7.2 Trigonometric Leveling

Key Takeaways

  • With zenith angle Z (0° at the zenith), vertical distance along the line of sight is VD = SD × cos(Z). Target elevation = elev of the instrument mark + h.i. + SD × cos(Z) − h.t.
  • The horizon form is equivalent: VA = 90° − Z and VD = SD × sin(VA) = HD × tan(VA), with HD = SD × sin(Z). Mixing zenith and horizon functions swaps VD with HD.
  • Worked one-way: mark 520.00 ft, h.i. = 5.28 ft, SD = 412.50 ft, Z = 82°15′30″, h.t. = 4.85 ft → VD = +55.57 ft → target mark 576.00 ft.
  • Worked depression: mark 318.75 ft, h.i. = 5.12 ft, SD = 287.40 ft, Z = 96°40′00″, h.t. = 6.00 ft → VD = −33.37 ft → target 284.50 ft. Cosine of Z > 90° is negative; do not drop the sign.
  • Reciprocal observations (A to B and B to A, then mean the two mark-to-mark differences) cancel most curvature and refraction on long sights. Use a level, not trig heights, when the job is a 0.01 ft pad with room for balanced setups.
Last updated: September 2026

7.2 Trigonometric Leveling

Quick Answer: Elev_target = elev_inst + h.i. + SD × cos(Z) − h.t., where Z is the zenith angle. The same vertical distance is SD × sin(VA) or HD × tan(VA) if VA is the vertical angle from the horizon (VA = 90° − Z). Reciprocal observations (both directions) cancel most curvature and refraction. Use trig heights on steep slopes, towers, and canyons; use a level when you can balance short sights and you need hundredths.

CES Domain II item B names trigonometric and differential leveling together. Differential leveling (§7.1) is the default elevation method. Trigonometric leveling is what a total station (or a theodolite plus EDM) does whenever the line of sight is not horizontal: you measure slope distance (SD) and a vertical pointing, reduce vertical distance (VD), then move from mark to mark with two tape heights.

Disambiguate names before you hit the calculator:

  • elev_inst — elevation of the station mark under the instrument (hub, monument, TP), not the elevation of the telescope.
  • h.i. — measured height from that mark up to the trunnion / horizontal axis. Typical values are 5.0–5.5 ft. This is not leveling HI (elevation of a horizontal line of sight).
  • h.t. — height of the target (prism, tape mark, or rod reading) above the target station mark.
  • Z — zenith angle, 0° straight up, 90° on the horizon, > 90° looking down. Same convention as Chapter 6.

Zenith Form

Along the line of sight:

VD = SD × cos(Z)
HD = SD × sin(Z) (needed only if you use the tan form below)

Then:

Elev_target = elev_inst + h.i. + VD − h.t.
Elev_target = elev_inst + h.i. + SD × cos(Z) − h.t.

Limits that catch a swapped sine/cosine: Z = 90° → cos(90°) = 0 → VD = 0 (purely horizontal). Z = 0° → VD = SD (looking at the zenith). When Z > 90°, cos(Z) is negative, so VD is negative and the target mark is below the trunnion; keep the sign instead of taking an absolute value and guessing “up or down.”

Horizon Form (Vertical Angle from the Horizon)

Some notes, theodolites, and clinometers store VA with 0° on the horizon, positive up, negative down.

VA = 90° − Z
VD = SD × sin(VA)
VD = HD × tan(VA)
Elev_target = elev_inst + h.i. + SD × sin(VA) − h.t.

sin(VA) = cos(Z) on the same pointing, so both forms agree. Feeding a zenith reading into sin(VA) or a horizon VA into cos(Z) returns something close to HD, which is the classic CES trap.

ConventionZeroVDWhen you see it
Zenith ZStraight upSD × cos(Z)Total station display
Horizon VAHorizonSD × sin(VA) = HD × tan(VA)Theodolite, clinometer, older books

Worked Example 1 — One-Way, Looking Up

Instrument over mark A, elev_A = 520.00 ft.
h.i. = 5.28 ft. Prism over mark B, h.t. = 4.85 ft.
SD = 412.50 ft. Z = 82°15′30″.

82°15′30″ = 82 + 15/60 + 30/3600 = 82.258333°
cos(82.258333°) = 0.134707
VD = 412.50 × 0.134707 = 55.57 ft (rounded to hundredths)

Horizon check: VA = 90° − 82.258333° = 7.741667° = 7°44′30″.
SD × sin(7.741667°) = 55.57 ft.
HD = 412.50 × sin(82.258333°) = 408.74 ft, and HD × tan(7.741667°) = 55.57 ft.

Elev_B = 520.00 + 5.28 + 55.57 − 4.85 = 576.00 ft.

Stack: 520.00 + 5.28 = 525.28 (elevation of the trunnion). 525.28 + 55.57 = 580.85 (elevation of the prism). 580.85 − 4.85 = 576.00 (mark B). Forgetting h.t. leaves you 4.85 ft high; forgetting h.i. leaves you 5.28 ft low. Using SD × sin(Z) as VD would give HD = 408.74 and a nonsense elevation near 929 ft.

At 412 ft, combined curvature and refraction is only about 0.0035 ft — smaller than a hundredth. Reciprocal observations are not what saves this short hillside shot; h.i., h.t., and the correct cosine are.

Worked Example 2 — One-Way, Looking Down (Z > 90°)

Instrument over mark C, elev_C = 318.75 ft. h.i. = 5.12 ft.
Prism over a streambed hub, h.t. = 6.00 ft. SD = 287.40 ft. Z = 96°40′00″ = 96.666667°.

cos(96.666667°) = −0.116092
VD = 287.40 × (−0.116092) = −33.37 ft

Elev_hub = 318.75 + 5.12 + (−33.37) − 6.00 = 284.50 ft.

Equivalent VA = 90° − 96.666667° = −6.666667° (depression). SD × sin(−6.666667°) = −33.37 ft. Taking the absolute value of VD and adding it because “the river is downhill” would put the hub at 318.75 + 5.12 + 33.37 − 6.00 = 351.24 ft — above the instrument, the wrong valley.

Worked Example 3 — Reciprocal Observations (Brief)

C+R and unmodeled refraction grow with distance squared. Across a canyon you cannot balance 40-ft level sights, so you occupy both ends.

Marks A and B, elev_A = 210.00 ft, SD = 3,850 ft, h.i. = h.t. = 5.40 ft (heights cancel in each one-way Δh).

A → B: Z = 88°12′00″.
VD_AB = 3,850 × cos(88.20°) = +120.93 ft.
One-way elev_B = 210.00 + 120.93 = 330.93 ft.

B → A (same day, similar atmosphere): Z = 91°48′20″.
VD_BA = 3,850 × cos(91.805556°) = −121.30 ft.
That is the height of A minus the height of B from B’s pointing, so the implied Δh_AB = +121.30 ft.
One-way elev_B from this pointing = 210.00 + 121.30 = 331.30 ft.

Mean Δh_AB = (120.93 + 121.30) / 2 = 121.115 → 121.12 ft.
Elev_B (reciprocal) = 210.00 + 121.12 = 331.12 ft.

The two one-ways differ by 0.37 ft. Combined C+R at 3,850 ft is about 0.31 ft (Chapter 9 develops 0.574 M² with M in miles: 3,850/5,280 ≈ 0.729 mi). Reciprocal averaging removes most of that systematic bend; a single long one-way does not. Heights that do not cancel (different h.i. and h.t. on the two occupations) stay in each one-way reduction and must still be taped carefully.

When Trig Leveling versus a Level

SituationPrefer
Pad, invert, or finished floor to 0.01 ft, with room to turnAutomatic or digital level, balanced BS/FS
Steep hillside where a level would need a dozen short setupsTrig heights from a total station
Top of a tower, bridge soffit, overhead pipeTrig, or an inverted rod on a level if the sight is short and nearly horizontal
River or canyon with no turning pointsReciprocal trig heights
Dense topo shots on a surfaceTotal-station trig or GNSS RTK (§7.3), not a level loop to every shot

A 3-minute vertical circle used as a “level replacement” over a 200 ft pad shot is the wrong tool; that geometry is in §7.3. Trig leveling is not sloppy leveling — it is a different measurement model. It still needs a checked vertical circle, a prism constant that matches the software, and h.i./h.t. measured to the same hundredth you want in the elevation.

Exam Traps

  1. Sine on a zenith angle — SD × sin(Z) is HD, not VD.
  2. Dropping the sign when Z > 90° — cosine is negative; the target is below the trunnion.
  3. Skipping h.i. or h.t. — a 5 ft tape error is 500 times a 0.01 ft spec.
  4. Calling leveling HI and total-station h.i. the same number — one is an elevation, the other is a 5-ft offset.
Loading diagram...
Trigonometric leveling: zenith or horizon VD, then h.i. and h.t.
Canyon example: one-way elevations of B versus reciprocal mean (feet)
Test Your Knowledge

A total station occupies a mark at 520.00 ft. Instrument height h.i. = 5.28 ft, slope distance to the prism = 412.50 ft, zenith angle Z = 82°15′30″, and target height h.t. = 4.85 ft. The elevation of the target mark is closest to which value?

A
B
C
D
Test Your Knowledge

A mark at 318.75 ft is occupied with h.i. = 5.12 ft. The pointing to a prism (h.t. = 6.00 ft) has SD = 287.40 ft and Z = 96°40′00″. The elevation of the target mark is closest to which value?

A
B
C
D
Test Your Knowledge

Why occupy both ends of a long canyon sight and mean the two trigonometric height differences?

A
B
C
D