Farmwithjunk
Super Member
Farmwithjunk said:Width in inches X MPH, divide by 10 will give you 10 hours worth of acreage. Divide by 10 AGAIN and get 1 hours worth. That equates to 1.4 acres per hour @ 5MPH and a 2X14" plow. That doesn't include "wasted motion" like turning and wheel spin. Factor in 10 to 15% for that.
After a good bit of searching in my office, I found a copy of a study done by Ohio State University AG Dept. back in the early 1970's. It gave their "magic formula" for figuring acreage covered by typical tillage implements per hour. The formula works with ANY sort of tractor drawn implements, not just tillage tools. I pulled this article out of a copy of "Prarie Farmer" magazine.
Width of implement (in inches) X MPH divided by 100 = per hour coverage. Wheel spin, turns, ect. reduce efficiency. Estimate percentage of wasted motion. Let's say it's 10%. With that, you take width (in inches) X MPH divide by 110. (100 + the "10%") It works out to a very accurate estimation in most cases. The only arbitrary figure is the percentage of wasted motion. Bigger fields (with less turns per acre) tend to lower the wasted time factor. Smaller implements on bigger tractors reduce the wheel spin percentage. In the case of implements that "overlap" previously worked ground, you need to use the width actually being worked, and NOT the width of the implement. (i.e. a 10' disc that's lapping over a foot into the last pass would use 9' (108") in the formula)
The purpose of their formula was to help determine that wasted motion percentage. By taking actual amount of acreage worked in an hour, and dividing it by (width X MPH) you can determine that final part of the equasion. Amount beyond 100 is the percentage of lost effort. The study was originally done to create a simple method of determining traction efficiency with different brands of tractor tires.
Fast, easy, and accurate formula.
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