Wheelchair ramp gradient and drain clearance
A community centre must install a wheelchair ramp from pavement point A(0,0) to a door platform at B(10.8,1.25), where coordinates are metres east and north. The site plan shows a drain cover centred at D(6.0,0.35) with a 0.30 m radius; the ramp must not pass within 0.20 m of the drain edge. The national guideline says a ramp's gradient must be no steeper than 1:12, and the ramp must have a level landing of at least 1.2 m at the top. The builder proposes a straight ramp from A to B. The centre manager is worried because the ramp looks steep and because the drain is close to the proposed centre-line. The available straight run from the building is 11.5 m, but the door platform itself occupies the last 1.2 m of that run, so the sloping part can use at most 10.3 m horizontally. The platform height is fixed at 1.25 m.
1. Calculate the gradient and angle of the proposed straight ramp AB. Decide whether it satisfies the 1:12 guideline. If not, calculate the maximum rise possible with the available 10.3 m sloping run at that guideline.
Read the worked answer
Rise=1.25 m, horizontal run=10.8 m. Gradient =1.25/10.8=0.1157, i.e. 1:8.64. The guideline maximum is 1/12=0.0833, i.e. 1:12. Since 0.1157>0.0833, the proposed ramp is too steep.
Angle: tan θ=0.1157, so θ≈6.6°, compared with the guideline angle tan⁻¹(1/12)≈4.8°.
To rise 1.25 m at 1:12 requires run=1.25×12=15.0 m. With only 10.3 m of sloping run, the maximum rise at 1:12 is 10.3/12=0.858 m, a shortfall of 1.25-0.858=0.392 m. A single straight ramp cannot meet both the height and the guideline on this site.
2. Determine the perpendicular distance from the drain centre D to the centre-line AB, and decide whether the 0.20 m edge clearance rule is met. State what the result means for the proposed route.
Read the worked answer
Line AB has direction (10.8,1.25), so a normal is (1.25,-10.8). Equation through A: 1.25x-10.8y=0.
For D(6,0.35): numerator |1.25(6)-10.8(0.35)|=|7.5-3.78|=3.72. Denominator √(1.25²+10.8²)=√118.2025≈10.872. Perpendicular distance ≈3.72/10.872=0.342 m.
The drain radius is 0.30 m, so clearance from the drain edge to the centre-line is 0.342-0.30=0.042 m. The rule requires 0.20 m, so the proposal fails by about 0.158 m. The centre-line must be shifted away from the drain or the drain relocated; simply keeping the straight AB route is not acceptable.