Solving the apparently slope/angle of a object, given camera rotations I'll preface my question with an apology - I'm not a regular speaker of math-lingo, so my question may come off as clumsy. I'm not particularly good or knowledgeable about math and don't often need to use or speak about it in depth. I use it in programming, but only at a elementary level for the most part.
I'm programming a plugin for the Unit engine that renders 3d objects into a series of images (sprites) and I need to be able to figure out the 'visible' slope and/or angle of an object's lines in order to figure out what angles to set my camera and the object.
The reason for this is that pixels rendered at clean slopes (2:1,3:1,1:1, etc) appear smoother than those rendered at fractional slopes (2.5:1,etc)
for example: the image in the picture below is just a cube. Its euler angles are (0,0,0) - no rotation. The camera is rotated at (45,35.8,0). In the resulting image, the lines appear to be at a 1:1 slope on one face and a 2:1 on the other (or 1:2?)

When the camera angle is changed to (45,-28.13,0) the resulting lines are at a non-smooth-appearing slope - as shown circled in red. The pixels move 1:1 for 3 pixels then 1:2 for 1, and 1:1 on another.

So, what im trying to figure out is: given the x and y angle of the camera (z always = 0) and the object's rotation, how can i find the 'apparent' slope/angle of cubes lines?
I need to use this information to then jump to the next "smooth" slope and adjust my object's rotation acoordingly.
If this is an obscenely complex question, any hints would be more than welcome - or at least telling me so, because I really have no idea on how to begin to apporach a solution or if one is actually possible.
note: the camera is orthographic - so no perspective distortion occurs.
 A: You should probably read a good book on computer graphics. I'd recommend my own (Computer Graphics: Principles and Practice, 3rd edition), but just about any good graphics book would let you get to the point of solving this. Roughly: you need to take the "view matrix" and apply it to ray-directions (i.e., vectors) in 3-space and see what the resulting slope in image space turns out to be. That's not hard if you can find the view matrix, but ...
A: You seem to have a line $l$ in 3D space which seems to get transformed to a line $T(l)$ on the 2D viewport plane, having some slope $m$ there. 
If the transformation is depending on some parameter vector $p$, you would like to calculate
$$
m = f(T_p(l)) \quad (*)
$$
to determine a close but better suited value $m'$ and then a corresponding $p'$ such that $m' = f(T_{p'}(l))$.
So we take two different points $u_1, u_2$ from $l$ and hope $T(u_1) \ne T(u_2)$ so we can calculate the slope
$$
m = \frac{T(u_2)_y - T(u_1)_y}{T(u_2)_x - T(u_1)_x}
$$
The transformation usually is affine and conveniently modeled by a $4\times 4$ matrix acting on homogeneous coordinate vectors $u = (u_x, u_y, u_z, u_w)^\top$ normalized to $u_w = 1$.
It will be a composition of rotations, translations and scalings, which can be performed as matrix product
$$
T = T_n T_{n-1} \dotsb T_1
$$
of the corresponding transformation matrices $T_i$. Have a look here to get an idea how these operations are modeled as matrices.
What these transformations are in your case depends on your 3D engine, you have to look at its documentation or experiment with it to find out.
In principle it should be possible to come up with an expression for equation $(*)$. If you can invert $m(p)$ into $p(m)$ for $m=m'$ outside a small neighbourhood of $m$ I can not tell, probably not.
Which means you will have to resort to trial and error, heuristics or machine learning. I expect this to be the tricky part.
For your cube your images seem to show that a decent transformation for one line of the object will result in good results for all lines of the object. I am not sure if that will be the case for different objects, e.g. a tetraeder.
