Rectangular Concrete Formula
Length in feet multiplied by width in feet multiplied by height in feet divided by 27 equals cubic yards needed.
When dealing in inches: convert inches to feet by dividing by 12.
For example: pour a 25 ft. sidewalk 5 ft. wide and four inches deep.
(25’ x 5 ‘ x 0.333’) / 27 = 1.54 cubic yards
Note* we achieved 0.333 ft. by taking four inches and dividing by twelve.
Another example: yardage for an 80 ft. driveway, 24 ft. wide and 5 inches deep.
(80’ x 24’ x 0.416’) / 27 = 29.58 cubic yards
Round Columns Formula
Area of a circle equals pie (3.14159) multiplied by the radius squared, multiplied by the height in feet, divided by 27 equals cubic yards needed.
Note* Radius equals diameter divided by 2.
For Example: A 2 ft. diameter light pole foundation 10 ft. tall.
(3.14159 x (1’ x 1’) x 10’ ) / 27 = 1.164 cubic yards
Another Example: A 4 ft. diameter foundation 4 ft. deep.
(3.14159 x (2’ x 2’) x 4‘ ) / 27 = 1.86 cubic yards
Courses of Block
Height in feet, divided by 2, multiplied by 3 equals number of courses.
For example: 8 ft. tall wall
(8’/2) * 3 = 12 courses
Number of Block per Course
Lineal feet of wall, divided by 4, multiplied by 3 equals number of block per course.
For example: 50 ft. long wall
(50’/4) * 3 = 37.5 block per course
Total Block for Project
Number of Block multiplied by number of course equals total number of block needed for project.
For example: 8 ft. tall wall by 50 ft. long
12 courses * 37.5 blocks per course = 450 total blocks
*Note: Block widths vary and do not affect formula
Stone
Length in Feet X Width in feet x depth in feet (inches divide by 12 to convert to feet) then divide by 27 to get cubic yardage. Approximate compaction factor 15% - multiply by 1.15 to get total yardage.
80 ' x 24 ' x 4 inches deep of 6AA stone
((80 x 24 x .333) / 27 ) x 1.15 = 27.26 yards
**Compaction factors are approximate and vary depending on material type and moisture content.
Rectangular Concrete Formula
Length in feet multiplied by width in feet multiplied by height in feet divided by 27 equals cubic yards needed.
When dealing in inches: convert inches to feet by dividing by 12.
For example: pour a 25 ft. sidewalk 5 ft. wide and four inches deep.
(25’ x 5 ‘ x 0.333’) / 27 = 1.54 cubic yards
Note* we achieved 0.333 ft. by taking four inches and dividing by twelve.
Another example: yardage for an 80 ft. driveway, 24 ft. wide and 5 inches deep.
(80’ x 24’ x 0.416’) / 27 = 29.58 cubic yards
Round Columns Formula
Area of a circle equals pie (3.14159) multiplied by the radius squared, multiplied by the height in feet, divided by 27 equals cubic yards needed.
Note* Radius equals diameter divided by 2.
For Example: A 2 ft. diameter light pole foundation 10 ft. tall.
(3.14159 x (1’ x 1’) x 10’ ) / 27 = 1.164 cubic yards
Another Example: A 4 ft. diameter foundation 4 ft. deep.
(3.14159 x (2’ x 2’) x 4‘ ) / 27 = 1.86 cubic yards
Courses of Block
Height in feet, divided by 2, multiplied by 3 equals number of courses.
For example: 8 ft. tall wall
(8’/2) * 3 = 12 courses
Number of Block per Course
Lineal feet of wall, divided by 4, multiplied by 3 equals number of block per course.
For example: 50 ft. long wall
(50’/4) * 3 = 37.5 block per course
Total Block for Project
Number of Block multiplied by number of course equals total number of block needed for project.
For example: 8 ft. tall wall by 50 ft. long
12 courses * 37.5 blocks per course = 450 total blocks
*Note: Block widths vary and do not affect formula
Note:These formulas are provided as a reference for the convenience of our site visitors. No allowances have been made for variations in grade thicknesses, waste, spillage, or shrinkage. Elmer's has no responsibility for or control over the resultant quantities using these formulas.
