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'CURVE TABLES. <br />Pdblished by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve, Tan. and <br />Ext.to any other radius maybe found nearly enough, bydividing the Tan. <br />or Ext opposite the given Central Angle by the given degree of curve. <br />..To find Deg. of Curve,having the Central Angle and Tangent: <br />Divide Tan, opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />Divide Ext. opposite the given Central Angle by the given External. <br />. To find- Nat. Tan: and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />or Ext.of twice the given angle divided by the radius of a I* curve will <br />be -the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE. <br />Wahted a Curve with 'an Ext. of about 12 ft. Angle <br />of `Intersection or I. P.=230 201 to- the R. at Station <br />542+72. <br />Ext in Tab. I opposite 23' 20'=120.87 <br />126.87 =12 = 10.07. Say a IOP'Curve. <br />Tan. in Tab. I opp. 230 20'=1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 23' 20' for a 100 Cur. 0.16 <br />-,118.31+0.16,=.118.47 = corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang. 23' 20'==M.313 =10=2.3333=L. C. <br />2* 19 def. for sta. 542 '1. P. sta. <br />2 542+72 <br />40 491;- +50 Tan. <br />2 1 .18.47 <br />it <br />7* 192it a"- 543 B. C.=sta. R21+53.53 <br />9049''-,'=`. " a. +50 <br />II* 40'= 543+ L. C. = 2 .33.33 <br />86.86 E. C. = Sta. 543+86.86 <br />100-53.53=46.47X3'(def-for I ft. of 10* Cur.)=129.41'= <br />M91 9. <br />�21, W=def. f.o� sta. 542 <br />50 ft. =2' 30' for a 10* Curve. <br />Def. for 36.86ft. =1*50iffor. alO*Curves- <br />r.P.An9.23-ZDI: <br />----------- A'- <br />e_ <br />.7& <br />9 <br />S_ 7 IC <br />S -z,3 7 <br />j <br />S, <br />/L <br />OAf-. <br />73 <br />S71, 7 -L <br />40 <br />5-1:S-3 <br />!r 3. <br />88 <br />3, 7 3 <br />r-3. I <br />F <br />0 .8 -7;, <br />p.4.3 <br />7//U-"/ <br />A) FS <br />3 S:8 <br />J0 <br />'CURVE TABLES. <br />Pdblished by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve, Tan. and <br />Ext.to any other radius maybe found nearly enough, bydividing the Tan. <br />or Ext opposite the given Central Angle by the given degree of curve. <br />..To find Deg. of Curve,having the Central Angle and Tangent: <br />Divide Tan, opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />Divide Ext. opposite the given Central Angle by the given External. <br />. To find- Nat. Tan: and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />or Ext.of twice the given angle divided by the radius of a I* curve will <br />be -the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE. <br />Wahted a Curve with 'an Ext. of about 12 ft. Angle <br />of `Intersection or I. P.=230 201 to- the R. at Station <br />542+72. <br />Ext in Tab. I opposite 23' 20'=120.87 <br />126.87 =12 = 10.07. Say a IOP'Curve. <br />Tan. in Tab. I opp. 230 20'=1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 23' 20' for a 100 Cur. 0.16 <br />-,118.31+0.16,=.118.47 = corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang. 23' 20'==M.313 =10=2.3333=L. C. <br />2* 19 def. for sta. 542 '1. P. sta. <br />2 542+72 <br />40 491;- +50 Tan. <br />2 1 .18.47 <br />it <br />7* 192it a"- 543 B. C.=sta. R21+53.53 <br />9049''-,'=`. " a. +50 <br />II* 40'= 543+ L. C. = 2 .33.33 <br />86.86 E. C. = Sta. 543+86.86 <br />100-53.53=46.47X3'(def-for I ft. of 10* Cur.)=129.41'= <br />M91 9. <br />�21, W=def. f.o� sta. 542 <br />50 ft. =2' 30' for a 10* Curve. <br />Def. for 36.86ft. =1*50iffor. alO*Curves- <br />r.P.An9.23-ZDI: <br />----------- A'- <br />