For smooth ceilings, all points on the ceiling shall have a detector within a distance equal to _____ times the selected spacing.

For smooth ceilings, all points on the ceiling shall have a detector within a distance equal to _____ times the selected spacing.

• A. 0.7 times the spacing
• B. 1.0 times the spacing
• C. 0.5 times the spacing
• D. 0.8 times the spacing

Answer: (A)

On smooth ceilings, detectors placed in a regular grid create circular coverage areas. The point farthest from any detector sits at the center of the square formed by four detectors. The distance from that center to a detector is S/√2, where S is the spacing between detectors, which is about 0.707 times the spacing. Designers round this to 0.7, ensuring every point on the ceiling lies within 0.7 times the spacing from a detector. For example, with a 20-foot spacing, every point would be within about 14 feet of a detector. The other fractions don’t match this geometric coverage: 1.0 would leave gaps, 0.5 would be more conservative than needed, and 0.8 would exceed the standard coverage allowance.

On smooth ceilings, detectors placed in a regular grid create circular coverage areas. The point farthest from any detector sits at the center of the square formed by four detectors. The distance from that center to a detector is S/√2, where S is the spacing between detectors, which is about 0.707 times the spacing. Designers round this to 0.7, ensuring every point on the ceiling lies within 0.7 times the spacing from a detector. For example, with a 20-foot spacing, every point would be within about 14 feet of a detector. The other fractions don’t match this geometric coverage: 1.0 would leave gaps, 0.5 would be more conservative than needed, and 0.8 would exceed the standard coverage allowance.