Fencing: Concrete Bags Needed For Post Holes

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Running out of concrete halfway through a fence line costs a trip to the store and a crew standing around. A post hole is a cylinder, so the volume is simple geometry: pi times the radius squared times the depth. This recipe converts that to cubic feet, multiplies by the number of posts and divides by the yield of one bag.


Quick formula: Cylinder volume in cubic inches, converted to cubic feet, times the post count, divided by bag yield:
=ROUNDUP(B2*PI()*(C2/2)^2*D2/1728/E2,0)

20 posts in 10-inch by 30-inch holes with a 0.6 cubic-foot bag need 46 bags.

Functions used (tap for the full reference guide):

The example

Three jobs with different hole sizes. Wider and deeper holes grow the bag count quickly because diameter is squared.

ABCDEFG
1CasePostsDia (in)Depth (in)Yield cu ftBagsPer post
2Backyard2010300.60462.3
3Pool fence1212360.60484.0
4Gate run88240.60101.3

The formula

The volume of one round hole, scaled by the number of posts and the bag yield, rounded up:

=ROUNDUP(B2*PI()*(C2/2)^2*D2/1728/E2,0) // bags needed; per post = bags / posts

How it works

Here is the logic, one piece at a time:

  1. PI()*(C2/2)^2 is the area of the circular hole in square inches; the radius is half the diameter.
  2. Multiplying by depth D2 gives cubic inches per hole, and dividing by 1728 converts to cubic feet.
  3. Multiplying by the post count in B2 and dividing by the bag yield in E2 gives the bags needed.
  4. ROUNDUP(...,0) rounds up because a partial bag still has to be bought.

Concrete bag yields are printed on the bag: roughly 0.6 cu ft for an 80 lb bag and 0.45 cu ft for a 60 lb bag.

Try it: interactive demo

Interactive

Enter the post count, hole size and bag yield.

Variations

Subtract the post itself

A 4x4 post displaces concrete. Subtract its footprint from the hole area for a tighter count.

=ROUNDUP(B2*(PI()*(C2/2)^2-3.5*3.5)*D2/1728/E2,0)

Add a spill allowance

Multiply by 1.05 inside ROUNDUP to cover waste and uneven holes.

=ROUNDUP(1.05*B2*PI()*(C2/2)^2*D2/1728/E2,0)

Pitfalls & errors

Diameter is squared, so a hole 2 inches wider than planned adds far more concrete than 2 inches of extra depth would.

Do not mix inches and feet. If depth is in inches the divisor is 1728; if depth is in feet, drop the 1728 and convert the diameter.

Gravel in the bottom of the hole displaces concrete too. Subtract the gravel depth from the depth cell if you use it.

Practice workbook

📊
Download the free Fencing: Concrete Bags Needed For Post Holes practice workbook
Edit the yellow post, hole and yield cells; bags needed recalculates.

Frequently asked questions

Why 1728?
There are 12 x 12 x 12 = 1,728 cubic inches in a cubic foot.
Does this work for square post holes?
Replace PI()*(C2/2)^2 with C2^2 for a square hole of side C2.

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Related formulas: Fencing: Posts and Panels From a Run · Fencing: Pickets Per Section · Masonry: Mortar Bags for a Block Wall

Function references: ROUNDUPPI