A Treatise on the Theory of Screws

311] THE GEOMETRICAL THEORY. 337 Let us now think of 0, as an instantaneous screw; it lies on the cylindroid and this cylindroid contains two principal screws of inertia. It follows from § 309 that the corresponding impulsive screw </>, lies on the same cylin- droid. That screw </>, can be determined by the construction there given. In like manner we construct on the other four cylindroids the screws </>2, </>3, (f>4, which are the impulsive screws coiresponding respectively to 02, 03, 04, 03, as instantaneous screws. Consider then the two pairs of corresponding impulsive screws and in- stantaneous (screws (r/, «) and (</>,, 02). We have arranged the construction so that 01 is reciprocal to r/. Hence, by the fundamental principle so often employed, a and 04 are conjugate screws of inertia, so that a must be reciprocal tO Øj. In like manner it can be proved that the instantaneous screw a for which we are in search must be reciprocal to </>3, </>4) </>5. We have thus dis- covered five screws, fy4, </>3, </>5, to each of which the required screw a must be reciprocal. But it is a fundamental point in the theory that the single screw reciprocal to five screws can be constructed geometrically (§ 25). Hence a is determined, and the geometrical solution of the problem is complete. It remains to examine the failure in this construction which arises when any one or more of the five screws </>, ... </>6 becomes indeterminate. This happens when y is reciprocal to two screws on the cylindroid in question. In this case t) is reciprocal to every screw on the cylindroid. Any one of such screws might be taken as the corresponding </>, and, of course, 0 would have been also indefinite, and a could not have been found. In this case 17 would have been reciprocal to the two principal screws of inertia, suppose j40, A4 which the cylindroid contained. Of course still more indeterminateness would arise if r/ had been also reciprocal to other screws of the series j40, J.,, _A2> As, A4, As. No screw could, however, be reciprocal to all of them. If 7/ had been reciprocal to five, namely, u41; A2, As, A4, As, then 77 could be no screw other than A, because the six principal screws of inertia are co- reciprocal ; would then be its own instantaneous screw, and the problem would be solved. We may therefore, under the most unfavourable conditions, take y to be reciprocal to four of the principal screws of inertia Ao, A1: A2, As, but not to or A5. We now draw the five cylindroids, 710214, A4A4, A2A4, A3A4, -40j45. We know that is reciprocal to no more than a single screw on each cylindroid. We therefore proceed to the construction as before, first finding 0, ... 05l one on each cylindroid ; then deducing <f>4 ... cf>5, and thus ultimately obtaining a. Thus the general problem has been solved. B. 22