Let $C_d$ be a unit edge-length cube in $d$ dimensions. I would like to orient it ("tilt" it) so that the vertical (last) coordinates of its $2^d$ vertices are maximally separated, in the sense that the minimum vertical distance between any two vertices is maximized over all orientations.

For $C_2$ (in standard orientation, edges parallel to Cartesian axes), tilting $\arctan \frac{1}{2} \approx 26.6^\circ$ separates the vertices by $\delta=1/\sqrt{5} \approx 0.447$:

For $C_3$ (in standard orientation), I believe that rotating the vector $(0,0,1)$ to lie on the vector $(\frac{1}{4},\frac{1}{2},1)$ results in a vertex separation of $\delta=1/\sqrt{21} \approx 0.218$:

My question is:

Q. What is the generalization to $C_d$ for $d>3$? What is the largest vertex separation $\delta$ achievable? Can one always achieve a uniform vertex separation (the same $\delta$ between each vertically adjacent pair), as in $C_2$ and $C_3$?