Get highest (Y-axis) vertices from primitive d20 dice in xna

Question

I have created a d20 dice (icosahedron) in xna and used BEPU physics to roll it. But now I'm trying to get the value from the dice after its rolled. I was thinking about looking for the 3 vertices that have the highest Y-axis coords after the dice rolled. so I could set something like this:

if (vertices[0].currentPosition.Y >= 2 && vertices[1].currentPosition.Y >= 2 && vertices[2].currentPosition.Y >= 2)
{
    diceValue = 5;
}

But how do i get the current position from the vertices? Or is there a better way to check what value is on top of the dice?

Answer 1

If you have the rotation matrix, I think the the easier is to get the cos between the normals and forget the vertex position.

The cos will be max and near 1 when the angle among the two vectors is near 0.

Each normal will have a face value so get the face value of the normal nearest to Vector3.UnitY is easy.

 Vector3 FaceNormals = new Vector3[] { Vector3.UnitX, -Vector3.UnitX, Vector3.UnitY, -Vector3.UnitY}
 int FaceValue = new int[] { 1,2,3,4 }

float cos = -1;
int faceValue = 0;
for (int n=0; n<FaceNormals.Length; n++) {
  var RotatedNormal = Vectro3.TransformNormal(FaceNormal[n], RotationMatrix);
  var newCos = Vector3.Dot(Vector3.UnitY , RotatedNormal );
  if (cos<newCos) {
     FaceValue = FaceValues[n];
     vod = newCos;
  }
}

where the Vector3.UnitY is fixed, and the face normal are rotated, so calculation is now in world space.

EDIT:

If you want to use vertices... calculate the center of each face, if the face has two triangles correlated:

var j =0;
Vector3[] Centers = new Vector3[indices.Lengt/6];
Vector3[] Normals= new Vector3[indices.Lengt/6];

for (int i =0; i<indices.lenght; i+=6)
{
      Vector3 center;
      for (int n=0; n<6;n++) {
          center+= vertices[indices[i+n]].Position;
      }
      center/=6;

      var A = vertices[indices[i+1]].Position - vertices[indices[i]].Position
      var B = vertices[indices[i+2]].Position - vertices[indices[i]].Position 
      normal = Vector3.Cross(A, B);

      Centers[j] = center;
      Normals[j] = normal;
      j++;
}

int[] FaceValues = { 1,2,3,4,5,6...}
float y = float.MinValue;
for (int n=0; n<Centers.Length; n++) {

    var rotated = Vector3.Transform(Centers[n], TransformMatrix);

    if (y < rotated.Y) {
        y = rotated.Y;
        FaceValue = FaceValues[n];
    }
}

You'll have the dice face value in "FaceValue"

Answer 2

Vertices are stored in memory using local coordinates. So no vertex has a "currentPosition," they're always positioned according to their offset from their original origin. In the "world," they are displayed according to some transformation: scale and rotation about their local origin, then translation with respect to the world origin. With the exception of static level geometry, vertices are almost never stored in a current position.

So, to get the "current" position of any vertex, you must apply a transformation to it that takes it from local to world space. Applied to all vertices, you need only find the vertices with the greatest Y coordinate.

It's been a while since I used XNA, but going off of the MSDN docs it should look something like this:

transform = Matrix.CreateFromQuaternion(dieRotation)
            * Matrix.CreateTranslation(diePosition);
var currentPositions = vertices
                       .Select(v => Vector3.Transform(v.Position, transform))
                       .OrderByDescending(p => p.Y)
                       .Take(numberOfVerticesPerFace)
                       .ToList();

Answer 3

You can calculate the orientations that the die will fall on, then test the current orientation against a pre-computed list of orientations to get the number.