A little background seems called for here, concerning Pluecker line coordinates, and why linear independence is relevant. The 3-space line L through points P, Q with homogeneous coordinates [P^0, P^1, P^2, P^3] and [Q^0, Q^1, Q^2, Q^3] has coordinate [L^1, L^2, L^3, L_3, L_2, L_1] defined by L^1 = P^0 Q^1 - P^1 Q^0, L^2 = P^0 Q^2 - P^2 Q^0, L^3 = P^0 Q^3 - P^3 Q^0, L_3 = P^1 Q^2 - P^2 Q^1, L_2 = P^3 Q^1 - P^1 Q^3, L_1 = P^2 Q^3 - P^3 Q^2. [For a finite point P, we may take P^0 = 1 and P^1, P^2, P^3 the familiar Cartesian x-, y-, z-components. The line meeting two planes has an analogous (dual) definition, with _ and ^ transposed --- the resulting coordinate is the same, up to an arbitrary nonzero scalar factor. Rescaling by a negative factor may be associated with reversal of orientation, where appropriate. Geometrically, L^i are proportional to the direction cosines, L_i components of the "moment" plane shared with the origin.] A line has freedom 4 and rescaling freedom 1, so the 6 components of a line must satisfy a single further constraint (Grassman): L_1 L^1 + L_2 L^2 + L_3 L^3 = 0. Any more general sextuplet which fails to satisfy this constraint may be interpreted as an impulsive "screw" (force along an axis combined with torque around it), or the derivative of a helical isometry, or an element of the Lie algebra of 3-space isometries. To mount the bicycle hub in a dynamically stable fashion relative to the rim requires that any applied screw should lie within the space spanned by the spokes; the rank of that space must therefore equal 6. Six spokes will in principle suffice provided that (1) the determinant of their coordinates is nonzero; and (2) some spoke is always acting in tension. [Maybe careful treatment of signs can also take care of the tension constraint (2)? In practice, rather more might be demanded: subject to the geometric constraints of the application (spokes meeting the rim, etc), the determinant might be maximised, in order to minimise the reactive tension in the spokes.] Let's apply this method to Dan's earlier suggestion of 3 radial spokes on each side [which we know already fails to oppose a torque around the hub]. It's easily shown that any 4 lines through the same point are dependent: therefore the axis of the hub lies in the subspace spanned by each set of 3. Since they have a 1-dimensional space in common, they span only 5 dimensions; therefore the 6 spokes are dependent, and rigidity fails along one (line-space) dimension. Finally, a confession: my triangle-in-square construction is less obvious than earlier claimed. With just the first four strings taut, the triangle could rotate also in its own plane; to verify the configuration would require checking that this motion also is restricted by the other two strings. Or we could simply apply the Pluecker determinant criterion ... But I'm tempted instead to investigate 3 spokes on each side of the hub, 2 spiralling one way and 1 in opposition. Fred Lunnon On 10/26/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
... I conjecture that a similar result applies to the bicycle spoke problem: 6 spokes locate the hub statically relative to the rim, provided they remain always in tension (that is, no spoke can shorten without causing some other to lengthen); and furthermore, they do so "rigidly" (transmitting all torques under tension) provided they are linearly independent. ...
On 10/24/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Suppose we idealise the wheel rim as a large square S [occasionally manifested in actual hardware!], the spokes as lengths of string [compliant in flection or compression], the hub as a small equilateral triangle T. The centres of S and T coincide, their planes are perpendicular; and one edge E of T is parallel to two opposite edges of S.
From each end of E in T, two strings run straight to nearby corners in S; from the third corner of T, two strings run straight to mid-points of edges parallel to E in S. Under tension, the motion of T wrt S is restricted by the former four strings to rotation about E; the tension is maintained, and the rotation prevented, by the latter two strings.
So pace Fuller and Coxeter, 6 spokes are both necessary and sufficient! The discrepancy between this result and Fuller's / Edmondson's 12 makes one wonder whether we're all discussing the same problem ... ...