Interpolating height for a point inside a grid based on a discrete height function

I have been wracking my brain to come up with a solution to this problem.

I have a lookup table that returns height values for various points (x,z) on the grid. For instance I can calculate the height at A, B, C and D in Figure 1. However, I am looking for a way to interpolate the height at P (which has a known (x,z)). The lookup table only has values at the grid intervals, and P lies between these intervals. I am trying to calculate values s and t such that:

A'(s) = A + s(C-A) B'(t) = B + t(P-B)

I would then use the these two equations to find the intersection point of B'(t) with A'(s) to find a point X on the line A-C. With this I can calculate the height at this point X and with that the height at point 开发者_如何学运维P.

My issue lies in calculating the values for s and t.

Any help would be greatly appreciated.

Try also bilinear interpolation or bicubic interpolation.

Depending on if you want to interpolate between ABC or ABCD the algorithm will change.

To interpolate between ABC (which I assume is what you want to do since you draw the diagonal) you will need to find the barycentric coordinates of P relative to ABC x and y positions then apply the barycentric coordinate to the height (z is assumed here) component of those triangles.

What about going this way: find u and v so that

  P = B + u(A-B) + v(C-B)

If you write this out, you'll see that this is a 2x2 linear system with unknowns u and v, so I guess you know how to go on from there.

Oh, and once you have u and v you use the same exact formula as above for the height, only this time A,B,C,P will be the heights at these points.

Considering points value are available at four corners of a square of unit length, interpolated value at any point(x,y) inside the square is given by:

f(x,y) = [ (1-y)f(0,0) + yf(0,1) ](1-x) + [ (1-y)f(1,0)+y(f(1,1)) ]x

If square has length other than 1,say L then f(x,y) is given by:

  f(x,y) = [ (L-y)f(0,0) + yf(0,L) ](L-x)/L^2 + [ (L-y)f(L,0)+y(f(L,L)) ]x/L^2

image

Here's an explicit example based on shape functions.

Consider the functions:

u1(x,z) = (x-x_b)/(x_c-x_b)

One has u1(x_b,z_b) = u1(x_a,z_a) = 0 (because x_a = x_b) and u1(x_c,z_c) = u1(x_d,z_d) = 1

u2(x,z) = 1 - u1(x,z)

Now we have u2(x_b,z_b) = u2(x_a,z_a) = 1 and u2(x_c,z_c) = u2(x_d,z_d) = 0

v1(x,z) = (z-z_b)/(z_a-z_b)

This function satisfies v1(x_a,z_a) = v1(x_d,z_d) = 1 and v1(x_b,z_b) = v1(x_c,z_c) = 0

v2(x,z) = 1 - v1(x,z)

We have v2(x_a,z_a) = v2(x_d,z_d) = 0 and v2(x_b,z_b) = v2(x_c,z_c) = 1

Now let's build new functions as follows:

S_D(x,z) = u1(x,z) * v1(x,z)

We get S_D(x_d, z_d) = 1 and S_D(x_a,z_a) = S_D(x_b,z_b) = S_D(x_c,z_c) = 0

S_C(x,z) = u1(x,z) * v2(x,z)

We get S_C(x_c, z_c) = 1 and S_C(x_a,z_a) = S_C(x_b,z_b) = S_C(x_d,z_d) = 0

S_A(x,z) = u2(x,z) * v1(x,z)

We get S_A(x_a, z_a) = 1 and S_A(x_b,z_b) = S_A(x_c,z_c) = S_A(x_d,z_d) = 0

S_B(x,z) = u2(x,z) * v2(x,z)

We get S_B(x_b, z_b) = 1 and S_B(x_a,z_a) = S_B(x_c,z_c) = S_B(x_d,z_d) = 0

Now define your interpolating function as

H(x,z) = h_a * S_A(x,z) + h_b * S_B(x,z) + h_c * S_C(x,z) + h_d * S_D(x,z),

where h_a is the heigh at point A, h_b is the height at point B, and so on.

You can easily verify that H is indeed an interpolating function:

H(x_a,z_a) = h_a, H(x_b,z_b) = h_b, H(x_c,z_c) = h_c and H(x_d,z_d) = h_d.

Now, in order to approximate the height at P, all you need to do is evaluate H at this point:

h_p = H(x_p, z_p)

The functions S are normally referred to as "shape functions". There's one such function for each node you want your interpolated value to depend on, and in this case they all satisfy Kronecker's delta property (they take the value one at one node and zero at all other nodes).

There are many ways to build shape functions for a given set of nodes. If I remember correctly, the construction of 2D shape functions by multiplication of 1D shape functions (as we've done in this case) is called "tensor product of functions" (easy in this case because the grid is rectangular). We have ended up with four functions (one per node), all of them linear combinations of {1, x, z, xz}.

If you want to use only three points for your interpolation, then you should be able to easily build three shape functions as linear combinations of {1, x, z} only, but you will loose a 25% of the height information provided by the grid and your interpolant will not be smooth inside the rectangle when h_b != h_d.