When going around a "small" path on a curved surface, the angle you come up short is given by the local Guass-Bonnet theorem. Usually the G-B theorem comes up in connection with topology, involving integrals over the whole manifold. But that is just one facet of it. The local version I find useful at the moment is:
$\iint_{enclosed} K dA = 2\pi - \sum \gamma_i $
Curvature $K$ is 1/(radius of curvature) per dimension, and we multiply together two dimensions. For a sphere, it's constant so the left hand side is just $KA$, $A$ being the area enclosed by your closed-path travel. $\gamma_i$ is the angle of turning at each point $i$ where you make a turn. (For a continuously curving path, there'd be an integral.) Going straight through a point $i$ is $\gamma_i=0$.
If you go around a square of side length L, with L much smaller than R, the area is $L^2$ (pretty close to it.) You think you'd have four turns of 90 degrees, adding to 360 (which is $2\pi radians), but Gauss-Bonnet says no. You'll make three 90 degree turns, and one that's not quite, if you travel a closed path ending at your starting point. Alternatively, you can make strict 90 degree turns four times, going exactly distance L for each side, and then measure how far away you end up from the starting point, and do some extra math. G-B still applies, just so not directly.
At this point, if you have a notion of how small an angular error you can measure, you can plug in numbers. $R_{Earth}$ is about 4000 miles (6400 km). Area $A$ is $L^2$ and $K=1/R^2$. On Earth, if we do this experiment on a reasonably human scale, we hope a few hundred meters or a few kilometers, we expect close to the same value for $A$. Say you can measure, or steer, to about half a degree accuracy. That's 1/100th radian, roughly. We already intend to make four ideal 90 degree turns (the $2\pi$ in the G-B formula) but expect to measure some small angular deficit
$\delta = 2\pi-\sum \gamma$.
when we finish the trip at the start. To get an idea what size trip we must make, just set $\delta = 0.04$ for four steering measurements good to .01 radians. (A naive estimate, not a proper error analysis.) Then,
$ {{L^2}\over{R^2}} = 0.04$
I'll let you grab a calculator and punch in the numbers, and try more accurate angle.
Note this is a rough "back-of-the-envolope" calculation. Better thought-out math would be needed for a real experiment.
The multi-layered structure protects against impact fracture.
If you hit an object very hard, you can create a crack; stresses will concentrate at that crack, and make it easier for the crack to propagate (think of the little notch in the ketchup packet: that's where you can tear the plastic...)
Now if you have a solid body (of anything), then that crack can continue to grow. But if you laminate, then the crack will hit the end of one lamina, and stop. That means that a laminated object will be much more impact resistant: it's easy to initiate a crack on the outermost surface (for example with a carbide-tipped object), but it's much harder to do so on an inner surface (which your tool cannot reach).
Best Answer
A parallelogram defined by its perimeter is unstable to shear. If the top corner is pushed it will scissor shut. A triangle defined by its perimeter is rigid toward shear. Solid planes cannot scissor shut. However, eccentricity of an axial force results in a bending moment acting on beam elements - buckling. Pattern the surface to resist other failure modes.