There is no conversion. "Cu ft" is a volume measurement, "kg" is a mass measurement.
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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
You need to know the density of the aggregate. Typically that will be in Imperial system (pounds per cubic feet) or in SI system (kilograms per cubic metre). Use Volume = Mass/Density to convert to volume in the same system. Then convert from SI to Imperial (or reverse) using 1 cubic metre = 35.315 cu ft or 1 cu ft = 0.02832 cubic metre.
Approximately 117.5 tons of dry sand. I willl use 1602 kg/m3 as the density of the sand. (See http://www.simetric.co.uk/si_materials.htm) We need common units, so first I will convert 1602 kg/m3 to lbs/cubic yard. 1602 kg/m3 = 1602 kg/m3 * (2.2 lb/kg) * (1m3/1.093 cu yds) = (1602*2.2)/1.093 = 2721 lbs/cu yard Now figure out the volume of sand you need (in cubic yards). volume = (14000 sq ft)*(2 in) = ((14000/9) sq yds)*((2/36) yds) = (14000/9)*(2/36) yds = 86.4 cu yds weight = density*volume=2721*86.4 lbs = 235094 lbs = 235094/2000 tons = 117.5 tons
1mtr 25x3 cu = 0.750 kg.
12 kg or 1/6th.