Ann: EXCEPTIONALLY HIGH GRADES TO 93.2g/t GOLD AT MYRTLEFORD, page-8

Nice hit of nearly 16ozs of gold per ton.

Let’s calculate the number of ounces of gold in an ore body with an intersection of 8.2 meters at a grade of 22.4 grams per tonne (g/t). As with the previous question, we need to estimate the tonnage, so I’ll use the same simplifying assumption: a 1-meter by 1-meter cross-section with a rock density of 2.7 tonnes per cubic meter (a typical value for gold-bearing rock).

Given:

• Intersection: 8.2 meters (m)
• Grade: 22.4 grams per tonne (g/t)

Step-by-Step Calculation:

1. Volume of the ore body:
   • Length = 8.2 m
   • Assume width = 1 m (simplification)
   • Assume height/thickness = 1 m (simplification)
   • Volume = 8.2 m × 1 m × 1 m = 8.2 cubic meters (m³)
2. Tonnage of the ore body:
   • Average density of rock = 2.7 tonnes/m³
   • Tonnage = Volume × Density = 8.2 m³ × 2.7 tonnes/m³ = 22.14 tonnes
3. Gold content in grams:
   • Grade = 22.4 g/t
   • Total gold = Tonnage × Grade = 22.14 tonnes × 22.4 g/t = 495.936 grams
4. Convert grams to ounces:
   • 1 troy ounce = 31.1035 grams (standard for precious metals)
   • Ounces of gold = 495.936 grams ÷ 31.1035 grams/ounce ≈ 15.94 ounces