- Asked 5/26
- 2077 Chaitra · 6 marks
Vertical curve -3.5% and +0.5%, IP 1300 m
Short answer
Equal-tangent vertical curve length 200 m. BVC at chainage 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m). Lowest point at chainage 2675 m, elevation 1300.438 m. Curve passes through 1301 m at the IP chainage.
A grade of (-)3.5% meets with another grade of 0.5%. The elevation and chainage of point of intersection are 1300 m and 2600 m respectively. Compute the suitable equal tangent vertical curve for full stations elevation assuming that vertical curve should pass through a point of elevation 2601 m at a chainage 1300 m. Take peg interval = 30 m.
In the question, the elevation and chainage of the given curve point appear to be interchanged (a chainage of 1300 m would be far away from the IP). The data are therefore taken as: IP at chainage 2600 m with RL 1300 m, and the curve passes through RL 1301 m at the IP chainage (2600 m).
An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate of the curve.
Step 1: Length of the curve
The curve is a sag curve, so the curve lies above the IP:
For an equal-tangent parabola, (grades in %), hence
Step 2: BVC, EVC and their levels
The curve is a sag (valley) curve with equal tangent lengths m.
Step 3: RL of curve points
Using the tangent-offset form of the parabola ( measured from BVC, grades in %):
with , and m, the offset is . The pegs are taken at multiples of 30 m.
| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 2500 | 0 | 1303.500 | 0.000 | 1303.500 |
| 2520 | 20 | 1302.800 | +0.040 | 1302.840 |
| 2550 | 50 | 1301.750 | +0.250 | 1302.000 |
| 2580 | 80 | 1300.700 | +0.640 | 1301.340 |
| 2610 | 110 | 1299.650 | +1.210 | 1300.860 |
| 2640 | 140 | 1298.600 | +1.960 | 1300.560 |
| 2670 | 170 | 1297.550 | +2.890 | 1300.440 |
| 2700 | 200 | 1296.500 | +4.000 | 1300.500 |
Check: the last curve RL (1300.5 m) equals the RL of EVC (1300.5 m), and at the IP chainage the curve lies 1 m above the IP, at RL 1301 m.
Step 4: Lowest point
At the lowest point the grade is zero, so m from BVC.
- Chainage = 2500 + 175 = 2675 m
- RL = 1303.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 1300.438 m
Answer: m, BVC at 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m), lowest point at chainage 2675 m (RL 1300.438 m); curve levels as tabulated.
✓ Answer checked by Claude Opus 5.5 (high effort)
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