Computing the normal from vertex positions is quite simple using the vector cross product.
The cross product of two vectors $u$ and $v$ (noted $u \times v$, or sometimes $u \wedge v$) is a vector perpendicular to $u$ and $v$, of length $||u \times v|| = ||u|| \cdot ||v|| sin(\theta)$, with $\theta$ the angle between $u$ and $v$. The direction of the vector will depend on the order of the multiplication: $u \times v$ is the opposite of $v \times u$ (the two directions perpendicular to the plane).
If you are not familiar with the cross product, I invite you to read about it and get comfortable with it. Normals will then seem simple.
Flat shading normals
If you have a triangle $ABC$, $AB \times AC$ is a vector perpendicular to the triangle and with a length proportional to its area. Since the normal is the unit vector perpendicular to the triangle's plane, you can get the normal with:
$N = \dfrac{AB \times AC}{||AB \times AC||}$
In code, this would look like n = normalize(cross(b-a, c-a)) for example. Just apply this over all your faces and you will have your normals per face.
For each triangle ABC
n := normalize(cross(B-A, C-A))
A.n := n
B.n := n
C.n := n
Note that this assumes vertices are not shared between triangles. I am not familiar with the Kinect API; it's quite possible that they are shared, in which case you would have to duplicate them, or move on to the next solution:
Smooth shading normals
After lighting with normals computed as above, you will notice that triangle edges are apparent. If this in not desirable, you can compute smooth normals instead, by taking into account all the faces that share a same vertex.
The idea is that if a same vertex is shared by three triangles $T1$, $T2$ and $T3$ for example, the normal $N$ will be the average of $N1$, $N2$ and $N3$. Moreover, if $T1$ is a big triangle and $T2$ is a tiny one, you probably want $N$ to be more influenced by $N1$ than by $N2$.
Remember how the cross product is proportional to the area? If you add up the cross products then normalize the sum, it will do exactly the weighted sum we want. So the algorithm becomes:
For each vertex
vertex.n := (0, 0, 0)
For each triangle ABC
// compute the cross product and add it to each vertex
p := cross(B-A, C-A)
A.n += p
B.n += p
C.n += p
For each vertex
vertex.n := normalize(vertex.n)
This technique is explained in longer detail this article by Iñigo Quilez: clever normalization of a mesh.
For more on normals, see also:
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