Mowing Time Calculator
Enter the lawn area, the deck width, how fast you walk or drive, the overlap between passes and a field efficiency. The calculator gives the area mowed per hour and the time the lawn takes.
Mowing time
0.169 hours
Mowing time in minutes
10.1 min
Area mowed per hour
59,136 sq ft/hr
Acres mowed per hour
1.358 acres/hr
Area per hour with no turns or stops
73,920 sq ft/hr
Width cut per pass
42 in
How it works
A mower cuts a strip as wide as its deck, less the overlap with the previous pass, and moves forward at its ground speed. Width times distance per hour is the area it could cut in an hour if it never stopped: the theoretical field capacity used in farm machinery management. At 1 mph a mower travels 5,280 ft in an hour.
No mower cuts all the time. Turns at the ends of each pass, trimming around trees and beds, and emptying the bag take time. Field efficiency is the share of the theoretical rate actually achieved. Iowa State University's Estimating the Field Capacity of Farm Machines (PM 696) lists median field efficiencies of 78–83% for rotary mower-conditioners; the 80% default here is in that range, but a small lawn with many obstacles can be well below it.
Mowing time is the lawn area divided by the effective rate.
Formula
width per pass (ft) = (deck width − overlap) (in) ÷ 12 theoretical (sq ft/hr) = width (ft) × speed (mph) × 5,280 effective (sq ft/hr) = theoretical × field efficiency time (hours) = area (sq ft) ÷ effective rate acres per hour = width (ft) × speed (mph) × efficiency ÷ 8.25
Example
A 42 in deck with the overlap left inside the efficiency figure cuts a 42 in (3.5 ft) strip. At 4 mph that is 3.5 × 4 × 5,280 = 73,920 sq ft an hour with no turns. At 80% efficiency, which covers overlap as well as turns and stops, it mows 59,136 sq ft (1.358 acres) an hour.
A 10,000 sq ft lawn takes 10,000 ÷ 59,136 = 0.169 hours, about 10.1 minutes.
Assumptions and limitations
- Field efficiency covers overlap between passes, turns, trimming round obstacles, emptying the bag and other stops, as Iowa State PM 696 (revised February 2026) defines it; that is why the overlap defaults to 0. An overlap entered alongside such a figure is counted twice; the overlap input is for an efficiency that covers only turns and stops. The 80% default is illustrative: PM 696 lists 78–83% for farm-scale rotary mower-conditioners, and home lawns with beds and trees are often lower.
- Ground speed is the average cutting speed, not the mower's top speed. The default and the walking range in the help text are typical figures, not manufacturer data.
- The time counts mowing only: not trimming and edging by hand, blowing clippings off paths, or getting the mower out and putting it away.
- The lawn is mowed in one direction in parallel passes; striping patterns, double cutting or slopes that force slower speeds take longer.
Frequently asked questions
Where does the 8.25 in acres per hour come from?
An acre is 43,560 sq ft and a mile is 5,280 ft, and 43,560 ÷ 5,280 = 8.25. So width in feet times speed in mph divided by 8.25 is acres per hour. Iowa State PM 696 gives the example of a 60 ft implement at 6 mph: 60 × 6 ÷ 8.25 = 43.6 acres per hour, or 37.1 at 85% efficiency.
How can I measure my mowing speed?
Iowa State PM 696 marks off 88 ft and times the run: speed in mph is 60 divided by the seconds taken. Covering 88 ft in 30 seconds is 2 mph; in 12 seconds, 5 mph.
More in Lawn & Garden calculators.