Drip irrigation calculator
This drip irrigation calculator turns emitter flow and spacing into an application rate: 0.3 gph emitters 12 in apart on beds 24 in wide apply about 0.24 in per hour, so 1 inch of water takes about 4 hours 9 minutes. It also gives the flow of a drip line.
Example: Units: US (gph, in, ft) · Emitter flow (gph or L/h) 0.3 · Emitter spacing (in or m) 12 · Line spacing or bed width (in or m) 24 · Water to apply (in or mm) 1 · Drip line length (ft or m) 100 → Run time: 0.241 in/h – 1 in takes 4 h 9 min. Source: FAO, Pressurized Irrigation Techniques, Kap. 4 „System design“, S. 4.1–4.2: Durchfluss der Leitung = Tropfer × Tropfermenge; Laufzeit = Gabe (mm) ÷ Beregnungsintensität (mm/h) (gelesen 10.10.2026). Updated: .
How it is calculated
Application rate (in/h) = emitter flow (gph) × 231 ÷ (emitter spacing in × line spacing in). Each emitter waters the rectangle between its neighbours and the next line; 231 cubic inches make one gallon. The University of Arkansas (FSA6174) uses 231.1 in this formula, Utah State University writes it as 1.6 × gph ÷ area in ft². The run time then follows the FAO rule: duration = water to apply ÷ application rate.
Run time for 1 inch of water
| Emitter flow | Emitter spacing | Line spacing | Inches per hour | Hours for 1 in |
|---|---|---|---|---|
| 0.3 gph | 12 in | 24 in | 0.24 | 4.16 |
| 0.5 gph | 12 in | 24 in | 0.40 | 2.49 |
| 0.5 gph | 12 in | 36 in | 0.27 | 3.74 |
| 1 gph | 18 in | 36 in | 0.36 | 2.81 |
Utah State’s example – 0.5 gph on 1 ft × 6 ft – gives 0.13 in per hour. In metric units the rule is simpler: 1 liter on 1 m² is 1 mm, so mm/h = L/h ÷ (spacing m × line spacing m).
Flow of a drip line
The FAO gives the flow of a lateral as the number of emitters times the emitter flow. A 100 ft tape with emitters every 12 in has 100 emitters; at 0.3 gph that is 30 gph or 0.5 gpm per 100 ft, as in the Arkansas example. Add up the lines that run at the same time to check that your water supply can deliver it.
How much and how often
FSA6174 notes that most fruit and vegetable crops need about 1 to 2 inches of water per week. The soil decides how that is split: according to FSA6174, clay soils absorb water slowly but hold it longer, and usually do best with lower emitter flows and less frequent irrigation cycles. Infiltration rate and water-holding capacity of your soil should guide flow, spacing and how long and how often you run the system.
Converting gallons and liters is quicker with the liters to gallons converter; for sprinklers on a lawn use the lawn watering calculator.
Frequently asked questions
How long should I run drip irrigation for 1 inch of water?
Divide 1 inch by the application rate. With 0.3 gph emitters 12 in apart on 24 in beds the rate is about 0.24 in/h, so about 4 hours 9 minutes.
How do I calculate inches per hour for a drip system?
Emitter flow in gph × 231 ÷ (emitter spacing × line spacing in inches). Utah State writes the same as 1.6 × gph ÷ area per emitter in ft².
How many gallons per hour does a 100 ft drip line use?
Number of emitters × emitter flow. With emitters every 12 in there are 100 emitters; at 0.3 gph that is 30 gph or 0.5 gpm.
How much water do vegetables need with drip?
The University of Arkansas gives about 1 to 2 inches per week for most fruit and vegetable crops; the right amount depends on crop, weather and soil.
Why does the rate depend on line spacing?
The water from each emitter is counted over the area it serves – emitter spacing times line spacing. Wider spacing spreads the same flow over more area, so the rate in inches per hour drops.
Sources and legal basis
- FAO, Pressurized Irrigation Techniques, Kap. 4 „System design“, S. 4.1–4.2: Durchfluss der Leitung = Tropfer × Tropfermenge; Laufzeit = Gabe (mm) ÷ Beregnungsintensität (mm/h) (gelesen 10.10.2026)
- University of Arkansas, FSA6174 „Drip Irrigation Fundamentals“: Drip Rate (in/h) = gph × 231,1 ÷ (Abstand in × Beetbreite in), Beispiel 0,3 gph / 12 in / 24 in, 0,5 gpm je 100 ft; Hinweise zu Boden und maximaler Laufzeit (gelesen 10.10.2026)
- Utah State University Extension, Drip Irrigation for Berries: Rate (in./hr.) = 1,6 × gph ÷ Fläche (ft²), Beispiel 0,5 gph auf 1 ft × 6 ft = 0,13 in/h (gelesen 10.10.2026)
- NIST SP 811, Appendix B.8: 1 gal (U.S.) = 3,785 412 L, 1 in = 2,54 cm (gelesen 10.10.2026)
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