Egan — Linear Complex, and a certain class of Twisted Curves. 55 



Hence 4<t = 4S' ={A- B cos \p)-, ds = del a = dip/ha = 4:d\p/h (A- B cos i/-)-. 

 Again z/s = const. = a/V{<i' + <t') = 1/ti. 



Hence 



h^]{A - Bcos^y 



Again, if *S' is the arc of the normal section of the cylinder, 

 dS = cdz/,7 = hdz = 4:bdxp/h'' (A - B cos i^)'. 

 Also X = Jcos \pdS, ?/ = fsin \pdS, 



hence the most general cyliiidrical helix whose tangents belong to a linear 

 complex is given by the equations 



z=N 



cos -ipdip 



y 



= Jf 



sin ^dip 



{A- B cos ipy 



(M, N const.) 



{A - B cos xpf 



• d^p^ 



(A -B cos )/,)'■ 

 Here Jf/iV = c/a. 



It is easily verified that these curves are algebraic only when A = ± B. 

 If A=B, we put i//' = i// + TT, which gives A'=-B. 

 Hence the algebraic curves of this type are given by 



2m 



cos i// d\p 



y = 2m 



sin i// dip 



(1 + cos xpy " """]{l + cos,iPf' 

 Putting t = tan ^\p, we obtain 



= H(1^ 



dip 



cos Tpy 



X = 771 



y = m 



m(a + <--), 



{l-t^)dt 



21 dt = m{b + t-], 



z = n\(l + t'^)dt = 71 Ic + t + 



Hence the only algebraic helices whose ta7igents belong to a linear complex are 

 certain twisted cubics,* 



3. Bertrand Curves. 



42. These curves are characterized by a linear relation between c and (t. 

 In this case equation (22) can be integrated by elliptic functions. In parti- 

 cular, we note that the circular helix is not the only curve of constant 

 curvature belonging to a linear complex. 



[ * Note added in the Press. — Another form of the equations of the complex helicea is given 

 by Keraval in Nouvelles Annales, 1909, pp. 42 sq.] 



