[Math] Determining which side of a 3D cube is facing the viewer

3dgeometrylinear algebra

Me and a friend are trying to render a rotating cube on a 2D plane(the screen) using java.
Here's the problem

The cube has 6 sides, each with a specific normal vector of the form $(0, 0, 1)$, $(0, -1, 0)$ and so on
The sides(and their normals) are rotated using a matrix as described on en.wikipedia.org/wiki/Perspective_projection#Perspective_projection
But unlike Wikipedia we project our rotated 3D-vectors on a plane situated on the positive X-axis using:

v = The 3D vector which has been rotated by the matrix = $(a, b, c)$

v2D = The 2D vector projected on the plane/screen = $(b, c)$

These new 2D vectors are then used to draw an image
To determine which of the six sides, that should be rendered we use the fact that the plane is located directly on the positive X-axis.

n = 3Dvector pointing directly outwards from the side

n2 = [rotation matrix]*n = The rotated normal pointing directly outwards from the rotated side = $(x, y, z)$

now if $x>0$ then the side is facing the same direction as the positive X-axis. And since the plane/screen to project on is located on the positive X-axis this means that the side is facing the screen and should be drawn on it.

So far, this is working perfectly and it looks like this:
Image1
But this cube could need some perspective and to get this we change the projection slightly:

v = The 3D vector which has been rotated by the matrix = $(a, b, c)$

$w=-a/700+1$;

v2D = The 2D vector projected on the plane/screen = $(b/w, c/w)$

This gives a cube where more distant objects are smaller
The only problem is that sometimes it shows sides that shouldn't be shown like:
Image3

So how should I compensate for the new perspective? It's clear that n2 is no longer pointing orthogonally from the surface of the sides. So that's what I need. A vector pointing orthogonally from the surface of the side, which has been distorted by w while n2 haven't

(edit) Here are some numbers.

The Cube has a width of $300$ pixels i e 150 pixels from origo to a side following it's normal
The plane/screen to project on is located on the positive X-axis at $(300, 0, 0)$
The viewer is located at $(1000, 0, 0)$ hence $700$ in the calculations above

Best Answer

When you weren't taking perspective into account (image 1), you were effectively viewing the scene from infinity, i.e. you replaced the parameter $700$ by infinity, resulting in a scaling factor $w=1$. When viewing from infinity, there is no difference between checking whether the normals point in a certain direction or checking on which side of the surface the viewer is, since the viewer is effectively infinitely far away and so which side she's on is determined exclusively by the direction of the normal.

But when you bring perspective into it, these two are no longer the same, and what you're actually interested in is not which direction the normal points, but which side of the surface the viewer is on. So you should figure out what viewer position your perspective projection corresponds to, and then instead of just testing the direction of the normals, form the scalar products of the normals with the viewing position, and check whether the result is greater or smaller than the corresponding product for points on the plane. I guess this can probably be rewritten as a transformation on the normals instead, but this to me seems a more natural way to think about it.

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[Math] A controlled trapezoid transformation with perspective projecton

Grabbed a pencil, piece of paper, and Maple and solved it by myself.

var eqRoot:Number = -Math.sqrt(newBottomWidth*newBottomWidth*oldHeight*oldHeight - newHeight*newHeight*oldWidth*oldWidth);
var focalLength:Number = Math.abs(eqRoot/(newBottomWidth - oldWidth));
var angle:Number = Math.atan2(eqRoot/(newBottomWidth*oldHeight), newHeight*oldWidth/(newBottomWidth*oldHeight));
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[Math] Finding the rotation angles of an plane

First off, there is no simple rule (at least as far as I know) to represent a rotation as a cascade of single axis rotations. One mechanism is to compute a rotation matrix and then compute the single axis rotation angles. I should also mention that there is not a unique solution to the problem you are posing, since there are different cascades of rotations that will yield the same final position.

Let's first build two references frames, $A$ and $B$.

$A$ will denote the standard Euclidian basis. $B$ will denote the reference frame which results from applying the desired rotation. I will use subscript notation when referring to a vector in a specific reference frame, so $v_A$ is a vector whose coordinates are defined with respect to the standard Euclidian basis, and $v_B$ is defined w.r.t. $B$.

All we know so far about $B$ is that the vector $(0,0,1)_B = \frac{1}{\sqrt{77}}(8, -3, 2)_A$. We have one degree of freedom to select how $(0,1,0)_B$ is defined in reference frame $A$. We only require that it be orthogonal to $\frac{1}{\sqrt{77}}(8, -3, 2)_A$. So we would like to find a unit vector such that

$$8x-3y+2z=0$$

One solution is $\frac{1}{\sqrt{6}}(1,2,-1)_A$, which you can verify. This forces our hand for how the vector $(1,0,0)_B$ is represented w.r.t. $A$, hence

$$(1,0,0)_B=\frac{1}{\sqrt{462}}(-1,10,19)_A$$

So the rotation matrix from $B$ to $A$ is

$$R_B^A=\left( \begin{array}{ccc} \frac{-1}{\sqrt{462}} & \frac{1}{\sqrt{6}} & \frac{8}{\sqrt{77}} \\ \frac{10}{\sqrt{462}} & \frac{2}{\sqrt{6}} & \frac{-3}{\sqrt{77}} \\ \frac{19}{\sqrt{462}} & \frac{-1}{\sqrt{6}} & \frac{2}{\sqrt{77}} \end{array} \right)$$

Given any vector with coordinates defined in the Euclidian basis (such as a point in the $xy$-plane), multiplying by the inverse this matrix will yield the representation in our new reference frame. If you want to search for the three axis rotations, I refer you to Wolfram MathWorld's page on Euler Angles.

Hope this helps! Let me know if anything is unclear or if I you think you see any mistakes.

Best Answer

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