Measuring Things — Question 10

Section 1-1 · Physics pages 30–31

Question 10 When the measuring tool changes the object A contact gauge measures a soft pad while applying force F. The intended measurand is the pad’s thickness before contact, at zero force. The following central readings are repeatable: At forces 0.50, 1.00, 1.50, and 2.00 N, the respective heights are 9.80, 9.60, 9.40, and 9.20 mm. F (N) h (mm) .5 1 1.5 2 9.2 9.4 9.6 9.8 10.0 Measured thickness versus contact force; the vertical axis starts at 9.0 mm. For parts (a) and (b), treat the central values as exact and use the explicitly assumed model h(F) = h0 −kF over 0 ≤F ≤2.00 N. No statistical uncertainty information is supplied. (a) Infer k and h0, including units. Check the model against all four central readings. (b) A technician reports 9.40 mm as the uncompressed thickness because the gauge is precise at 1.50 N. Evaluate the claim quantitatively. (c) Explain why the plotted data alone do not establish the zero-force thickness. Design a valida- tion protocol that addresses both extrapolation and contact-induced deformation, and identify an observation that would challenge the assumed model.
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Question 10 - Solution When the measuring tool changes the object F (N) h (mm) .5 1 1.5 2 9.2 9.4 9.6 9.8 10.0 Extrapolated zero-force thickness (a) Determine the conditional linear model. ∆h ∆F = 9.20 −9.80 2.00 −0.50 = −0.40 mm/N. Thus k = 0.40 mm/N and h0 = h + kF = 9.80 + (0.40)(0.50) = 10.00 mm . The model predicts 9.80, 9.60, 9.40, and 9.20 mm at the four listed forces, respectively. The negative slope matches compression under increasing load. (b) Identify the changed measurand. h(1.50 N) −h0 h0 = 9.40 −10.00 10.00 = −0.060 = −6.0%. At 1.50 N, 9.40 mm is the loaded thickness. Relative to the model’s zero-force value, it is smaller by 0.60 mm, or 6.0%. Excellent repeatability at that force does not make it a measurement of uncompressed thickness. (c) Test the extrapolation and the mechanism. No point was measured below 0.50 N. Agreement over the measured range does not establish linearity down to zero; the open point and dashed segment represent extrapolation. One suitable protocol is to calibrate a noncontact optical thickness measurement using a rigid reference and measure the unloaded pad under controlled orientation and temperature. Then collect lower- force gauge readings and repeat an increasing/decreasing force sequence after a defined recovery time. Compare the optical value and near-zero trend with 10.00 mm using defensible measurement bounds. Curvature at low force, different thicknesses on loading and unloading, or an optical result incompatible with the extrapolated value would challenge the model. Repeatable central readings alone cannot supply an uncertainty for h0.

Question 10

When the measuring tool changes the object

A contact gauge measures a soft pad while applying force FF. The intended measurand is the pad’s thickness before contact, at zero force. The following central readings are repeatable: At forces 0.500.50, 1.001.00, 1.501.50, and 2.00N2.00\ \mathrm{N}, the respective heights are 9.809.80, 9.609.60, 9.409.40, and 9.20mm9.20\ \mathrm{mm}.

Diagram shown in the original SVG worksheet.

For parts (a) and (b), treat the central values as exact and use the explicitly assumed model h(F)=h0−kFh(F)=h_0-kF over 0≤F≤2.000\leq F\leq2.00 N. No statistical uncertainty information is supplied.

  1. Infer kk and h0h_0, including units. Check the model against all four central readings.

  2. A technician reports 9.409.40 mm as the uncompressed thickness because the gauge is precise at 1.501.50 N. Evaluate the claim quantitatively.

  3. Explain why the plotted data alone do not establish the zero-force thickness. Design a validation protocol that addresses both extrapolation and contact-induced deformation, and identify an observation that would challenge the assumed model.

Question 10 - Solution

When the measuring tool changes the object

Diagram shown in the original SVG worksheet.

(a) Determine the conditional linear model.

ΔhΔF=9.20−9.802.00−0.50=−0.40mm/N.\frac{\Delta h}{\Delta F}=\frac{9.20-9.80}{2.00-0.50}=-0.40\ \mathrm{mm/N}. Thus k=0.40mm/N\boxed{k=0.40\ \mathrm{mm/N}} and h0=h+kF=9.80+(0.40)(0.50)=10.00mm.h_0=h+kF=9.80+(0.40)(0.50)=\boxed{10.00\ \mathrm{mm}}. The model predicts 9.809.80, 9.609.60, 9.409.40, and 9.209.20 mm at the four listed forces, respectively. The negative slope matches compression under increasing load.

(b) Identify the changed measurand.

h(1.50N)−h0h0=9.40−10.0010.00=−0.060=−6.0%.\frac{h(1.50\ \mathrm N)-h_0}{h_0}=\frac{9.40-10.00}{10.00}=-0.060=-6.0\%.

At 1.501.50 N, 9.409.40 mm is the loaded thickness. Relative to the model’s zero-force value, it is smaller by 0.600.60 mm, or 6.0%6.0\%. Excellent repeatability at that force does not make it a measurement of uncompressed thickness.

(c) Test the extrapolation and the mechanism.

No point was measured below 0.500.50 N. Agreement over the measured range does not establish linearity down to zero; the open point and dashed segment represent extrapolation. One suitable protocol is to calibrate a noncontact optical thickness measurement using a rigid reference and measure the unloaded pad under controlled orientation and temperature. Then collect lower-force gauge readings and repeat an increasing/decreasing force sequence after a defined recovery time. Compare the optical value and near-zero trend with 10.0010.00 mm using defensible measurement bounds. Curvature at low force, different thicknesses on loading and unloading, or an optical result incompatible with the extrapolated value would challenge the model. Repeatable central readings alone cannot supply an uncertainty for h0h_0.