Laserfiche WebLink
I <br />...CURVE TABLES <br />Published by KEUFFEL & ESSER CO... <br />NOW TO USE CURVE TABLES <br />Cable I. contains Tangents and Externals to a 1° curve. Tan. and <br />to any other' radius maybe found nearly enough; by dividing tbeTan. <br />it. opposite the given Central Angle by the given degree of curve. <br />Co find Deg; of Curve, having the Central Angle and Tangent: <br />ie Tan. opposite the given Central Ankle by'the given Tangent: <br />o find Deg. of Curve, having the Central'Angle,.and External: <br />ie Ext. opposite the given Central Angle by the given External. <br />Co find.Nat..Tam and Nat. Ex. Sec. for any angle by Table I.-.'Tan. <br />it. of twice the given angle divided'by the radius -of a 1 curve will <br />Ge Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Eiit. of about 12 ft. Angle <br />of Intersection or L P. =23°' 20' t-o—the R:' at Station <br />542 +72. <br />Ext. in Tab. 1. opposite 230 20'=120.87 - <br />120.87+12-10.07. Say a 10° Curve. . i' <br />Tan: in Tab. I opp. 23' 20'= 1183.1 <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 in same way) <br />. Ang..23° 20'=23.33b+10=2.3333=L. C. <br />2° 19 def. for sta. 542 L P.=sta. 542-72 <br />4° 49j''= " " +50' Tan. = 1 .18.47 <br />7" 19}'= " « 543 <br />9° 49#,'= " " " . I B. C.=sta. 541 -53.53 <br />11 ° 40f— " " '" 543+ L. C. = 2 .33.33 <br />8$.8fi E: C.=Sta. 543+86.86 <br />100-53.53=46.47XW(def. for 1 ft. of 10' Cur.)=139.41'= <br />2° 19' I'= for sta. 542. <br />Def. for 50 ft. =2' 30' for a"10* Curve. <br />Def. for 36.86 ft. =1° 50V for a 10° Curve.. <br />g y_ <br />4 �# <br />IZ q _- <br />'•r° 3--� <br />�ao <br />r� . <br />OE 4e <br />JS,a �. <br />` <br />sa <br />s�- <br />i E <br />��)ivi <br />• <br />- <br />livi <br />2 L-` <br />�• <br />Z$ �o .�Yy 3.35`- 3 <br />z�, <br />2•Sp <br />�l7I-moi Ai� ri q8 <br />e t <br />.' <br />t' <br />i <br />/1O°� o 5�_cfq 3- nq <br />5 <br />emp 9 <br />7LIPC 1596�_ <br />"i o r <br />n. Z fs Y 2 1 <br />zs <br />s i. <br />19,14 1 _0MALL ant <br />' <br />4 <br />l': 1 <br />tt <br />N w, <br />44o <br />t ' <br />. <br />.. <br />ar <br />Uri It, <br />Qi. �¢rru34Va'r, <br />Laft <br />' �aL i Z Z I <br />...CURVE TABLES <br />Published by KEUFFEL & ESSER CO... <br />NOW TO USE CURVE TABLES <br />Cable I. contains Tangents and Externals to a 1° curve. Tan. and <br />to any other' radius maybe found nearly enough; by dividing tbeTan. <br />it. opposite the given Central Angle by the given degree of curve. <br />Co find Deg; of Curve, having the Central Angle and Tangent: <br />ie Tan. opposite the given Central Ankle by'the given Tangent: <br />o find Deg. of Curve, having the Central'Angle,.and External: <br />ie Ext. opposite the given Central Angle by the given External. <br />Co find.Nat..Tam and Nat. Ex. Sec. for any angle by Table I.-.'Tan. <br />it. of twice the given angle divided'by the radius -of a 1 curve will <br />Ge Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Eiit. of about 12 ft. Angle <br />of Intersection or L P. =23°' 20' t-o—the R:' at Station <br />542 +72. <br />Ext. in Tab. 1. opposite 230 20'=120.87 - <br />120.87+12-10.07. Say a 10° Curve. . i' <br />Tan: in Tab. I opp. 23' 20'= 1183.1 <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 in same way) <br />. Ang..23° 20'=23.33b+10=2.3333=L. C. <br />2° 19 def. for sta. 542 L P.=sta. 542-72 <br />4° 49j''= " " +50' Tan. = 1 .18.47 <br />7" 19}'= " « 543 <br />9° 49#,'= " " " . I B. C.=sta. 541 -53.53 <br />11 ° 40f— " " '" 543+ L. C. = 2 .33.33 <br />8$.8fi E: C.=Sta. 543+86.86 <br />100-53.53=46.47XW(def. for 1 ft. of 10' Cur.)=139.41'= <br />2° 19' I'= for sta. 542. <br />Def. for 50 ft. =2' 30' for a"10* Curve. <br />Def. for 36.86 ft. =1° 50V for a 10° Curve.. <br />