Acres per Hour Calculator
Work out how fast you cover ground. Enter implement width in feet, ground speed in mph, and a field efficiency percentage to get acres per hour — plus how long a field of a given acreage will take.
Example: with Implement width (ft) 15 · Ground speed (mph) 5 · Field efficiency (%) 80 · Field size (acres) 40 → Work rate: 7.27 acres per hour.
- Time for the field5.5 hr (5 hr 30 min) for 40 acres
- Ground covered316,800 sq ft per hour
Computed by the calculator below using its default values. Change any input to see your own numbers.
Field capacity: acres/hr = width (ft) × speed (mph) × efficiency ÷ 8.25. The 8.25 comes from 43,560 sq ft per acre ÷ 5,280 ft per mile.
Where the 8.25 comes from
An implement moving at 1 mph sweeps a strip 5,280 ft long each hour. Multiply by its width in feet and you get square feet per hour; divide by 43,560 sq ft per acre and the constants collapse to a single divisor: 43,560 ÷ 5,280 = 8.25. That is the standard field capacity formula in every agricultural engineering handbook — acres/hr = width × mph ÷ 8.25 at perfect efficiency.
Nothing runs at perfect efficiency. Headland turns, overlap between passes, refills, and clogs all steal time, which is what the efficiency percentage models.
Picking an honest efficiency
ASABE machinery-management data puts typical field efficiency at 70-90% for most operations: mowing and tillage usually run about 80-90%, planters 65-85%, and small odd-shaped fields with lots of turning sit at the low end. Overlap is the quiet killer — running a 15 ft bush hog with a half-foot of overlap is really a 14.5 ft machine before any turning losses. If you track your own hours per field, back-solve your true efficiency and use that.
How it’s calculated
Acres per hour = width (ft) × speed (mph) × (efficiency ÷ 100) ÷ 8.25, where 8.25 = 43,560 sq ft per acre ÷ 5,280 ft per mile. Time = field acres ÷ acres per hour. Square feet per hour = acres per hour × 43,560.
Efficiency is a single flat percentage — real losses vary with field shape, moisture, and operator, so treat times as planning estimates, not promises.
Acres per hour at 100% efficiency
| Width | Speed | Acres/hr |
|---|---|---|
| 5 ft (finish mower) | 3 mph | 1.8 |
| 6 ft (bush hog) | 4 mph | 2.9 |
| 10 ft (disc) | 5 mph | 6.1 |
| 15 ft (batwing) | 5.5 mph | 10.0 |
| 20 ft (field cultivator) | 6 mph | 14.5 |
Computed with acres/hr = width × mph ÷ 8.25; multiply by your efficiency (typically 0.7-0.9).
Common mistakes
- Using the mower's advertised size instead of its effective cut — overlap between passes trims 5-10% off the working width.
- Running the math at 100% efficiency; turns and refills mean real-world output is usually 70-90% of theoretical.
- Mixing units: width goes in feet and speed in mph. A 72 in deck is 6 ft.
- Quoting PTO or road speed instead of actual working speed under load, which is often 1-2 mph slower.
Frequently asked questions
What is the acres per hour formula?
Acres per hour = width in feet × speed in mph × field efficiency ÷ 8.25. The 8.25 constant is 43,560 sq ft per acre divided by 5,280 ft per mile.
How many acres can I mow per hour with a 6 ft bush hog?
At 4 mph and 80% efficiency: 6 × 4 × 0.8 ÷ 8.25 ≈ 2.3 acres per hour. A 10 acre field is about a 4.3 hour job.
What field efficiency should I use?
80% is a fair default for mowing and tillage in decent-sized fields. Drop to 65-75% for small or irregular fields with lots of headland turns; rise toward 90% for long straight passes with minimal overlap.
Why is my real time worse than the calculator says?
Usually overlap and turning. Effective width is what you actually cut per pass, not the machine's rating, and every end-row turn is time with zero acres covered — both belong in your efficiency number.
Does this work for seeding, spraying, or plowing?
Yes — the formula is generic field capacity. Just use that implement's working width and realistic speed; sprayers often run 8-12 mph while moldboard plows crawl at 4-5.