. American engineer and railroad journal . o 19 00 sin. tang. kllAC-AHC A C: whileand Substituting these values in the above proportion, wehave 2 R : g: : tang, k (180 - H C A) : tang. J B A C. But the tang. A (180° - H C .4) = tang. 90 - k H C A,and from the proportion we have the equation tang, go - i H CA — 2 R X tang. kBA C But since the tangent of 90° minus a given angle isequal to the cotangent of the given angle, we have, fromthe above, the equation , . „^ . 2 R X tang. iBACcot. i H C A = 5—: (,g^ This analysis (as well as that for Eq. 27) is somewhat CB A = 164° 28 42 ar. co
. American engineer and railroad journal . o 19 00 sin. tang. kllAC-AHC A C: whileand Substituting these values in the above proportion, wehave 2 R : g: : tang, k (180 - H C A) : tang. J B A C. But the tang. A (180° - H C .4) = tang. 90 - k H C A,and from the proportion we have the equation tang, go - i H CA — 2 R X tang. kBA C But since the tangent of 90° minus a given angle isequal to the cotangent of the given angle, we have, fromthe above, the equation , . „^ . 2 R X tang. iBACcot. i H C A = 5—: (,g^ This analysis (as well as that for Eq. 27) is somewhat CB A = 164° 28 42 ar. comp. .j/>/. • -(- ^^= Subtracting from this result the value of the k gauge =, we have as the radius of the turnout 380 002. THE CHORD DISTANCE. Having thus found the radius of the turnout curve, thechord distance //./, fig. 14, may be readily found. The angle // A .4 = 180 —CB .4, the angle .1/ B A =i H B A. and B A M = 180 — {B A M -f M B A).Thus in the right-angled triangle ABM we have thethree angles and the
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