Hall Ramp Gradient and Accessibility
A community hall has a door threshold 0.72 m above the concrete path. The committee plans a straight ramp from the path to the door. The straight distance from the footpath edge to the wall is 8.60 m. Safety rules require a flat landing 1.20 m long at the top of the ramp and a flat area 0.90 m long at the bottom, so these cannot be part of the sloping run. The guideline for an unassisted wheelchair ramp is a maximum gradient of 1:12, meaning 1 unit of rise for every 12 units of horizontal run. The proposed ramp will use the remaining horizontal distance. A builder says the ramp is 'close enough' because the angle is under 5 degrees. The committee wants to know whether the ramp meets the 1:12 guideline and, if not, what the largest door threshold height would be that still meets it with the available run.
1. Calculate the available horizontal run and the proposed gradient. Compare it with the 1:12 guideline.
Read the worked answer
Available horizontal run = total straight distance - top landing - bottom flat area = 8.60 - 1.20 - 0.90 = 6.50 m.
Proposed gradient = rise : run = 0.72 : 6.50. Divide both parts by 0.72: 1 : 6.50/0.72 = 1 : 9.03, to 2 decimal places. As a decimal, gradient = 0.72 / 6.50 = 0.1108.
The guideline maximum is 1:12, which is 0.72 / 8.64 = 0.0833 as a decimal. The proposed ramp is steeper than the guideline because 1:9.03 is a larger rise for the same run than 1:12. It does not meet the guideline.
2. Calculate the angle of the proposed ramp. Evaluate the builder's claim that it is 'close enough' because the angle is under 5 degrees.
Read the worked answer
tan(theta) = rise / run = 0.72 / 6.50 = 0.11077. theta = tan^-1(0.11077) = 6.32 degrees, to 2 decimal places.
The builder's claim is factually wrong even on its own terms: 6.32 degrees is not under 5 degrees. It is also not appropriate to judge accessibility by angle alone when the guideline is written as a maximum gradient. A gradient of 1:12 corresponds to tan(theta) = 1/12, so theta = 4.76 degrees. The proposed 6.32-degree ramp is steeper than the guideline, so 'close enough' is not a safe judgement.
3. Find the largest door threshold height that would meet the 1:12 guideline with the available 6.50 m run. Recommend what the committee should do.
Read the worked answer
For a 1:12 gradient, maximum rise = run / 12 = 6.50 / 12 = 0.5417 m, or about 54.2 cm.
The current threshold is 0.72 m, so it is too high by 0.72 - 0.5417 = 0.178 m, about 17.8 cm.
To meet the guideline with the current 0.72 m rise, the required horizontal run is 12 x 0.72 = 8.64 m. The available run is only 6.50 m, so the ramp is short by 8.64 - 6.50 = 2.14 m.
The committee should not accept the proposed straight ramp. Options are to lower the door threshold by about 18 cm, extend the available run by about 2.14 m (for example by redesigning the path or using a switchback), or seek a professionally designed alternative. If the threshold cannot be lowered, the straight-ramp plan cannot meet 1:12 with the space available.