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Measuring Things — Question 5

Section 1-1 · Physics pages 20–21

Question 5 Can the display establish a difference? An ideal digital comparator rounds to the nearest 0.02 mm. Assume rounding is its only error. To avoid a tie-breaking convention, use the conservative closed interval display d =⇒ x ∈[d −0.01, d + 0.01] mm. Two objects are measured separately. The indicated sizes are dA = 2.36 mm and dB = 2.38 mm. mm 2.35 2.36 2.37 2.38 2.39 2.40 2.41 A B B (alternative) Closed rounding bins: a display does not locate the value within its bin. (a) Write the allowed interval for each true size and the worst-case interval for xB −xA. (b) Do these bounds guarantee that B is strictly larger than A? Explain why “different displays” is not sufficient justification under the prescribed convention. (c) Suppose B instead displays 2.40 mm. What changes in your conclusion? (d) A student repeats the original readings 100 times, always obtaining the same two displays, then reports xB −xA = 0.02000 mm with negligible error. Is that justified? State what kind of additional information would be needed to support finer resolution.
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Question 5 - Solution Can the display establish a difference? mm 2.35 2.36 2.37 2.38 2.39 2.40 2.41 A B B (alternative) (a) Work with allowed values. xA ∈[2.35, 2.37], xB ∈[2.37, 2.39] (mm). For a difference, the lower bound uses the smallest B and largest A, and the upper bound uses the largest B and smallest A: xB −xA ∈[2.37 −2.37, 2.39 −2.35] = [0, 0.04] mm . (b) A boundary value matters. IB −IA = [Bmin −Amax, Bmax −Amin]. The conservative interval includes zero, so strict inequality is not guaranteed by these bounds. The displayed difference is 0.02 mm, but it is not an exact true difference. A specified deterministic tie rule could refine the endpoint issue; no such refinement is licensed here. (c) A larger displayed separation. If B displays 2.40 mm, its interval is [2.39, 2.41] mm. Therefore xB −xA ∈[0.02, 0.06] mm . Every permitted difference is positive, so B is strictly larger under the assumptions. (d) Repetition does not create sub-display information. An unchanged object can repeatedly fall anywhere within the same rounding bin. Repeating the identical quantized output does not shrink its guaranteed interval. Finer discrimination requires, for example, a calibrated higher-resolution instrument or a separately validated measurement model that supplies information within the bin. Averaging identical display values does not supply that information.

Question 5

Can the display establish a difference?

An ideal digital comparator rounds to the nearest 0.020.02 mm. Assume rounding is its only error. To avoid a tie-breaking convention, use the conservative closed interval display d⇒x∈[d−0.01,d+0.01]mm.\text{display }d\quad\Longrightarrow\quad x\in[d-0.01,d+0.01]\ \mathrm{mm}. Two objects are measured separately. The indicated sizes are dA=2.36d_A=2.36 mm and dB=2.38d_B=2.38 mm.

Diagram shown in the original SVG worksheet.

  1. Write the allowed interval for each true size and the worst-case interval for xB−xAx_B-x_A.

  2. Do these bounds guarantee that B is strictly larger than A? Explain why “different displays” is not sufficient justification under the prescribed convention.

  3. Suppose B instead displays 2.402.40 mm. What changes in your conclusion?

  4. A student repeats the original readings 100 times, always obtaining the same two displays, then reports xB−xA=0.02000x_B-x_A=0.02000 mm with negligible error. Is that justified? State what kind of additional information would be needed to support finer resolution.

Question 5 - Solution

Can the display establish a difference?

Diagram shown in the original SVG worksheet.

(a) Work with allowed values.

xA∈[2.35,2.37],xB∈[2.37,2.39](mm).x_A\in[2.35,2.37],\qquad x_B\in[2.37,2.39]\quad(\mathrm{mm}). For a difference, the lower bound uses the smallest B and largest A, and the upper bound uses the largest B and smallest A: xB−xA∈[2.37−2.37,2.39−2.35]=[0,0.04]mm.\boxed{x_B-x_A\in[2.37-2.37,\,2.39-2.35]=[0,0.04]\ \mathrm{mm}}.

(b) A boundary value matters.

IB−IA=[Bmin−Amax,Bmax−Amin].I_B-I_A=[B_{\min}-A_{\max},\ B_{\max}-A_{\min}].

The conservative interval includes zero, so strict inequality is not guaranteed by these bounds. The displayed difference is 0.020.02 mm, but it is not an exact true difference. A specified deterministic tie rule could refine the endpoint issue; no such refinement is licensed here.

(c) A larger displayed separation.

If B displays 2.402.40 mm, its interval is [2.39,2.41][2.39,2.41] mm. Therefore xB−xA∈[0.02,0.06]mm.\boxed{x_B-x_A\in[0.02,0.06]\ \mathrm{mm}}. Every permitted difference is positive, so B is strictly larger under the assumptions.

(d) Repetition does not create sub-display information.

An unchanged object can repeatedly fall anywhere within the same rounding bin. Repeating the identical quantized output does not shrink its guaranteed interval. Finer discrimination requires, for example, a calibrated higher-resolution instrument or a separately validated measurement model that supplies information within the bin. Averaging identical display values does not supply that information.

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