Width of a tilted cube I'm projecting a cube onto a hexagon (for RGB to HSL conversion) and I want to calculate the width of the hexagon. Cube is, first, tilted by 45° ccw on the x axis, and then, tilted by 35.26° ccw on the y axis. Black corner is at the bottom and the white corner is at the top. Both (black and white) corners pass through the z axis. ![image]( I have four values at hand, but don't know what to do with them, atm. I can get the width of the hexagon with this 3D software, but I need to be able to calculate it mathematically.

Based on your technical drawing with measurements posted in it, the width $H$ of the hexagon can be deduced with simple trigonometry: $$ H = 360.62\cos\,(54.74^\circ) + 255\cos\,(35.26^\circ). $$

And in general, for any angle $\theta$ that you tilt the cube of side length $w$ and face-diagonal length $\sqrt 2 \cdot w$ over, the width $H$ of the hexagon will be $$ H = \sqrt 2 \cdot w\cos\,(\pi/2 - \theta) + w\cos\,\theta. $$