An 8-foot wide by 90-foot long by 8-inch depth requires a minimum of 2,565.5 cubic yards (22.8 tons) of gravel.
Can't answer because with cubic yards you need a third dimension.
I don't think you can convert a cubic dimension into a squared dimension as a cubic dimension applies to volume and a squared dimension applies to area. Length X Width = Area (squared dimension); Length X Width X Depth = Volume (cubic dimension). One square yard = nine square feet; One cubic yard = 27 cubic feet.
Density of sand can vary depending on the grain size and moisture content and how tightly it is compactedIf you dont need very accurate densities the following should helpSand, wet 1922 Sand, wet, packed 2082 Sand, dry 1602 Sand, loose 1442 Sand, rammed 1682 Sand, water filled 1922 Sand with Gravel, dry 1650 Sand with Gravel, wet 2020density of sand will vary depending upon the condition.that is for wet,dry,gravel..
Not enough info to determine an answer. "Cubic inches" is a measurement of Volume with 3 dimensions. A "24 inch Pipe" can mean either a Length of 24 inches or a Diameter of 24 inches. To calculate a Volume of a pipe, you need the Diameter (or Radius) and the Length.
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You cannot make a direct conversion. You need to specify how deep you want the gravel to be in your area that needs covering.
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You will need 0.309 cubic yards (about 600 pounds) of gravel.
You will need 20 cubic yards or 25.7 tons of gravel.
You will need about 3.7 cubic yards (around 4.75 tons) of gravel.
You would need about 13 cu yards or about 15 tons.
You will need 65.93 cubic yards or 84.6 tons of gravel.
Approximately 24.7 cubic yards or 31.7 tons of gravel.
You will need about 65.185 cubic yards (83.6 tons) of gravel for this area.
You'll need about 28.704 cubic yards (37 tons) of gravel.