Linear interpolation of three 3D points in 3D spac

Question

I have three 3D points like p1(x1,y1,z1), p2(x2,y2,z2), p3(x3,y3,z3). I have another point, but I know only x, y value of that point like p4(x4,y4,Z), in which Z is the value I like to compute.

I am sure p4(x4,y4) point is inside a triangle formed by p1(x1,y1), p2(x2,y2), p3(x3,y3) by checking with delaunay triangulation approach. How can I compute Z value of point p4? I like to implement it in C programming. Actually I am trying to implement griddata in MATLAB.

Thanks

Answer 1

This is to support both MBo's and Konstantin's answers. Please don't accept this question, but one of the others.

This is how you would implement a solution in MATLAB:

%// Your known 3 points
p1 = [ 1 10  0]';
p2 = [-1 10 10]';
p3 = [ 0  0 10]';

%// your 4th target point
p4 = [0 5  NaN]';

%// Difference matrix/vector
A = [p2-p1  p3-p1];
b = p4-p1;

%// Compute solution
p4(end) = p1(end) + A(3,:)*(A(1:2,:)\b(1:2));

Now, in C++, the mere fact of including the relevant eigen libraries blows up the executable size rather spectacularly. What eigen is capable of is complete overkill for this simple 2x2 system.

So I wouldn't go as far as resort to eigen, unless you have tons of other linear algebra things to do. It is a simple 2x2 system, which is easy enough to solve by hand.

Just KISS it; see DanielKO's answer :)

Answer 2

You can express P4 coordinates in the P1P2P3 vector basis.

x4 = x1 + A * (x2 - x1) + B * (x3 - x1)
y4 = y1 + A * (y2 - y1) + B * (y3 - y1)

This is easy-to-solve linear equation system. You have to find A and B coefficients, then use them to calculate z-coordinate

z4 = z1 + A * (z2 - z1) + B * (z3 - z1)

Answer 3

The mathematical problem here is to solve the following system of equations

p1 + a * (p2 - p1) + b * (p3 - p1) = (x4, y4, z4)

or equivalently

 (x1,y1,z1) + a * (x2-x1, y2-y1, z2-z1) + b * (x3-x1, y3-y1, z3-z1) = (x4, y4, z4)

for a, b, and z4.

To solve it in C/C++, you could either implement the Gauss algo (see also the Numerical Recipes book, it is available online), or use Linear Algebra libraries, such as Eigen, or others.

Remark: the approach is the same regardless if the point (x4, y4) lies within the triangle (x1, y1), (x2, y2), (x3, y3), or not.

Answer 4

p1, p2, p3 define a plane. You can represent it by a point and a normal. For instance, P=p1, N=(p2-P) x (p3-P) (that is, N = cross product of p1p2 and p1p3).

Now for p4 to be in the same plane, it satisfies the plane equation:

  (p4-P) · N = 0  %// dot product
⇒ (x4-x1)*N.x + (y4-y1)*N.y + (z4-z1)*N.z = 0

Re-arranging:

z4 = z1 - ((x4-x1)*N.x + (y4-y1)*N.y)/ N.z

No linear system to solve, you just need a cross product.