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i jos <br /> DIVZGE.NN's RAILROAD CURVE, <br /> AND <br /> REDUCTION TABLES 4 <br /> . �g 3/ t_ S' a .._ . Nsw York to <br /> I I <br /> PC 7' P <br /> S°L <br /> _ r p <br /> Z= a K / s� <br /> G <br /> •��-_--- — _ — -- — CURVE. FORMULAS <br /> Radius=R=r 0 (1)Degree of Curve=D and sin.P=L,*(2) <br /> 16 Tangent=T=RtanA (3)Length of Curve=L=100$(4) <br /> )Mdd1a ordinate=M=R(1--cm j)(5) =Rvera A(6) r <br /> a t�F Z7 Q Fiternal=E=Tran+(7)=R+cos.n}-t(8)—Rexsec#(9)' <br /> �= �5. Long Chord=C=2 R sin.4,k(10),&=Central Angle <br /> EXPLANATION AND USE OF TABLES <br /> Statlons.—Given P. I.—Sts. 161+80.35 to find Stu. of P.•'C. <br /> �osf a"an;c" and P.T. 0=62° 10' D=8"20'. From Table IV for r curve T <br /> 8454.1 and+SM-414.49 ft. From Table V oorrection.%36 or T <br /> -- -- 414.85 €t. P. C.-Sts. P.L—T-157+45.50. Also from (4) L <br /> 746.00 and P.T.u-Sts.P.C.+1.---164+91.50. <br /> jz n9'c ffeat o <br /> Offsets.—Tan eets vary <br /> /g"vaiC i Lo" nA C 8 (aPpro+sixnately) dire* 'With <br /> D and with square of the distance. Thus tangent offset for Sta. <br /> 159 an above curve is 2.16 1L found as follows. From Table III tangent <br /> w i t for.100 ft.-7.27 ft. Distanoe-15"ta.P. C...64ZOR hence <br /> offs4.7.27 (54.50+100?-2.16 ft. Also square of any <br /> J i divided by twice the radiusala roximately)the dlstartee fin <br /> ' tangent to curvb. Thus (54�0p+688.26}•2.16 it. <br /> DdWetiona.--Dedeetion angle—I3 �y ft <br /> D for 100 ft., D f*b0 ., <br /> ft <br /> a c o For c n minutes).3 z C t or�deR.for 1 Ott.from Taiste <br /> C C. For'Sts►. 158 e4 above curve*..3 x 54.6 x 8 X136.2' or <br /> i$:2',.a�1 Wx 54.5.136.2'fmm'table M. For Sta. 169 4fleo. <br /> Own angle--r 16.2'}8°20Y+2=-W 26.21,etc. <br /> or curve <br /> bo 891 o Table manner to tante, 'thus <br /> r a V for i°curve,�99p.6 <br /> T e V co1'recdoai�-.10 or <br /> 1.87 ft. Or suppose ,p�2°and 8 i,►�and#nand to be <br /> .421t. What is DY From Table IV FPPWU and+42=-&S or D— <br /> Z, <br /> ELD 1300K <br />