Conduit / Tray Fill
Define the conduit (or tray) and the cables with their outer diameter and I compute the fill percentage, the NEC limit (40% for 3+ cables) and how many more cables fit.
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How it is computed · NEC Ch.9 Table 1 fill
1. Area of each round cable:
π × (OD/2)², with the OD (outer diameter) in mm.
2. Duct area: round conduit π × (Ø/2)²;
rectangular tray width × height, in mm².
3. Fill:
fill% = Σ(cable area × qty) / duct area × 100.
4. NEC limit by total number of conductors:
1 cable → 53%, 2 cables → 31%,
3 or more → 40%.
5. Extra cables that fit:
floor((duct area × limit − used area) / one-cable area),
using the largest-diameter cable as the reference (conservative).
6. Real fill also depends on the bend radius,
elbows and pull ease. These figures are for sizing,
not a replacement for the manufacturer's table or the local electrical code.
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How it works
The calculator computes the fill percentage of a rectangular raceway (inner width × height) or a round conduit (inner diameter) from the cables inside it. Each cable is modeled as a circle of area π × (OD/2)² using its outer diameter in mm, and fill is fill% = Σ(area × quantity) / duct area × 100.
The result is compared against the fill limit of the NEC, Chapter 9, Table 1: 53 % with 1 conductor, 31 % with 2 and 40 % with 3 or more. It also calculates how many extra cables fit under that limit, using the largest-diameter cable on the list as reference (a conservative criterion). These are sizing values: the actual pull also depends on bends, bend radii and the manufacturer tables.
Example: how many Cat6 cables fit in a 40 × 25 mm raceway?
- Duct area:
40 × 25 = 1000 mm². Area of one Cat6 with 6 mm OD:π × 3² ≈ 28.3 mm². - With 10 cables:
283 mm²used → 28.3 % fill. - The NEC limit for 3+ conductors is 40 % (400 mm²): it complies with room to spare.
- Extra cables:
floor((400 − 283) / 28.3) = 4more Cat6 before reaching the limit.