Checking a Ramlila Ground Bandobast Against the Numbers
A district Ramlila committee expects 24,000 people at a ground measuring 9,000 square metres. The accepted planning norm for a standing religious or cultural crowd in this district is 2.5 persons per square metre. The ground has five entry gates, each with one metal detector and two frisking staff; the measured screening rate is 25 persons per minute per gate. Roughly 70 per cent of the crowd arrives between 4:30 p.m. and 6:00 p.m., when the performance begins. The whole ground is served by one 4 metre exit lane, and traders' stalls line the inside of that lane. The committee's written plan reads: “पाँच गेट पर्याप्त हैं क्योंकि मैदान बड़ा है।” The additional superintendent of police must clear or amend the bandobast plan by 2 p.m.
1. Work out the screening arithmetic for the 4:30–6:00 p.m. window. What happens at 6:00 p.m., and how many gates are actually needed?
Read the worked answer
Arrivals in the window: 70 per cent of 24,000 = 16,800 persons in 90 minutes, i.e. 16,800 ÷ 90 = 187 persons per minute.
Installed capacity: 5 gates × 25 per minute = 125 per minute. In the 90 minutes available the gates clear 125 × 90 = 11,250 persons.
At 6:00 p.m., therefore, 16,800 − 11,250 = 5,550 people are still outside in the queue, on the road, with no screening and no cover. Clearing that backlog at 125 per minute takes a further 44 minutes, so the last of the 4:30–6:00 arrivals is inside at about 6:44 p.m. - that is 16,800 ÷ 125 = 134 minutes from 4:30.
Gates needed: 187 ÷ 25 = 7.5, so 8 gates as the bare minimum, and 9 if a 10 per cent reserve is kept for a detector failure or a slow bag check. The committee's sentence is wrong in its reasoning, not just its number: a big ground does not screen anybody. Gate capacity, not area, sets the queue.
2. Is the ground itself adequate for 24,000 at the stated norm? What is the minimum correction in area or in numbers?
Read the worked answer
Compare holding capacity, not the crowd's expectation.
Permissible at the norm: 9,000 m² × 2.5 persons/m² = 22,500 persons.
Expected: 24,000. The ground is over by 1,500 persons, i.e. the crowd would stand at roughly 24,000 ÷ 9,000 = 2.67 persons/m², about 7 per cent above the norm - which sounds small on paper and is not small in a crowd, because density above the comfortable figure degrades movement, and movement is what fails first.
Minimum correction, either way: add 1,500 ÷ 2.5 = 600 m² of usable ground, or cap admission at 22,500 and hold the balance outside with a screen and an announcement.
The number that should worry the additional SP more than either is the single 4 metre exit lane, lined with stalls. Screening over capacity produces a slow queue, which is an inconvenience; a single obstructed lane produces a crush, which is a disaster. The plan must name a second exit and clear the stalls from the lane before anything else is approved.
3. The committee offers one fix: three more gates from a reserve company. Is that the single most effective measure? Justify your choice.
Read the worked answer
Three more gates take screening to 8 × 25 = 200 per minute, which clears the 187 per minute arrival rate and empties the queue on time. It solves the queue. It does not solve the ground.
After the fix the same 24,000 people stand inside at 2.67 per square metre, and they must all leave through one 4 metre lane with stalls along it. So the binding constraint has simply moved: throughput is a 45-minute problem that gets worse gradually, while density plus single-exit is an instantaneous problem that gets worse suddenly. In crowd safety, the instantaneous problem governs.
The most effective single measure is therefore to hold admission at 22,500 and open a second exit with the lane cleared - with the three gates accepted as well, since they are free capacity and reduce the outside queue. If the committee will only accept one, take the cap and the second exit. Approving the gates alone, because they are visible and easy to count, is exactly how a plan passes review and still fails on the night.