Subtense bar and tacheometer gradient A-B

Reading of the data: the subtense bar at B, observed from A, gives the horizontal distance AB. The tacheometer at R ($K=100$, $C=0$, staff vertical) gives the distance RB and the heights. For A only the axial reading (1.100 m) is recorded, so RA cannot be found from stadia readings; it is found from the triangle RAB (AB and RB known, angle ARB known from the bearings). The instrument axis at R is the same for both sights, so only the readings and vertical components are needed for the difference in level.

Step 1: Distance AB from the subtense bar (angle at A)

The bar is out of normal by $4^\circ$, so its effective length is $l\cos4^\circ=3\cos4^\circ=2.9927$ m.

$$AB=\frac{l'}{2}\cot\frac{\beta}{2}=\frac{2.9927}{2}\cot\frac{0^\circ50'20''}{2}=204.396\ \text{m}$$

Step 2: Distance RB and vertical component from R

Zenith angle $83^\circ45'$ gives $\theta=+6^\circ15'$.

Point B (bearing 210°00', angle of elevation $6^\circ 15'$):

$$\begin{aligned} s&=1.850-0.650=1.200\ \text{m}\\ D&=100\times1.200\times\cos^2 6^\circ 15'=118.578\ \text{m}\\ V&=\tfrac12\times100\times1.200\times\sin 2(6^\circ 15')=12.986\ \text{m} \end{aligned}$$

Step 3: Distance RA from the triangle RAB

Bearings from R: RA $=315^\circ$ and RB $=210^\circ$, so angle ARB $=105^\circ$. By the cosine rule:

$$AB^2=RA^2+RB^2-2\,RA\,RB\cos105^\circ$$ $$RA^2-2(118.578)\cos105^\circ\,RA+(118.578^2-204.396^2)=0\ \Rightarrow\ RA=138.599\ \text{m}$$

(the other root is negative).

Step 4: Vertical component for A

Zenith angle $98^\circ45'$ gives $\theta=-8^\circ45'$ (depression):

$$V_A=RA\tan\theta=138.599\times\tan(-8^\circ45')=-21.332\ \text{m}$$

Step 5: Difference in level and gradient

$$\begin{aligned} \text{RL}_A&=\text{RL}_{axis}+V_A-1.100=\text{RL}_{axis}-21.332-1.100\\ \text{RL}_B&=\text{RL}_{axis}+V_B-1.250=\text{RL}_{axis}+12.986-1.250\\ \Delta H&=\text{RL}_B-\text{RL}_A=(12.986-1.250)-(-21.332-1.100)=+34.169\ \text{m} \end{aligned}$$ $$\text{Gradient}=\frac{\Delta H}{AB}=\frac{+34.169}{204.396}=0.16717$$

Answer: B is 34.169 m higher than A; AB = 204.396 m; the gradient from A to B is rising 1 in 6.0 (16.72 %).