Rigger 2
Advanced Rigging Formulas, Weight Calculations, and OSHA/ASME Standards for NCCCO Certification
What is Rigger Level 2?
Rigger Level 2 (Advanced Rigger) goes beyond basic rigging to include weight calculations, center of gravity determination, sling tension analysis, D/d ratios, and advanced hitch configurations. A qualified Rigger 2 can calculate the weight of any object, determine the proper sling configuration, and verify that all equipment is rated for the load.
This guide covers the key OSHA and ASME standards, the essential formulas with worked examples and diagrams, a standard weights chart, the top 10 NCCCO exam questions, and a 20-question quiz — half general knowledge, half formula-based math with detailed pictures.
Key OSHA & ASME Standards
Top 10 NCCCO Exam Questions
- 1How do you calculate the weight of an object? Formula: Weight = Volume × Unit Weight (density). You must know the material's density (e.g., steel = 490 lbs/ft³) and the object's volume.
- 2What is the D/d ratio and why is it important? Formula: D/d = Pin or Hook Diameter ÷ Rope Diameter. A D/d ratio of 25:1 or greater maintains 100% of the sling's rated capacity. Lower ratios reduce capacity.
- 3What is the minimum recommended sling angle from horizontal? 30 degrees is the absolute minimum per ASME B30.9. 60 degrees is the preferred minimum for most lifts.
- 4How does a choker hitch affect sling capacity? A choker hitch reduces capacity to approximately 75% of the vertical rating. Formula: Choker Capacity = Vertical WLL × 0.75.
- 5What is the design factor for wire rope slings? 5:1 per ASME B30.9 — the breaking strength is 5 times the rated WLL.
- 6How do you find the center of gravity of a multi-object system? Formula: CG = (W1×D1 + W2×D2 + ...) ÷ (W1 + W2 + ...). The CG must be directly below the hook for a balanced lift.
- 7What happens to tension as the sling angle decreases? Tension increases dramatically. At 30°, each leg carries 100% of the load; at 60°, each leg carries ~58%.
- 8What are the inspection criteria for removing a sling from service? Missing/ illegible tag, broken wires, kinks, birdcaging, corrosion, damaged fittings, or any condition that compromises the sling's integrity.
- 9What is the formula for sling tension using the Tension Factor? Tension per Leg = (Load ÷ Number of Legs) × Tension Factor. TF from chart: 90°=1.000, 60°=1.155, 45°=1.414, 30°=2.000.
- 10How do you calculate the volume of a cylindrical tank? Formula: V = π × r² × h. For a rectangular tank: V = L × W × H. Multiply by the material's unit weight to get total weight.
Essential Rigger 2 Formulas
Weight of an Object
Weight = Volume × Unit Weight
Calculate the object's volume, then multiply by the material's unit weight (density) from the standard weights chart.
Volume of a Cylinder / Tank
V = π × r² × h
r = radius, h = height. For diameter: V = π × (d÷2)² × h. Multiply by unit weight for total weight.
Volume of a Rectangular Tank
V = L × W × H
L = length, W = width, H = height. Multiply by unit weight for total weight.
Sling Tension (Tension Factor Method)
Tension per Leg = (Load ÷ Number of Legs) × Tension Factor
TF from chart: 90°=1.000, 60°=1.155, 45°=1.414, 30°=2.000. If angle not on chart: TF = Sling Length ÷ Sling Height.
Center of Gravity (Multi-Object)
CG = (W1×D1 + W2×D2 + ...) ÷ (W1 + W2 + ...)
W = weight of each object, D = distance of each object from a reference point. The hook must be positioned above the CG.
D/d Ratio
D/d = Pin or Hook Diameter ÷ Rope Diameter
D = diameter of the pin, hook, or edge the rope bends around. d = diameter of the wire rope. D/d ≥ 25:1 maintains 100% capacity.
Choker Hitch Capacity
Choker Capacity = Vertical WLL × 0.75
A choker hitch reduces the sling's rated capacity to approximately 75% of its vertical WLL due to bending stress at the choke point.
Standard Weights Chart
Use these unit weights (density) to calculate the weight of any object: Weight = Volume × Unit Weight.
| Material | Unit Weight (lbs/ft³) | Notes |
|---|---|---|
| Steel | 490 | Structural steel, plate, bar |
| Cast Iron | 450 | Engine blocks, machinery |
| Aluminum | 170 | Lightweight metal |
| Brass | 526 | Fittings, valves |
| Copper | 556 | Pipe, wire |
| Lead | 710 | Counterweights, shielding |
| Concrete (reinforced) | 150 | Slabs, beams, columns |
| Water | 62.4 | Tanks, pipes (per ft³) |
| Wood (Oak) | 62 | Hardwood lumber |
| Sand (dry) | 100 | Bulk material |
| Gravel | 120 | Bulk material |
| Glass | 162 | Plate glass |
⚡ EXAM TIP: Steel = 490, Concrete = 150, Water = 62.4 — memorize these three; they appear on almost every NCCCO exam.
Worked Examples with Diagrams
Example 1: Weight of a Steel Cylinder
Formula
Weight = π × r² × h × Unit Weight
A steel cylinder has a radius of 2 ft and a height of 10 ft. Steel weighs 490 lbs/ft³.
Volume = π × r² × h = π × (2)² × 10 = π × 4 × 10 = 125.66 ft³
Unit Weight of Steel = 490 lbs/ft³
Weight = 125.66 × 490 = 61,574 lbs
Result: The steel cylinder weighs approximately 61,574 lbs (30.8 tons).
Example 2: Volume of a Cylindrical Tank
Formula
V = π × r² × h
A water tank has a radius of 3 ft and a height of 8 ft. What is its volume?
V = π × r² × h
V = π × (3)² × 8 = π × 9 × 8 = 226.19 ft³
Volume ≈ 226.19 cubic feet
Result: The tank holds approximately 226.19 ft³ of water (about 1,693 gallons).
Example 3: Weight of a Concrete Block
Formula
Weight = L × W × H × Unit Weight
A concrete block measures 4 ft × 3 ft × 2 ft. Concrete weighs 150 lbs/ft³.
Volume = L × W × H = 4 × 3 × 2 = 24 ft³
Unit Weight of Concrete = 150 lbs/ft³
Weight = 24 × 150 = 3,600 lbs
Result: The concrete block weighs 3,600 lbs (1.8 tons).
Example 4: Sling Tension at 60°
Formula
Tension per Leg = (Load ÷ Number of Legs) × Tension Factor
A 12,000 lb load is lifted with a two-leg bridle at 60°. TF at 60° = 1.155.
Load = 12,000 lbs, Legs = 2, Angle = 60°
Tension Factor (from chart) at 60° = 1.155
Tension per leg = (12,000 ÷ 2) × 1.155 = 6,000 × 1.155
Tension per leg = 6,930 lbs
Result: Each sling leg carries 6,930 lbs — about 58% of the load.
Example 5: Center of Gravity
Formula
CG = (W1×D1 + W2×D2) ÷ (W1 + W2)
Object 1 weighs 2,000 lbs at 4 ft from the left end. Object 2 weighs 3,000 lbs at 10 ft from the left end.
CG = (2,000 × 4 + 3,000 × 10) ÷ (2,000 + 3,000)
CG = (8,000 + 30,000) ÷ 5,000
CG = 38,000 ÷ 5,000 = 7.6 ft
Result: The center of gravity is 7.6 ft from the left end. The hook must be positioned above this point.
Example 6: D/d Ratio
Formula
D/d = Pin Diameter ÷ Rope Diameter
A wire rope sling with a diameter of 1 inch is used over a pin with a diameter of 6 inches.
D (pin diameter) = 6 inches
d (rope diameter) = 1 inch
D/d = 6 ÷ 1 = 6:1
Result: The D/d ratio is 6:1. Since this is below 25:1, the sling's capacity is reduced. Consult the manufacturer's reduction chart.
Example 7: Choker Hitch Capacity
Formula
Choker Capacity = Vertical WLL × 0.75
A wire rope sling has a vertical WLL of 10,000 lbs. It is used in a choker hitch.
Vertical WLL = 10,000 lbs
Choker reduction factor = 0.75 (75%)
Choker Capacity = 10,000 × 0.75 = 7,500 lbs
Result: The choker hitch capacity is 7,500 lbs — 25% less than the vertical rating.
Test Your Rigger 2 Knowledge
Take our 20-question quiz — 10 general knowledge questions on OSHA/ASME standards, and 10 formula-based math questions with detailed diagrams covering weight calculations, tank volume, sling tension, center of gravity, D/d ratio, and choker hitch capacity.