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CURVE AND REDUCTION TABLES <br /> ST YY RT_y`ie.0 -T-- 'Pubifabed b7 Nu/ens Dfetagen Co. <br /> 25 - 1424. <br /> e z <br /> f _ � <br /> N V <br /> I <br /> CURVE FORMULAS <br /> 50 <br /> 1. Radius R= <br /> y sin D/2 <br /> / 2. Degree of Curve: D—I00 L• Also,sin D/2=50 <br /> / <br /> 1/z Also,T=T for 1 curve+C. <br /> { 3. Tangent T=R tan L D <br /> All'i <br /> 4. Length of Curve: L=100 D <br /> 5. Long Chord L. C.=2$sin%I. <br /> 1 /ofd.' oras 6. Middle Ordinate: M—R (1—ens I) <br /> 0 7. External E=7R R. Also Ea <br /> 5z o o cos <br /> r �§I— T ► 3a L <br /> 5� 30 23 3 <br /> EXPL&RATION AND USE OF TABLES <br /> Given P.I.Sts.83+40.7.1-45-20'and D=6'30'find: <br /> Stations-P.C.=P.I.-T. T- for i°Curve <br /> �N' C 2392.8 D + C. From Tables V and VI <br /> y ;l <br /> 9 t 2S 310 T a 6.5 +.197=368.32=3+68.82. Sts. P. C.-83+40.7-(3+68.32)-79+72.38. <br /> 8.38 <br /> 1 N 30 Z N P.T.=P.C.+L,and L=100 I 4-$=100 6.5 =697.38 Therefore,P.T.=(79+72.38) <br /> +(6+97.38)=86+69.76. <br /> G Offteb-Tangent oSeete v�q (a pmateiq) dineotly with D and <br /> with the <br /> square of the�istanoe.From Tahle III�angent Offset for 100 feet=5.669 feet.Distsnoe <br /> =80-13ta.P.C."07.62.Hence odest=5.66 X 27.82 ! <br /> i \ 100) =482 ft.Also,&q.of any <br /> distance, equahi(approximately)the distance from tangent <br /> � JJ to curve. h 6*s+(2Xf .432 h. <br /> F le, 9 .t7`3 Defleotfons-Degeotion angle=3!D for 100 ft..X D for 50 ft.,etc.For"X"ft, <br /> x/ Deflection Angle(in minutes)=.3 XX XD.For ata.80 of-above curve Deflection Angle <br /> =.3 X27.62 X6.5=53.86'.Also 3=tion Angle=dfl,for 1 ft.from Table III XX=1.95 <br /> 7ctJa �7 �rr2 /!.E iJ/S/S��C}� �' /C�rt X27.62=53.86'. ForSt&.181 Deflection <br /> l / Angle=53.88'+6��•4" <br /> d .e f,;1^ - T GVOC/ lkternals-From Table V fqt P curve.with central angle of 45`201,E=479.6. <br /> ,/ Therefore.for 6.30rcurve,E:470 6 l M from Table VI= <br /> N„I,e,!Al cz ,� /r 1 P((� 6 d M" _ "^'n 7.378+.039=7.417. <br />