Changing Units — Question 7

Question 7

Exact conversion does not manufacture precision

A part’s length is reported as $L=(2.000\pm0.002)\ \mathrm{in}$. Here the $\pm$ specifies a closed hard bound: every value from $1.998$ to $2.002\ \mathrm{in}$ is allowed. Use the exact definition $1\ \mathrm{in}=25.4\ \mathrm{mm}$.

For this exercise, a part is certified to meet a dimensional requirement only if its entire allowed interval lies within the inclusive specification $[50.70,50.90]\ \mathrm{mm}$.

1. Convert the central length, half-width, and both endpoints to millimetres. Does the part meet the stated certification rule?

2. An estimator uses $1\ \mathrm{in}\approx25\ \mathrm{mm}$, treating the approximation as exact. What central value and interval result? How does this alter the decision, and what central-value error does it introduce?

3. A colleague proposes reporting $50.800000\ \mathrm{mm}$ with no bound because the factor 25.4 is exact. Explain why this is misleading. Give a shorter enclosing interval with endpoints rounded to hundredths of a millimetre.

4. Compare the relative half-width before and after the exact conversion. Explain why the result should be unchanged for a positive purely multiplicative conversion.

Question 7 - Solution

Exact conversion does not manufacture precision

(a) Apply the same positive factor to all interval data.

$$L_0=(2.000)(25.4)\ \mathrm{mm}=50.800\ \mathrm{mm},\qquad \delta L=(0.002)(25.4)\ \mathrm{mm}=0.0508\ \mathrm{mm}.$$ The exact transformed interval is $$\boxed{L\in[50.7492,\ 50.8508]\ \mathrm{mm}}.$$ Both endpoints lie inside $[50.70,50.90]\ \mathrm{mm}$, so the part passes the specified certification rule. This conclusion relies on the given bound being valid.

(b) Quantify the approximation error.

$$\Delta L_{\mathrm{bias}}=(25.0-25.4)\frac{\mathrm{mm}}{\mathrm{in}}(2.000\ \mathrm{in})=-0.800\ \mathrm{mm}.$$

Using 25 instead gives a centre $50.000\ \mathrm{mm}$, half-width $0.050\ \mathrm{mm}$, and interval $[49.950,50.050]\ \mathrm{mm}$. That interval lies below the specification, producing a false rejection under this calculation.

The central-value error is $(50.000-50.800)\ \mathrm{mm}=\boxed{-0.800\ \mathrm{mm}}$, much larger than the true converted half-width. A rough estimation factor is unsuitable for this decision; the omitted conversion error is not part of the original measurement bound.

(c) Keep the bound and round outward.

An exact conversion factor adds no uncertainty but removes none from the original measurement. Appending decimal zeros is not evidence of finer knowledge, and a bare central value suppresses the stated bound.

A shorter enclosing report is $\boxed{L\in[50.74,\ 50.86]\ \mathrm{mm}}$. The lower endpoint is rounded down and the upper endpoint up. Nearest rounding to $[50.75,50.85]\ \mathrm{mm}$ would exclude some allowed values and is not an enclosing interval.

(d) The common factor cancels.

$$\frac{0.002}{2.000}=0.001=0.1\%,\qquad \frac{0.0508}{50.800}=0.001=0.1\%.$$ For $L'=kL$ with exact $k>0$, $\delta L'/L'_0=k\delta L/(kL_0)$. The relative half-width is therefore invariant under this conversion.