ is calculated as

\[\begin{align*} \int\limits_{S_i} \frac{\mathbf{r} \times \boldsymbol\gamma}{r^3} \,\mathrm{d} S_i' = z b_{00} \mathbf{c}_1 \underbrace{ \int\limits_{S_i} \frac{1}{r^3} \,\mathrm{d} S_i' }_{H_{00}} - z a_{00} \mathbf{c}_2 \underbrace{ \int\limits_{S_i} \frac{1}{r^3} \,\mathrm{d} S_i' }_{H_{00}} + b_{00} \mathbf{c}_3 \underbrace{ \int\limits_{S_i} \frac{u}{r^3} \,\mathrm{d} S_i' }_{H_{10}} - a_{00} \mathbf{c}_3 \underbrace{ \int\limits_{S_i} \frac{v}{r^3} \,\mathrm{d} S_i' }_{H_{01}} .\end{align*}\]

The integral terms $H_{00}$, $H_{10}$, and $H_{01}$ are given in Pate 2017, Appendix A.4:

\[\begin{align*} & H_{00} = \frac{1}{h} \tan^{-1}\left.\left( \frac{al}{ a^2 + h^2 + h\sqrt{l^2 + a^2 + h^2}} \right)\right\rvert_{l_1}^{l_2} \\ & H_{10} = \left. \left[ \frac{l}{\sqrt{l^2 + a^2}} \ln\left( \sqrt{l^2 + a^2 + h^2} + \sqrt{l^2 + a^2} \right) - \ln\left( l + \sqrt{l^2 + a^2 + h^2} \right) - \frac{l \ln h }{\sqrt{l^2 + a^2}} \right] \right\rvert_{l_1}^{l_2} \\ & H_{01} = \left. \left\{ \frac{a}{\sqrt{l^2 + a^2}} \left[ \ln h - \ln\left( \sqrt{l^2 + a^2 + h^2} + \sqrt{l^2 + a^2} \right) \right] \right\} \right\rvert_{l_1}^{l_2} \end{align*}\]

where

\[\begin{align*} & a = \mathbf{q}_1 \cdot \mathbf{c}_1 \\ & l_1 = \mathbf{q}_1 \cdot \mathbf{c}_2 \\ & l_2 = \mathbf{q}_2 \cdot \mathbf{c}_2 \\ & h = \sqrt{z^2 + \delta^2} \\ & z = \left( \mathbf{x} - \mathbf{p}_1 \right) \cdot \hat{\mathbf{n}} \end{align*}\]

NOTE: $H_{10}$ and $H_{01}$ becomes undefined when the projection of $\mathbf{x}$ onto the panel plane lays on one of the vertices, leading to the a division of zero by zero in $\frac{l}{\sqrt{l^2 + a^2}}$ and $\frac{a}{\sqrt{l^2 + a^2}}$. Both $H_{10}$ and $H_{01}$ actually converge to 0 in that situation, and can be manually prescibed to that value.

Finally, the velocity induced by the sheet is evaluated as

\[\begin{align*} \boxed{ \mathbf{u}(\mathbf{x}) = \frac{1}{4\pi} \int\limits_S \frac{\mathbf{r} \times \boldsymbol\gamma}{\Vert \mathbf{r} \Vert^3} \,\mathrm{d} S' = \frac{1}{4\pi} \sum\limits_i \left[ z b_{00} H_{00} \mathbf{c}_1 - z a_{00} H_{00} \mathbf{c}_2 + \left(b_{00} H_{10} - a_{00} H_{01}\right) \mathbf{c}_3 \right] \, } ,\end{align*}\]

To speed up computation, we can use trigonometric and logarithmic identities to rewrite the formulas given by Pate as

\[\begin{align*} & H_{00} = \frac{1}{h} \tan^{-1}\left( \frac{m_2 - m_1}{1 + m_1 m_2} \right) \qquad \text{ where } \quad m_i = \frac{al_i}{ a^2 + h^2 + h\sqrt{l_i^2 + a^2 + h^2}} \\ & H_{10} = \ln\left( \frac{m_2}{m_1} \right) \qquad \text{ where } \quad m_i = \frac{1}{l_i + \sqrt{l_i^2 + a^2 + h^2}} \left( \frac{\sqrt{l_i^2 + a^2 + h^2} + \sqrt{l_i^2 + a^2}}{h} \right)^\frac{l_i}{\sqrt{l_i^2 + a^2}} \\ & H_{01} = - \ln \left( \frac{m_2}{m_1} \right) \qquad \text{ where } \quad m_i = \left( \frac{\sqrt{l_i^2 + a^2 + h^2} + \sqrt{l_i^2 + a^2}}{h} \right)^\frac{a}{\sqrt{l_i^2 + a^2}} ,\end{align*}\]

thus reducing the number of $\log$ and $\tan^{-1}$ evaluations.