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jr <br />� ,q CURVE, TAB -LES <br />Published by KEUFFFL 8 ESSER CO. + g I= <br />78 <br />HOW TO USE CURVE TABLES <br />Table T. contains Tangents and Externals to a I° curve. Tan. and <br />or Ext. to t. o y other radius may be found nearly enough, by dividing the Tan. `$ <br />PPosite the <br />v <br />r <br />0 <br />_ <br />Q <br />�1 <br />given Central Angle by the given degree of curve. <br />To <br />Divide Tad Deg. of Curve, having the Central Angle _and Tangent:'°_ <br />PPosite the given Central Angle by the given Tangent. <br />To find Deg., <br />o£ Curve, having the Central Angle and External: <br />Divide Ext. opposite the given Central Angle by the given External <br />To find Nat. Tan. <br />.: <br />and Nat. Ex. Sec, for any an by Table L: Tan. A <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. <br />See. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about I2 ft. Angle <br />of Intersection or f. P. =23° 20' to the R. <br />542 X72. at Station <br />Lxt. in Tab. 1 0 PPosite 23° 20' =120.87 <br />120.87 : 12=10.07. Say a 10° Curve. I <br />t <br />Tan, in Tab. I opp. 23' 20'= 1183.1 <br />` <br />.; <br />r <br />- + <br />� <br />1I83:1=10=11$.31. <br />Coerection for A. 23° 20' for a 10° Cur. =0.16 ! <br />118.31. --{-0.16 =118,47 =corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang. 23° 20'=23.33 X14 <br />v (�� <br />-2.3333-L, C. <br />2°-192'=def. forsta_ 542 L P.= y <br />4° 492 sta. 542+72 <br />+50 Tan. = 1.1$.47 <br />7° 192'= .0 u v 543 <br />914911= « u B. C.=sta. 541-1-53.53 <br />i-50, � <br />I 1' 401 = " 543+ L. C. = 2.33.33 <br />86.86 E. C. = Sta. , 543-}-86.86 <br />—71 100-53.53=46,47x3'(def.-for 1 ft. of 10° Cur.) =139.41'= <br />2° 19z'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 it. =1.° 50;' for a 10° Curve. <br />iOle, <br />i <br />� ff d IP. Ai, y.23.2D' <br />a <br />jr <br />