In 3D, given three points $P_1$, $P_2$, and $P_3$ spanning a non-degenerate triangle $T$. How to determine if the projection of a point $P$ onto the plane of $T$ lies within $T$?
2 Answers
The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle
There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.
Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as
- $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
- $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
- $\alpha=1-\gamma-\beta$
The coordinates of the projected point is
- $P'=\alpha P_1+\beta P_2 +\gamma P_3$
The point $P'$ lies inside $T$ if
- $0\leq\alpha\leq 1$,
- $0\leq\beta\leq 1$, and
- $0\leq\gamma\leq 1$.
-
$\begingroup$ Are there any corner cases where this should fail? The paper doesn't say anything to it $\endgroup$ Nov 29, 2017 at 11:41
-
$\begingroup$ Doesn't this only work if point P already lies on the plane created by the triangle [P1,P2,P3]? $\endgroup$– MasterHDOct 25, 2018 at 8:59
-
$\begingroup$ @MasterHD The point is projected onto the plane, so it does not have to lie on the plane. $\endgroup$ Oct 25, 2018 at 9:11
-
$\begingroup$ Quick question... in the equations for gamma and beta, what on earth is this "[" and this "]". I imagine it must mean "signed magnitude" and explicitly not just magnitude (for the author would have written | or || , plus Craig's answer below codes it as such), but I can't find something online to confirm what [ and ] are meaning here. $\endgroup$– XenialDec 8, 2019 at 5:07
-
1$\begingroup$ @Xenial They are just the ordinary grouping brackets. Note that they are not necessary in this case. $\endgroup$ Dec 8, 2019 at 8:10
Thanks to Håkon Hægland for asking and answering this question, and especially for providing the key information from the hard-to-obtain paper: Wolfgang Heidrich, 2005, Computing the Barycentric Coordinates of a Projected Point, Journal of Graphics Tools, pp 9-12, 10(3).
I was web searching for exactly this information. Among the top hits were a question on the gamedev.stackexchange.com site, and this one. After implementing the technique presented here by Håkon, I posted that code and a link to this question at the game dev site: Easy way to project point onto triangle (or plane). Later it occurred that the C++/Eigen implementation might be useful to readers of this question:
bool pointInTriangle(const Eigen::Vector3f& query_point,
const Eigen::Vector3f& triangle_vertex_0,
const Eigen::Vector3f& triangle_vertex_1,
const Eigen::Vector3f& triangle_vertex_2)
{
// u=P2−P1
Eigen::Vector3f u = triangle_vertex_1 - triangle_vertex_0;
// v=P3−P1
Eigen::Vector3f v = triangle_vertex_2 - triangle_vertex_0;
// n=u×v
Eigen::Vector3f n = u.cross(v);
// w=P−P1
Eigen::Vector3f w = query_point - triangle_vertex_0;
// Barycentric coordinates of the projection P′of P onto T:
// γ=[(u×w)⋅n]/n²
float gamma = u.cross(w).dot(n) / n.dot(n);
// β=[(w×v)⋅n]/n²
float beta = w.cross(v).dot(n) / n.dot(n);
float alpha = 1 - gamma - beta;
// The point P′ lies inside T if:
return ((0 <= alpha) && (alpha <= 1) &&
(0 <= beta) && (beta <= 1) &&
(0 <= gamma) && (gamma <= 1));
}
-
2$\begingroup$ This can be done more easily without cross products. See gamedev.stackexchange.com/questions/23743/… $\endgroup$ Jan 27, 2018 at 22:17