Constant-Strength Doublet (Vortex Ring)

Recall that the potential field induced by doublets is given by

\[\begin{align*} \phi_\mu(\mathbf{x}) = % \frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \mu \hat{\mathbf{n}} \cdot \nabla \left(\frac{1}{r}\right) \,\mathrm{d}S ,\end{align*}\]

while the component induced by sources is

\[\begin{align*} \phi_\sigma(\mathbf{x}) = % -\frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \sigma \frac{1}{r} \,\mathrm{d}S .\end{align*}\]

Assuming constant-strength panels, each strength comes out of the integrals and we rewrite the above equations as

\[\begin{align*} \phi_\mu(\mathbf{x}) = % \mu G_\mu (\mathbf{x}) \quad , \quad G_\mu (\mathbf{x}) \equiv \frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \hat{\mathbf{n}} \cdot \nabla \left(\frac{1}{r}\right) \,\mathrm{d}S ,\end{align*}\]

and

\[\begin{align*} \phi_\sigma(\mathbf{x}) = % \sigma G_\sigma (\mathbf{x}) \quad , \quad G_\sigma (\mathbf{x}) \equiv -\frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \frac{1}{r} \,\mathrm{d}S ,\end{align*}\]

Notice that computation of the potential field induced by the doublet element is similar to the computation of the normal velocity induced by the source element:

\[\begin{align*} \hat{\mathbf{n}} \cdot \nabla\phi_\sigma = -\sigma \frac{f_{\tiny \partial V_b}}{\pi } \int\limits_{\partial V_b} \hat{\mathbf{n}} \cdot \nabla \left( \frac{1}{r} \right) \,\mathrm{d}S = -\sigma G_\mu. \end{align*}\]

Hence, we can simply reuse the computation of the source-induced velocity to calculate the potential of the doublet element as

\[\begin{align*} \phi_\mu = -\mu \frac{\hat{\mathbf{n}} \cdot \mathbf{u}_\sigma}{\sigma} ,\end{align*}\]

where $\hat{\mathbf{n}} \cdot \mathbf{u}_\sigma = U_z$ with $U_z$ as defined in the previous section.

In order to calculate the velocity field induced by the doublet, instead of calculating $\nabla\phi_\mu$ we take advantage of the fact that the velocity induced by a constant-strength doublet panel is the same than the velocity induced by a vortex ring (see Katz and Plotkin Sec. 10.4.3),

\[\begin{align*} \mathbf{u}_\mathrm{ring} \left( \mathbf{x} \right) = \frac{\Gamma}{4\pi} \sum\limits_{i,j\in A} \frac{\mathbf{r}_i \times \mathbf{r}_j}{ \Vert \mathbf{r}_i \times \mathbf{r}_j \Vert^2} \mathbf{r}_{ij} \cdot \left( \frac{\mathbf{r}_i}{r_i} - \frac{\mathbf{r}_j}{r_j} \right) ,\end{align*}\]

when $\Gamma = \mu$, and where $\mathbf{r}_{ij} = \mathbf{p}_j-\mathbf{p}_i$, $\mathbf{r}_i = \mathbf{x} - \mathbf{p}_i$, $\mathbf{r}_j = \mathbf{x} - \mathbf{p}_j$, $A = \{(1,2),\,\dots,\,(n-1, n),\,(n, 1) \}$ and $n$ the number of vertices that make the panel, each with position $\mathbf{p}_k$.

NOTE: For the same reasons explained in the previous section, $\phi_\mu$ poses a discontinuity at the surface since $\lim\limits_{z\rightarrow \pm 0} \phi_\mu (0, 0, z) = -\frac{\mu}{\sigma} \lim\limits_{z\rightarrow \pm 0} U_z (0, 0, z) = \mp \frac{\mu}{2}$. In FLOWPanel, we let $\phi_\mu (0, 0, 0) = 0$, but we also have shifted all control points slightly in the direction of $\hat{\mathbf{n}}$. Remembering that $\hat{\mathbf{n}}_\mathrm{HS} = -\hat{\mathbf{n}}$, the control points are thus shifted in the $-z$ direction, effectively obtaining $\boxed{\phi_\mu (\mathbf{x}_\mathrm{cp}) \approx \frac{\mu}{2}}$.

OBSERVATIONS

• The velocity induced by a segment of the vortex ring shown above becomes singular when the denominator terms $r_i$, $r_j$, or $\Vert \mathbf{r}_i \times \mathbf{r}_j \Vert$ approach $0$. For this reason, FLOWPanel adds a small epsilon to each of these terms, while also defining a cutoff threshold for $\Vert \mathbf{r}_i \times \mathbf{r}_j \Vert$ under which the velocity induced becomes $0$ (thus, self-induced velocity is forced to be zero).

NOTE: The small offset added to the denominator terms $r_i$, $r_j$, and $\Vert \mathbf{r}_i \times \mathbf{r}_j \Vert$ corresponds to body.kerneloffset, while the cutoff threshold for self-induced velocity corresponds to body.kernelcutoff.

The potential and velocity field of a doublet panel of unitary strength ($\mu=1$) is shown below

$\nabla \phi$ = $\mathbf{u}$ verification