Powder Factors

Area Calculations

Basic Formulas

Area = Length x Width

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

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

Volume Calculations

Basic Formulas

Volume = Length x Width x Height

Remember: 27 ft³ = 1 yd³

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

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

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)

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

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

Blast Volume Calculations

Imperial Formula

Blast Volume = (L x W x H) / 27 yd³

Metric Formula

Blast Volume = L x W x H m³

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

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

Rock Density

Determining Rock Density

Measurements needed:

• Dry weight of rock sample
• Volume with rock
• Volume without rock (water displacement method)

Density = Mass / Volume

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

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

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

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

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

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

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

Powder Factor Expressions

Imperial Units

Metric Units

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

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

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

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

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

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

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

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?

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?

Quick Reference Formulas

Conversion Factors

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