$C = w\,S\,e$ — read aloud C equals w S e. $C$ is the effective field capacity, the ground actually covered per hour; $w$ is the working width of the machine — the swath it really cuts, not the width of the frame; $S$ is the field speed while working; and $e$ is the field efficiency, a bare fraction.

The metric bridge is the same 10,000 m² as everything else here: a metre of width travelled at a kilometre an hour sweeps 1,000 m²/h, so ha/h = m × km/h ÷ 10. Its imperial twin divides by 8.25, for the same reason and with different definitions: 43,560 ft² per acre over 5,280 ft²/h per foot-mph.

The honest part of that equation is $e$, and it is defined by measurement: $e = \dfrac{C_e}{C_t}$, the effective capacity over the theoretical one, where theoretical means width times speed with no time lost at all. Efficiency absorbs the headland turns, the refills, the unblocking, the overlap between passes and the operator's lunch. Field-work planning figures run about 70 to 85% for tillage and seeding, lower for anything that stops to unload. Quoting the theoretical capacity is how a day's work becomes two.

Then $t = \dfrac{A}{C}$ — hectares over hectares-per-hour leaves hours. $A$ is the field, $C$ the capacity, and $C$ must be the effective one. Apply efficiency a second time here and you are discounting the same lost time twice.

The reason this lesson exists is that it is almost never asked as arithmetic. It is asked as: the rain arrives in nine hours — do we start this field or the other one? The hours are the working, and the answer is a decision. Speed is a real lever on it — capacity is proportional to speed — but every implement has a speed past which work quality goes, and the losses efficiency measures grow with it too.