Position and Displacement — Question 6

Question 6

Three relative-position measurements must agree with one geometry

Three stationary markers A, B, and C lie on one east-positive line. Define $$d_{AB}=x_B-x_A,\quad d_{BC}=x_C-x_B,\quad d_{AC}=x_C-x_A.$$ All comparisons refer to the same markers, instant, orientation, and metre scale. A survey reports The reported signed comparisons are $d_{AB}=+8\ \mathrm{m}$, $d_{BC}=-3\ \mathrm{m}$ and $d_{AC}=+6\ \mathrm{m}$. Each true comparison initially lies within $0.20\ \mathrm{m}$ of its report. These are hard bounds, not standard deviations. Before imposing geometric consistency, every combination of values inside the intervals is allowed.

1. Derive the identity the three comparisons must obey. Show that the central reported values violate it.

2. Find the allowed interval of the residual $r=d_{AB}+d_{BC}-d_{AC}$. Can any values within the stated bounds represent one simultaneous configuration?

3. Would changing the common origin repair this discrepancy? Explain using the coordinate differences.

4. As a separate hypothetical case, enlarge each bound to $0.40\ \mathrm{m}$. Construct one consistent set of comparison values and one set of coordinates producing them. Does admitting a consistent configuration establish that this enlarged bound is experimentally justified?

Question 6 - Solution

Three relative-position measurements must agree with one geometry

(a) Eliminate the intermediate coordinate.

$$r=d_{AB}+d_{BC}-d_{AC}=8-3-6=-1\ \mathrm{m}.$$

$$d_{AB}+d_{BC}=(x_B-x_A)+(x_C-x_B)=x_C-x_A=d_{AC}.$$ Every consistent configuration therefore has $\boxed{r=0}$. The reported centres give $8.0-3.0-6.0=\boxed{-1.0\ \mathrm{m}}$, not zero. Using only A–B and B–C would predict A–C as 5.0 m rather than 6.0 m.

(b) Check all allowed error extremes.

$$|\delta r|\leq|\delta d_{AB}|+|\delta d_{BC}|+|\delta d_{AC}|\leq0.60\ \mathrm{m};\quad0\notin[-1.6,-0.4].$$

The initial intervals are $[7.8,8.2]$, $[-3.2,-2.8]$, and $[5.8,6.2]\ \mathrm{m}$. Thus $$r_{\min}=7.8-3.2-6.2=-1.6\ \mathrm{m},\qquad r_{\max}=8.2-2.8-5.8=-0.4\ \mathrm{m}.$$ So $\boxed{r\in[-1.6,-0.4]\ \mathrm{m}}$, excluding zero. No single configuration can satisfy all three stated bounds and comparison assumptions. One should investigate errors, labels, timing, or conventions rather than assign inconsistent marker coordinates.

(c) A common origin shift cancels in every comparison.

Replacing every coordinate by $x'=x-a$, with the same fixed $a$, leaves each difference unchanged. The residual therefore stays unchanged. An origin choice cannot repair inconsistent relative separations. This is distinct from accidentally using different origins for different reports, which would violate the problem’s assumptions.

(d) Exhibit feasibility under the hypothetical larger bounds.

With bounds of $\pm0.40\ \mathrm{m}$, choose $$\boxed{d_{AB}=8.4\ \mathrm{m},\quad d_{BC}=-2.8\ \mathrm{m},\quad d_{AC}=5.6\ \mathrm{m}}.$$ These lie within the new intervals and satisfy $8.4-2.8=5.6$. One realization is $x_A=0$, $x_B=8.4$, $x_C=5.6\ \mathrm{m}$. Any common fixed shift gives another coordinate description of the same configuration.

This construction proves mathematical compatibility with the larger bounds. It does not justify enlarging an instrument’s error allowance without supporting evidence.