On a smooth ceiling with a listed spacing of 50′, what is the maximum distance the detector can be from any point?

On a smooth ceiling with a listed spacing of 50′, what is the maximum distance the detector can be from any point?

• A. 20 ft
• B. 28 ft
• C. 35 ft
• D. 42 ft

Answer: (C)

On a ceiling laid out with detectors spaced at 50 ft, the worst-case point is at the center of the square formed by four neighboring detectors. The distance from that center to any one detector is the half-diagonal of the square, which is (spacing/2) × sqrt(2). With a 50 ft spacing, half of that is 25 ft, and 25 × sqrt(2) ≈ 25 × 1.414 ≈ 35.4 ft. So the maximum distance from any point to the nearest detector is about 35 ft, which rounds to 35 ft.

On a ceiling laid out with detectors spaced at 50 ft, the worst-case point is at the center of the square formed by four neighboring detectors. The distance from that center to any one detector is the half-diagonal of the square, which is (spacing/2) × sqrt(2). With a 50 ft spacing, half of that is 25 ft, and 25 × sqrt(2) ≈ 25 × 1.414 ≈ 35.4 ft. So the maximum distance from any point to the nearest detector is about 35 ft, which rounds to 35 ft.