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I` <br /> DIETZCEN,5 RAILROAD CURVE, <br /> IFIELDBOOK VU AND <br /> REDUCTION 'SABLES <br /> EOaWd&t,2*14,by Euzem Dktsgm Cb: Now Yak CNAW I- <br /> PInpje <br /> i <br /> ` CURVE FORMULAS <br /> Rbtllva=R=mgt(1)Degree of Curve=D and sin.D`s(21t <br /> i Tangent=T=Rtan� (3)Length of Curve=L=100g(4) <br /> x J Middle ordinate=M=R(1—cos. (5) sRvere (6) <br /> External=E=Ttan�-(7)=R+cos.Jj-R(8)�Rexsecg(9) <br /> j Long Chord=C=2 R sin.9(10)o-Central Angle <br /> i I EXPLANATION ANIS USE OF TABLES <br /> Sttliaas.—Given P. I.—Sta. 161+W.35 to find Sta. of P. C. <br /> and P.T. A--W I0' D--8.20'. From Table IV for 1• curve T= <br /> 8"1 ond-f-8M=414.49 ft. From Table V correction-mm,36 or T <br /> 414,86 <br /> i ft .TC�?,PCI-I-Twai--16X547+4Alm from (4) L— <br /> MOA <br /> j OdJObr-Tangent ofleets varyr (appro�dmately) dtredi wi£h <br /> D mi►d with square of the diste"aee. us tangent offset or Sta. <br /> 158 an above curve is2.16 ft.found as follows. From Table III tangent <br /> offset for:100 ft.**1.27 it. Distanoe-a-15"ta.P. Q m54.50t hence <br /> offso -7.27 (54.50+100)0 .16 ft. Also square of any distance <br /> divided by Was tae adiv$ nail(approximately)the distance frc u <br /> tangent to outve. ue (54� +(2 x 688.26)2.16 ft. <br /> eta For offf:A -(in o; 3 i�D for 100 <br /> °for'1�tDfro�oa�JA, <br /> ! III i C. For f1tai. 158 of above cuurvem-.3 x 54.5 x 8 136.2' or <br /> P i I r 16.21,ori 60 x 64.5-136.2'fromable M. For Sti. 159 dellec- <br /> tion ongie--•12°1"i 8°201+2==W 262,etc. <br /> $stands+-W y be found is's laa.mtmner to tangents. Thus 1 <br /> k tot curve above is 91.$7. For from Table IV for 1°curve R=960.6 <br /> i for d° 70'x4160.6+8 x91. from Table V correction--10 or <br /> I E-41.87 ft. Or suppose o d B L measured and found to be <br /> 42 ft. Whet b Dt Frons Table IV Emm=.9 and+42=5.5 or Da <br /> 5-30'. <br /> • f <br /> ? • ,0 . <br />