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w i <br /> UG rJ RAILROAD CURVE. - <br /> 4, <br /> O *14,*44.W .)*Iqm Co..Naw'ta*ter Z 7^0 . . <br /> / mle- -- .S <br /> /G Z /G• <br /> / _. <br /> -/ CuRVt FORMULAS <br /> I;aditn=lt= (1)Degree of Cows=D and sin.s=s M <br /> Nv <br /> - TAngMtq-T'R.ta nAy (a)Length of•Curve=I,=100J(4) <br /> * --cos. (8) �Bvere (B) <br /> -A$sm l=E=Tt4u!k(7)=R ^ca.#-R(8'=AexseC- #(O <br /> - Long Chord=C=2 R sin.s(10)4=Central Angle <br /> EXPLANATION AND USE Of TABLES <br /> - - S Hosts.--Given.P. I.-=Sta. 161+60.35 to find Sta. of P. C. <br /> A=62° 10' Dom$° 201. From Table IV for 1° curve T= <br /> 8454.1�aad+83§=414.49 ft. From Table V oorrRAion=.36 or T= <br /> 414.85 'tC' P. C.—Sta. P.I.—T�157+45.50. Also from (4) LF-- <br /> 746.00. <br /> F746.00 apd P.T.-Sta.P.C.+D*164+91.50. <br /> Qfisetsr--Tangent offsets vary (apprommately) directly with <br /> D end with squamof the distance. Thus tangent offset for Sta. <br /> 158 on above curve is 2.16 ft.found as follows. From Table iIItangent: <br /> offset for 100 lt.-7.27 ft. Distance=15"ta.P. C.=54.50t hence ' <br /> o .27.(54.60+100)2=2.16 ft. Also square of any distance <br /> f di by twice the radius equals( roximately the distance from <br /> app <br /> • r tangent to onrve. Thus (b4.50)z+(2 x 688,26)= 18-ft: 1 <br /> Defiecdms.--Dellectign angle=34 D for 100 ft., ;/D-for 50 ft., <br /> ete. For o ft q* .3 x C x ll°or efi.,for 1 it.from Table <br /> III% tibove •curve—.3 x 54.5 x 8H=138.2' or <br /> 2° 18 ,ot�+R lip 36.2'l�notn TableiIL For Sta. 159 deflec- <br /> tion ' <br /> . td 1b d udlar manner to tangents. Thus <br /> . E h'ot'ftm*4IV for 1°curve E=98U.6 <br /> for td iso lablo V correction=--.10 or y' <br /> 46. me <br /> ip,measured and F. 30.9 and t 42 5.6 or d to be a <br /> _ r 4242 ft._ 9Ybiltet.1 _ - 4'f! <br /> 5"30'. <br /> 6.< <br /> e <br /> Y <br /> . e <br />