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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.

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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:

Alloy SeriesDensity (lbs/in³)Density (g/cm³)
1xxx pure aluminum0.09752.698
3xxx (common sheet/coil)0.09752.700
5xxx (marine)0.0975–0.09802.700–2.710
6xxx (structural)0.0975–0.09802.700–2.710
2xxx (aerospace)0.10082.790
7xxx (high strength)0.10222.830

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)

Gauge (Ga)Thickness (inches)Thickness (mm)
280.0126″0.320
260.0159″0.404
240.0201″0.511
220.0253″0.643
200.0320″0.813
180.0403″1.024
160.0508″1.290
140.0641″1.628
120.0808″2.052
100.1019″2.588

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