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Point height on triangle in 3D space.


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#1 jajcek   Members   -  Reputation: 274

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Posted 02 January 2013 - 03:54 AM

Hello,

 

I believe there is a lot of topics about it, but I just wrote my function that returns the height. However I have strage differences in height values when moving between cells (even triangles) and I just can't realise after two days what is wrong. I am writing the code on the grounds of http://www.toymaker.info/Games/html/terrain_follow.html

 

Here is my structure which contains the terrain (the size of terrain is 256x256; the cell is a square with size of 1.0):

 

terrainstruct.png

 

here is the function:

 

 

 

float Terrain::getHeight( float x, float z ) {
    if( !m_heightMap ) return -1;


    // which cell are we in
    int _cellX = ( int )( x - m_heightMap[0].x );
    int _cellZ = ( int )( z - m_heightMap[0].z );


    // to determine which triangle we are in
    //  _____
    // |\    |
    // |  \  |
    // |____\|
    float _relativeXPos = x - _cellX;
    float _relativeZPos = z - _cellZ;


    float _height = 0.0f;
    float _vertex1[3], _vertex2[3], _vertex3[3], _vector1[3], _vector2[3];
    VectorType _normals;

    // if we are in the left triangle
    if( _relativeXPos <= _relativeZPos ) {
        // get three vertices from the triangle
        _vertex1[0] = m_heightMap[_cellX+256].x;
        _vertex1[1] = m_heightMap[_cellX+256].y;
        _vertex1[2] = m_heightMap[_cellX+256].z;


        _vertex2[0] = m_heightMap[_cellX].x;
        _vertex2[1] = m_heightMap[_cellX].y;
        _vertex2[2] = m_heightMap[_cellX].z;


        _vertex3[0] = m_heightMap[_cellX+1].x;
        _vertex3[1] = m_heightMap[_cellX+1].y;
        _vertex3[2] = m_heightMap[_cellX+1].z;


        // calculate two vectors for this triangle
        _vector1[0] = _vertex2[0] - _vertex1[0];
        _vector1[1] = _vertex2[1] - _vertex1[1];
        _vector1[2] = _vertex2[2] - _vertex1[2];
        _vector2[0] = _vertex3[0] - _vertex1[0];
        _vector2[1] = _vertex3[1] - _vertex1[1];
        _vector2[2] = _vertex3[2] - _vertex1[2];


        // calculate the normal vector of the triangle
        _normals.x = ( _vector1[1] * _vector2[2] ) - ( _vector1[2] * _vector2[1] );
        _normals.y = ( _vector1[2] * _vector2[0] ) - ( _vector1[0] * _vector2[2] );
        _normals.z = ( _vector1[0] * _vector2[1] ) - ( _vector1[1] * _vector2[0] );


        // normalize
        float _length = sqrt( ( _normals.x * _normals.x ) + ( _normals.y * _normals.y ) + ( _normals.z *  _normals.z ) );
        _normals.x /= _length;
        _normals.y /= _length;
        _normals.z /= _length;


        //P.y = V0.y+ (N.x * dx + N.z * dz ) / -N.y


        _height = _vertex1[1] + ( _normals.x * _relativeXPos + _normals.z * _relativeZPos ) / ( -_normals.y );


    } else {
        // get three vertices from the triangle
        _vertex1[0] = m_heightMap[_cellX+1].x;
        _vertex1[1] = m_heightMap[_cellX+1].y;
        _vertex1[2] = m_heightMap[_cellX+1].z;


       _vertex2[0] = m_heightMap[_cellX+257].x;
       _vertex2[1] = m_heightMap[_cellX+257].y;
       _vertex2[2] = m_heightMap[_cellX+257].z;


       _vertex3[0] = m_heightMap[_cellX+256].x;
       _vertex3[1] = m_heightMap[_cellX+256].y;
       _vertex3[2] = m_heightMap[_cellX+256].z;


       // calculate the two vectors for this triangle
       _vector1[0] = _vertex2[0] - _vertex1[0];
       _vector1[1] = _vertex2[1] - _vertex1[1];
       _vector1[2] = _vertex2[2] - _vertex1[2];
       _vector2[0] = _vertex3[0] - _vertex1[0];
       _vector2[1] = _vertex3[1] - _vertex1[1];
       _vector2[2] = _vertex3[2] - _vertex1[2];


       // calculate the normal vector of the triangle
       _normals.x = ( _vector1[1] * _vector2[2] ) - ( _vector1[2] * _vector2[1] );
       _normals.y = ( _vector1[2] * _vector2[0] ) - ( _vector1[0] * _vector2[2] );
       _normals.z = ( _vector1[0] * _vector2[1] ) - ( _vector1[1] * _vector2[0] );


      // normalize
       float _length = sqrt( ( _normals.x * _normals.x ) + ( _normals.y * _normals.y ) + ( _normals.z *     _normals.z ) );
       _normals.x /= _length;
       _normals.y /= _length;
       _normals.z /= _length;


       //P.y = V0.y+ (N.x * dx + N.z * dz ) / -N.y
       _height = _vertex1[1] + ( _normals.x * _relativeXPos + _normals.z * _relativeZPos ) / ( -_normals.y );
}

    return _height;
}

 

for example I have bigger height difference between position (2.99, 0.5) and (3.01, 0.5), but the vertices of those triangles have almost the same height and the camera jumps strangely.

 

I'd really appreciate some help, please.

 

Thank you.



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#2 Bacterius   Crossbones+   -  Reputation: 9304

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Posted 02 January 2013 - 06:15 AM

Did you check your normal is pointing in the right direction? The cross product may return two different normals depending on winding, if your winding isn't consistent your normals will not be either.

 

That said I'm not sure I understand what you want to do. What do you mean by height, do you mean the height of the mesh represented by your triangulated heightmap at any (x, z) position even between cells? If so, isn't it just a matter of finding the triangle you're in (which you are doing) and interpolating height from the triangle's vertices?


The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.

 

- Pessimal Algorithms and Simplexity Analysis


#3 jajcek   Members   -  Reputation: 274

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Posted 02 January 2013 - 06:44 AM

The cells are connected, you can't be "between cells". I'd like to achieve simple effect of moving my camera on the terrain as it was moving a character.

PS How to interpolate the height in 3D triangle? Maybe it is what I am looking for.

// EDIT

okay the interpolation was what I was looking for, thank you, but still I don't understand why the code above doesn't work...

Edited by jajcek, 02 January 2013 - 07:42 AM.


#4 Bacterius   Crossbones+   -  Reputation: 9304

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Posted 02 January 2013 - 08:09 AM

The cells are connected, you can't be "between cells". I'd like to achieve simple effect of moving my camera on the terrain as it was moving a character.

PS How to interpolate the height in 3D triangle? Maybe it is what I am looking for.

// EDIT

okay the interpolation was what I was looking for, thank you, but still I don't understand why the code above doesn't work...

I meant inside a cell (not just the corners of each cell, so being able to "sample" the heightmap at any point, not just the vertices).

 

 

But what I don't understand is how your code gets the vertices from the heightmap - it can't reach all the cells with the current code. In a 2D array represented as a 1-dimensional array, the (x, z) element is:

 

m_heightMap[z * Xwidth + x]

 

In your case, Xwidth = 256. I don't see this happening anywhere in the code, in fact you're not even using _cellZ. I think this is what's wrong with the code.

 

----------

 

If this wasn't the problem, then maybe the algorithm is wrong, but it seems way too complicated for something this simple. This is what I would do:

- find the triangle in which your (x, z) coordinate is in

- you will notice the triangle surface is planar so the dh/dx and dh/dz gradient (change in height with respect to change in x and z coordinates ) is constant, calculate those quantities by taking two vertices v1 and v2 of the triangle (this can be precomputed):

 

dh/dx = (v2.y - v1.y) / (v2.x - v1.x)

dh/dz = (v2.y - v1.y) / (v2.z - v1.z)

 

- now, find the x and z distance from your point P to any vertex V (or any point on the triangle, for that matter) of the triangle, and use the gradient to find its height:

 

h = V.Y + (dh/dx) * (P.X - V.X) + (dh/dz) * (P.Z - V.Z)

 

Think of it as starting at point V, moving up or down the triangle's surface in a straight line to align yourself with point P on the x-axis by moving on the z-axis, and then going to point P again in a straight line, still remaining on the triangle's surface, travelling along the x-axis.


The slowsort algorithm is a perfect illustration of the multiply and surrender paradigm, which is perhaps the single most important paradigm in the development of reluctant algorithms. The basic multiply and surrender strategy consists in replacing the problem at hand by two or more subproblems, each slightly simpler than the original, and continue multiplying subproblems and subsubproblems recursively in this fashion as long as possible. At some point the subproblems will all become so simple that their solution can no longer be postponed, and we will have to surrender. Experience shows that, in most cases, by the time this point is reached the total work will be substantially higher than what could have been wasted by a more direct approach.

 

- Pessimal Algorithms and Simplexity Analysis





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