Pg 82 - Laserfiche WebLink

Home Browse Search

CURVE' -TABLES.- Published by KEUFFEL & ESSER C.Q. HOW' TO USE CURVE TABLES. Table I. contains Tangents and Externals to a I' curve. tan. and Ext. to any, other radius may befound nearly enough, by dividing the Tan. or Ext. opposite the given Central -Angle by the given degree of curve. To find Deg. of Curve, haVing the Central Angle.and.. Tangent: Divide Tan. opposite the given Central Angle by the given- Tangent. To find - Deg. of Curve,, having -the Central. Angle and External: Divide Ekt. opposite the given iCefitral.Angle by the given External. .To find Nat. Tan:and Nat. Ex. See. for any angle by Table I.: Tan. or Ext. of twice the giVen. angle divid6&byith6 iadids of a_1 curve will be the Nat. Tan'. or Nat. Ex. Sec.' - EXAMPLE;-�:. Wanted a Curve with an Ext. of about 12 ft. Angle of Intersection or,- 1. P. =230,20' to - the R. "at 'Station 542+72. Ext. in Tab. I opposite 23°:20' = 120.87 120.87 +12 10.07. Say a IOP Curve. Tan. in Tab. I opp. 23° 20'=1183.1 1183.1=10=118.31.- Correction 183.1+10=118.31.- Correction for A. 23* 20' for a 10° Cur. 0.16. 118.31+0.16 =118.47 =corrected Tangent. �(If corrected Ext. is required find in same way) Ang. 23* 20'=23.33 10 = 2.3333 = L. C. 2* 1914'= defJor sta. 542. T. P. =sta. 542+72 4° 497', 'i +50- Tan. 1 .18.47 '543 7* 1.92 9* 497, a a +50 B. C. sta. .541+53.53 - I , 11* 40'=. a +" C. 33.33 a,. a. 543 -. L : 86.86 E. C —Sta- 543+86.A6 7 100-53.53-46.47XX(def. f6r.l.ft.-of 10° Cur.')=139.41'= V.Al. . M ' for tai .542.. Def. for 50 ft=2* 30' for a- 10° Curve.- Def.jor 96.86 ft.= V 507' for a 10° Curve. - To, 4:; ry A/- ------------ .4 J So •¢4 .63.oe 47,04 417-e,.3 42, Z8 47 2rly-p CSW 47- — 2 1- 42WALG 0 &S, Z all- A-1 0 -, 4%L4 47,78 47.33 s- 4,70 C, 4 .14 8 r z :`0.36 JO 0 ?47 4' 93 zs- .5"o045 A- 2/ e 41 V CURVE' -TABLES.- Published by KEUFFEL & ESSER C.Q. HOW' TO USE CURVE TABLES. Table I. contains Tangents and Externals to a I' curve. tan. and Ext. to any, other radius may befound nearly enough, by dividing the Tan. or Ext. opposite the given Central -Angle by the given degree of curve. To find Deg. of Curve, haVing the Central Angle.and.. Tangent: Divide Tan. opposite the given Central Angle by the given- Tangent. To find - Deg. of Curve,, having -the Central. Angle and External: Divide Ekt. opposite the given iCefitral.Angle by the given External. .To find Nat. Tan:and Nat. Ex. See. for any angle by Table I.: Tan. or Ext. of twice the giVen. angle divid6&byith6 iadids of a_1 curve will be the Nat. Tan'. or Nat. Ex. Sec.' - EXAMPLE;-�:. Wanted a Curve with an Ext. of about 12 ft. Angle of Intersection or,- 1. P. =230,20' to - the R. "at 'Station 542+72. Ext. in Tab. I opposite 23°:20' = 120.87 120.87 +12 10.07. Say a IOP Curve. Tan. in Tab. I opp. 23° 20'=1183.1 1183.1=10=118.31.- Correction 183.1+10=118.31.- Correction for A. 23* 20' for a 10° Cur. 0.16. 118.31+0.16 =118.47 =corrected Tangent. �(If corrected Ext. is required find in same way) Ang. 23* 20'=23.33 10 = 2.3333 = L. C. 2* 1914'= defJor sta. 542. T. P. =sta. 542+72 4° 497', 'i +50- Tan. 1 .18.47 '543 7* 1.92 9* 497, a a +50 B. C. sta. .541+53.53 - I , 11* 40'=. a +" C. 33.33 a,. a. 543 -. L : 86.86 E. C —Sta- 543+86.A6 7 100-53.53-46.47XX(def. f6r.l.ft.-of 10° Cur.')=139.41'= V.Al. . M ' for tai .542.. Def. for 50 ft=2* 30' for a- 10° Curve.- Def.jor 96.86 ft.= V 507' for a 10° Curve. - To, 4:; ry