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r 8o FiEIL.D B OK 230 j 11 DIETZG .N'S RAILROAD CURVE AND RE.DUCTION_ TABLES 3111 CbWrisbt,1914,by Eugene Dietsm Co.,New York Ciiy 2°O o Lri P.G T r 2 540 OSS L' � � Iz 7.2o y A �S� � - � � CURVE FORMULAS Radius=R= ao L)=!()ry z / '<° I „n (i)Degree of Curve=D and sin.2.a(2) _53 7 Tangent=T=Rtan6 (3)Length of Curve=L=100,,(4) s3s n/ = f/, Middle ordinate=M=R(1-cos. (5) =Avers�(6) f PL/ -4 =Ttan� (7)_$'cOs4-R(8)=Rexsec (9) y z 3 {4 3 s f /a5' Long Chord=0=2 R sin. ((10),&=Central Angle s-'3 ° `�, I C. I EXPLANATION AND USE OF TABLES i -- y j7 Stations.-Given P. I. ta. 161+110,35 to find Sta. of P. C. f�z f� E / 7 ,f G 11 and P. T. 0=i2o10' D-8* 20'. From Table IV for le curve T= v, - 3454.1 and+8%=414.49 ft. From Table V correction=.36 or T= o£' 7 414.85 ft. P. C.-Sta. P.I.-T=157+45.50. Also from (4) L= 7 I 746.00 and P.T.==Sta.P.C.+L-=164+91.50. Offsets.-Tangent ofliaets vary (a proximately) directly with D sad with square of the distance. Thus tangent offset for Sta. 158 on above curve is 2.18 ft.found as follows. From Table III tangent offset for 100 ft.7.27 ft. Distance=158--Sta.P. C.=54.50,hence W 21 r' offset=7.27 (54.50+100)2=2.16 ft. Also square of any distance i ! divided by twice the radius equals(approximately)the distance from -3 f..i I tangent to curve. Thus (54.50)2+( x 688.26)--2.16 ft. ` Deflections.-Deflection angle=-M D for 100 ft., %D for 50 ft., S a 54 G eo „1 f etc. For c ft.=(in minutes).3 x C x W or=xiefl.for 1 ft.from Table III x C. For Sta. 158 of above curve-.3 x 54.5 x 8}J-136.2' or 20 16.21,or=.2.50 x 54.5 136.2'from Table III. For fits. 159 deflec- tion angle-2°162'+8°201+2==W 262',etc. Esternals.-May be found in similar manner to tangents. Thus 6 s� E for curve above is 91.87, For from Table IV for 10 curve E=980.6 S for 8°.20'=960.6+8Y3�9).27 and from Table V correction-.10 or ��9�1 E=91.87 ft. Or suppcee p==3r and E is measured and found to be 42 ft. What is D? From Table IV E-230.9 and+42=5.5 or D_- So 30'. v .3 4,