# Powder Factors

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## Area Calculations

### Basic Formulas

**Area = Length x Width**

| Unit | Calculation |
|------|-------------|
| 1 square yard | 1 yd x 1 yd = 1 yd² = 3 ft x 3 ft = 9 ft² |
| 1 square meter | 1 m x 1 m = 1 m² |

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### Area Problem - Imperial

**Problem:** What is the area of a slab of concrete with a length of 8 feet and a width of 6 feet?

**Solution:**
Area = 8 ft x 6 ft = **48 ft²**

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### Area Problem - Metric

**Problem:** What is the area of an asphalt parking space with a length of 4.5 m and a width of 3 m in square meters?

**Solution:**
Area = 4.5 m x 3 m = **13.5 m²**

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## Volume Calculations

### Basic Formulas

**Volume = Length x Width x Height**

**Remember:** 27 ft³ = 1 yd³

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### Volume Problem - Imperial

**Problem:** If a 48 ft² slab is 1 foot thick, what is its volume?

**Solution:**
- (a) In cubic feet: 8 ft x 6 ft x 1 ft = **48 ft³**
- (b) In cubic yards: 48 ft³ ÷ 27 = **1.78 yd³**

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### Volume Problem - Metric

**Problem:** If the asphalt of a 13.5 m² parking space is 0.300 m thick, what is its volume in cubic meters?

**Solution:**
Volume = Length x Width x Height = 4.5 m x 3 m x 0.3 m = **4.05 m³**

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## Rock Volume Calculations

### Imperial Formula

**Rock Volume = (B x S x H) / 27 yd³**

Where:
- B = Burden (feet)
- S = Spacing (feet)
- H = Bench Height (feet)
- 27 ft³ = 1 yd³

### Metric Formula

**Rock Volume = B x S x H m³**

Where:
- B = Burden (meters)
- S = Spacing (meters)
- H = Bench Height (meters)

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### Rock Volume Problem - Imperial

**Problem:** A borehole has 12 foot burden and a 15 foot spacing on 30 foot bench. How many cubic yards are there in this borehole?

**Solution:**
Rock Volume = (B x S x H) / 27 = (12 x 15 x 30) / 27 = 5400 / 27 = **200 yd³**

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### Rock Volume Problem - Metric

**Problem:** A borehole has 3.2 m burden and a 5 m spacing on a 10 m bench. How many cubic meters are there per borehole?

**Solution:**
Rock Volume = B x S x H = 3.2 x 5 x 10 = **160 m³**

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## Blast Volume Calculations

### Imperial Formula

**Blast Volume = (L x W x H) / 27 yd³**

### Metric Formula

**Blast Volume = L x W x H m³**

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### Blast Volume Problem - Imperial

**Problem:** A shot is 150 feet long by 50 feet wide with a bench height of 60 feet. How many cubic yards are there in the shot?

**Solution:**
Blast Volume = (150 x 50 x 60) / 27 = 450,000 / 27 = **16,667 yd³**

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### Blast Volume Problem - Metric

**Problem:** A shot is 30 meters long and 15 meters wide with a bench height of 9 meters. How many cubic meters are there in the shot?

**Solution:**
Blast Volume = 30 x 15 x 9 = **4,050 m³**

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## Rock Density

### Determining Rock Density

**Measurements needed:**
- Dry weight of rock sample
- Volume with rock
- Volume without rock (water displacement method)

**Density = Mass / Volume**

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### Rock Density Conversion - Imperial

**Problem:** How many tons per cubic yard are there if the rock weight is 168 lbs per cubic foot?

**Solution:**
168 lbs/ft³ x 27 ft³/yd³ = 4,536 lbs/yd³
4,536 lbs/yd³ ÷ 2,000 lbs/ton = **2.27 tons/yd³**

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### Rock Density Conversion - Metric

**Problem:** How many tonnes per cubic meter are there if the rock weight is 1,211 kg per cubic meter?

**Solution:**
1,211 kg/m³ ÷ 1,000 kg/tonne = **1.211 tonnes/m³**

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## Tonnage Calculations

### Imperial Tonnage Formula

**Tonnage = Volume x Weight/Unit Volume**

**Problem:** A borehole has a 12 foot burden and a 15 foot spacing on a 30 foot bench. The rock weighs 2.16 tons per cubic yard. How many tons are there per borehole?

**Solution:**
1. Volume = (B x S x H) / 27 = (12 x 15 x 30) / 27 = 200 yd³
2. Tonnage = 200 yd³ x 2.16 tons/yd³ = **432 tons**

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### Metric Tonnage Formula

**Tonnage = Volume x Weight/Unit Volume**

**Problem:** A borehole has a 4 m burden and a 5 m spacing on a 9 m bench. The rock weighs 2 tonnes per cubic meter. How many tonnes are there per borehole?

**Solution:**
1. Volume = B x S x H = 4 x 5 x 9 = 180 m³
2. Tonnage = 180 m³ x 2 tonnes/m³ = **360 tonnes**

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## Charge Weight Calculations

### Metric Formula

**Weight/meter = (0.785 x De² x d) / 1000**

Where:
- 0.785 and 1000 are conversion factors
- De = Explosive Diameter in millimeters
- d = Explosive Density in g/cc

### Imperial Formula

**Weight/foot = 0.34 x De² x d**

Where:
- 0.34 is a conversion factor
- De = Explosive Diameter in inches
- d = Explosive Density in g/cc

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### Charge Weight Problem - Imperial

**Problem:** How many pounds per foot are there in a 6" borehole bulk loaded with ANFO with a density of 0.88 g/cc?

**Solution:**
Weight/foot = 0.34 x 6² x 0.88 = 0.34 x 36 x 0.88 = **10.77 lbs/ft**

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### Charge Weight Problem - Metric

**Problem:** How many kilograms per meter are there in a 152 mm borehole bulk loaded with ANFO with a density of 0.88 g/cc?

**Solution:**
Weight/meter = (0.785 x 152² x 0.88) / 1000
= (0.785 x 23,104 x 0.88) / 1000
= **15.96 or 16 kg/m**

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## Powder Factor Expressions

### Imperial Units

| Expression | Abbreviation |
|------------|-------------|
| Pounds per cubic yard | lb/yd³ |
| Cubic yards per pound | yd³/lb |
| Pounds per ton | lb/ton |
| Tons per pound | ton/lb |

### Metric Units

| Expression | Abbreviation |
|------------|-------------|
| Kilograms per cubic meter | kg/m³ |
| Cubic meters per kilogram | m³/kg |
| Kilograms per tonne | kg/t |
| Tonnes per kilogram | t/kg |

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## Powder Factor Calculation Process

1. Calculate the **VOLUME** of material to be blasted
   **OR**
   Calculate the **WEIGHT** of material to be blasted

2. **AND**
   Calculate the explosive **WEIGHT**

3. **THEN**
   Calculate **POWDER FACTOR** in the desired units

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### Powder Factor Problem - Imperial (Volume)

**Problem:** If 9,800 pounds of explosives are used to break 10,000 cubic yards of rock, what is the powder factor expressed in pounds per cubic yard?

**Solution:**
Powder factor = explosive weight / yardage
= 9,800 lb / 10,000 yd³
= **0.98 lb/yd³**

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### Powder Factor Problem - Metric (Volume)

**Problem:** If 4,250 kilograms of explosive are used to break 4,000 cubic meters of rock, what is the powder factor expressed in kilograms per cubic meter?

**Solution:**
Powder factor = explosive weight / volume
= 4,250 kg / 4,000 m³
= **1.063 kg/m³**

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### Powder Factor Problem - Imperial (Tonnage)

**Problem:** If 1,200 lbs of explosives are used to break 3,000 tons of rock, what is the powder factor expressed in tons per pound?

**Solution:**
Powder factor = tonnage / weight
= 3,000 tons / 1,200 lbs
= **2.5 tons/lb**

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### Powder Factor Problem - Metric (Tonnage)

**Problem:** If 750 kg of explosive are used to break 2,400 tonnes of rock, what is the powder factor expressed in tonnes per kilogram?

**Solution:**
Powder factor = tonnage / weight
= 2,400 tonnes / 750 kg
= **3.2 tonnes/kg**

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## Shot Tonnage Calculations

### Imperial Example 1

**Problem:** A shot is 50 feet wide by 150 feet long with a bench height of 60 feet. The rock weighs 2.2 tons/yd³. How many tons are there in this blast?

**Solution:**
1. Shot volume = (50 x 150 x 60) / 27 = 16,667 yd³
2. Shot tonnage = 16,667 yd³ x 2.2 tons/yd³ = **36,667 tons**

### Imperial Example 2

**Problem:** A shot is 50 feet wide by 150 feet long with a bench height of 60 feet. The rock weighs 2.3 tons/yd³. How many tons are there in this blast?

**Solution:**
1. Shot volume = (50 x 150 x 60) / 27 = 16,667 yd³
2. Shot tonnage = 16,667 yd³ x 2.3 tons/yd³ = **38,334 tons**

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### Metric Example

**Problem:** A shot is 15 m wide by 50 m long with a bench height of 20 m. The rock weighs 2.6 tonnes/m³. How many tonnes are there in this blast?

**Solution:**
1. Shot volume = 15 x 50 x 20 = 15,000 m³
2. Shot tonnage = 15,000 m³ x 2.6 t/m³ = **39,000 tonnes**

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## Charge Weight Tables

### Using Charge Weight Tables

**Example:** 4" = 101.6 mm hole with bulk product density of 1.0 g/cc

**What is the explosive's column build rate?**

| Units | Calculation | Result |
|-------|-------------|--------|
| Imperial | 5.45 x 1.0 | **5.45 lb/ft** |
| Metric | 8.1 x 1.0 | **8.1 kg/m** |

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### Adjusted for Different Density

**Same hole:** 4" = 101.6 mm with bulk product density of 0.84 g/cc

**What is the explosive's column build rate?**

| Units | Calculation | Result |
|-------|-------------|--------|
| Imperial | 5.45 x 0.84 | **4.6 lb/ft** |
| Metric | 8.1 x 0.84 | **6.8 kg/m** |

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## Quick Reference Formulas

| Calculation | Imperial | Metric |
|-------------|----------|--------|
| Area | L x W (ft²) | L x W (m²) |
| Volume | L x W x H (ft³) | L x W x H (m³) |
| Rock Volume | (B x S x H) / 27 (yd³) | B x S x H (m³) |
| Charge Weight | 0.34 x De² x d (lb/ft) | (0.785 x De² x d) / 1000 (kg/m) |
| Powder Factor (vol) | Weight / Volume | Weight / Volume |
| Powder Factor (ton) | Tonnage / Weight | Tonnage / Weight |
| Tonnage | Volume x Density | Volume x Density |

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## Conversion Factors

| From | To | Factor |
|------|-----|--------|
| ft³ | yd³ | ÷ 27 |
| lbs | tons | ÷ 2,000 |
| kg | tonnes | ÷ 1,000 |

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*This document is the property of Maritime Blasting Services Ltd., Moncton, NB*