📐 Essential Formula Sheet
Comprehensive Quick Reference Guide for California Contractor License Exam
📏 Area & Volume Calculations
Rectangle/Square Area
Area = Length × Width
Example: A room 12 ft × 15 ft = 180 square feet
Note: Always ensure both measurements are in the same units before multiplying.
Triangle Area
Area = ½ × Base × Height
Example: Triangle with base 10 ft and height 8 ft = ½ × 10 × 8 = 40 sq ft
Note: Height must be perpendicular to the base for accurate calculation.
Circle Area
Area = π × r² (where r = radius, π ≈ 3.14159)
Example: Circle with radius 5 ft = 3.14 × 5² = 78.5 sq ft
Note: Radius is half the diameter. If given diameter, divide by 2 first.
Circle Circumference
Circumference = 2 × π × r OR π × d
Example: Circle with radius 3 ft = 2 × 3.14 × 3 = 18.84 ft
Trapezoid Area
Area = ½ × (Base₁ + Base₂) × Height
Example: Trapezoid with bases 8 ft and 12 ft, height 5 ft = ½ × (8 + 12) × 5 = 50 sq ft
Parallelogram Area
Area = Base × Height
Example: Parallelogram with base 15 ft and height 8 ft = 120 sq ft
Volume of Rectangular Space
Volume = Length × Width × Height
Example: Room 10 ft × 12 ft × 8 ft = 960 cubic feet
Cylinder Volume
Volume = π × r² × Height
Example: Tank with radius 3 ft and height 10 ft = 3.14 × 9 × 10 = 282.6 cu ft
Sphere Volume
Volume = (4/3) × π × r³
Example: Sphere with radius 4 ft = (4/3) × 3.14 × 64 = 268.1 cu ft
Cone Volume
Volume = (1/3) × π × r² × Height
Example: Cone with radius 3 ft and height 9 ft = (1/3) × 3.14 × 9 × 9 = 84.78 cu ft
🏗️ Material Calculations
Concrete Volume (Cubic Yards)
Cubic Yards = (Length × Width × Depth in feet) ÷ 27
Example: Slab 20 ft × 30 ft × 0.5 ft = 300 ÷ 27 = 11.11 cubic yards
Note: Always order 10% extra for waste and spillage.
Concrete for Circular Slab
Cubic Yards = (π × r² × Depth in feet) ÷ 27
Example: Circular patio, radius 10 ft, depth 4 inches (0.33 ft) = (3.14 × 100 × 0.33) ÷ 27 = 3.84 cu yds
Board Feet
Board Feet = (Thickness in inches × Width in inches × Length in feet) ÷ 12
Example: 2" × 6" × 10 ft board = (2 × 6 × 10) ÷ 12 = 10 board feet
Number of Boards Needed
Boards = Total Board Feet Required ÷ Board Feet per Board
Example: Need 500 board feet, each board is 10 bf = 500 ÷ 10 = 50 boards
Paint Coverage
Gallons Needed = Total Square Feet ÷ Coverage per Gallon
Example: 1,000 sq ft ÷ 350 sq ft per gallon = 2.86 gallons (buy 3)
Note: Rough surfaces require more paint. Smooth drywall: 350-400 sq ft/gal. Textured: 250-300 sq ft/gal.
Wall Area for Paint (with openings)
Area = (Perimeter × Height) - (Window & Door Areas)
Example: Room 12×15 ft, 8 ft high, two 3×5 ft windows, one 3×7 ft door = [(54×8) - (30 + 21)] = 381 sq ft
Drywall Sheets Needed
Sheets = Total Wall Area ÷ Sheet Size (typically 32 sq ft for 4×8 sheet)
Example: 800 sq ft of wall ÷ 32 = 25 sheets (add 10% for waste = 28 sheets)
Brick/Block Count
Units = Wall Area ÷ (Unit Length × Unit Height)
Example: 100 sq ft wall, standard brick (8" × 2.67") with mortar = 100 ÷ 0.89 = 112 bricks
Note: Standard calculation: 7 bricks per sq ft including mortar joints.
Mortar for Brick/Block
Cubic Feet = (Wall Area × Mortar Joint Thickness) × 1.15
Example: 100 sq ft of brick wall = approximately 5 cubic feet of mortar
Tile Required
Tiles = (Total Area ÷ Tile Size) × 1.10
Example: 150 sq ft floor, 12"×12" tiles = (150 ÷ 1) × 1.10 = 165 tiles
Grout Coverage
Pounds = (Length + Width) ÷ (Length × Width) × Area × Depth × 0.0625
Example: For 12"×12" tiles, 1/8" joints, 100 sq ft ≈ 7.5 lbs grout
Carpet/Flooring Square Yards
Square Yards = Square Feet ÷ 9
Example: 450 sq ft room = 450 ÷ 9 = 50 square yards
⚡ Electrical Formulas
Ohm's Law (Voltage)
Voltage (V) = Current (I) × Resistance (R)
Example: 10 amps × 12 ohms = 120 volts
Ohm's Law (Current)
Current (I) = Voltage (V) ÷ Resistance (R)
Example: 120 volts ÷ 12 ohms = 10 amps
Ohm's Law (Resistance)
Resistance (R) = Voltage (V) ÷ Current (I)
Example: 120 volts ÷ 10 amps = 12 ohms
Power (Watts)
Power (W) = Voltage (V) × Current (I)
Example: 120 volts × 15 amps = 1,800 watts
Total Amperage
Amperage = Watts ÷ Voltage
Example: 2,400 watts ÷ 120 volts = 20 amps
Total Wattage
Wattage = Voltage × Amperage
Example: 240 volts × 30 amps = 7,200 watts
Three-Phase Power
Power (W) = √3 × Voltage × Current × Power Factor
Example: 1.732 × 208V × 50A × 0.9 = 16,229 watts
Voltage Drop
Vd = (2 × K × I × L) ÷ CM
Example: K=12.9 (copper), 20A, 100ft run, 12AWG (6530 CM) = (2×12.9×20×100)÷6530 = 7.9V drop
Note: NEC recommends maximum 3% voltage drop for branch circuits, 5% total.
Wire Size Calculation
CM = (2 × K × I × L) ÷ Allowable Vd
Example: For 20A, 150ft, max 3.6V drop = (2×12.9×20×150)÷3.6 = 21,500 CM (use 6 AWG)
Energy Consumption (kWh)
kWh = (Watts × Hours Used) ÷ 1,000
Example: 1,500W heater running 8 hours = (1,500 × 8) ÷ 1,000 = 12 kWh
💰 Cost & Bidding Calculations
Markup Percentage
Markup = (Selling Price - Cost) ÷ Cost × 100
Example: Sell for $15,000, cost $10,000 = ($5,000 ÷ $10,000) × 100 = 50% markup
Profit Margin (Gross)
Profit Margin = (Selling Price - Cost) ÷ Selling Price × 100
Example: Sell for $15,000, cost $10,000 = ($5,000 ÷ $15,000) × 100 = 33.3% margin
Selling Price from Markup
Selling Price = Cost × (1 + Markup %)
Example: Cost $8,000, want 40% markup = $8,000 × 1.40 = $11,200
Selling Price from Margin
Selling Price = Cost ÷ (1 - Desired Margin %)
Example: Cost $8,000, want 30% margin = $8,000 ÷ 0.70 = $11,428.57
Labor Cost
Labor Cost = Hours × Hourly Rate × Number of Workers
Example: 40 hours × $35/hr × 3 workers = $4,200
Unit Price
Unit Price = Total Cost ÷ Number of Units
Example: $5,000 for 250 sq ft = $20 per square foot
Total Job Cost
Total = Materials + Labor + Equipment + Overhead + Profit
Example: $3,000 + $4,000 + $500 + $1,200 + $1,800 = $10,500
Overhead Rate
Overhead % = Total Overhead Costs ÷ Total Revenue × 100
Example: $50,000 overhead, $250,000 revenue = ($50,000 ÷ $250,000) × 100 = 20%
Break-Even Point
Break-Even = Fixed Costs ÷ (Price - Variable Cost per Unit)
Example: $10,000 fixed costs, sell at $100, variable cost $40 = $10,000 ÷ $60 = 167 units
Labor Burden Rate
Burden = (Base Wage + Benefits + Taxes + Insurance) ÷ Base Wage
Example: $25/hr base, $8/hr burden = ($25 + $8) ÷ $25 = 1.32 or 32% burden
Cost Per Hour (True Labor Rate)
True Rate = Hourly Wage × (1 + Burden %) + Overhead Allocation
Example: $25/hr wage, 32% burden = $25 × 1.32 = $33/hr true cost
📐 Slope & Grade Formulas
Roof Slope (Pitch)
Slope = Rise ÷ Run × 12
Example: 4 ft rise over 12 ft run = 4 ÷ 12 × 12 = 4:12 pitch
Roof Slope in Degrees
Degrees = arctan(Rise ÷ Run)
Example: 4:12 pitch = arctan(4 ÷ 12) = 18.43 degrees
Rafter Length
Rafter = √(Run² + Rise²)
Example: 12 ft run, 4 ft rise = √(144 + 16) = √160 = 12.65 ft
Grade Percentage
Grade % = (Rise ÷ Run) × 100
Example: 2 ft rise over 20 ft run = (2 ÷ 20) × 100 = 10% grade
Stair Rise & Run
Total Rise ÷ Number of Risers = Individual Riser Height
Example: 108 inches total rise ÷ 14 risers = 7.71" per riser (acceptable: 7" to 7.75")
Note: Building code requires risers 4-7.75", treads minimum 10". Total of riser + tread should be 17-18".
Pythagorean Theorem
c² = a² + b² (where c = hypotenuse)
Example: Sides 3 ft and 4 ft = √(9 + 16) = 5 ft diagonal
Note: Use 3-4-5 rule for checking square corners (or multiples: 6-8-10, 9-12-15).
Diagonal of Rectangle
Diagonal = √(Length² + Width²)
Example: 12 ft × 16 ft room = √(144 + 256) = 20 ft diagonal
Excavation Cut/Fill
Volume = Length × Width × Average Depth
Example: 30ft × 40ft area, 3ft average depth = 3,600 cu ft ÷ 27 = 133.3 cu yds
🌡️ HVAC Formulas
BTU Requirement (Basic)
BTUs = Square Feet × 20 (for standard 8-9 ft ceilings)
Example: 1,500 sq ft × 20 = 30,000 BTUs needed
BTU Requirement (Detailed)
BTUs = (Sq Ft × 25) + (500 per person) + (1,000 per kitchen)
Example: 1,200 sq ft, 4 people, kitchen = (30,000 + 2,000 + 1,000) = 33,000 BTUs
Tons of Cooling
Tons = BTUs ÷ 12,000
Example: 36,000 BTUs = 36,000 ÷ 12,000 = 3 tons
Note: One ton = 12,000 BTUs per hour of cooling capacity.
Airflow (CFM)
CFM = (Room Volume in cubic feet) ÷ Minutes per air change
Example: 2,000 cu ft room ÷ 4 minutes = 500 CFM
CFM for Cooling
CFM = (BTU/hr × 1.08) ÷ Temperature Difference
Example: 36,000 BTU system, 20°F temp difference = 36,000 ÷ (1.08 × 20) = 1,667 CFM
Duct Velocity
Velocity (FPM) = CFM ÷ Duct Area (sq ft)
Example: 1,000 CFM through 1.5 sq ft duct = 1,000 ÷ 1.5 = 667 FPM
Duct Sizing
Area = CFM ÷ Desired Velocity (typically 700-900 FPM)
Example: 800 CFM at 800 FPM = 800 ÷ 800 = 1 sq ft (12" × 12" duct)
Heat Loss (Basic)
Heat Loss = Area × U-Value × Temperature Difference
Example: 200 sq ft wall, U=0.08, 40°F difference = 200 × 0.08 × 40 = 640 BTU/hr loss
SEER to EER Conversion
EER ≈ SEER × 0.875
Example: 16 SEER unit ≈ 16 × 0.875 = 14 EER
💧 Plumbing Formulas
Water Pressure (PSI)
PSI = Height in feet × 0.433
Example: 50 ft height × 0.433 = 21.65 PSI
Note: Each foot of elevation = 0.433 PSI. Conversely, 1 PSI = 2.31 feet of head.
Pressure to Height
Height (feet) = PSI × 2.31
Example: 40 PSI × 2.31 = 92.4 feet of head
Pipe Volume (Gallons)
Gallons = π × r² × Length (in inches) ÷ 231
Example: 2" diameter pipe, 100 ft long = 3.14 × 1² × 1,200 ÷ 231 ≈ 16.3 gallons
Pipe Volume (Cubic Feet)
Cu Ft = π × r² × Length (all in feet)
Example: 3" (0.25 ft) diameter, 50 ft long = 3.14 × 0.125² × 50 = 2.45 cu ft
Flow Rate (GPM)
GPM = (Volume in gallons) ÷ Time in minutes
Example: 50 gallons filled in 5 minutes = 50 ÷ 5 = 10 GPM
Water Velocity in Pipe
Velocity (FPS) = (0.408 × GPM) ÷ (Diameter in inches)²
Example: 10 GPM through 2" pipe = (0.408 × 10) ÷ 4 = 1.02 FPS
Drain Slope
Slope = 1/4" per foot (minimum) = 2% grade
Example: 20 ft drain run × 0.25" = 5 inches total drop minimum
Water Heater Recovery
Recovery (GPH) = (BTU Input × 0.7) ÷ 8.33 ÷ Temperature Rise
Example: 40,000 BTU heater, 70°F rise = (40,000 × 0.7) ÷ 8.33 ÷ 70 = 48 GPH
Fixture Unit Calculation
Total Fixture Units = Sum of all fixtures
Example: Toilet (3 FU) + Sink (1.5 FU) + Shower (2 FU) = 6.5 FU
Note: Use DFU (Drainage Fixture Units) to size drain pipes per plumbing code.
🔧 Structural & Load Calculations
Live Load
Load (lbs) = Area (sq ft) × Load per sq ft
Example: 200 sq ft floor, 40 PSF live load = 200 × 40 = 8,000 lbs
Note: Residential floors typically 40 PSF live load, 10 PSF dead load.
Beam Load Capacity
Capacity = (Fiber Stress × Section Modulus) ÷ Safety Factor
Example: Complex calculation requiring engineering tables and lumber grade specs
Point Load Distribution
Load per Support = Total Load ÷ Number of Supports
Example: 10,000 lb load on beam with 4 posts = 10,000 ÷ 4 = 2,500 lbs per post
Joist Spacing Maximum
Spacing depends on: Joist size + Span + Load + Species
Example: 2×10 joists, 12 ft span, 40 PSF = 16" spacing (refer to span tables)
Footing Size
Area = Total Load ÷ Soil Bearing Capacity
Example: 12,000 lb load, 2,000 PSF soil = 12,000 ÷ 2,000 = 6 sq ft footing needed
Snow Load
Load = Snow Depth (inches) × Snow Density (typically 0.5 PSF/inch)
Example: 24" snow × 0.5 PSF/inch = 12 PSF additional load
📏 Unit Conversions
Quick Reference Conversions
| From | To | Multiply By |
|---|---|---|
| Inches | Feet | ÷ 12 |
| Feet | Inches | × 12 |
| Feet | Yards | ÷ 3 |
| Square Feet | Square Yards | ÷ 9 |
| Cubic Feet | Cubic Yards | ÷ 27 |
| Gallons | Cubic Feet | ÷ 7.48 |
| Cubic Feet | Gallons | × 7.48 |
| PSI | Feet of Head | × 2.31 |
| Feet of Head | PSI | × 0.433 |
| Horsepower | Watts | × 746 |
| BTU/hour | Watts | × 0.293 |
| Tons (cooling) | BTU/hour | × 12,000 |
Temperature Conversions
°F = (°C × 9/5) + 32 | °C = (°F - 32) × 5/9
Example: 20°C = (20 × 1.8) + 32 = 68°F | 75°F = (75 - 32) × 0.556 = 23.9°C
🔨 Specialty Calculations
Rebar Spacing
Number of Bars = (Length - 2 × Edge Distance) ÷ Spacing + 1
Example: 20 ft slab, 3" edges, 18" spacing = (240 - 6) ÷ 18 + 1 = 14 bars
Insulation R-Value Total
Total R-Value = R₁ + R₂ + R₃ + ... (additive)
Example: R-13 batts + R-5 foam + R-1 drywall = R-19 total
Window U-Factor to R-Value
R-Value = 1 ÷ U-Factor
Example: Window with U-0.30 = 1 ÷ 0.30 = R-3.33
Retaining Wall Pressure
Pressure (PSF) = 30 × Height (ft) × 1.5 (safety factor)
Example: 6 ft wall = 30 × 6 × 1.5 = 270 PSF at base
Window Rough Opening
R.O. = Window Width + 2" to 3" (Height + 2" to 3")
Example: 36" × 48" window = 38-39" × 50-51" rough opening
Door Rough Opening
R.O. Width = Door Width + 2.5" | Height = Door Height + 2.5"
Example: 36" × 80" door = 38.5" × 82.5" rough opening
Adhesive Coverage
Gallons = Area ÷ Coverage Rate (varies by product)
Example: 500 sq ft, adhesive covers 200 sq ft/gal = 2.5 gallons
Sealant Calculation
Tubes = (Linear Feet × Bead Width × Bead Depth) ÷ 77
Example: 100 ft, 1/4" wide, 1/4" deep = (100 × 0.25 × 0.25) ÷ 77 = 0.08 tubes (≈ 1 tube)