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l - IMO TAB AND t :5" INFORMATION 79 � �f0 l �d T �. �c fj a r �� j-2 , v < ? , of CURVE FORMULAS y.Z Radi y(1) Degree of Curve=D and sin.P2_3X0(2) t T — ttana°—(3) Length of Curve=L=10 D MVd .ohlinste=M� (4) Mid a ohdinste=M=R(i—cos.a) (5)=Rvers a (6) J External--PP-Ttau�°� M( 11—cos. R(8)=Rexsec s(9) Long Chord—C==2 R sin-s (10) Q=,entral Angle t' - S� EXPLANATION AND USE OF TABLES Status.—Given P. I.==Sta. 161+60.35 to find Sta. of P. C. and P. T. p==62° 10' Dom* 20'.. From Table IV for 1° curve T= andd�,-t t-414.49 ft. From Table V correction=.36 or-T= y3 i`3 414,85 ft. f'. =Sta. P.I--T=157 +45.50. Also from (4) 1— 746.00 and P.T.=-&a.P.C.+L=164{-91.50. Offsets.—Tangent offsets vary (approximately) directly with D and with Thus tangent offset for Sta. j-, square of the distance. y 158 on above curve is 2.16 ft.found as follows. From Table III tangent offset for 100 ft.=7.27 ft. Distance=15"ta. P. C-=54.50 hence r , offset-7.27 (54.50—. 1002-2.16 ft. Also square of any aistance cF 6• ° — - divided by twice the radius equals (approximately) the distance from y / tangent to curve. Thus(54.50=+(2 x 688.26) 2.16 ft. — ` DeSectiona.—Deflection angle---% D for 100 ft., /D for 50 It., y1 vF 3 9� etc. For a ft.==Cm minutes) .3 x C x D°of=deft.for 1 1t.from Table -7 3 �.� f III x C. Fonts. 158 of above curve=•3 x 54.8 x 8%§ 136:2' or y3•y J ` 2° 16.2',or=2.50x54.5=136.2'from Table IIL For Sta. 159 deflec- tion angle=2°16.2' -}-8°20' +2=6°26.21,etc. be -0 9 7 E for o aboveis 115.37. or ro�m Table IV fomannerr curve Thus 7 ° for 8° 20'=980.6+8%—I15.27 and from Table V correction—.10 or y y y E 118.37 Et. Or su A=32°and E is measured and found to be 42 ft. What is DY Prom Table IV Fo=W&9 and +426.8 or D- 5°30'.