Following observations were taken using a Tacheometer with staff held vertical. The additive and multiplying constants of the instrument are 0 and 100, respectively. The reduced level (RL) of the staff station R, in $m$, is . (rounded off to two decimal places)
Instrument station Staff station Staff reading (m) Vertical angle Remarks
P Q 1.4, 2.7, 3.9 7° RL of Q = 102 m
P R 2.1, 2.8, 3.6 5°

Question

The primary purpose of mine ventilation is to:

Answer 1

Instead of first isolating the elevation of the instrument's line of sight, let's work directly with the difference in RL between R and Q, since both sightings share the same instrument and the same line of collimation.

For any staff station, $\text{RL} = H + V - \text{middle reading}$, where $H$ is the (unknown but common) elevation of the line of sight. Since $H$ is the same for both Q and R taken from station P, it cancels out when we subtract: $\text{RL}_R - \text{RL}_Q = (V_R - m_R) - (V_Q - m_Q)$, where $m_Q$ and $m_R$ are the middle hair readings.

For Q: staff intercept $s_Q = 3.9 - 1.4 = 2.5\ \text{m}$, and $V_Q = \frac{100 \times 2.5}{2}\sin(14^\circ) = 125 \times 0.2419 = 30.24\ \text{m}$. Middle reading $m_Q = 2.7\ \text{m}$, so $V_Q - m_Q = 30.24 - 2.7 = 27.54\ \text{m}$.

For R: staff intercept $s_R = 3.6 - 2.1 = 1.5\ \text{m}$, and $V_R = \frac{100 \times 1.5}{2}\sin(10^\circ) = 75 \times 0.17365 = 13.02\ \text{m}$. Middle reading $m_R = 2.8\ \text{m}$, so $V_R - m_R = 13.02 - 2.8 = 10.22\ \text{m}$.

Now take the difference: $\text{RL}_R - \text{RL}_Q = 10.22 - 27.54 = -17.32\ \text{m}$.

Since RL of Q is given as 102 m, $\text{RL}_R = 102 - 17.32 = 84.68\ \text{m}$.

Let's summarize:

The elevation of the line of sight is common to both sightings taken from P, so it cancels out when comparing R and Q directly.
Each sighting's own vertical intercept minus its middle hair reading gives that station's height relative to the line of sight.
Subtracting these two relative heights gives the RL difference between R and Q directly.

So the reduced level of station R works out to about 84.68 m.

Answer 2

Instead of first isolating the elevation of the instrument's line of sight, let's work directly with the difference in RL between R and Q, since both sightings share the same instrument and the same line of collimation.

For any staff station, $\text{RL} = H + V - \text{middle reading}$, where $H$ is the (unknown but common) elevation of the line of sight. Since $H$ is the same for both Q and R taken from station P, it cancels out when we subtract: $\text{RL}_R - \text{RL}_Q = (V_R - m_R) - (V_Q - m_Q)$, where $m_Q$ and $m_R$ are the middle hair readings.

For Q: staff intercept $s_Q = 3.9 - 1.4 = 2.5\ \text{m}$, and $V_Q = \frac{100 \times 2.5}{2}\sin(14^\circ) = 125 \times 0.2419 = 30.24\ \text{m}$. Middle reading $m_Q = 2.7\ \text{m}$, so $V_Q - m_Q = 30.24 - 2.7 = 27.54\ \text{m}$.

For R: staff intercept $s_R = 3.6 - 2.1 = 1.5\ \text{m}$, and $V_R = \frac{100 \times 1.5}{2}\sin(10^\circ) = 75 \times 0.17365 = 13.02\ \text{m}$. Middle reading $m_R = 2.8\ \text{m}$, so $V_R - m_R = 13.02 - 2.8 = 10.22\ \text{m}$.

Now take the difference: $\text{RL}_R - \text{RL}_Q = 10.22 - 27.54 = -17.32\ \text{m}$.

Since RL of Q is given as 102 m, $\text{RL}_R = 102 - 17.32 = 84.68\ \text{m}$.

Let's summarize:

The elevation of the line of sight is common to both sightings taken from P, so it cancels out when comparing R and Q directly.
Each sighting's own vertical intercept minus its middle hair reading gives that station's height relative to the line of sight.
Subtracting these two relative heights gives the RL difference between R and Q directly.

So the reduced level of station R works out to about 84.68 m.

Show Hint

Correct Answer: 84.68

Solution and Explanation

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