smstonypoint
Super Member
- Joined
- Oct 13, 2009
- Messages
- 5,820
- Location
- SC (Upstate) & NC (Piedmont)
- Tractor
- NH TN 55, Kubota B2320 & RTV 900, Bad Boy Outlaw ZTR
I use the mower width times MPH divided by 10 as my estimate for mowing time. This is about 81% efficient so if you know the size of field you see how well you keep to speed, overlap and turns. 5280ft/Hr x 10ft/43560ftft x 1 MPH = 1.21 ac / hr This would be for a 10ft mower at 1 MPH
ksbrdmkr's formula to obtain theoretical efficiency (acres/hour) can be simplified to (speed in mph * mower width in feet)/8.25. Dividing by 10 rather than 8.25 allows for overlap, turning, etc. and results in a field efficiency of 83%.
http://www.caes.uga.edu/departments/bae/extension/handbook/documents/capacity.pdf gives typical field efficiencies for various implements. The reported values for mowers range from 75-85%.
TractorData.com - Mowing with tractors has a mowing time calculator. That calculator assumes >90% efficiency -- too high in my opinion.
eXmark provides a table showing acres per hour mowing times for alternative mower width and speeds at 100% and 80% efficiencies (Mower Productivity - Exmark | The Efficiency of Our Lawn Mowers and notes that the 80% data are more representative of actual practice.
Steve
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