Center Pivot Rotation Time & Application Depth

I tell you how long a center pivot needs to run to apply a target depth of water — or how much depth it will lay down in a given run time.

Calculator

T = 452.6 × A × d × P / (Q × E)
Solve for
acres
gpm
gpm/acre

= 780 gpm total at the current irrigated area

%

Below 100% only if a wiper/GPS switch shuts part of the circle off (e.g. to avoid a road or farmstead). Leave at 100% for a full-circle pivot.

%
in
hr
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About this calculator

For a center pivot, the time needed to complete one pass while applying a given depth of water depends on the irrigated area, the flow rate delivered at the pivot point, how much of the full circle is actually being irrigated, and how efficiently that water is applied:

T = 452.6 × A × d × P / (Q × E)

where T is rotation time (hours), A is irrigated area (acres), d is the depth applied (inches), P is the fraction of the full circle being irrigated (decimal), Q is the flow rate at the pivot (gpm), and E is irrigation application efficiency (decimal). The constant 452.6 converts an acre-inch of water into gallons per minute per hour (1 acre-inch ≈ 27,154 gallons, and 27,154 ÷ 60 ≈ 452.6).

Pick whether you want to solve for rotation time or application depth; the other quantity becomes an input slider. This is the same relationship behind WSU's 1" Application Time calculator, generalized here to any target depth instead of only 1 inch, and solvable in either direction. Flow rate can be entered directly in gpm, or as a design rate in gpm/acre (a common way pivots are sized) — check the box under the flow field and the tool multiplies it by the irrigated area for you.

Worked example: A 130-acre full-circle pivot (P = 100%) delivering 800 gpm at 85% application efficiency needs about 86.5 hours (3.6 days) to lay down 1 inch of water. Push the depth slider down to 0.5 in and the rotation time drops to about 43.3 hours — half the depth, half the time.