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Vertical curve chainage and elevations, R 8000 m

Short answer

Length L = N×R = 0.07×8000 = 560 m. BVC chainage 0+945, RL 865.12 m. EVC chainage 1+505, RL 866.52 m. Highest point at 1+245, RL 870.75 m. At 90 m left of PVI, RL = 869.99 m.

An ascending gradient of 3.75% meets with descending gradient of 3.25%. Calculate the chainage and elevation of beginning of the curve, end of the curve, highest point of the curve and 90 m left from the point of vertical intersection if the chainage and elevation of PVI are 1+225.00 and 875.62 m respectively. The radius of curve provided is 8000 m.

An ascending 3.75% grade followed by a descending 3.25% grade gives a summit curve; N=0.0375+0.0325=0.07N=0.0375+0.0325=0.07.

Length and chainages

L=NR=0.07(8000)=560 m,L2=280 mL=NR=0.07(8000)=560\ \text{m},\qquad \frac{L}{2}=280\ \text{m}

PVI chainage = 1+225.00 = 1225 m.

  • Beginning of curve (BVC) = 1225 − 280 = 945 m (0+945)
  • End of curve (EVC) = 1225 + 280 = 1505 m (1+505)

Levels of BVC and EVC

RLBVC=875.62−0.0375(280)=865.120 m,RLEVC=875.62−0.0325(280)=866.520 mRL_{BVC}=875.62-0.0375(280)=865.120\ \text{m},\qquad RL_{EVC}=875.62-0.0325(280)=866.520\ \text{m}

Highest point

x=g1R=0.0375(8000)=300 m from BVCx=g_1R=0.0375(8000)=300\ \text{m from BVC}

Chainage =945+300=1245=945+300=\mathbf{1245} m (1+245).

RLmax=RLBVC+g12R2=865.120+0.03752(8000)2=870.745 mRL_{max}=RL_{BVC}+\frac{g_1^2R}{2}=865.120+\frac{0.0375^2(8000)}{2}=870.745\ \text{m}

90 m left of PVI

Chainage =1225−90=1135=1225-90=1135 m, distance from BVC x=280−90=190x=280-90=190 m:

y=RLBVC+g1x−x22R=865.120+0.0375(190)−19022(8000)=869.989 my=RL_{BVC}+g_1x-\frac{x^2}{2R}=865.120+0.0375(190)-\frac{190^2}{2(8000)}=869.989\ \text{m}

Answer: BVC at 0+945 (RL 865.12 m); EVC at 1+505 (RL 866.52 m); highest point at 1+245 (RL 870.75 m); 90 m left of PVI RL = 869.99 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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