Posted 9/18/2024 10:29 (#10894948) Subject: Harvesting field efficiency
If I run at 2.5 mph with a 35-foot header, I found an equation from the internet: 2.5 × 35 ÷ 8.25 = 10.6. This gives me my acres per hour. However, I am not sure how to find my field efficiency to determine my final acres per hour. Could you please help me with this?
Harvesting field efficiency
