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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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