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 ... ...

On 10/28/10, Fred lunnon <fred.lunnon@gmail.com> wrote:

... A more plausible construction has 3 spokes spiralling one way (tangentially to the hub flanges) on one side, with 3 spiralling the other way on the other side. This configuration is independent, and I simply cannot visualise how it might be deformed by compressing any combination of spokes ...

Obviously the hub can screw in both directions along its axis: in fact, with the hub in position, it is impossible actually to tighten the strings (tension the spokes) at all!

... However 7 rams in tension are (necessary and) sufficient for a robot in general: for example, the Stewart platform with all 6 legs equally extended is opposed by a single extra ram, mounted vertically upward from the platform centre. This presumably constituted the other half of Coxeter's argument, and serves to demolish Edmondson's.

For 7-ram sufficiency, the platform configuration (hub location relative to rim) must be given: if it varied generally, 12 rams (in opposing pairs say) would be required for rigidity. This evidently explains Edmondson's confusion --- if it leads only to Fuller's geodesic domes being over-engineered, maybe that's no bad thing!

... but I don't at this point actually have a copperfastened 7-spoke configuration to propose.

My earlier proposal for 4 spokes in opposed pairs spiralling on one side, 3 spokes radial on the other side, still looks convincing to me. To prove it would involve partitioning line space into 7 subsets, corresponding to the 7 bases comprising 6 from 7 spokes (appropriately signed), such that always some component with respect to the corresponding basis is positive (denoting tension). Well, that's perfectly feasible, I suppose ... Fred Lunnon

Theorem: _________________________________________________________________ 7 spokes in tension are necessary and sufficient to mount rigidly a hub in a rim. _________________________________________________________________ Necessity of 7 is essentially a consequence of the sufficiency method, there being no linear relation possible between 6 members of a basis. Sufficiency follows the same strategy as earlier for 8 spokes: the production of linear relations, involving all the spokes, whose coefficients are all non-negative. This configuration involves synchronised regular 12-gons of holes P[k], R[k], Q[k], on parallel planes for left, right flange, rim, with radii q, q, r, flanges at distance +p, -p along the common central hub (z-axis). The lacing connects on the left side (radially) L[1] from P[0] to Q[0] L[2] from P[4] to Q[4] L[3] from P[8] to Q[8] and on the right side (spirally in opposing pairs) L[4] from R[1] to Q[5] L[5] from R[5] to Q[1] L[6] from R[7] to Q[11] L[7] from R[11] to Q[7]; the squared lengths of the spokes are respectively s^2, t^2 = p^2 + (r-q)^2, p^2 + (r+q)^2. The determinant of the first 6 spokes (before normalisation by s^3 t^4) is 9 sqrt3 (r - q) p^3 q r^4, establishing spanning of line-space; and the single relation between the spokes is 4*(L[1] + L[2] + L[3]) + 3*(L[4] + L[5] + L[6] + L[7]), establishing reaction under tension. QED It would be interesting to know how much of this Coxeter actually knew (off the top of his head?), or whether he just guessed lucky (I reckon I'm scoring around 50% here --- about as bad as could be!) Since Fuller & co. were apparently unaware of this technique (er --- the screw theory, that is), it presumably is not common knowledge amongst structural engineers --- which I find surprising. There's obviously more questions that could be asked at this juncture --- such as, what hole layout and spoke lacing minimises the maximum spoke tension in response to a unit applied screw? [Note that the relevant metric appears to be the purely directional sqrt( (L^1)^2 + (L^2)^2 + (L^1)^2 + (L^1)^3 ), being also the length of a line segment between normalised points.] But perhaps I'm going to be able to put this beauty and myself to bed now (doubtless I'm not alone). Nice problem, Dan! Dan? Are you still there? Fred Lunnon [30/10/10] On 10/30/10, Fred lunnon <fred.lunnon@gmail.com> wrote:

... Analysis of a symmetrical 8-spoke wheel lacing:

The rim comprises n = 8 (nipple) points at vertices Q[k] of a regular octagon of circumradius r; the hub flanges comprise 8 (elbow) points at vertices P[k] of pair of squares of circumradius q, lying in parallel planes at distance p on either side of the rim, and rotated by pi/4 with respect to one another.

With p = hub semi-axis, q = hub radius, r = rim radius, the flange and rim points have projective coordinates [P^0, P^1, P^2, P^3] etc as follows: P[k] = [1, q sin(k pi/4), q cos(k pi/4), p (-1)^k]; Q[k] = [1, r sin(k pi/4), r cos(k pi/4), 0]. for k = 0,...,7.

The spokes L[k] connect points as follows L[1] from P[0] to Q[2] L[2] from P[1] to Q[7] L[3] from P[2] to Q[0] L[4] from P[3] to Q[5] L[5] from P[4] to Q[6] L[6] from P[5] to Q[3] L[7] from P[6] to Q[4] L[8] from P[7] to Q[1]

with Pluecker coordinates [L^1, L^2, L^3, L_3, L_2, L_1] L[1] = [r, -q, -p, -q r, +p r, 0] L[2] = [(-r-q)c, (+r-q)c, p, +q r, +p r c, +p r c] L[3] = [-q, r, -p, +q r, 0, -p r] L[4] = [(-r-q)c, (-r+q)c, p, -q r, +p r c, -p r c] L[5] = [-r, q, -p, -q r, -p r, 0] L[6] = [(+r+q)c, (-r+q)c, p, +q r, -p r c, -p r c] L[7] = [q, -r, -p, +q r, 0, +p r] L[8] = [(+r+q)c, (+r-q)c, p, -q r, -p r c, +p r c] where c = 1/sqrt(2). It may be verified that spokes have length t = sqrt(p^2 + q^2 + r^2); also that their coordinates satisfy the Grassmann relation.

Before normalisation by 1/t^8, a sample 6x6 determinant is | L[1] L[2] L[3] L[4] L[6] L[7] | = 16*(2*r^2/q^2 - 1)*(p*q*r)^3 which (except in the unlikely event that r/q = c) is nonzero; hence they together span line space, and any screw applied to the configuration would be resisted if the spokes were replaced by rams.

The spokes satisfy the additional subspace relations L[3] + L[4] + L[7] + L[8] = 0 L[1] + L[2] + L[5] + L[6] = 0 Immediately from these it follows that a screw which compresses any combination of spokes is equivalent to one which tenses some other combination, and so can be resisted by tension alone. [In computing line vectors, it is important that the points are taken consistently in order (say) hub followed by rim, so that the signs are meaningful!]

Fred Lunnon [30/10/10]