These are first-party reproducible checks, not independent certification, structural design, or supplier quotes.
Validation checks the published quantity math. Verify measurements, selected-product specifications, supplier minimums, and local requirements before ordering.
/// Start from the review question
Choose a reproducible case
| Review task | Published case | Calculator |
|---|---|---|
| Does a 10 ft × 10 ft × 4 in slab with 10% waste produce 1.3580 yd³ and 62 bags? | Concrete slab volume and 80 lb bag count | Open calculator |
| Does an 8 ft × 5 ft × 3 ft box contain 120 ft³ and about 4.4444 yd³? | Rectangular cubic feet and equivalent units | Open calculator |
| Does one 20 ft × 1.5 ft × 0.75 ft footing produce 0.8333 yd³ net and 42 bags after 10% waste at 0.60 ft³ per bag? | Concrete footing volume and entered bag yield | Open calculator |
| Does a 12 ft × 10 ft deck using 5.5 in boards, 0.125 in gaps, and 12 ft stock plan 25 boards with 10% waste? | Deck board rows and whole stock boards | Open calculator |
| Does a 30 ft × 20 ft footprint at 6:12 pitch and 10% waste produce about 738 ft², 7.38 squares, and 23 bundles? | Simple sloped roof area, squares, and bundles | Open calculator |
| Does a 12 ft × 10 ft surface using 12 in square tile and 10 tiles per box require 132 tiles and 14 boxes with 10% waste? | Tile pieces and full boxes | Open calculator |
| Does a 20 ft × 10 ft bed at 3 in deep contain about 1.8519 yd³ before waste and 2.0370 yd³ with 10%? | Mulch bed volume before and after allowance | Open calculator |
/// Case 01
Concrete slab volume and 80 lb bag count
Question: Does a 10 ft × 10 ft × 4 in slab with 10% waste produce 1.3580 yd³ and 62 bags?
| Checkpoint | Published value |
|---|---|
| Slab | 10 ft × 10 ft × 4 in |
| Waste allowance | 10% |
| Selected package | 80 lb QUIKRETE Concrete Mix No. 1101 |
| Published package yield | Approximately 0.60 ft³ |
Formula
Convert depth to feet; multiply length × width × depth; divide ft³ by 27 for yd³; multiply by 1.10; divide adjusted ft³ by 0.60 ft³ per selected bag.
| Checkpoint | Published value |
|---|---|
| Base volume | 33.3333 ft³ = 1.2346 yd³ |
| Adjusted volume | 36.6667 ft³ = 1.3580 yd³ |
| Unrounded packages | 61.1111 bags |
Rounding and expected result
Display volume to four decimals and round the final package quantity up to the next whole bag.
| Checkpoint | Published value |
|---|---|
| Base volume | 1.2346 yd³ |
| Volume with 10% waste | 1.3580 yd³ |
| 80 lb bags | 62 bags |
The package result is scoped to the named QUIKRETE No. 1101 80 lb yield. Field conditions, delivery, placement, and other products require separate verification.
Related sources: nist-sp811-foot-metre · quikrete-1101-80-yield
/// Case 02
Rectangular cubic feet and equivalent units
Question: Does an 8 ft × 5 ft × 3 ft box contain 120 ft³ and about 4.4444 yd³?
| Checkpoint | Published value |
|---|---|
| Shape | Rectangular box |
| Dimensions | 8 ft × 5 ft × 3 ft |
| Quantity | 1 |
Formula
Multiply length × width × height for cubic feet, then divide by exactly 27 for cubic yards and use the exact international-foot relationship for metric equivalents.
| Checkpoint | Published value |
|---|---|
| Cubic feet | 8 × 5 × 3 = 120 ft³ |
| Cubic-yard divisor | 27 ft³ per yd³ |
Rounding and expected result
Keep the engine value at full precision and display the cubic-yard equivalent to four decimals.
| Checkpoint | Published value |
|---|---|
| Volume | 120 ft³ |
| Cubic yards | 4.4444 yd³ |
This is geometric volume only. It does not add material density, compaction, waste, package coverage, shipping capacity, or price.
Related sources: nist-sp811-foot-metre
/// Case 03
Concrete footing volume and entered bag yield
Question: Does one 20 ft × 1.5 ft × 0.75 ft footing produce 0.8333 yd³ net and 42 bags after 10% waste at 0.60 ft³ per bag?
| Checkpoint | Published value |
|---|---|
| Footing | 1 × 20 ft × 1.5 ft × 0.75 ft |
| Waste allowance | 10% |
| Entered bag yield | 0.60 ft³ per bag |
Formula
Multiply footing quantity × length × width × depth; divide ft³ by 27 for yd³; multiply by 1.10; divide adjusted ft³ by the entered 0.60 ft³ yield and round up.
| Checkpoint | Published value |
|---|---|
| Net volume | 22.5 ft³ = 0.8333 yd³ |
| Volume with waste | 24.75 ft³ = 0.9167 yd³ |
| Unrounded packages | 41.25 bags |
Rounding and expected result
Display volume to four decimals and round the final package quantity up to the next whole bag.
| Checkpoint | Published value |
|---|---|
| Net volume | 0.8333 yd³ |
| Volume with waste | 0.9167 yd³ |
| Bags | 42 bags |
The 0.60 ft³ bag yield is an entered example and must be replaced with the selected product's label or current technical data. The calculator does not size or approve a footing.
Related sources: nist-sp811-foot-metre
/// Case 04
Deck board rows and whole stock boards
Question: Does a 12 ft × 10 ft deck using 5.5 in boards, 0.125 in gaps, and 12 ft stock plan 25 boards with 10% waste?
| Checkpoint | Published value |
|---|---|
| Deck | 12 ft long × 10 ft wide |
| Board face and gap | 5.5 in + 0.125 in |
| Stock length | 12 ft |
| Cutting allowance | 10% |
Formula
Rows = ceil((deck width in inches + one gap) ÷ (board face width + gap)); adjusted board length = rows × deck length × 1.10; stock boards = ceil(adjusted length ÷ stock length).
| Checkpoint | Published value |
|---|---|
| Board rows | 22 rows |
| Board length before allowance | 264 linear ft |
| Board length with allowance | 290.4 linear ft |
Rounding and expected result
Round the row count up first, retain length precision, then round the final stock-board count up.
| Checkpoint | Published value |
|---|---|
| Stock boards | 25 boards |
The case uses actual board face width and an entered gap. It excludes picture framing, breaker boards, stairs, fascia, diagonal layouts, and structural framing.
Related sources: nist-sp811-foot-metre
/// Case 05
Simple sloped roof area, squares, and bundles
Question: Does a 30 ft × 20 ft footprint at 6:12 pitch and 10% waste produce about 738 ft², 7.38 squares, and 23 bundles?
| Checkpoint | Published value |
|---|---|
| Horizontal footprint | 30 ft × 20 ft |
| Pitch | 6 in rise per 12 in run |
| Waste allowance | 10% |
| Product coverage | 3 bundles per square |
Formula
Slope factor = √(1 + (6 ÷ 12)²); adjusted roof area = footprint × slope factor × 1.10; squares = adjusted area ÷ 100 ft²; bundles = ceil(squares × 3).
| Checkpoint | Published value |
|---|---|
| Horizontal footprint | 600 ft² |
| Slope factor | 1.118034 |
| Base sloped roof area | 670.82 ft² |
| Adjusted roof area | 737.90 ft² |
| Bundle quotient before rounding | 22.137 bundles |
Rounding and expected result
Retain area and square precision for display, then round only the final bundle order up to a whole bundle.
| Checkpoint | Published value |
|---|---|
| Roof area with waste | About 738 ft² |
| Roofing area | 7.38 squares |
| Bundles | 23 bundles |
The footprint model covers one simple pitch and one product coverage set. Mixed pitches or products, starters, caps, valleys, flashing, manufacturer coverage, access, and safety require separate takeoffs and review.
Related sources: arma-residential-manual-roofing-square
/// Case 06
Tile pieces and full boxes
Question: Does a 12 ft × 10 ft surface using 12 in square tile and 10 tiles per box require 132 tiles and 14 boxes with 10% waste?
| Checkpoint | Published value |
|---|---|
| Surface | 12 ft × 10 ft |
| Tile face | 12 in × 12 in |
| Cutting allowance | 10% |
| Package count | 10 tiles per box |
Formula
Adjusted area = length × width × 1.10; individual tiles = ceil(adjusted area ÷ tile face area); boxes = ceil(individual tiles ÷ 10).
| Checkpoint | Published value |
|---|---|
| Base surface | 120 ft² |
| Area with allowance | 132 ft² |
| Tile face area | 1 ft² |
Rounding and expected result
Round individual pieces up before dividing by package count, then round the package result up again.
| Checkpoint | Published value |
|---|---|
| Individual tiles | 132 tiles |
| Full boxes | 14 boxes |
The case is face-area based. Pattern matching, grout joints, borders, breakage, attic stock, trim, substrate, and the selected carton label remain separate checks.
Related sources: nist-sp811-foot-metre
/// Case 07
Mulch bed volume before and after allowance
Question: Does a 20 ft × 10 ft bed at 3 in deep contain about 1.8519 yd³ before waste and 2.0370 yd³ with 10%?
| Checkpoint | Published value |
|---|---|
| Bed | 20 ft × 10 ft |
| Placed depth | 3 in |
| Allowance | 10% |
Formula
Convert 3 in to 0.25 ft; multiply length × width × depth; divide cubic feet by 27; multiply the base volume by 1.10.
| Checkpoint | Published value |
|---|---|
| Placed volume | 50 ft³ |
| Cubic-yard divisor | 27 ft³ per yd³ |
Rounding and expected result
Retain engine precision and display cubic-yard quantities to four decimals; do not round to bags or loads.
| Checkpoint | Published value |
|---|---|
| Base volume | 1.8519 yd³ |
| Volume with allowance | 2.0370 yd³ |
The case estimates placed geometry only. Product type, settling, existing grade, density, bag coverage, supplier increments, delivery, and installation remain separate decisions.
Related sources: nist-sp811-foot-metre