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FI�LD aoo� 1031 DIETZGEWRAILROADIAND 3R , I REDUCTION TABLES CkWrigbt.1914.by Eugene Dietago!Co.,New York all i I I y C _ a R CURVE FORMULAS Radius=R �(1)Degree of Curve=D and sin.D=s(2) Tangent=T=R.tan (3)Length of Curve=L=100 (4) i Middle ordinate=M=R(1—cos. 3)(5) ( ) =livers� 6 External=E=Ttan4-(7)=R+cos.Q R(8)-R(8) ! song Chord=C=2 R sin.s(10)A=Central Angle w EXPLANATION AND USE, OF TABLES 3 124° I `3 Stations.—Given P: I—Staff 161+60.35 to find Sta. of P. C. �9 } -, % » IJ L ^� 397, �. and P. T. 0&==620 101D==r 20'. 1 From Table IV for curve Q— V/ Ie sF 3454.1 and+83 =414.49 ft. From Table V correction—.36 or T= 414.85 ft. P. C.=-Sta. P.I.—T=157+45.50. Also from (4) L= 746.00 1 and P.T.—Sta.P.C.+Lm164+91.50. Offsets,—Tangent offsets vary (spproxunately) directly with D and with square of the distance. Thus tangent offset for Sta. s 158 on above curve is 2.16 ft.found as follows. From Table III tangent offset for 100 ft.7.27 ft. Distance-158--Sta,P. C.=540,hence offset=-7.27 (54.50+100)22.16 ft. Also squareof any "distance divided by twice the radius equals(approximately)the distance from tangent to curve. Thus (54.50)2+(2 x 688:26}=2.16 ft. Deflections:Deflection angle--%y D for.IW ft., Y.D for W ft., etc. For c ft.—(in minutes).3 x C x D°or—defl.for 1 ft.from Table III x C. For Sta. 158 of above curve==.3 x 54.5 x 8 136.2' or � c,-,;T, qo - 20 16.2,or=2.50x54.5=136.2 from Table III, For Sia. 159 deflec- -" 4-1 so I tion angle-2°16.2'+8°20'+2-----V 26.2',eta 6 II{ Externals.--May be found in similaa manner to tangents. Thus 5o E for curve above is 91.37. For from Table 1V for 1°curve E=960.6 + for 80 20'=960.8+8%y 91.27 and from Table V correction—.10 or x✓ E--91.37 ft. Or suppose x=32°and E is measured and found to be 42 ft. What is D7 From Table IV F-230.9 and+42=5.5 or D-- . - 5°'n,• � z