---
title: "Subtense bar, elevation difference and gradient P-Q"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:subtense-staff-elevation-difference-gradient-pq"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/subtense-staff-elevation-difference-gradient-pq
site: BCT Engineering
---

# Subtense bar, elevation difference and gradient P-Q

Surveying II · Chapter 2: Tacheometry

- Asked in: 2075 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/subtense-staff-elevation-difference-gradient-pq
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

Calculate the elevation difference and gradient between station P and Q from the data given below, which are observed by a theodolite from station A with tachometric constant 100 and 0. The staff was held at station Q and subtense bar at station P. The subtended angle between the instrument and 3 m long bar was 00°17'40".

| Inst. St'n | Target St'n | Azimuth | Vertical angle | Staff reading (m) | Subtends bar height | Remarks |
|---|---|---|---|---|---|---|
| A | P | 37°45' | (-) 10°31' | × | 1.75 m | |
| | Q | 112°15' | (+) 7°15' | 0.56, 1.61, 2.66 | × | Staff vertical |

## Short answer

Subtense bar: D=1.5cot(0°8'50")=583.767 m, V=-108.37 m, RL_P-RL_axis=-110.12 m. Stadia: s=2.10 m, H=206.656 m, V=26.29 m, RL_Q-RL_axis=24.68 m. Elevation difference = 134.80 m (Q higher), PQ = 564.81 m, gradient = 1 in 4.19.

## Answer

Assumptions: the height of instrument at A is not given, so it cancels in the RL difference. "Subtense bar height 1.75 m" is taken as the height of the bar (the point sighted) above the ground at P. Staff is vertical; $k = 100$, $c = 0$.

#### Station P (subtense bar, b = 3 m, θ = 17'40")

$$
D_{AP} = \frac{b}{2}\cot\frac{\theta}{2} = 1.5\cot(0°08'50") = 583.767\ \text{m}
$$

$V_P = D_{AP}\tan(-10^\circ31') = -108.370$ m

Level of P above the axis of instrument: $RL_P - RL_{axis} = V_P - 1.75 = -110.120$ m

#### Station Q (staff, vertical angle +7°15')

- $s = 2.66 - 0.56 = 2.100$ m
- $H_{AQ} = k s\cos^2\theta = 100\times 2.10\times\cos^2 7^\circ15' = 206.656$ m
- $V_Q = \frac{ks}{2}\sin 2\theta = 50\times 2.10\times\sin 14^\circ30' = 26.290$ m
- $RL_Q - RL_{axis} = V_Q - 1.61 = 24.680$ m

#### Elevation difference

$$
RL_Q - RL_P = 24.680 - (-110.120) = 134.800\ \text{m}
$$

Q is higher than P by 134.800 m.

#### Distance PQ

Angle PAQ = $112^\circ15' - 37^\circ45' = 74^\circ30'$.

$$
PQ = \sqrt{583.767^2 + 206.656^2 - 2(583.767)(206.656)\cos 74^\circ30'} = 564.811\ \text{m}
$$

#### Gradient

$$
\frac{134.800}{564.811} = 1\text{ in }4.19
$$

**Answer: elevation difference = 134.80 m (Q higher than P); PQ = 564.81 m; gradient = 1 in 4.19 rising from P to Q.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Distance to bar at P (3 m bar, angle θ):
   A. 1.5 cot(θ/2)
   B. 3 tanθ
   C. 300 s
   D. 3 cot θ
   Answer: A. D = (b/2) cot(θ/2). (Key point: Distance AP)

2. Vertical component to P with D and depression:
   A. V = D tan(-10°31')
   B. V = D sin2θ
   C. V = Ks
   D. V = D cos θ
   Answer: A. The subtense distance is horizontal, so V = D tanθ. (Key point: V to P)

3. RL of P relative to axis is:
   A. V_P - 1.75
   B. V_P + 1.75
   C. 1.75 - V_P
   D. V_P
   Answer: A. The bar height 1.75 m is subtracted like h. (Key point: Level of P)

4. Angle PAQ is:
   A. 112°15' - 37°45' = 74°30'
   B. Sum of azimuths
   C. Sum of vertical angles
   D. 90°
   Answer: A. Difference of azimuths. (Key point: Angle PAQ)

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