The equation of a quadric surface has 10 coefficients, so the freedom of the one-sheet hyperboloid equals 9; subtracting 6 for spatial isometries leaves freedom 3 for the shape/size of the surface. [This is easier to visualise for the 3 orthogonal axes of an ellipsoid.] Since the joints must also have freedom 3, the rods must not only (1) rotate around one another without changing their relative angles, but also (2) rotate "along" each other orthogonally to the previous; and finally (3) slide along one another. Any doubly-ruled surface is a quadric, so this model generates any and only one-sheet hyperboloids. JA's model generates a freedom-1 subset: in particular, its cross-sections are circular rather than elliptical. I think that joints with at most two of the freedoms (1)-(3) restrict the surface to that in which they were assembled; in which case the C & R construction, as reported by TP, appears to be incorrect. Fred Lunnon