Acres Per Hour Calculator

Estimate how many acres a field implement covers per hour from its working width, ground speed, and field efficiency.

Quick Facts

Formula
Acres/hr = width(ft) × speed(mph) × efficiency ÷ 8.25
The 8.25 converts ft² to acres (43,560 ft²/acre ÷ 5,280 ft/mi).

Your Results

Calculated
Effective field capacity
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Acres covered per hour
Theoretical capacity
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At 100% efficiency
Hours per 100 acres
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Time to cover 100 ac

Ready

Enter width, speed, and efficiency, then calculate.

About the Acres Per Hour Calculator

Acres per hour — known formally as effective field capacity — measures how much land a piece of equipment can work in one hour. It is the core number behind planning tillage, planting, spraying, mowing, and harvesting operations, because it tells you how long a field will take and how much fuel, labor, and daylight the job will consume. The calculation combines how wide the implement is, how fast it travels, and how much of the theoretical maximum is actually achieved once turns and overlaps are counted.

The formula

Acres per hour is calculated as:

Acres/hour = (Working width in feet × Speed in mph × Field efficiency) ÷ 8.25

The constant 8.25 is a unit conversion, not an empirical guess. In one hour at S miles per hour you travel S × 5,280 feet. Multiplied by the width W in feet, the swept area is W × S × 5,280 square feet. Dividing by 43,560 square feet per acre gives (W × S × 5,280) ÷ 43,560, and 5,280 ÷ 43,560 equals exactly 1 ÷ 8.25 = 0.12121. So the swept area per hour is W × S ÷ 8.25 acres. Field efficiency (entered as a decimal, e.g. 0.80 for 80%) scales that theoretical figure down to the coverage you actually get.

Why field efficiency matters

A machine almost never covers 100% of its theoretical capacity. Time is lost turning at headlands, overlapping previous passes, refilling seed or spray tanks, unplugging, and adjusting. Field efficiency captures all of that as a single percentage. Typical ranges published by agricultural engineering references (such as ASABE Standard D497) are:

  • Moldboard and chisel plowing, tillage: roughly 80–90%
  • Planting and drilling: roughly 55–80%, depending on refill frequency
  • Spraying: roughly 55–80%
  • Mowing and mower-conditioning: roughly 75–85%
  • Combining (grain harvest): roughly 60–75%

Smaller, irregular, or hilly fields sit toward the low end because turns and point rows eat a larger share of the time; large rectangular fields sit toward the high end.

Worked example

A 30-foot implement travelling at 5 mph at 80% efficiency covers (30 × 5 × 0.80) ÷ 8.25 = 120 ÷ 8.25 = 14.55 acres per hour. Its theoretical maximum at 100% efficiency would be (30 × 5) ÷ 8.25 = 18.18 acres per hour. A 100-acre field would therefore take 100 ÷ 14.55 ≈ 6.9 hours of actual field time.

Frequently Asked Questions

What is the formula for acres per hour?
Effective field capacity in acres per hour equals working width in feet times ground speed in mph times field efficiency, divided by 8.25. The 8.25 converts square feet to acres: one acre is 43,560 ft² and one mph is 5,280 ft/hr, so width × speed × 5,280 ÷ 43,560 reduces to width × speed ÷ 8.25.
Where does the constant 8.25 come from?
It is 43,560 ÷ 5,280 = 8.25. Because one acre is 43,560 square feet and travelling at 1 mph covers 5,280 feet per hour, a 1-foot-wide pass at 1 mph sweeps 5,280 ft² per hour, which is 5,280 ÷ 43,560 = 0.12121 acre, or 1 ÷ 8.25. Dividing width-feet times speed-mph by 8.25 gives acres per hour directly.
What field efficiency should I use?
Use the operation-specific ranges: about 80–90% for tillage, 55–80% for planting and spraying, 75–85% for mowing, and 60–75% for combining. Pick the lower end for small, irregular, or hilly fields and the higher end for large, square fields. If you have measured a job before, dividing acres actually done by hours actually worked, then back-solving, gives your own real efficiency.
Do I use width in inches or feet?
Feet. The 8.25 constant is built for width in feet and speed in miles per hour. If your implement width is given in inches, divide by 12 first (a 360-inch header is 30 feet). Metric users should instead compute hectares per hour as width(m) × speed(km/h) × efficiency ÷ 10.

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