Hallway Curves. 



125 



investigation as regards a compound curve, there being not the same 

 variety of appearance in the figures assumed by this as by the less 

 " prehensible " and eel-like S curve, to all of which, however, the 

 same trigonometrical principles apply. We have only to be attentive 

 to the sign positive or negative of the gonometrical ratios, and for 

 that purpose, and that only, to estimate all the angles as opening on 

 the same hand right or left ; and as to their angular quantity 

 (as of a and B), to take them simply as they appear by their necessary 

 designation in the figures. The repetition of this precept may be 

 necessary. 



To prove tliat the locus of osculating point is a circular arc. 



Note — That a and B are the angles of intersection of the radii of 

 the two arcs, at D and B, with BD. 



Fig. M. 



The |_ OAD = half supt. of O 

 or, OAD = t!_ a = ADB + 8 



2 



(by Ext. |_— two inter. & opposite) 

 V ADB=^ 



— 2 



and the L PB A = half supt. of P 

 or, PBA=E±i=ABD + /S 



ABD: 



8-B 



and '. " The angle BAD = supt. Z. " 



Fig. N. 



p„ J- are parallels to < p^ 



(Either an S or C.) 



The L OAD = half supt. of O 



or, OAD = ^L S = ADB + g 



(for ADB = AD'P; and OAD is 

 the exterr. \_ of the A, AD'P ; and 

 AD'P and 8 are the interr. and 

 opposite [_cs.) 



a-8 



ABP = PAB : 



•.• ADB = 



MPB B- 8 



(as shown by construction of the 



figure) 

 MPB or 8 - abeing exterr. [_ of the 

 A, ABP and A and B being equals. 



But ABP = fi-DBA = fi~ g 



•■■DBA = ^ 



— 2 



and • . • The angle BAD = supt. of 



