We have h=Rθ(1−cosα), where Rθ is the radius of curvature orthogonal to the meridian. Approximating cosα ∼1−α2/2 and rearranging, we find α=(2h/Rθ)1/2. The length element is thus dρ=Rθα=(2hRθ)1/2 and the area is: A=dsdρ=ds(2hRθ)1/2. As Rθ ...
The convex part is bulged in the direction of the concave part, and has a radius of curvature lower than that of the arch so as to define a space between the convex part and a metatarsus of the foot at or near a front end (31) of the convex part....