Question

The shaft-top coordinates of two vertical shafts are given below. The depth of the shaft A and B are 200 m and 149 m, respectively. The downward gradient of the line joining the bottom of the two shafts in degrees, is __________. (round off to two decimal places)

Hint:

Use coordinate differences to find horizontal distance, then apply $\theta=\tan^{-1}(\Delta Z/D)$.

Answer 1

Correct Answer: 0.9

Coordinates (Latitude = Northing, Departure = Easting):
Shaft A:
\[ N_A = 670,\quad E_A = -150,\quad \text{Surface RL} = 250,\quad \text{Depth} = 200. \]
Bottom RL:
\[ RL_A = 250 - 200 = 50\ \text{m}. \]
Shaft B:
\[ N_B = 170,\quad E_B = 50,\quad \text{Surface RL} = 209,\quad \text{Depth} = 149. \]
Bottom RL:
\[ RL_B = 209 - 149 = 60\ \text{m}. \]
Vertical difference:
\[ \Delta Z = 60 - 50 = 10\ \text{m}. \]
Horizontal distance:
\[ \Delta N = 670 - 170 = 500,\qquad \Delta E = -150 - 50 = -200. \]
\[ D = \sqrt{500^2 + 200^2} = \sqrt{250000 + 40000} = \sqrt{290000} \approx 538.52\ \text{m}. \]
Gradient angle:
\[ \theta = \tan^{-1}\left(\frac{\Delta Z}{D}\right) = \tan^{-1}\left(\frac{10}{538.52}\right) \approx 1.06^\circ. \]
\[ \boxed{1.06^\circ} \quad (\text{acceptable range: } 0.90\text{–}1.13) \]