malachi constant
Senior Member
- Location
- Minneapolis
210.65 Meeting Rooms (B)(2) Floor Outlets states:
A meeting room with any floor dimension that is 12 ft or greater in any direction and that has a floor area of at least 215 SF shall have at least one floor receptacle outlet, or at least one floor outlet to serve receptacles, located at a distance not less than 6 ft from any fixed wall for each 215 SF or fraction thereof.
^^^ note that per 210.65(A) the above does not apply for rooms over 1000 SF.
What the Sam Hill does that mean? Let's use a 20' x 15' meeting room as an example.
* Room is 300 SF, so we are in the 215 < SF < 1000 sweet spot where "floor receptacle outlets" are required.
* I will be installing a floor BOX, quantities as required until code is satisfied.
* Assume no whips to furniture - the receptacles will be mounted within the floor box.
* I need one floor box for sure, since we are over 215 SF.
* It/they will need to be at least six feet away from all walls.
* Do I need two floor boxes? One "outlet" for the first 215 SF, and a second "outlet" for the next 85 SF (the "fraction thereof").
* The rub is, if it is 214 SF you don't need a floor outlet. If it is 215 SF you need one floor outlet. If it is 216 SF you need...two? (One for the first 215 SF and a second for the 216th SF / the "fraction thereof" of the other half of the room?)
I have taken the approach that:
*<215 SF = zero
* exactly 215 SF = 1
* exactly 430 clearly requires two
* there is some gray area as to when you have to go from one to two, as fraction thereof sounds like you should be rounding up.
* Surely "216 SF means two" is not the common sense answer (because 214->0, 215->1, 216->2 is stupid).
* So, arbitrarily, I'll do 1 from 215-250. Then use round numbers: 2 from 250-400; 3 from 401-600; 4 from 601-800; and 5 from 801-1000 SF. (1001 = zero required.)
* But I feel like I could get busted by an inspector for only putting one in a 245 SF room. "Sure, you have one, but that only covers your first 215 SF, what about the second for the "fraction thereof" of the rest of the room?"
Aside from the "how many floor receptacle outlets" are required there is the question of "does a floor box with a quad count as two floor receptacle outlets?" Or do you need two distinct floor boxes to get to two floor receptacle outlets? That doesn't really clarify the math logic of how many are required, but putting multiple duplexes in one floor box at least minimizes the cost of rounding up.
I understand the intent of the code but I don't understand the details. Has anyone else had to put enough thought into this where you can be of assistance here? I looked in NEC 2026 and it doesn't get clarified there. Thanks in advance.
A meeting room with any floor dimension that is 12 ft or greater in any direction and that has a floor area of at least 215 SF shall have at least one floor receptacle outlet, or at least one floor outlet to serve receptacles, located at a distance not less than 6 ft from any fixed wall for each 215 SF or fraction thereof.
^^^ note that per 210.65(A) the above does not apply for rooms over 1000 SF.
What the Sam Hill does that mean? Let's use a 20' x 15' meeting room as an example.
* Room is 300 SF, so we are in the 215 < SF < 1000 sweet spot where "floor receptacle outlets" are required.
* I will be installing a floor BOX, quantities as required until code is satisfied.
* Assume no whips to furniture - the receptacles will be mounted within the floor box.
* I need one floor box for sure, since we are over 215 SF.
* It/they will need to be at least six feet away from all walls.
* Do I need two floor boxes? One "outlet" for the first 215 SF, and a second "outlet" for the next 85 SF (the "fraction thereof").
* The rub is, if it is 214 SF you don't need a floor outlet. If it is 215 SF you need one floor outlet. If it is 216 SF you need...two? (One for the first 215 SF and a second for the 216th SF / the "fraction thereof" of the other half of the room?)
I have taken the approach that:
*<215 SF = zero
* exactly 215 SF = 1
* exactly 430 clearly requires two
* there is some gray area as to when you have to go from one to two, as fraction thereof sounds like you should be rounding up.
* Surely "216 SF means two" is not the common sense answer (because 214->0, 215->1, 216->2 is stupid).
* So, arbitrarily, I'll do 1 from 215-250. Then use round numbers: 2 from 250-400; 3 from 401-600; 4 from 601-800; and 5 from 801-1000 SF. (1001 = zero required.)
* But I feel like I could get busted by an inspector for only putting one in a 245 SF room. "Sure, you have one, but that only covers your first 215 SF, what about the second for the "fraction thereof" of the rest of the room?"
Aside from the "how many floor receptacle outlets" are required there is the question of "does a floor box with a quad count as two floor receptacle outlets?" Or do you need two distinct floor boxes to get to two floor receptacle outlets? That doesn't really clarify the math logic of how many are required, but putting multiple duplexes in one floor box at least minimizes the cost of rounding up.
I understand the intent of the code but I don't understand the details. Has anyone else had to put enough thought into this where you can be of assistance here? I looked in NEC 2026 and it doesn't get clarified there. Thanks in advance.