Sand & Gravel Calculator

Sand & Gravel is evaluated from Material, Length and Width. The calculation reports Cubic Feet, Cubic Yards and Estimated Tons.

Results

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About the Sand & Gravel Calculator

Sand & Gravel is treated here as a quantitative relation between Material, Length, Width and Depth and Cubic Feet, Cubic Yards, Estimated Tons and With 10% Waste.

The calculator uses a multi formula configuration. Each reported value is read as a direct evaluation of the stored rules with the declared field formats and units.

Formula basis:
Volume (ft^3) = length x width x (depth / 12)
Volume (yd^3) = ft^3 / 27
Weight (tons) = yd^3 x material density (tons/yd^3)
Add 10% for compaction, settling, and waste.
Note: excavated material weighs ~20 - 30% less than compacted material. When ordering fill, calculate compacted volume needed and order 20% extra to account for settling.

Interpret the outputs in the order shown by the result fields. Optional inputs affect only the outputs that depend on those variables.

Formula & How It Works

The calculation applies the following relations exactly as recorded in the metadata:

Volume (ft^3) = length x width x (depth / 12)
Volume (yd^3) = ft^3 / 27
Weight (tons) = yd^3 x material density (tons/yd^3)
Add 10% for compaction, settling, and waste.
Note: excavated material weighs ~20 - 30% less than compacted material. When ordering fill, calculate compacted volume needed and order 20% extra to account for settling.

Each output field is produced by substituting the supplied inputs into the relevant relation and then applying the declared rounding or text format.

Worked Examples

Example 1: Gravel driveway: 40 ft × 12 ft × 4 in base

Inputs

material: Crushed Stone #411 length: 40 width: 12 depth: 4

Cubic Feet: 160 ft^3. Cubic Yards: 5.93 yd^3. Estimated Tons: 8.3 tons. With 10% Waste: 6.52 yd^3

With Material = Crushed Stone #411, Length = 40, Width = 12 and Depth = 4 as the stated inputs, the result is Cubic Feet = 160 ft^3, Cubic Yards = 5.93 yd^3 and Estimated Tons = 8.3 tons. Each value corresponds to the declared output fields.

Example 2: Sandbox: 8 ft × 8 ft × 6 in sand

Inputs

material: Mason Sand length: 8 width: 8 depth: 6

Cubic Feet: 32 ft^3. Cubic Yards: 1.19 yd^3. Estimated Tons: 1.6 tons. With 10% Waste: 1.3 yd^3

With Material = Mason Sand, Length = 8, Width = 8 and Depth = 6 as the stated inputs, the result is Cubic Feet = 32 ft^3, Cubic Yards = 1.19 yd^3 and Estimated Tons = 1.6 tons. Each value corresponds to the declared output fields.

Example 3: Patio base: 14 ft × 16 ft × 4 in gravel

Inputs

material: Crushed Stone #57 length: 14 width: 16 depth: 4

Cubic Feet: 74.67 ft^3. Cubic Yards: 2.77 yd^3. Estimated Tons: 3.87 tons. With 10% Waste: 3.04 yd^3

With Material = Crushed Stone #57, Length = 14, Width = 16 and Depth = 4 as the stated inputs, the result is Cubic Feet = 74.67 ft^3, Cubic Yards = 2.77 yd^3 and Estimated Tons = 3.87 tons. Each value corresponds to the declared output fields.

Example 4: Fill dirt: 25 ft × 20 ft × 12 in low spot

Inputs

material: Fill Dirt length: 25 width: 20 depth: 12

Cubic Feet: 500 ft^3. Cubic Yards: 18.52 yd^3. Estimated Tons: 21.3 tons. With 10% Waste: 20.37 yd^3

With Material = Fill Dirt, Length = 25, Width = 20 and Depth = 12 as the stated inputs, the result is Cubic Feet = 500 ft^3, Cubic Yards = 18.52 yd^3 and Estimated Tons = 21.3 tons. Each value corresponds to the declared output fields.

Common Use Cases

Calculate gravel needed for a driveway base
Estimate sand for a sand box or play area
Determine fill material for a landscaping project