Context: K-coverage provides fault tolerance - if k=3, you can tolerate 2 simultaneous sensor failures at any location before losing monitoring capability. This formula calculates the minimum sensors needed for complete k-coverage of an area.
Formula: \[
N_{sensors} \geq K \times \frac{A}{\pi R_s^2} \times \alpha
\]
Where: A = area (m²), Rs = sensing radius (m), K = coverage level, α = overlap factor (typically 1.2-1.4)
Worked Example: A chemical plant safety system monitors a 10,000 m² production floor. Sensors have Rs = 15m. Safety regulations require k=3 coverage (tolerate 2 failures).
Step 1 – Calculate single sensor coverage: \[\text{Coverage per sensor} = \pi R_s^2 = \pi \times 15^2 = 707 \text{ m}^2\]
Step 2 – Minimum sensors for 1-coverage (theoretical): \[N_1 = \frac{10,000}{707} = 14.1 \text{ sensors}\]
Step 3 – Apply overlap factor (α = 1.3 for realistic deployment): \[N_1 = 14.1 \times 1.3 = 18.3 \approx 19 \text{ sensors}\]
Step 4 – Scale for k=3 coverage: \[N_3 = 3 \times 19 = 57 \text{ sensors}\]
Cost Analysis:
- 1-coverage: 19 sensors × $85 = $1,615
- 3-coverage: 57 sensors × $85 = $4,845
Key Insight: k=3 costs 3x the baseline but provides critical redundancy - if 2 sensors fail at any location, the third continues monitoring. For life-safety applications, this $3,230 premium ($4,845 - $1,615) is trivial compared to the cost of an undetected chemical leak ($500K+ in cleanup and regulatory fines).