A level does one thing: it gives a perfectly horizontal line of sight. Everything else is bookkeeping, and the bookkeeping has exactly two signs in it.

$E_2 = E_1 + \text{BS} - \text{FS}$ — E-two equals E-one plus B S minus F S. $E_1$ is the elevation of a point you already know, usually a benchmark, in metres. BS is the backsight: the rod reading taken on that known point. FS is the foresight: the rod reading on the new point. $E_2$ is the new point's elevation, and it is what you are solving for.

The words mislead every year, so say it plainly: backsight and foresight have nothing to do with which way the telescope faces. A backsight is taken on a point of known elevation; a foresight on a point whose elevation is being established. The instrument may well be pointing forward down the route when it takes a backsight.

The signs come from a quantity that then vanishes. Add BS to the known elevation and you have the height of instrument — the elevation of the line of sight itself. Subtract FS and you come back down to the ground at the new point. The instrument's own height never appears in the answer, which is why the level need not be set up anywhere in particular, or even level with anything.

Stadia: $D = K s + C$. The telescope carries two extra horizontal hairs above and below the crosshair. $s$ is the rod intercept — upper hair reading minus lower hair reading, in metres. $K$ is the stadia interval factor, a bare number, 100 on virtually every instrument ever made. $C$ is the additive constant in metres: zero on modern internal-focusing instruments, a few tenths on older external-focusing ones. $D$ is the horizontal distance. It is similar triangles and nothing more: the fixed angle between the hairs covers more staff the further away the staff is.

And when a parcel's corners have been coordinated, its area comes straight off the reduced field book by the shoelace: $A = \tfrac{1}{2}\left[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right]$, where each $(x, y)$ is a corner's easting and northing in metres. List the corners counter-clockwise and the answer comes out positive; clockwise and it comes out negative, same size. The $\tfrac{1}{2}$ is the one people drop, and dropping it sells a parcel at twice its area.