MIL-HDBK-1012/3

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EXAMPLE TWO (Continued)

Example: If the width of the distribution zone is 100 ft. and

the quantity of feeder ducts required is 13 then:

Length of feeder duct = (100 ft.) x 13

4

= 25 ft. x 13

= 325 ft.

(2) For a telecommunications closet located toward

either edge of the distribution zone, divide the width of the

distribution zone by 2 and multiply by the number of feeder ducts

required:

Total length of feeder duct = (width) x no. of ducts

=

2

Example: If the width of the distribution zone is 100 ft. and

the quantity of feeder ducts required is 13 then:

Length of feeder duct = (100 ft.) x 13

2

= 50 ft. x 13

= 650 ft.

d) In accordance with NFPA 70, Articles 300 and 354

and EIA/TIA 569, par. 4, in-floor duct systems shall be

installed:

(1) Under 21 mm to 25 mm (3/4 in. to 1 in.) of

concrete cover depending on the size of the raceway or flush with

the floor and covered with floor covering;

(2)

With trench cover, where applicable, flush

with the floor;

(3) Maintaining a maximum 40 percent fill ratio;

BICSI also recommends an additional 15 percent reduction in fill

for each 90-degree bend. Consequently, a 40 percent fill with

two 90-degree bends would be reduced to a 28 percent fill.

(4) Laid in straight lines with feeder ducts or

trenches perpendicular to the distribution ducts;

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