. Elements of railroad track and construction . In Fig. 69, assuming 0 D as equal to 0 A or 0 B,the difference being assumed small, we have from thetriangle A 0 B, 1 = 2 (R - I G) tan i <^ = 2 (R - i G)^ = 2 GN - ^ (18) 116 RAILROAD TRACK AND CONSTRUCTION. If the last term in (18) be dropped, 1 = 2 G N, the sameas in the turnout from a straight track. Computationswill show that (18), (16), and (17) give results, in manycases, that vary materially from the true values from(12), (15), and (16), and are given here only becausethey are given in many Handbooks. 117. Turnout from Convex Side of M
. Elements of railroad track and construction . In Fig. 69, assuming 0 D as equal to 0 A or 0 B,the difference being assumed small, we have from thetriangle A 0 B, 1 = 2 (R - I G) tan i <^ = 2 (R - i G)^ = 2 GN - ^ (18) 116 RAILROAD TRACK AND CONSTRUCTION. If the last term in (18) be dropped, 1 = 2 G N, the sameas in the turnout from a straight track. Computationswill show that (18), (16), and (17) give results, in manycases, that vary materially from the true values from(12), (15), and (16), and are given here only becausethey are given in many Handbooks. 117. Turnout from Convex Side of Main Curve.—Given the radius, R, of the main curve and the frog-number, to find the radius, R2,and the lead, 1, of the the triangle A 0 B, ,OAB + OBA =18O°-0,OAB-OBA=F (this isproved by drawing the tangentsBx and A x, then 0 A B = 90°+ B A X, 0 B A = A B X +X B 0, and since B A x = A B x,0 A B - 0 B A = 90° + B Ax-ABx-xBO =90°,-xBO= F), 0 A = R - i G, and0 B = R + i G, then from trig-onometry, 1 G) tan I (180 - ). (R + ^ G) + (R tan^F 2^taiiiF = J^ (19). (R + i G) - (R - i G) from which cot t - Q In the triangle 0 A B, drawing the line 0 D, bisectingthe angle A 0 B and the line A B,AB = 2 OA sin ^ AOB, or 1 = 2(R + |G)sin|0 (20). CIRCULAR TURNOUTS. 117 Also assuming OD=OA=OB, the difference beingassumed negligible, i = 2 (R + I G) tan I </» = 2GN + ^ (21),and neglecting the last term in (21), we have 1 = 2 ON (22). To iand R2: In the triangle 0 d B, d 0 B = ^S isknown from (19), and since F and R are given, we have ^ ^ OB sin OiOB^^ = sinOOiB substituting and transposing as in ^ 116, we have ^^ = (^^ + ^^)si^#^)-^^(23). In the triangle d A B, d A B = 180° - 0 A B,OAB =90° -1^, /. OiAB = 180° - (90° -J^)= 90° + 19^; Oi B A = 0 A B - A Oi B, A d B =F - ^, .-. d B A = 90° - i 0 - (F - 0 ) = 90° -F + i (^; and from trigonometry, OiB + OiA ^ tan | (OiAB + OiBA) ^ tan f (180 - (F - 0))OiB - OiA tan ^ (OiAB - OiBA) tan | F combining and transposing, G
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