Question

Three simple curve are connected to each other such that the 1st and the 2nd form a compound curve while the second and the 3rd form a reversed curve. The distance between the point of curvature and the point of tangency of the compound curve which is also the point of reversed curvature of the reversed curve is 485 meters. If I 450, Rı = 190.99 m, /2 = 60°, I3 20°, R3 = 163.7 m and the stationing of PC is 3+720. Determine the stationing of the PCC Determine the stationing of the PRC Determine the stationing of the PT of the reversed curve 3.

Answer 1

we will dnaw the outlay of problem At first Solution PCE Poimt of curvatue Point ot tomgena o PRC = tue compmd cuve amd also tne poimt ot nerersed cunvadwre Pec LI LC 2 I2/2 R3=163.7 485 PRe 22 L3 LRE 20 PT R-190.99m

Stohimin of Pc Now, As fo curve CuNe length of Stea. Pc t Sta.Pec= Sta. PC t Li RI 190.099m Now, I- 4C T1XPII 150m. LI there fore, Sta Pec Sta. pr tL S0 +150 3720 +150: 3870 Sta. 3+8 70 So, sta. Pce- As to ouhiomimg we will comsider f PRC toiangle the Pc -Pet- PRe 2 o Lz LCI 127.50 485m Long Chond = 2RSn I/, we know 4here tosre Le 2R,simT/, () 2X190.99 x Sim 146.18m.

Sine ne ot iongle, 485 Lei L Sim s6 sim 日 Simg LL2 Now, 488 485 Lei 0gr SIM127.5 Sm 129.5 485 o, Le 146.18 85 Sim127-5 ±9.82415 Le=381.98 m 13.83 G+ ダ+8=180 2nd For Cunve Ll2= 2R2 381.18 03 122- 2xsin 60 R2 381.98m TXR2 TX 381-の8×めo L2= ナhere fore 7 180 180 400m. L2 S0 Sta. PR= Stea. Pcc + Sta +400 11 4270 3 870 +400 SU Sta, PR Sta. 4+2to

As os steoutioning ot Now PT Sta. PRC Lg Sta Pt Sta Cur ne 3, R3 163.7 m T320 TA R3AT3 these fore 17 18 0 HX 163.7 x 20 17 1 & o 7.-142m there tore Ste PT- 4240ナ57、142 4327142 So, Sta. Pt -Sta. 4 + 32t.(42