The last thing a field calculation does is put a number on a delivery docket, and each material is bought in its own currency. Getting the currency wrong is a worse error than getting the arithmetic slightly wrong, because nobody catches it until the truck arrives.

Concrete, by volume: $V = L\,W\,T\left(1 + \dfrac{w}{100}\right)$ — V equals L W T, times one plus w over one hundred. $L$, $W$ and $T$ are the length, width and thickness of the pour, all in metres by the time they reach the formula, and $V$ is in cubic metres. $w$ is the waste allowance in percent — 5 to 10 % is normal, covering over-dig, spillage, an uneven subgrade and the wash-out at the end. It is not padding. Concrete cannot be topped up an hour later; a slab short by a wheelbarrow gets a cold joint it will keep for fifty years.

The unit trap here is the thickness. Drawings give slabs in millimetres and plan dimensions in metres, so 150 mm must become 0.15 m before it multiplies. Leave it as 150 and the order is a thousand times too large, which at least gets a phone call from the batching plant.

Asphalt, by mass: $M = A\,t\,\rho$ — M equals A t rho. $A$ is the paving area in square metres, $t$ the compacted lift thickness in metres, and $\rho$ — rho — the compacted density of the mix, around 2300–2400 kg/m³ for a dense-graded surface course. The product is a mass in kilograms; divide by a thousand for tonnes, which is how the plant sells it. Two words carry the whole risk: compacted thickness. Behind the paver the mat is roughly a quarter thicker than it will be after the roller, and ordering on the loose depth buys a quarter more mix than the job needs.

A stockpile, by geometry: $V = \dfrac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 A_2}\right)$, the truncated pyramid. $A_1$ is the base area, $A_2$ the flat top area, both in square metres, and $h$ is the vertical height — straight up, not along the batter face. The middle term, the geometric mean of the two areas, is what makes it work: a taper narrows linearly in its side lengths, so its areas narrow as the square, and a straight average of the ends would over-count the middle. Watch the formula do the whole family: set $A_2$ to zero and it collapses to $\tfrac{1}{3}A_1 h$, a full pyramid; set $A_2 = A_1$ and it becomes $A_1 h$, a prism.