Railway Curves. 



129 



(In this investigation, let the centre of the arc that is "without" 

 the arc of contact be designated by O, and its chd. by AD, 

 and let a be at that end of BD, as in all the figures.) 



In all cases of $ curves, 

 L OAQ=(A + a)-90° 

 PAQ = 270°-(A + a) 



And in compound curves, 

 OAQ and PAQ are the 



same |_ = 90° + a — A* 



and r. arc of contact is common to the two As, OAQ, PAQ 



Then, 



If ?■', the radius OA, be given, or If r", the radius AP, be given, 



Then in A OAQ - Given two 

 sides, viz., r' and r. arc of con- 

 tact, and included [_ OAQ — 

 To find [ AOQ, the |_ of theseg- 

 ment opposite to AHD (being 

 half the [ at the centre O). 



Then in A PAQ — Given two 

 sides, viz., r" and r. arc of con- 

 tact, and included [_ PAQ 



To find [_ APQ, whostTsupt. 



is APV, the |_ of the segment 

 . opposite to AGB 



(by the well-known rule a + b : a — b : : tan. 



A+B : tan. A-B 



and thus finding half-difference of the unknown [_es, and knowing 

 half their sum, we get AOQ) which is the half sampltts half difference 

 if its opposite side (r. arc of contact) is greater than r' or r" respec- 

 tively— (and minus, if less.) 



At this L AOQ, 



or 



At [_ APY, 



The chd. AD must be deflected \ 

 from the By. tangent. The length > 

 of chd. AD = 2 r' sin. AOQ. j 



( The chd. AB must be deflected 

 < from the Ey. tangt. The length 

 { of chd. AB = 2 r" sin. APV. 



* InFigs. M, N, ODQ = OAQ = a + compt. of supt. of A ) . _ AO 



In Fig. O =a+ compt. A ' / =A + °~ yo 



and PAO = supt. of OAQ = 270° - (A + a) 



In a compound curve (Fig. P) ODQ = OAQ= a-comp. sunt. 

 A= a + 90°-A. 



s 



