Step 1: Determine Number of Passes
\[ \text{Number of passes} = \frac{\text{Panel width}}{\text{Web depth}} = \frac{200}{0.6} = 333.\overline{3} \approx 334 \text{ passes} \] Step 2: Calculate Time per Complete Cycle
Cutting time: \( \frac{1200}{5} = 240 \, \text{minutes} = 4 \, \text{hours} \)
Retreat time: \( \frac{1200}{10} = 120 \, \text{minutes} = 2 \, \text{hours} \)
Operational delay: 2 hours
\[ T_{\text{cycle}} = 4 \, \text{hours} + 2 \, \text{hours} + 2 \, \text{hours} = 8 \, \text{hours} \] Step 3: Determine Productive Capacity
Only one shift per day is considered productive for unidirectional cutting.
Available time per day: \( 8 \, \text{hours} \)
Cycles per day: \( \frac{8 \, \text{hours}}{8 \, \text{hours}} = 1 \, \text{complete cycle} \)
Productive passes per day: 1 (since each cycle completes one pass)
Step 4: Calculate Total Extraction Time
Accounting for operational efficiency of 90%:
\[ \text{Effective passes per day} = 0.9 \] \[ \text{Days required} = \frac{334}{0.9} \approx 371 \, \text{days} \quad \Rightarrow \quad \boxed{370} \, \text{days} \]
| Feature in mine plan | Symbol | ||
| P | shaft | 1 | ![]() |
| Q | staple shaft | 2 | ![]() |
| R | Abandoned shaft | 3 | ![]() |
| S | Abandoned staple shaft | 4 | ![]() |
