How to compute a point on a line in CGAL
Asked Answered
S

2

6

Given a 3D line in CGAL, how do I compute a point on that line that is some known distance from an endpoint?

Schreibman answered 17/8, 2010 at 18:26 Comment(0)
J
4

If you have two points P0 and P1, you can make a vector V = P1 - P0.

Given distance D from P0, you can get the resulting point R = P0 + (D ÷ ||V||) ⋅ V.

(Linearly interpolate between the lines, changing D into a percentage by dividing by the full length of the line.)


I don't know CGAL (and the documentation kind of sucks), but I assume it'd be something like this:

Line_3<K> l = /* ... */;
Vector_3<K> v = l.to_vector();
Point_3<K> r = l.p + (d * d / v.squared_length()) * v;

Note I can't even find a way to get the starting point of a line, so that one is up to you. (The l.p part is made up.)

Jello answered 17/8, 2010 at 19:2 Comment(3)
To get the two points in a line: typedef Kernel::Line_3 Line; Line l = Line(point1, point2); cout << l.point(0) << l.point(1) << "\n";Murderous
Well, it won't compile - CGAL doesn't define one of the operators for points and vectors. I don't really know which one, because gcc is the worst compiler known to man. But it's very close, so you get a check :)Murderous
@Max: Ah, if you post the errors I'm sure we could help. (Edit the question, I mean.) And thanks.Jello
L
0

You can just subtract the points you're interested in to get a slope vector and then walk along it. A MWE is below:

// Compile with: clang++ -DBOOST_ALL_NO_LIB -DCGAL_USE_GMPXX=1 -O2 -g -DNDEBUG -Wall -Wextra -pedantic -march=native -frounding-math main.cpp -lgmpxx -lmpfr -lgmp

#include <CGAL/Exact_predicates_exact_constructions_kernel.h>

using K = CGAL::Exact_predicates_exact_constructions_kernel;
using Point_3 = K::Point_3;

class InterPointInterpolator {
 public:
  InterPointInterpolator(const Point_3 &a, const Point_3 &b) : a(a), b(b) {}
  // Returns points interpolated from a at t=0 to b at t=1
  Point_3 operator()(const double t) const {
    const auto m = b - a;
    return a + t * m;
  }
 private:
  Point_3 a;
  Point_3 b;
};

int main(){
  InterPointInterpolator ipi(Point_3(0, 0, 0), Point_3(10, 5, 20));
  for(int i=0;i<=10;i++){
    const auto interpolated_point = ipi(i/10.0);
    std::cout<<interpolated_point<<std::endl;
  }

  return 0;
}

Output:

0  0   0
1  0.5 2
2  1   4
3  1.5 6
4  2   8
5  2.5 10
6  3   12
7  3.5 14
8  4   16
9  4.5 18
10 5   20
Lavonlavona answered 17/8, 2010 at 18:26 Comment(0)

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