How many sausages would you need to throw into the ocean to raise sea level one inch?
Assumptions I’ll use:
Ocean area ≈ 3.61×10^8 km² = 3.61×10^14 m².
Raise by 1 inch = 0.0254 m.
A typical sausage ≈ cylinder 10 cm long × 2.5 cm diameter (0.10 m × 0.025 m).
Sausage volume = π·r²·h, density ≈ 1000 kg/m³ (water-like).
Calculation (numbers rounded sensibly):
Extra ocean volume = area × height = 3.61×10^14 m² × 0.0254 m ≈ 9.1694×10^12 m³.
Sausage volume = π·(0.0125 m)²·0.10 m ≈ 4.9087×10^-5 m³.
Number of sausages = 9.1694×10^12 m³ ÷ 4.9087×10^-5 m³ ≈ 1.87×10^17 sausages.
So you’d need on the order of 1.9×10^17 sausages (about 190 quadrillion sausages).
Mass equivalent (for scale):
Mass per sausage ≈ 0.0491 kg → total mass ≈ 9.17×10^15 kg ≈ 9.17×10^12 metric tonnes.
Short takeaway: unimaginably huge — hundreds of quadrillions of sausages, weighing trillions of tonnes. Even with very different sausage sizes you’d still need on the order of 10^15–10^18 sausages.
