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0 <br />CURIE TABLES. <br />Published by KEUFFEL & ESSER CO'. <br />HOW T® USE CURVE TABLES. <br />Tahle I. contains Tangents and Externals to a 10 curve. Tan. and <br />Ext. to any other radius may be f ound nearly enough, by dividing theTan. <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 />l. <br />Divide Ext. opposite the given Central Angle by the e by ablExte .- T <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table L: Tan. <br />or Ext. of twice the given angle divided by the radius of a 1° curve will <br />'be the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of *about 12 ft. Angle <br />of Intersection or I. R=9_30 20' to the R. at Station <br />542-1-72. <br />]sxt. in Tah. I opposite 23° 20 =120.87 ' 174. 0 9 <br />1213,87-19-10.07. Say a 10° Curve. <br />Tan. in Tab. i opp. 23° 20'=1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 113.31+0.16 118.47 =co,rect r a d Tangent, . <br />(If corrected Ext.. is required find in same wav) <br />Ang. 23° 20'= 23.33'-10 = 2.3333 = L. C. <br />2° 192'=def . for sta. 542 1. P. =Eta. 542+72 <br />40 40" _ cccc It +50 Tan. _ <br />1 .18.47 <br />7° 191'= r. « « 543 B..C.=sta. 541-1-53.53 <br />90 492' _ « « « +50 2 .33.33 <br />543 -1 L. C. _ <br />8611.86 E. C. =Sta. 543+903.86 <br />100 — 53.53 = 46.47 X T(d ef . for 1 ft. of 10° Curl =139.41'= <br />21 19;'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 1.0° Curve. <br />Def. for 36.86 ft. =1° 50l' for a 10° Curve. <br />J <br />- <br />3 <br />.27. �_ s <br />4 73 <br />2 <br />.71 <br />TT`f-8- <br />i <br />tl <br />17 2 <br />7 <br />82 <br />CURIE TABLES. <br />Published by KEUFFEL & ESSER CO'. <br />HOW T® USE CURVE TABLES. <br />Tahle I. contains Tangents and Externals to a 10 curve. Tan. and <br />Ext. to any other radius may be f ound nearly enough, by dividing theTan. <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 />l. <br />Divide Ext. opposite the given Central Angle by the e by ablExte .- T <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table L: Tan. <br />or Ext. of twice the given angle divided by the radius of a 1° curve will <br />'be the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of *about 12 ft. Angle <br />of Intersection or I. R=9_30 20' to the R. at Station <br />542-1-72. <br />]sxt. in Tah. I opposite 23° 20 =120.87 ' 174. 0 9 <br />1213,87-19-10.07. Say a 10° Curve. <br />Tan. in Tab. i opp. 23° 20'=1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 113.31+0.16 118.47 =co,rect r a d Tangent, . <br />(If corrected Ext.. is required find in same wav) <br />Ang. 23° 20'= 23.33'-10 = 2.3333 = L. C. <br />2° 192'=def . for sta. 542 1. P. =Eta. 542+72 <br />40 40" _ cccc It +50 Tan. _ <br />1 .18.47 <br />7° 191'= r. « « 543 B..C.=sta. 541-1-53.53 <br />90 492' _ « « « +50 2 .33.33 <br />543 -1 L. C. _ <br />8611.86 E. C. =Sta. 543+903.86 <br />100 — 53.53 = 46.47 X T(d ef . for 1 ft. of 10° Curl =139.41'= <br />21 19;'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 1.0° Curve. <br />Def. for 36.86 ft. =1° 50l' for a 10° Curve. <br />J <br />