How much gravel would be needed to cover an area 30 feet by 9 feet to a depth of 5 inches?

Answer 1

1ft = 12 in

→ volume required = 30 ft x 9 ft x 5 in = (30 x 12) in x (9 x 12) in x 5 in = 194400 cu in

1 in = 2.54 cm (exactly)

1 m = 100 cm

→ 1 cu m = 1c00 cm x 100 cm x 100 cm = 1,000,000 cu cm

→ 1 cu in = 1 in x 1 in x 1 in = 2.54 cm x 2.54 cm x 2.54 cm = 16.387064 cu cm

→ 194400 cu in = 194400 x 16.387064 cu cm ≈ 3,185,645 cu cm ≈ 3.186 cu m

Depending upon the gravel used (using coverage figures from a supplier):

20mm gravel:

one 850 kg bag covers aprox 0.6 cu m → 1 kg covers approx 600,000 / 850 cu cm ≈ 705.88 cu cm

→ require approx 3,185,645 / 705.88 kg = 4513 kg = 4.513 tonnes

Pea shingle:

one 25 kg bag covers approx 0.02 cu m → 1 kg covers approx 20,000 / 25 cu cm = 800 cu cm

→ require approx 3,185,645 /800 kg =3982 kg = 3.982 tones

Depending upon the gravel used, you will need about 4 to 5 tonnes