How to Calculate Aluminum Coil Weight, Length, and Core Dimensions
Learn how to calculate aluminum coil weight from dimensions, how to find coil length from weight and gauge, and how to estimate running length for any aluminum coil stock.
Related Calculators
Aluminum Coil Dimensions
An aluminum coil is characterized by: - OD: Outer diameter (outside of the coil) - ID: Inner diameter (core/mandrel diameter) - W: Width of the strip - t: Strip thickness (gauge)
Coil Weight Formula
Weight = π × (OD² − ID²) / 4 × W × density
In consistent units (all in inches, density in lbs/in³): Weight (lbs) = π/4 × (OD² − ID²) × W × ρ
Aluminum density by alloy:
Most standard coil stock (3003, 3105, 5052) uses ρ = 0.0975 lbs/in³.
Worked Example: Coil Weight Calculation
Given: OD = 48 in, ID = 20 in, Width = 24 in, 3105 aluminum
Weight = π/4 × (48² − 20²) × 24 × 0.0975 = 0.7854 × (2,304 − 400) × 24 × 0.0975 = 0.7854 × 1,904 × 24 × 0.0975 = 0.7854 × 1,904 × 2.340 = 3,500 lbs (approximate)
Coil Length Formula
Length = Weight / (W × t × density)
Or from geometry: Length = π × (OD² − ID²) / (4 × t)
Example: From the coil above (t = 0.019 in) Length = π × (48² − 20²) / (4 × 0.019) = π × 1,904 / 0.076 = 5,982 / 0.076 = 78,700 inches = 6,558 feet
Gauge to Thickness Conversion (Aluminum)
Note: Aluminum gauge differs from steel gauge — always confirm which standard applies.
Pounds Per Thousand Square Feet (PSFT)
A common coil industry metric: PSFT = Thickness (in) × 0.0975 × 1,000 × 144 = Thickness × 14,040
For 0.019-inch (18 ga) aluminum: PSFT = 0.019 × 14,040 = 266.8 lbs/MSF
This means 1,000 square feet of 18-gauge aluminum weighs ~267 lbs.
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Frequently Asked Questions
- How do I calculate the weight of an aluminum coil?
- Weight = π/4 × (OD² − ID²) × W × density. For 3003 or 3105 aluminum, density = 0.0975 lbs/in³. Example: OD = 36 in, ID = 16 in, W = 12 in: Weight = 0.7854 × (1,296 − 256) × 12 × 0.0975 = 0.7854 × 1,040 × 12 × 0.0975 = 0.7854 × 1,216.8 = 956 lbs approx.
- How do I find the running length of an aluminum coil?
- Length = π × (OD² − ID²) / (4 × thickness). Or: Length = Weight / (Width × Thickness × Density). Example: 500 lb coil, 12-inch wide, 0.020-inch thick: Length = 500 / (12 × 0.020 × 0.0975) = 500 / 0.0234 = 21,368 inches = 1,781 feet.
- What is the density of aluminum in lbs per cubic inch?
- Most common aluminum alloys (3003, 3105, 5052, 6061) have a density of approximately 0.0975 lbs/in³ (2.70 g/cm³). Aerospace alloys (2024, 7075) are slightly denser at 0.101–0.103 lbs/in³. For standard sheet metal and coil work, 0.0975 lbs/in³ is the industry standard value.
- How many square feet does an aluminum coil cover?
- Coverage = Length × Width. If a coil has 2,000 feet of running length at 24-inch width: Coverage = 2,000 ft × 2 ft = 4,000 sqft. To find length from weight: Length = Weight / (Width in inches × Thickness in inches × 0.0975 × 12). At 24-inch width, 0.019-inch gauge: 1,000 lbs covers approximately 22,700 sqft.
Last updated 7/28/2026