A Treatise on the Theory of Screws

236] FREEDOM OF THE FIFTH ORDER. 257 In like manner for the pectenoid relating to 0 and ß, we have (pe + Pß) M' -k'N’ = 0, where M' = 0, N’ = 0 also represent planes passing through 0, whence eliminating pg we have (— Pß) AIM' — kNM' + k'N'M = 0. The equation of the pectenoid can also be deduced directly as follows. Let five screws of the five-system be given. Take a point 0 on one of these screws (a) and through 0 draw four screws ß, y, 8, e which belong to the four-system defined by the remaining four screws of the original five. Let there be any three rectangular axes drawn through 0. Let alt a2, a3 be the direction cosines of a and let ßlt ß2, ß3 be the direction cosines of ß, and similarly for y, 8, e. Let 6 be some other screw of the five-system which passes through 0 and let be its direction cosines, then if twists of amplitudes a, ß', y', 8', e, 0' neutralize we must have a' ax + ß'ßi + 7'71 + S'&i + e'ej + 6'6, = 0, a'a2 + ß’ß3 + y'y2 + O, + e'e2 + 6'62 = 0, a'a3 + ß'ß3 + y'y3 + 8'83 + e'es + 6'6:1 = 0, because the rotations neutralize, and also apacii + ß’pßßi + y’Py'P + S'pA. + ep^ + 6'pe 6r = 0, a.'paa2 + ß’pßßi + y'PvVs + %pA + e'pee2 + 0'pe02 = 0, apaa3 + ß'pßß3 + TPtYs + + e'P& + O'Pe^s = °> whence by elimination of a, ß',... O', we have ax 7i ei = 0. a2 ß2 72 e2 6, a3 ß3 y3 S3 63 03 p««i pßß, pyy2 ps8i pA Pa«2 Pßßs Pyy-2 PA P^3 PA paa3 pßß3 pyy3 p$8s p„es pA This equation has the form ia+mø2+nø3 LA + M'02 + N'63 Let pe = p-h and 0, = cc + p, 02 = y + p, 03 = z + p then by reduction and transformation of axes we obtain pz = ay, where y and z are planes at right angles and a is constant. This is the equation of the pectenoid.