I have some data that I am using to generate a triangulation of a surface without boundary in $\mathbb{R}^3$. The triangulation gives each triangle a unit normal pointing outwards. I need to find some point in the interior of the surface.

I will outline my general strategy. I let $V$ denote the set of vertices of the triangulation and compute a number that I will call the fineness $$ \delta = \min_{x, y \in V \mid x\neq y} \lVert x-y\rVert. $$ I pick some vertex $v$ and consider the point $$ p = v -\frac{\delta}{4}u, $$ where $u$ is some unit vector. It is possible to show that $p$ is closer to $v$ than any other vertex in the triangulation, so it suffices to choose $u$ such that it lies below all of the triangles containing $v$.

If $i$ indexes the triangles containing $v$ as a vertex and $n_i$ is the outer normal of the triangle $i$, choosing $u$ such that $\langle u, n_i\rangle >0$ for all $i$ will put $p$ below the surface.

We can map this onto a linear algebra problem by writing $u$ as a weighted combination of the outer normals: $u = \sum_i w_i n_i$. If all of the inner products of the outer normals were non-negative, we could solve this by taking the $w_i$ to be the components of the Perron–Frobenius eigenvector of the matrix $A_{i,j} = \langle n_i, n_j\rangle$. However, the surface can have regions of high curvature, so $A$ is not always non-negative.

A natural notion of the normal at $v$ is the angle weighted averaged of the normals. This lead me to try taking $w_i = \theta_i$, where $\theta_i$ is angle of the triangle $i$ at $v$. However, this alone does not guarantee $\langle u, n_i \rangle > 0$ for all $i$. This seems like some iterative approach is needed, where the weights corresponding to positive inner products should be decreased and those corresponding to non-positive ones should increase. I think one strategy should be to add to the weights of non-positive terms and then normalize all of the weights so that $\sum_i w_i = 1$. If all inner products are positive, we terminate the iterative procedure. Of course, we would have to make $u$ a unit vector at the end of this.

My questions are as follows. Am I on the right track? Is there a simpler approach that I am missing? If I am on the right track, what update rule should I choose for the weights that guarantees convergence?

Edit: The shapes that I am working with are definitely not convex. The data points are from intestinal organoids, which can roughly be shaped like octopi holding broccoli.

I have a triangulation of a surface without boundary in $\mathbb{R}^3$. The triangulation gives a unit normal pointing outwards for each triangle. I need to find some point in the interior of the surface.

I will outline my general strategy. I let $V$ denote the set of vertices of the triangulation and compute a number that I will call the fineness $$ \delta = \min_{x, y \in V \mid x\neq y} \lVert x-y\rVert. $$ I pick some vertex $v$ and consider the point $$ p = v -\frac{\delta}{4}u, $$ where $u$ is some unit vector. It is possible to show that $p$ is closer to $v$ than any other vertex in the triangulation, so it suffices to choose $u$ such that it lies below all of the triangles containing $v$.

If $i$ indexes the triangles containing $v$ as a vertex and $n_i$ is the outer normal of the triangle $i$, choosing $u$ such that $\langle u, n_i\rangle >0$ for all $i$ will put $p$ below the surface.

We can map this onto a linear algebra problem by writing $u$ as a weighted combination of the outer normals: $u = \sum_i w_i n_i$. If all of the inner products of the outer normals were non-negative, we could solve this by taking the $w_i$ to be the components of the Perron–Frobenius eigenvector of the matrix $A_{i,j} = \langle n_i, n_j\rangle$. However, the surface can have regions of high curvature, so $A$ is not always non-negative.

A natural notion of the normal at $v$ is the angle weighted averaged of the normals. This lead me to try taking $w_i = \theta_i$, where $\theta_i$ is angle of the triangle $i$ at $v$. However, this alone does not guarantee $\langle u, n_i \rangle > 0$ for all $i$. This seems like some iterative approach is needed, where the weights corresponding to positive inner products should be decreased and those corresponding to non-positive ones should increase. I think one strategy should be to add to the weights of non-positive terms and then normalize all of the weights so that $\sum_i w_i = 1$. If all inner products are positive, we terminate the iterative procedure. Of course, we would have to make $u$ a unit vector at the end of this.

My questions are: Am I on the right track? Is there a simpler approach? If I am on the right track, what update rule for the weights will guarantee convergence?

Edit: The shapes that I am working with are definitely not convex. The data points are from intestinal organoids, which can roughly be shaped like octopi holding broccoli.

I have some data that I am using to generate a triangulation of a surface without boundary in $\mathbb{R}^3$. The triangulation gives each triangle a unit normal pointing outwards. I need to find some point in the interior of the surface.

I will outline my general strategy. I let $V$ denote the set of vertices of the triangulation and compute a number that I will call the fineness $$ \delta = \text{min}_{x, y \in V| x\neq y} ||x-y||. $$ I pick some vertex $v$ and consider the point $$ p = v -\frac{\delta}{4}u, $$ where $u$ is some unit vector. It is possible to show that $p$ is closer to $v$ than any other vertex in the triangulation, so it suffices to choose $u$ such that it lies below all of the triangles containing $v$.

If $i$ indexes the triangles containing $v$ as a vertex and $n_i$ is the outer normal of the triangle $i$, choosing $u$ such that $\langle u, n_i\rangle >0$ for all $i$ will put $p$ below the surface.

We can map this onto a linear algebra problem by writing $u$ as a weighted combination of the outer normals: $u = \sum_i w_i n_i$. If all of the inner products of the outer normals were non-negative, we could solve this by taking the $w_i$ to be the components of the Perron-Frobenius eigenvector of the matrix $A_{i,j} = \langle n_i, n_j\rangle$. However, the surface can have regions of high curvature, so $A$ is not always non-negative.

A natural notion of the normal at $v$ is the angle weighted averaged of the normals. This lead me to try taking $w_i = \theta_i$, where $\theta_i$ is angle of the triangle $i$ at $v$. However, this alone does not guarantee $\langle u, n_i \rangle > 0$ for all $i$. This seems like some iterative approach is needed, where the weights corresponding to positive inner products should be decreased and those corresponding to non-positive ones should increase. I think one strategy should be to add to the weights of non-positive terms and then normalize all of the weights so that $\sum_i w_i = 1$. If all inner products are positive, we terminate the iterative procedure. Of course, we would have to make $u$ a unit vector at the end of this.

My questions are as follows. Am I on the right track? Is there a simpler approach that I am missing? If I am on the right track, what update rule should I choose for the weights that guarantees convergence?

Edit: The shapes that I am working with are definitely not convex. The data points are from intestinal organoids, which can roughly be shaped like octopi holding broccoli.

I have some data that I am using to generate a triangulation of a surface without boundary in $\mathbb{R}^3$. The triangulation gives each triangle a unit normal pointing outwards. I need to find some point in the interior of the surface.

I will outline my general strategy. I let $V$ denote the set of vertices of the triangulation and compute a number that I will call the fineness $$ \delta = \min_{x, y \in V \mid x\neq y} \lVert x-y\rVert. $$ I pick some vertex $v$ and consider the point $$ p = v -\frac{\delta}{4}u, $$ where $u$ is some unit vector. It is possible to show that $p$ is closer to $v$ than any other vertex in the triangulation, so it suffices to choose $u$ such that it lies below all of the triangles containing $v$.

If $i$ indexes the triangles containing $v$ as a vertex and $n_i$ is the outer normal of the triangle $i$, choosing $u$ such that $\langle u, n_i\rangle >0$ for all $i$ will put $p$ below the surface.

We can map this onto a linear algebra problem by writing $u$ as a weighted combination of the outer normals: $u = \sum_i w_i n_i$. If all of the inner products of the outer normals were non-negative, we could solve this by taking the $w_i$ to be the components of the Perron–Frobenius eigenvector of the matrix $A_{i,j} = \langle n_i, n_j\rangle$. However, the surface can have regions of high curvature, so $A$ is not always non-negative.

A natural notion of the normal at $v$ is the angle weighted averaged of the normals. This lead me to try taking $w_i = \theta_i$, where $\theta_i$ is angle of the triangle $i$ at $v$. However, this alone does not guarantee $\langle u, n_i \rangle > 0$ for all $i$. This seems like some iterative approach is needed, where the weights corresponding to positive inner products should be decreased and those corresponding to non-positive ones should increase. I think one strategy should be to add to the weights of non-positive terms and then normalize all of the weights so that $\sum_i w_i = 1$. If all inner products are positive, we terminate the iterative procedure. Of course, we would have to make $u$ a unit vector at the end of this.

My questions are as follows. Am I on the right track? Is there a simpler approach that I am missing? If I am on the right track, what update rule should I choose for the weights that guarantees convergence?

Edit: The shapes that I am working with are definitely not convex. The data points are from intestinal organoids, which can roughly be shaped like octopi holding broccoli.