Skip to main content

Vertical curve radius 5000 m, BVC and highest point

Short answer

Summit curve length L = N R = 0.05 × 5000 = 250 m. BVC RL = 310.83 m, EVC RL = 311.46 m. Highest point is at 137.5 m from BVC with RL = 312.72 m.

An ascending gradient of 2.75% meets with descending gradient of 2.25%. The radius of curve is 5000 m. If the reduced level of the curve at a distance of 60 m from BVC is 312.12 m, find the reduced level of BVC, EVC and highest point of the curve.

Ascending 2.75% meeting descending 2.25% gives a summit curve. N=0.0275+0.0225=0.05N=0.0275+0.0225=0.05.

Length of curve

L=NR=0.05(5000)=250 mL=NR=0.05(5000)=250\ \text{m}

RL of BVC

Parabolic summit curve, xx measured from BVC: y=RLBVC+g1x−x22Ry=RL_{BVC}+g_1x-\dfrac{x^2}{2R}. At x=60x=60 m, y=312.12y=312.12:

312.12=RLBVC+0.0275(60)−6022(5000)=RLBVC+1.65−0.36312.12=RL_{BVC}+0.0275(60)-\frac{60^2}{2(5000)}=RL_{BVC}+1.65-0.36 RLBVC=310.830 mRL_{BVC}=310.830\ \text{m}

RL of EVC

RLEVC=RLBVC+g1L−L22R=310.830+0.0275(250)−250210000=311.455 mRL_{EVC}=RL_{BVC}+g_1L-\frac{L^2}{2R}=310.830+0.0275(250)-\frac{250^2}{10000}=311.455\ \text{m}

Check through the PVI: RLPVI=310.830+0.0275(125)=314.268RL_{PVI}=310.830+0.0275(125)=314.268 m, and RLPVI−0.0225(125)=311.455RL_{PVI}-0.0225(125)=311.455 m. It agrees.

Highest point

The highest point is where the slope is zero: g1−xR=0g_1-\dfrac{x}{R}=0.

x=g1R=0.0275(5000)=137.5 m from BVCx=g_1R=0.0275(5000)=137.5\ \text{m from BVC} RLmax=RLBVC+g12R2=310.830+0.02752(5000)2=312.721 mRL_{max}=RL_{BVC}+\frac{g_1^2R}{2}=310.830+\frac{0.0275^2(5000)}{2}=312.721\ \text{m}

Answer: L = 250 m; RL of BVC = 310.83 m; RL of EVC = 311.46 m; highest point at 137.5 m from BVC with RL = 312.72 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Transportation Engineering I chapter 3 questions →