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,o <br />'CURVE TABLES, <br />47L,4$ <br />Published by KEUFFEL & ESSER CO. f.�t <br />HOW TO USE CURVE TABLES 7 , <br />Table J. contains Tangents and Externals to a 1° curve. Tari: and <br />Ext. to any other radius may be found nearly enough, by dividing 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 />F: or Ext. of twice the given angle- divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. Sec.: <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. •P.=23°-20' to- the R: at' Station <br />542+72. <br />Ext. in Tab:.I opposite 230 20'=2,120.87 <br />120.877-12=10.07. Say, a 10° Curve. <br />Tan. in Tab. I opp. 23° 20'= 1'183.1 <br />1183.1=10 =118.31: <br />- Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31+0.16=118.47=corrected Tangent.. ;- <br />(If corrected Ext. is required find i6 same way) <br />A rig. 23° 20'=23.33 =10 = 2.3333 = L. C. <br />20 19;'=def. for sta. 542 L P. =sta. 542+72 <br />4° 49"= " +50 Tan.,= 1 .18.47 '. <br />70 191' _ " " " 543 <br />9°'491,'= " " +50 B. C.=sta. 541+53.53 <br />11 ° 40, _ « • « « .. 543+ L. C. = • t 2 .33.33 <br />86.86 E. C.=Sta: 543+86.86 <br />100-53.53=46.47X3'(def. for 1 -ft. of 10°.Cur:) =139.41'= <br />2° 19.2'=def. for sta. 542. <br />Def: for 50 ft. =2°,30' for a 10°•Curve: - <br />Def. for 36.86 ft.=1° 50;' for a 10° Curve. <br />2%� <br />S /Z MIA <br />s, ¢s. <br />FTr�, <br />Nl <br />.5 <br />98'.86 <br />ZIZc <br />�e%,rrvv� <br />r <br />79 <br />O,g ass <br />c� <br />s' 3 G <br />881,33 <br />a a <br />860.9._- <br />G 5-,4B <br />¢ <br />y�i� <br />6 kWegG <br />,7o <br />s7,89 <br />G <br />.siZ,2.o <br />min a Irv, <br />cao� s <br />,o <br />'CURVE TABLES, <br />47L,4$ <br />Published by KEUFFEL & ESSER CO. f.�t <br />HOW TO USE CURVE TABLES 7 , <br />Table J. contains Tangents and Externals to a 1° curve. Tari: and <br />Ext. to any other radius may be found nearly enough, by dividing 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 />F: or Ext. of twice the given angle- divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. Sec.: <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. •P.=23°-20' to- the R: at' Station <br />542+72. <br />Ext. in Tab:.I opposite 230 20'=2,120.87 <br />120.877-12=10.07. Say, a 10° Curve. <br />Tan. in Tab. I opp. 23° 20'= 1'183.1 <br />1183.1=10 =118.31: <br />- Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31+0.16=118.47=corrected Tangent.. ;- <br />(If corrected Ext. is required find i6 same way) <br />A rig. 23° 20'=23.33 =10 = 2.3333 = L. C. <br />20 19;'=def. for sta. 542 L P. =sta. 542+72 <br />4° 49"= " +50 Tan.,= 1 .18.47 '. <br />70 191' _ " " " 543 <br />9°'491,'= " " +50 B. C.=sta. 541+53.53 <br />11 ° 40, _ « • « « .. 543+ L. C. = • t 2 .33.33 <br />86.86 E. C.=Sta: 543+86.86 <br />100-53.53=46.47X3'(def. for 1 -ft. of 10°.Cur:) =139.41'= <br />2° 19.2'=def. for sta. 542. <br />Def: for 50 ft. =2°,30' for a 10°•Curve: - <br />Def. for 36.86 ft.=1° 50;' for a 10° Curve. <br />2%� <br />