Length — Question 7

Question 7

A distant target and two different distances

A surveyor sets out a perpendicular baseline $b=30.0\ \mathrm{m}$ as shown. The observed angular quantity is $u=\tan\theta=0.0250$. You may use the supplied right-triangle relations $$D=\frac{b}{u},\qquad S=\sqrt{D^2+b^2}.$$ No trigonometric calculation of $\theta$ is needed.

1. Find the central values of $D$ and $S$. Is the quotient $b/u$ the distance from B to the target? Explain the geometric distinction.

2. Treat $b$ as exact and take $u\in[0.0245,0.0255]$ as a closed hard bound. Find the resulting interval for $D$ and explain which endpoint of $u$ produces the greatest distance.

3. The baseline is extended to $60.0\ \mathrm{m}$, still perpendicular to AT. The same target then gives $u\in[0.0495,0.0505]$. Find the new distance interval. Explain why this improves the inference when the absolute bound on $u$ remains the same.

Question 7 - Solution

A distant target and two different distances

(a) Identify the side computed by each relation.

$$D=\frac{30.0}{0.0250}\ \mathrm{m}=\boxed{1200\ \mathrm{m}},\qquad S=\sqrt{1200^2+30^2}\ \mathrm{m}\approx\boxed{1200.375\ \mathrm{m}}.$$ $D$ is AT, the perpendicular range from A. B-to-target distance is $S$, the hypotenuse, about $0.375\ \mathrm{m}$ longer. A small numerical difference does not make the two measurands identical.

(b) Invert the bounds in the correct order.

For positive $u$, increasing $u$ decreases $b/u$. Thus $$D\in\left[\frac{30.0}{0.0255},\frac{30.0}{0.0245}\right]\ \mathrm{m} \approx\boxed{[1176.47,\ 1224.49]\ \mathrm{m}}.$$ The smallest allowed angle quantity produces the greatest inferred distance. The exact fractions specify the hard interval; displayed decimal endpoints are approximate.

(c) A larger baseline gives a larger angular signal.

$$D_{\max}-D_{\min}=\frac{b}{u_0-\delta u}-\frac{b}{u_0+\delta u}=\frac{2b\delta u}{u_0^2-(\delta u)^2}.$$

$$D\in\left[\frac{60.0}{0.0505},\frac{60.0}{0.0495}\right]\ \mathrm{m} \approx\boxed{[1188.12,\ 1212.12]\ \mathrm{m}}.$$ Both central estimates are 1200 m, but the interval width decreases from about $48.02\ \mathrm{m}$ to $24.00\ \mathrm{m}$. The same absolute bound $\pm0.0005$ is a smaller fraction of 0.0500 than of 0.0250. The improvement assumes the baseline itself remains accurately known and the stated geometry and angular bounds remain valid.