The Effective Field Capacity Calculator helps you find out how much land agricultural equipment can cover in a given time. It measures the real working efficiency of machinery like tractors, seeders, or harvesters in field operations. While theoretical capacity assumes non-stop perfect operation, this calculator adjusts for practical factors such as turning time, machine overlap, and downtime.
Farmers, agricultural planners, and machinery dealers use this tool to estimate labor time, optimize machinery usage, and improve cost-efficiency on the farm. This calculator belongs to the Agricultural Equipment Efficiency Calculators category.
formula of Effective Field Capacity Calculator
For Metric Units (hectares per hour):
EFC = (S * W * E) / 10
For Imperial Units (acres per hour):
EFC = (S * W * E) / 8.25
Variables:
EFC (Effective Field Capacity):
The actual field area a machine covers per hour, accounting for all working conditions. Units: hectares/hour (ha/h) or acres/hour (ac/h).
S (Operating Speed):
The speed at which the equipment moves through the field. Units: kilometers per hour (km/h) or miles per hour (mph).
W (Working Width):
The width of the implement that actively works the soil or crops. Units: meters (m) or feet (ft).
E (Field Efficiency):
A decimal representing real-life efficiency, including losses from turns, refueling, or adjustments. For example, 80% efficiency = 0.8.
Use the metric version for metric units and the imperial version for imperial units.
Reference Table for Common Agricultural Inputs
| Operating Speed (S) | Working Width (W) | Field Efficiency (E) | EFC (ha/h or ac/h) |
|---|---|---|---|
| 5 km/h | 2.5 m | 0.85 | 1.06 ha/h |
| 6 km/h | 3.0 m | 0.80 | 1.44 ha/h |
| 4 mph | 10 ft | 0.75 | 3.64 ac/h |
| 5 mph | 12 ft | 0.80 | 5.82 ac/h |
This table helps you quickly estimate the output without needing to perform full calculations every time.
Example of Effective Field Capacity Calculator
Problem:
A tractor operates at 6 km/h with a plow that has a 2.5 m working width. The field efficiency is 0.85.
Step 1: Apply the formula
EFC = (6 * 2.5 * 0.85) / 10
EFC = (12.75) / 10 = 1.275 ha/h
Result:
The tractor effectively covers 1.275 hectares per hour under these conditions.
Most Common FAQs
A: The divisor adjusts for unit conversions. In the metric system, dividing by 10 converts from km·m to hectares. In the imperial system, dividing by 8.25 converts mph·feet to acres.
A: Typical field efficiency ranges from 0.70 to 0.90 depending on the operation. Seeding or spraying may have higher efficiency, while tillage operations often have lower values due to turning and overlaps.
A: You can increase operating speed, widen the implement, or improve field efficiency by reducing downtime and overlap during operations.