{
  "id": "acres-per-hour",
  "version": "41ecc3e89012",
  "status": "published",
  "name": "Acres Per Hour Calculator",
  "question": "How many acres per hour?",
  "summary": "Computes the acres per hour (effective field capacity) of a tractor, mower, sprayer or planter from its working width, ground speed and field efficiency, with hectares per hour and the hours a field takes.",
  "category": "construction",
  "url": "https://www.acalculator.org/construction/acres-per-hour-calculator",
  "markdown": "https://www.acalculator.org/construction/acres-per-hour-calculator.md",
  "kind": "function",
  "method": "Acres per hour = speed × width × field efficiency ÷ 43,560 sq ft; in US units, mph × ft × efficiency ÷ 8.25. Hours = field size ÷ acres per hour.",
  "assumptions": [
    "The width is the effective width, without overlap; the speed is the working speed in the field.",
    "Field efficiency covers time lost to turns, overlap, filling, emptying and small repairs; Iowa State’s example uses 83%.",
    "1 acre = 43,560 sq ft and 1 mile = 5,280 ft, so mph × ft ÷ 8.25 gives acres per hour."
  ],
  "inputs": {
    "$schema": "https://json-schema.org/draft/2020-12/schema",
    "type": "object",
    "properties": {
      "w": {
        "title": "Working width",
        "description": "The width the implement covers in one pass, without overlap (a 60-inch mower deck is 5 ft).",
        "type": [
          "number",
          "string"
        ],
        "x-quantity": "length",
        "minimum": 0.01,
        "maximum": 100
      },
      "v": {
        "title": "Ground speed",
        "description": "The working speed in the field.",
        "type": [
          "number",
          "string"
        ],
        "x-quantity": "speed",
        "minimum": 0.01,
        "maximum": 50
      },
      "e": {
        "title": "Field efficiency",
        "description": "The share of time spent working at full width, after turns, overlap, filling and stops.",
        "type": "number",
        "x-unit": "percent",
        "minimum": 1,
        "maximum": 100
      },
      "field": {
        "title": "Field size",
        "description": "The area to cover, to work out the hours. Leave empty to skip.",
        "type": [
          "number",
          "string"
        ],
        "x-quantity": "area",
        "exclusiveMinimum": 0,
        "maximum": 1000000000
      }
    }
  },
  "outputs": {
    "acres": {
      "label": "Field capacity (acres per hour)",
      "description": "Speed × width × field efficiency ÷ 43,560 sq ft.",
      "format": "number"
    },
    "hectares": {
      "label": "Field capacity (hectares per hour)",
      "description": "The same capacity in hectares per hour.",
      "format": "number"
    },
    "theoretical": {
      "label": "Theoretical capacity (acres per hour)",
      "description": "Speed × width ÷ 43,560 sq ft, at 100% field efficiency.",
      "format": "number"
    },
    "hours": {
      "label": "Hours for the field",
      "description": "The field size ÷ the field capacity.",
      "format": "number"
    }
  },
  "defaultAnswer": {
    "inputs": {
      "w": "20 ft",
      "v": "5 mph",
      "e": 80
    },
    "outputs": {
      "acres": 9.696969696969697,
      "hectares": 3.9242244096,
      "theoretical": 12.121212121212121
    },
    "text": "You can cover about 9.7 acres per hour."
  },
  "examples": [
    {
      "given": {
        "w": 9.144,
        "v": 2.2352,
        "e": 83
      },
      "expect": {
        "acres": 15.090909090909092,
        "theoretical": 18.181818181818183
      },
      "source": "Iowa State University Extension, Ag Decision Maker A3-24, Estimating the Field Capacity of Farm Machines (field capacity in acres per hour = speed in mph × width in feet × field efficiency ÷ 8.25, where 8.25 = 43,560 ÷ 5,280; a 30-foot implement at 5.0 mph and 83% field efficiency covers about 15 acres per hour), https://www.extension.iastate.edu/AGDM/crops/html/a3-24.html (retrieved 2026-10-03); hand calculation in content.mdx: 30 × 5 × 0.83 ÷ 8.25 = 15.09 acres per hour"
    },
    {
      "given": {
        "w": 6.096,
        "v": 2.68224,
        "e": 80,
        "field": 647497.027584
      },
      "expect": {
        "acres": 11.636363636363637,
        "hours": 13.75
      },
      "source": "Iowa State University Extension, Ag Decision Maker A3-24, Estimating the Field Capacity of Farm Machines (field capacity in acres per hour = speed in mph × width in feet × field efficiency ÷ 8.25, where 8.25 = 43,560 ÷ 5,280; a 30-foot implement at 5.0 mph and 83% field efficiency covers about 15 acres per hour), https://www.extension.iastate.edu/AGDM/crops/html/a3-24.html (retrieved 2026-10-03); hand calculation in content.mdx: 20 × 6 × 0.80 ÷ 8.25 = 11.64 acres per hour; 160 ÷ 11.64 = 13.75 h"
    },
    {
      "given": {
        "w": 6,
        "v": 2.7777777777777777,
        "e": 75
      },
      "expect": {
        "hectares": 4.5,
        "acres": 11.11974216602244
      },
      "source": "Iowa State University Extension, Ag Decision Maker A3-24, Estimating the Field Capacity of Farm Machines (field capacity in acres per hour = speed in mph × width in feet × field efficiency ÷ 8.25, where 8.25 = 43,560 ÷ 5,280; a 30-foot implement at 5.0 mph and 83% field efficiency covers about 15 acres per hour), https://www.extension.iastate.edu/AGDM/crops/html/a3-24.html (retrieved 2026-10-03); hand calculation in content.mdx: 6 m × 10,000 m per hour × 0.75 = 45,000 m² = 4.5 ha per hour"
    }
  ],
  "sources": [
    "Iowa State University Extension, Ag Decision Maker A3-24, Estimating the Field Capacity of Farm Machines (Hanna; Rosentrater, updated February 2026): field capacity = speed × width × field efficiency ÷ 8.25, with 8.25 = 43,560 ÷ 5,280; a 30-foot implement at 5.0 mph and 83% field efficiency covers about 15 acres per hour. https://www.extension.iastate.edu/AGDM/crops/html/a3-24.html (retrieved 2026-10-03)"
  ],
  "related": [
    "acreage",
    "corn-yield",
    "speed"
  ],
  "changelog": []
}