Wheelchair ramp: gradient, angle and cost
A school in Warangal plans to build a straight wheelchair ramp to a classroom door whose sill is 0.75 m above the ground. The available horizontal space along the wall is 9.6 m, but the design rules require a flat landing of 1.2 m at the top of the ramp and a maximum gradient of 1 in 12 (vertical rise to horizontal run). The ramp will be 1.5 m wide. The contractor proposes to use the whole 9.6 m as sloping run and put the landing beyond it, which the school rejects because a boundary wall is there. Cement work for the ramp surface costs ₹850 per square metre. The school also wants to know the angle the ramp makes with the ground, to the nearest degree, for the safety notice. The physical education teacher suggests that if a straight ramp does not fit, a switchback ramp with two flights and a landing could be used in the same 9.6 m.
1. Does the proposed straight ramp meet the 1-in-12 rule? Determine the minimum sloping run needed and the total horizontal length, and state whether the 9.6 m space is enough.
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The rule is rise/run ≤ 1/12, so run ≥ 12 × rise. For a 0.75 m rise the minimum sloping run is 12 × 0.75 = 9.0 m. A compliant ramp also needs the 1.2 m top landing, so the minimum horizontal length is 9.0 + 1.2 = 10.2 m. Only 9.6 m is available, so the straight ramp as proposed fails. If the contractor tries to fit the landing inside the 9.6 m, the sloping run becomes 9.6 − 1.2 = 8.4 m, giving a gradient of 0.75/8.4 = 1 in 11.2, which is steeper than the permitted 1 in 12. The proposal cannot be approved as drawn.
2. For a compliant straight ramp, find the angle of inclination to the nearest degree and the area of the sloping surface for a 1.5 m width; hence find the cement cost.
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tan θ = rise/run = 0.75/9.0 = 0.0833, so θ = tan⁻¹(0.0833) ≈ 4.76°, i.e. 5° to the nearest degree. The sloping length is the hypotenuse: L = √(9.0² + 0.75²) = √81.5625 ≈ 9.03 m. Surface area = 9.03 × 1.5 ≈ 13.55 m². Cost = 13.55 × ₹850 ≈ ₹11,510 (about ₹11,500). Note that the landing area is extra and is not included in this figure.
3. If the school insists on using the 9.6 m wall and a switchback, explain with numbers how two flights plus a 1.2 m landing could work, and state one practical disadvantage.
Read the worked answer
Split the 0.75 m rise into two flights of 0.375 m. Each flight then needs a run of 12 × 0.375 = 4.5 m, with a 1.2 m landing at the turn. Because the two flights are parallel and stacked, the length used along the wall is about 4.5 + 1.2 = 5.7 m (plus the top landing at the door), which fits within 9.6 m. The problem moves to width: two 1.5 m flights plus a gap need roughly 3.2 m of floor width, and the landing must be at least that wide. Practical disadvantages: it blocks corridor space, requires users to turn 180°, needs handrails on both sides, and gives a longer travel distance than a straight ramp.