Center of a irregular hexahedra given the vertices

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In summary: This means that the center must be one of the four new vertex coordinates.In summary, the center of an irregular convex hexahedron is found by averaging the coordinates of the vertices.
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pyroknife
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I am trying to compute the center of irregular convex hexahedras (6 faced volume).

I have the 8 indices that define the hexahedra. I was reading that the center is simply the sum of the indices divided by the number of indices and this is true for all polygons and polyhedras. Is this true?
 
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How do you define "center" in an irregular shape?
 
  • #3
I am asking myself that same question. I believe I am referring to the geometrical/spatial center, but I am not positive.

Basically I have a 3-D structured mesh grid, with cells. And I am trying to find the "center" of that cell.
I believe it is a volume-based center?

Hmmmm, if it is a volume based center, then an averaging of the vertices would not yield the center of volume.
 
  • #4
What do you mean by "geometrical/spatial center"?

There is a center of mass with a clear definition, for example. There might be other well-defined points, but you have to decide what you are actually looking for. "The center" is too unclear.
 
  • #5
I will look into this and figure out what it is that I actually want.What "type of center" does an averaging of all the nodal points provide?
 
  • #6
I'm not sure if that point has any special meaning.
 
  • #7
The "centroid" of a triangle has coordinates equal to the average of the coordinates of the vertices. Other figures, such as a hexagon, can be divided into triangles. The coordinates of the centroid of such a figure is the weighted average of the coordinates of the centroids of the triangles, weighted by area.
 
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pyroknife said:
I am trying to compute the center of irregular convex hexahedras (6 faced volume).

I have the 8 indices that define the hexahedra. I was reading that the center is simply the sum of the indices divided by the number of indices and this is true for all polygons and polyhedras. Is this true?

Start with a cube. Take four of the corners and bring them close together so that the cube becomes something close to a four sided pyramid in shape. The average of the eight corner z coordinates is one 1/2 half of the height, but the center of mass is only 1/4 the height.
 
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What is the center of an irregular hexahedra?

The center of an irregular hexahedra is the point that is equidistant from all of its vertices. It is often referred to as the centroid or geometric center.

How do you find the center of an irregular hexahedra?

To find the center of an irregular hexahedra, you can use the midpoint formula to calculate the average of the coordinates of its vertices in each dimension. The resulting point will be the center.

Is the center of an irregular hexahedra always located within the shape?

Yes, the center of an irregular hexahedra is always located within the shape. This is because it is the average of all the vertices, which are all located within the shape.

What is the significance of the center of an irregular hexahedra?

The center of an irregular hexahedra is a key point for understanding the shape's symmetry and balance. It can also be used in calculations for determining the volume and surface area of the shape.

Can the center of an irregular hexahedra be located outside the shape?

No, the center of an irregular hexahedra must always be located within the shape. If it appears to be outside the shape, it is likely due to a miscalculation or mistake in determining the coordinates of the vertices.

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