CURVE AND REDUCTION TABLES Published by Eugene Dietzgen Co. CURVE FORMULAS 1. Radius : R=50/sin D/2 2. Degree of Curve: D=100 I/L Also, sin D/2=50/R 3. Tangent : T=R tan 1/2 I. Also, T=T for 1° curve/D + C. 4. Length of Curve: L=100 I/D 5. Long Chord : L.C.=2R sin 1/2 I. 6. Middle Ordinate: M=R (1-cos 1/2 I) 7. External : E=R/cos 1/2 I - R. Also, E=T tan 1/4 I. EXPLANATION AND USE OF TABLES Given P.I. Sta. 83+40.7, I=45° 20' and D=6°30' find: Stations—P.C.=P.I.-T. T=T for 1° Curve/D + C. From Tables V and VI T=2392.8/6.5+.197=368.32=3+68.32. Sta. P. C.=83+40.7-(3+68.32)=79+72.38. P. T.=P. C.+L, and L=100 I/D =100 45.33/6.5 =697.38 Therefore, P. T.=(79+72.38) +(6+97.38)=86+69.76. Offsets—Tangent offsets vary (approximately) directly with D and with the square of the distance. From Table III Tangent Offset for 100 feet =5.669 feet. Distance =80-Sta. P. C.=27.62. Hence offset=5.66×(27.62/100)²=.432 ft. Also, square of any distance, divided by twice the radius equals (approximately) the distance from tangent to curve. Thus (27.62)²÷(2×881.95)=.432 ft. Deflections—Deflection angle=1/2 D for 100 ft., 1/4 D for 50 ft., etc. For "X" ft., Deflection Angle (in minutes)=.3×X×D. For Sta. 80 of above curve Deflection Angle =.3×27.62×6.5=53.86'. Also Deflection Angle=dfI. for 1 ft. from Table III×X=1.95 ×27.62=53.86'. For Sta. 181 Deflection Angle =53.86'+6° 30'/2 =4° 8.86'. Externals—From Table V for 1° curve, with central angle of 45° 20', E=479.6. Therefore, for 6° 30'curve, E=479.6/6.5 + Correction from Table VI =7.378+.039=7.417. 1