地形一般是用网格再从高度图里读取每个顶点的高度来生成。若进一步，想实现摄像机在地形上行走的效果，就需要算出地形上任意一点的高度。总结了3个方法：
一：
《Introduction to 3D Game Programming With Directx 9.0》这本书里介绍的，利用向量来算。
float Terrain::GetHeight(float x, float z)
{
// Translate on xz-plane by the transformation that takes
// the terrain START point to the origin.
x = ((float)_width / 2.0f) + x;
z = ((float)_depth / 2.0f) - z;
// Scale down by the transformation that makes the
// cellspacing equal to one. This is given by
// 1 / cellspacing since; cellspacing * 1 / cellspacing = 1.
x /= (float)_CellSpacing;
z /= (float)_CellSpacing;
// From now on, we will interpret our positive z-axis as
// going in the 'down' direction, rather than the 'up' direction.
// This allows to extract the row and column simply by 'flooring'
// x and z:
float col = ::floorf(x);
float row = ::floorf(z);
// get the heights of the quad we're in:
//
// A B
// *---*
// | / |
// *---*
// C D
float A = GetHeightMapEntry(row, col);
float B = GetHeightMapEntry(row, col+1);
float C = GetHeightMapEntry(row+1, col);
float D = GetHeightMapEntry(row+1, col+1);
//
// Find the triangle we are in:
//
// Translate by the transformation that takes the upper-left
// corner of the cell we are in to the origin. Recall that our
// cellspacing was nomalized to 1. Thus we have a unit square
// at the origin of our +x -> 'right' and +z -> 'down' system.
float dx = x - col;
float dz = z - row;
// Note the below compuations of u and v are unneccessary, we really
// only need the height, but we compute the entire vector to emphasis
// the books discussion.
float height = 0.0f;
if(dz < 1.0f - dx) // upper triangle ABC
{
float uy = B - A; // A->B
float vy = C - A; // A->C
// Linearly interpolate on each vector. The height is the vertex
// height the vectors u and v originate from {A}, plus the heights
// found by interpolating on each vector u and v.
height = A + Lerp(0.0f, uy, dx) + Lerp(0.0f, vy, dz);
}
else // lower triangle DCB
{
float uy = C - D; // D->C
float vy = B - D; // D->B
// Linearly interpolate on each vector. The height is the vertex
// height the vectors u and v originate from {D}, plus the heights
// found by interpolating on each vector u and v.
height = D + Lerp(0.0f, uy, 1.0f - dx) + Lerp(0.0f, vy, 1.0f - dz);
}
return height;
}
用到的2个函数
float Terrain::Lerp(float a, float b, float t) //一个插值函数
{
return (a - (a*t) + (b*t));
}
int Terrain::GetHeightMapEntry(int row, int col) //读取高度函数
{
return _heightmap[row * _numVertsPerRow + col]; // 高度图数据存在_heightmap里
}
二：
先计算出摄像机所在三角形的平面方程，然后带入摄像机的X，Z坐标，即可得高度Y
具体实现过程见[url]http://creatorchen1984.spaces.live.com/[/url] 《获取地形上某一点高度》，写的十分详细
三：
网上找的一个方法
假设你的地形为terrain[][];用下面的函数求出地形上点(x,z);的y值,将人物的高度加上这个y值即可.
float GetHeight(GLfloat x, GLfloat z)
{
float h=0;
float Xb,Yb;
int Xa,Ya;
Xa=(int)x;
Ya=(int)z;
Xb=x-Xa;
Yb=z-Ya;
float a=terrain[Xa][Ya].y;
float b=terrain[Xa+1][Ya].y;
float c=terrain[Xa][Ya+1].y;
float d=terrain[Xa+1][Ya+1].y;
h=(a*(1-Xb)+b*Xb)*(1-Yb)+(c*(1-Xb)+d*Xb)*Yb;
return h;
} [/img]..
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