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.���' �,�ia�k 7 DILT;GG1'S RAILROAD CURVE <br /> `sC"�l�cr AND <br /> CO C /{r/ph Fa unc/) <br /> FIE j K 199 REDUCTION TABLES <br /> CoW40e ,1914,by Euoga Dietum Co.,Now Y4xk Cop <br /> 7'P . <br /> 3 t a <br /> c <br /> j E �x <br /> R <br /> ° <br /> oun _ CURVE FORMULAS <br /> iYCo�Lot B/ac% 9 Rediae=A=sin.60 (1)Degree of Curve=D and ain.P=V <br /> r D/Z '2'g <br /> i Tangent=T=Rtan' (3)Length of Curve=L=100$(4) <br /> Middle ordinate=M=R(I—cos. b(5) =Rvers�(6) <br /> I ' I AO= <br /> External=E=Ttan - 7 <br /> 4 R=cos. a-j—R(8)=Rexaec0) <br /> Long Chord=C=2 R sin.s(10)4,cCentral Angle <br /> EXPLANATION AND USE OF TABLES <br /> i Stations.- Given P. I.==Sts. 161+60.35 to Lind Sta. of P. C. <br /> y� mod P. T. 4,=82° 10' D=8° 20'. From Table IV for 11' curve Ta <br /> h i 8454.1 and+ .%-:V 49 ft. From Table V ppr etion�38 or T= <br /> 414.85 ft. P. C.=Sts. P.I.—T=157+45.50. Also from (4) L= <br /> 748.00 awed P.T.�ta.P.C.+L�264+91,50. <br /> Q if�>to/','T P T/!A/e r,9 hrr Offsets.Tangent offsets vary (a�ppmximatel di y yet <br /> 0 and with square of the distance. `1'�us tangent)offeetfor Sta <br /> W on above carve is 2.16 ft.found as follows. From Table'$II tangent <br /> W7.27 <br /> r 100 ft,=7.27 ft. Distance=158—Sta.P. C.•Q54.W hence <br /> (54,50+100y4�1.18 ft. Also square of sup distance <br /> i by twice the radius equals(approximately)the diawsaee from <br /> soagont to curve. Thus (54.50)4-t(2x688.28)-2.18 ft. <br /> Deliections.—Deflection angle— D for 100 ft., �/D for 50 ft., <br /> C Fora ft.=(in minutes).3 tt C x�or=defl.for 1,t.from Tabic <br /> i '1�C. For Sta, 158 of above ourve-A x 54.5 x 8�-136.2' or <br /> .2',X2.50 x 54.136.2'from Table III. For Sta. 159 deflec- <br /> tionangff e�2°16.2'+8 201+2=6°26.2',etc. <br /> 'PZ9� _ I o I -Altnual0•—May be found ini4milar manner to tangents. Thus <br /> !J �3.3 J 4 E for curve above is 91.37. For from Table IV for 1°.carve 6 <br /> for �° 21Y=980.8+Sj��1.27 sad from Table V <br /> i—.10 or <br /> E9187 ft. Or suppose 4,=32°and E is measured and found to be <br /> j 42 ft. Whet js DY From Table IV E�230.9 and+42=5.5 or Dm. <br /> 5°30'. <br /> l <br /> 10 <br />