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I it -�-- � �-� � _ � - � ��1• DIETZGEN'S RAILROAD CURVE I I AND REDUCTION TABLES F" I ,D B 0;'._'K ' 02 ! QWri&t.1914,by Lupw Dietiaon Co..New York CC by 11 PC T Mp c i I WR.rf 5 a O - i rU 7 S, �5 �M'7oPc Iv O1✓ 2 CURVE FORMULAS i W¢soao _D=ao Radius=R=50(1)Degree of Curve=D and bin. x(2) Tangent=T=Rtan5' (3)Length of Curve=L=100$(4) i �, I l l 2 .`i 4• ( Middle ordinate=M=R(1—oos. f)(5) =Rvers (6) Clio F•�t.; - .`1"Z 4 2 �101f,.G'i__ External=E=TtanA-(7)=R+cos.#—R(8)=Rexsec#(9) • 11 3; ,S 44Long Chord=C=2 R sin. (10)A=Central Angle L°"S EXPLANATION AND USE OF TABLES Stations.—Given P. I.aSta. 161+60.35 to find Sta. of P:C. and P.T. n=62° 10' D-4°20'. From Table IV for lo curve T— war+r,dF 3454.1 and-e•8X=414.49 ft. From Table V correction=-.36 or T .6 O 0. L- 414.85 ft. P. C.—Sta. P.I:Tm157+45.50. Also from (4) T- 746.00 and P T.—Sta.'P.C.+L-164+91.50 1 4 1 Offseb.—Tangent offsets vary (appro3nmately) direetl with D and with square of the distance. Thus tangent offset for Sta. 13 5,1-+ )1 0 o? 158 on above curve is 2.16 it.found as follows. From Table III tangent offset for 100 ft.-7.27 ft. Distanoes158—&a.P. C.—.54,50 hence offset,-7.27 (54.50+100".16 ft. Also square of any distance divided by twice the radiusuals(approximately)the distance fraon i 3 2 tangent to curve. Thus (54-50)s+(2 z 688.26}-2.18 ft. 1 Deflections.—Deflection angle-31 D for 100 ft., X D for 50 ft., 5` 1 eta For o ft.-m(m minutes).8 x C x D*or—deft.for 1 ft.from Table 1 III x C. For Sta. 158 of above 0urve--3 x 54.5 x 8X==136.2' or r 16.2',ors2.50 x 54.5136.2'from Table M. For Sta. 159 deflec- tion angle--4°16.2'+9*20'+2==S°26.2,etc. j Externals.—May be found in similar manner to tangents. Thus: I E for curve above is 91.87. For from Table IV for 1°curve E=980.6 i for 8° 20'-960.6+8%-91.27 and from Table V correctionum:10 or E-91.37 ft. Or suppose &-32°and E is measured and found to be 42,ft. What is D? From Table IV F-230.9 and+42 .8 or D. M'. to r s l 4 { a I o /l?. 73 e V