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I am quite new to 3D graphical programming and thus far only understand that normal somehow define the direction in which a vertex faces and therefore the direction in which light is reflected. I have now idea how they are calculated though, only that they are defined by a Vector3.

For a visualizer that I am creating I am importing a bunch of coordinate which represent layer upon layer of line segments. At the moment I am only using a vertex buffer and adding the start and end point of each line and then rendering a linelist.

The thing is now that I need to calculate the normal for the vertices of these line segments so that I can get some realistic lighting. I have no idea how to calculate these normal but I know they all face sideways and not up or down. To calculate them all I have are the start and end positions of each line segment. The below image is a representation of what I think I need to do in the case of an example layer:

enter image description here

The red arrows represent the normal that should be calculates, the blue text represent the coordinates of the vertices and the green numbers represent their indices.

I would greatly appreciate it if someone could please explain to me how I should calculate these normal.

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  • \$\begingroup\$ do you want to calculate normals for faces? or for line segments ? is it in 2D or 3D ? \$\endgroup\$
    – concept3d
    Commented Nov 3, 2013 at 11:43
  • \$\begingroup\$ I am working in 3d space with a collection of layers of line segments, the line segments therefore also have z coordinates. I do not have any faces though and want to calculate normals for the line segments. \$\endgroup\$
    – Gerharddc
    Commented Nov 3, 2013 at 12:17
  • \$\begingroup\$ as I said in my answer a 3d line segment have infinitely many normals. So I don't know what you want should mean? A normal is a geometric attribute which a 3D line doesn't have. maybe you should post an image so we can understand what are you working on. \$\endgroup\$
    – concept3d
    Commented Nov 3, 2013 at 13:02
  • \$\begingroup\$ is the Z always 0 ? \$\endgroup\$
    – concept3d
    Commented Nov 3, 2013 at 13:15
  • \$\begingroup\$ No, the z increases for every layer \$\endgroup\$
    – Gerharddc
    Commented Nov 3, 2013 at 13:17

1 Answer 1

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The red arrows represent the normal that should be calculates.

2D is the only case that those red lines (or flipped) are the normals. for calculating a normal for 2D line segment (which is the title indicate), there are only two vectors perpendicular to any given vector, hence you can do the following:

 V = {x, y} 
 Vperp = {-y,x}

[Edit based on comments] Your case is simply 2D in 3D space you only need to deal with Z as constant. (Which is actually the case of 2D where is always Z=0) I am assuming the layer are in plane Z = const value. This is also equivalent to rotating the vector by 90 degrees.

 V = {x, y, Z} // Z is const regardless of the value
 Vperp = {-y,x, Z}

In case Z was not const for each layer you only need to multiply the vector by a rotation matrix, but you need to figure out the axis that you should rotate around.

Otherwise for calculating a normal for 3D line segment, Well there are infinitely many perpendicular lines on a 3D line segments.

For calculating the normal for a face (which is I think is what the question wants) you can do the following:

Given a triangle with A,B,C, you need to take the cross product for any two non-parallel edges and then normalize it. This will give you a face normal. In order to calculate the normal for each vertex you need to average normal of each face that shares the vertex.

Vec3 a,b,c;
Vec3 faceNormal = (a-b) cross (c-a);
faceNormal = normalize(faceNormal);

for_each Face that share vertex:
    vertexNormal += FaceNormal;

vertexNormal = normalize(vertexNormal);   
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  • \$\begingroup\$ I am working in 3d space with a collection of layers of line segments, the line segments therefore also have z coordinates. I do not have any faces though and want to calculate normal for the line segments, how should I do that? \$\endgroup\$
    – Gerharddc
    Commented Nov 3, 2013 at 12:19

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