Geometrically construct sides on cube

Question

How many degrees would I skew a circle to display it on each side (3) of a cube (die) perfectly centered from the viewpoint of the camera?

Answer 1

The question as posed isn't completely clear to me (i.e. 'how many degrees'), but I assume you need to know how a circle drawn on a side of a cube looks, if we observe it in the direction of the main diagonal of the cube.

Let $u,v,w$ be the Cartesian axes going from a corner along the edges of our cube. Projected onto an $x,y$ plane perpendicular to the main diagonal, they will look like this:

For the three circles on the sides of the cube the equations in $u,v,w$ would look like this:

$$(u-u_0)^2+(v-v_0)^2=R^2 \\ (u-u_0)^2+(w-w_0)^2=R^2 \\ (v-v_0)^2+(w-w_0)^2=R^2$$

From elementary geometry we can find for any point in the plane with (projected) $u,v,w$ coordinates the related $x,y$ coordinates, depending on the sector (side of the cube) the point belongs to:

$$\begin{cases} u=x+\frac{1}{\sqrt{3}}y \\ v= \frac{2}{\sqrt{3}}y \end{cases} \tag{1}$$

$$\begin{cases} u=x-\frac{1}{\sqrt{3}}y \\ w= -\frac{2}{\sqrt{3}}y \end{cases} \tag{2}$$

$$\begin{cases} v=-x+\frac{1}{\sqrt{3}}y \\ w= -x-\frac{1}{\sqrt{3}}y \end{cases} \tag{3}$$

Now simply substitute the above expressions into each circle equation, and we obtain for circles centered at their respective $u_0,v_0,w_0$ and of radius $R$:

$$ \left( x+\frac{1}{\sqrt{3}}y-u_0 \right)^2+\left( \frac{2}{\sqrt{3}}y-v_0 \right)^2=R^2$$

$$ \left( x-\frac{1}{\sqrt{3}}y-u_0 \right)^2+\left( \frac{2}{\sqrt{3}}y+w_0 \right)^2=R^2$$

$$ \left( x-\frac{1}{\sqrt{3}}y+v_0 \right)^2+\left(x+ \frac{1}{\sqrt{3}}y+w_0 \right)^2=R^2$$

Here's an example for $u_0=v_0=w_0=5$ and $R=3$: