Measuring Things — Question 4

Question 4

A damaged ruler and a useful difference

A ruler’s zero end is damaged. The intact scale indicates the two ends of a rigid bar at $x_L=13.4$ cm and $x_R=28.7$ cm. Each endpoint has a reading-error bound of $\pm0.1$ cm. No statistical independence is assumed: use worst-case bounds.

1. Find the indicated bar length and its worst-case reading-error bound.

2. Suppose every ruler coordinate has the same unknown additive offset $z=+0.6$ cm. Show algebraically whether this changes the endpoint difference.

3. Now consider a separate calibration model, $r=1.010x+z$, where $r$ is the indicated coordinate and $x$ the true coordinate in centimeters. Find the corrected central length and corrected worst-case bound. Does subtraction eliminate this scale-factor error?

Treat $1.010$ as exact for part (c).

Question 4 - Solution

A damaged ruler and a useful difference

(a) Subtract the endpoints and bound the difference.

$$|\delta L|=|\delta r_R-\delta r_L|\le |\delta r_R|+|\delta r_L|\le0.20\ \mathrm{cm}.$$

$$L_r=28.7-13.4=15.3\ \mathrm{cm}.$$ The smallest possible difference is $28.6-13.5=15.1$ cm; the largest is $28.8-13.3=15.5$ cm. Thus $$\boxed{L_r=15.3\pm0.2\ \mathrm{cm}\quad\text{(worst-case bound)}}.$$

(b) A common additive offset cancels.

If $r_R=x_R+z$ and $r_L=x_L+z$, then $$r_R-r_L=(x_R+z)-(x_L+z)=x_R-x_L.$$ The $+0.6$ cm offset does not alter the difference. The remaining endpoint reading errors are still bounded as above.

(c) A scale-factor error does not cancel.

With $r=1.010x+z$, subtraction instead gives $\Delta r=1.010\Delta x$. Hence $$\Delta x=\frac{15.3}{1.010}=15.1485\ldots\ \mathrm{cm},\qquad \delta(\Delta x)=\frac{0.2}{1.010}=0.1980\ldots\ \mathrm{cm}.$$ The exact transformed interval is $[15.1/1.010,\,15.5/1.010]$ cm, approximately $[14.9505,15.3465]$ cm. The central estimate is about $15.15$ cm; the bound is about $0.20$ cm. Subtraction removes a shared offset, not a multiplicative error.