Field Efficiency

Also known as machine efficiency · field time efficiency · effective versus theoretical capacity

e=CeCte = \frac{C_e}{C_t}

Worked example: 3.2 of a theoretical 4.0 ha/h → 80% — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

Field work rates →

UniversityApplied Field Engineering

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning.

See your Report Card
Compete with your friends
share your results
Learning zone

Field Efficiency explained

CtCee

Field efficiency is the difference between what a machine could do and what it does. Theoretical capacity is width times speed with no time lost at all — a machine that starts at one corner, never turns, never fills, never unloads and never stops. Effective capacity is what the field actually gets. The ratio is typically between one half and nine tenths, and where it falls in that range is more about the field than the machine.

What lives inside the number is worth enumerating, because each item is attacked differently. Turning at the headlands is usually the largest single loss and scales with how many turns the field shape forces. Filling a planter or a sprayer, and unloading a combine, is the next, and it is a function of tank size against the area a tank covers. Then come overlap between passes, adjustment stops, and the ordinary business of clearing a blockage. Published ranges reflect the mix: tillage runs 70–90% because a tillage tool stops for almost nothing, combines 65–80%, planters 55–75%, and sprayers 50–70% because a sprayer empties a tank every few hectares.

Because the losses are dominated by field geometry, the same machine has a different efficiency on every farm and sometimes on every field. A long rectangular field with a headland at each end is the best case. A small field, an awkward shape, a wet spot to drive round, or a set of point rows against a diagonal boundary all raise the number of turns per hectare, and the efficiency falls even though nothing about the machine has changed. This is the honest reason a contractor's quoted work rate does not reproduce on a farm of small fields, and it is not a complaint about the contractor.

The classic mistake is treating the figure as a property of the machine and carrying a book value between farms. The right move is to measure it: record the area covered and the clock time for a real day, divide to get effective capacity, and divide again by width × speed. That measured figure is worth more than any table, and it is the one that should go into a machinery costing. A measured value above 90% is usually a sign that the theoretical capacity was computed from too narrow a width or too low a speed, not that the operator has beaten the published range.

Field Efficiency formula

e=CeCte = \frac{C_e}{C_t}
Where
  • ee= Field efficiency (%)
  • CeC_e= Effective field capacity (ha/h)
  • CtC_t= Theoretical field capacity (ha/h)

Missing one of these? Work it out first, then come back