Smoke detectors

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JDB3

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I am drawing a blank right now. What is the general requirement for how far smoke detectors can be apart or for the maximum square footage (room size) you can go before requiring placement of a smoke detector?

Thanks
 
JDB3 said:
I am drawing a blank right now. What is the general requirement for how far smoke detectors can be apart or for the maximum square footage (room size) you can go before requiring placement of a smoke detector?

Thanks
Click to expand...

Thirty foot spacing or a 900 sq ft area. Technically it's a circle with a 21 foot radius from the center of the detector.
 
gadfly56 said:
Thirty foot spacing or a 900 sq ft area. Technically it's a circle with a 21 foot radius from the center of the detector.
Click to expand...

Technically it isn't. Technically it is a circle with a 30 foot diameter, and a 15 foot radius (for a normal height ceiling). No point may be more than .7 times the diameter of the coverage, which allows on detector to cover a 30 x 30 room or longer spacing in a narrow corridor.
 
Strathead said:
Technically it isn't. Technically it is a circle with a 30 foot diameter, and a 15 foot radius (for a normal height ceiling). No point may be more than .7 times the diameter of the coverage, which allows on detector to cover a 30 x 30 room or longer spacing in a narrow corridor.
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Technically, it is. 30/0.7 = 42 (plus or minus). A square with 30 foot sides is circumscribed by a circle with a radius of 21 feet (diameter 42 feet).

From NFPA 72-2013:
17.7.3.2.3.1* In the absence of specific performance-based de-
sign criteria, one of the following requirements shall apply:

(1) The distance between smoke detectors shall not exceed a
nominal spacing of 30 ft (9.1 m) and there shall be detectors
within a distance of one-half the nominal spacing,
measured at right angles from all walls or partitions extending
upward to within the top 15 percent of the ceiling
height.​
(2)*All points on the ceiling shall have a detector within a
distance equal to or less than 0.7 times the nominal 30 ft
(9.1 m) spacing (0.7S).​

See also figure A17.6.3.1.1(h) clearly showing the use of the 21 foot radius for determining coverage for an irregular space.
 
gadfly56 said:
Technically, it is. 30/0.7 = 42 (plus or minus). A square with 30 foot sides is circumscribed by a circle with a radius of 21 feet (diameter 42 feet).

From NFPA 72-2013:
17.7.3.2.3.1* In the absence of specific performance-based de-
sign criteria, one of the following requirements shall apply:
(1) The distance between smoke detectors shall not exceed a
nominal spacing of 30 ft (9.1 m) and there shall be detectors
within a distance of one-half the nominal spacing,
measured at right angles from all walls or partitions extending
upward to within the top 15 percent of the ceiling
height.​
(2)*All points on the ceiling shall have a detector within a
distance equal to or less than 0.7 times the nominal 30 ft
(9.1 m) spacing (0.7S).​

See also figure A17.6.3.1.1(h) clearly showing the use of the 21 foot radius for determining coverage for an irregular space.
Click to expand...

I stand by my statement. From the perspective of the code you quoted 30/.7 has no place. I expect that you know exactly what you mean, but the way you are presenting is, a room that was 30 feet wide by 42 feet long would only require one smoke detector. That is not true. The cited code is 30*.7= 21 as the maximum distance for oblique angles. In no case can you draw a 42 foot diameter circle and state that one detector covers the ENTIRE area within it.
 
Strathead said:
In no case can you draw a 42 foot diameter circle and state that one detector covers the ENTIRE area within it.
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The code quoted says "one of the following shall apply". Item (2) says you can draw a 42 foot diameter circle around each detector, and as long as all those circles cover the entire ceiling, you are good to go.

Cheers, Wayne

P.S. I can't think of a situation where (2) is more restrictive than (1), so it seems like they could just drop (1) and use (2). Of course, people like to think in terms of rectangles, and (1) is geared towards that.
 
wwhitney said:
The code quoted says "one of the following shall apply". Item (2) says you can draw a 42 foot diameter circle around each detector, and as long as all those circles cover the entire ceiling, you are good to go.

Cheers, Wayne

P.S. I can't think of a situation where (2) is more restrictive than (1), so it seems like they could just drop (1) and use (2). Of course, people like to think in terms of rectangles, and (1) is geared towards that.
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All points need to be within .7D, yes, but this part of (1) requires at least one point on each wall to be within .5D.
there shall be detectors
within a distance of one-half the nominal spacing,
measured at right angles from all walls or partitions
Click to expand...

So the 42 foot long room needs more than one detector unless there are no end walls on the room (e.g. carport).

OOPS. Missed the "one or the other" part.
Still, I suspect my interpretation is what was intended, the code just did not word it correctly for what the CMP wanted.

However, that provision does seem to prevent putting the one detector at one end of a 30 foot room when applying (1).
 
Since it says 42 foot diameter circle, rather than 21 foot radius, does this mean the smoke detector can be anywhere within that 42 foot circle?
 
Strathead said:
I stand by my statement. From the perspective of the code you quoted 30/.7 has no place. I expect that you know exactly what you mean, but the way you are presenting is, a room that was 30 feet wide by 42 feet long would only require one smoke detector. That is not true. The cited code is 30*.7= 21 as the maximum distance for oblique angles. In no case can you draw a 42 foot diameter circle and state that one detector covers the ENTIRE area within it.
Click to expand...

Yes, I should have multiplied. If I have a smoke detector, every point within 42 feet of the detector is covered. That would cover an area of 1,385 square feet. Conversely, a square 30 feet on a side (900 square feet) will fit inside this circle. The prescriptive spacing of 30 feet is intended to accommodate typical human construction which involves square and rectangular spaces (on the whole).

A room that is 30 feet wide by 42 feet long has portions that DO NOT fit into the circle. In fact, no space that is greater than 0 by 42 feet is going to fit inside that circle. However, if you look at the 2013 edition, you'll see that you could use 41 foot spacing in a corridor that's not more than 10 feet wide. See figure A17.6.3.1.1(g).
 
GoldDigger said:
I do not see any reference to a 42 foot diameter. Where do you see that?
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The prescriptive spacing is 30 center to center and 15 feet from a wall. Or you can use 0.7 times the prescriptive spacing (30*.07=21) and satisfy the criteria that all points on a smooth ceiling have to be no farther than this distance from the smoke detector. That then is the radius of the circle (21 feet) with the smoke detector at the center. Or a diameter of 42 feet.
 
gadfly56 said:
Yes, I should have multiplied. If I have a smoke detector, every point within 42 feet of the detector is covered.
Click to expand...

That is what we get for typing quickly. I meant to say a 42 foot circle it a square room. If you had a 42 foot diameter round room then you could use one detector. When I was typing I was not envisioning a round room.
 
gadfly56 said:
The prescriptive spacing is 30 center to center and 15 feet from a wall. Or you can use 0.7 times the prescriptive spacing (30*.07=21) and satisfy the criteria that all points on a smooth ceiling have to be no farther than this distance from the smoke detector. That then is the radius of the circle (21 feet) with the smoke detector at the center. Or a diameter of 42 feet.
Click to expand...


then we can throw in slopes.;)
 
gadfly56 said:
However, if you look at the 2013 edition, you'll see that you could use 41 foot spacing in a corridor that's not more than 10 feet wide. See figure A17.6.3.1.1(g).
Click to expand...
I don't have access to that document, but a 41' x 10' rectangle doesn't quite fit inside a 42' diameter circle. The diagonal of the rectangle is sqrt(41^2 + 10^2) = 42.2'. A 40' x 10' rectangle fits, however, as does a 41' x 9' rectangle.

Cheers, Wayne
 
Dennis Alwon said:
My understanding, and this may vary depending on manufacturer, is that a 31 foot hallway needs 2 sd
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It depends on the width of the hallway. Using figure A17.6.3.1.1(g) from NFPA 72-2013, if your hallway was 31 feet long it could be covered by a single detector as long as it was under 28'-4" in the other dimension. I don't believe this depends on the manufacturer because no manufacturer uses a listed spacing for smoke detectors. They all use the prescriptive spacing in NFPA 72 (30 feet).
 
Dennis Alwon said:
My understanding, and this may vary depending on manufacturer, is that a 31 foot hallway needs 2 sd
Click to expand...

No Dennis, that isn't correct. You would have to use trig to figure it out with math, but in a five foot wide hallway the corners perpendicular to the tip of the circle are not much further from a smoke detector in the center of the hallway than the tip. As the above code cites, any point on the ceiling that is less than 21 feet from the detector is an acceptable placement.
 
Strathead said:
No Dennis, that isn't correct. You would have to use trig to figure it out with math, but in a five foot wide hallway the corners perpendicular to the tip of the circle are not much further from a smoke detector in the center of the hallway than the tip. As the above code cites, any point on the ceiling that is less than 21 feet from the detector is an acceptable placement.
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Here is the instructions for BRK. I don't see where all this math stuff is necessary

More specifically, install Smoke Alarms:
• On every level of your home, including finished attics and basements.
• Inside every bedroom, especially if people sleep with the door partly or completely
closed.
• In the hall near every sleeping area. If your home has multiple sleeping areas, install a
unit in each. If a hall is more than 40 feet (12 meters) long, install a unit at each end.
• At the top of the first-to-second floor stairway, and at the bottom of the basement stairway.
Click to expand...
 
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