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moo <br />CURVE AND REDUCTION_. TABLES <br />Published by Eugene Dietagen Co. <br />I <br />1' <br />1 <br />/ <br />CURVE FORMULAS <br />L Radiu3 R= <br />sin D/2 <br />2. Degree of Curve: D=100 L• Also, sin D/2=R <br />Tforl.°curve <br />3. Tangent T=R tan % I. Also, T= D +C. <br />4. Length of Curve: L=100 D <br />5. Long Chord L. C.= 2R sin !4 I. <br />6. Middle Ordinate: M=R (1—cos % I) <br />7. External E= COS % I—R. Also, E=T tan % I. <br />EXPLANATION AND USE OF TABLES <br />Given P.I. Sta.83+40.7, I =45° 20' and D =6°30' find: <br />Stations -P. C. = P. I. - T. T- T for 1 ° Curve <br />D C. From Tables V and VI <br />T=2-3 ' F.47=368.32=3+68.32. Eta. P. C.=83+40.7-(3+68.32)=79+72.38.. <br />P. T. =P. C.+L, and L =100D =100 li 3 = 697.38 Therefore, F. T. _ (79+72.38) <br />6.+(6+97.38)=66+69.76, <br />Ofiaeta-Tangent offsets vary (?proximately) directly with D and with the <br />square of the distance! From Table IhTangent Offset for 100 feet =5.669 feet. Distance <br />=80 -Sts. P.C.=27.62. Hence offset=5.66X(2�Op)2=.432 ft. Also, square of any <br />distance, divi6dby twice the radius equals (approximately) the distance from tangent <br />to curve. Thus V.62)1=(2X881.95)=.432 ft. <br />Deflectfou-Deflection angle =!4 D for 100 ft.. Y, D for 50 ft., etc. For "X" ft.,: <br />Deflection Ange (in minutes) =.3 XX XD. For Sta. 80 of above curve Deflection Angle' <br />=.3 X27.62)<65=53.86', Also Deflection Angle =df1, for L ft. from Table III XX =1.95 <br />X 27.62 - 53.66'. For Sta, 181 Deflection Angle =53.86'+6-3a =4° 8.86'. <br />2 <br />Exter=ab-From Table V for 1' curve, with central angle of 45' 201, E=479.6. <br />Therefore, for6'30'curve, E =479'6+ Correction from Table VI =7.378+.039=7.417. <br />6.5 <br />1 11 <br />\1 <br />