Length — Question 9

Section 1-4 · Physics pages 88–89

Question 9 Accurate chords can still underestimate a curved length A circular boundary has radius R = 1.000 m. A survey approximates its circumference using N equal straight chords joining equally spaced points. Take the chord-sum model and exact circum- ference to be LN = 2NR sin(180◦/N), C = 2πR. You may use sin 30◦= 0.5, sin 15◦≈0.258819, and sin 7.5◦≈0.130526. These approximations are sufficient for the requested comparisons; no trigonometric derivation is required. R curved boundary straight chord Six equal straight chords approximate a circular boundary. The solid circle is the boundary whose length is sought. (a) Compute LN for N = 6, 12, 24 and the percentage shortfall 100(C −LN)/C. Which of these choices meets a requirement of less than 1% geometric error? (b) Explain why even perfectly measured chords underestimate this boundary’s length. Would a finer ruler alone correct the discrepancy at fixed N? (c) A different, unknown boundary is sampled at successively finer spacing, giving increasing chord sums. Does that observation alone prove the length is infinite, or establish that the latest sum is within 1% of the true length? Explain the missing evidence.
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Question 9 - Solution Accurate chords can still underestimate a curved length R curved boundary straight chord (a) Compare the same target length at each sampling density. LN = 2NR sin(π/N), εN = 100 ( 1 −LN 2πR ) %. The circumference is C = 2π m ≈6.283185 m. The supplied sine values give: L6 = 6.000 m, ε6 ≈4.51%, L12 = 6.211656 m, ε12 ≈1.14%, L24 = 6.265248 m, ε24 ≈0.285%. Of the listed choices, 24 chords meet the less-than-1% requirement. Rounding the sine values as supplied does not alter this decision. (b) Improve the geometric approximation, not just the readout. sin z < z (0 < z < π/2) =⇒LN < 2NR π N = 2πR = C. A straight chord is shorter than the circular arc connecting its endpoints. Adding precisely measured chords therefore still omits curved distance. A finer ruler can reduce chord-measurement error but does not replace chords by arcs. Increasing N, or applying a justified model-based correction, addresses this separate geometric discrepancy. (c) More samples do not supply a universal convergence guarantee. The circle itself is a counterexample to the claim of infinity: its chord sums increase yet approach the finite value 2πR. For an unknown boundary, unresolved bends may add further length. A few increasing sums do not bound that missing contribution. A defensible accuracy claim needs additional information, such as a validated shape or curvature bound, an appropriate convergence/error model, and a clear definition of the spatial scale at which the boundary is to be measured. The latest chord sum alone establishes neither infinite length nor a 1% error limit.

Question 9

Accurate chords can still underestimate a curved length

A circular boundary has radius R=1.000mR=1.000\ \mathrm{m}. A survey approximates its circumference using NN equal straight chords joining equally spaced points. Take the chord-sum model and exact circumference to be LN=2NRsin⁡(180∘/N),C=2πR.L_N=2NR\sin(180^\circ/N),\qquad C=2\pi R. You may use sin⁡30∘=0.5\sin30^\circ=0.5, sin⁡15∘≈0.258819\sin15^\circ\approx0.258819, and sin⁡7.5∘≈0.130526\sin7.5^\circ\approx0.130526. These approximations are sufficient for the requested comparisons; no trigonometric derivation is required.

Diagram shown in the original SVG worksheet.

  1. Compute LNL_N for N=6,12,24N=6,12,24 and the percentage shortfall 100(C−LN)/C100(C-L_N)/C. Which of these choices meets a requirement of less than 1%1\% geometric error?

  2. Explain why even perfectly measured chords underestimate this boundary’s length. Would a finer ruler alone correct the discrepancy at fixed NN?

  3. A different, unknown boundary is sampled at successively finer spacing, giving increasing chord sums. Does that observation alone prove the length is infinite, or establish that the latest sum is within 1%1\% of the true length? Explain the missing evidence.

Question 9 - Solution

Accurate chords can still underestimate a curved length

Diagram shown in the original SVG worksheet.

(a) Compare the same target length at each sampling density.

LN=2NRsin⁡(π/N),εN=100(1−LN2πR)%.L_N=2NR\sin(\pi/N),\qquad\varepsilon_N=100\left(1-\frac{L_N}{2\pi R}\right)\%.

The circumference is C=2πm≈6.283185mC=2\pi\ \mathrm{m}\approx6.283185\ \mathrm{m}. The supplied sine values give: L6=6.000m,ε6≈4.51%,L12=6.211656m,ε12≈1.14%,L24=6.265248m,ε24≈0.285%.\begin{aligned}L_6&=6.000\ \mathrm m,&\varepsilon_6&\approx4.51\%,\\L_{12}&=6.211656\ \mathrm m,&\varepsilon_{12}&\approx1.14\%,\\L_{24}&=6.265248\ \mathrm m,&\varepsilon_{24}&\approx0.285\%.\end{aligned} Of the listed choices, 24 chords meet the less-than-1%1\% requirement. Rounding the sine values as supplied does not alter this decision.

(b) Improve the geometric approximation, not just the readout.

sin⁡z<z(0<z<π/2)⇒LN<2NRπN=2πR=C.\sin z<z\ (0<z<\pi/2)\ \Longrightarrow\ L_N<2NR\frac{\pi}{N}=2\pi R=C.

A straight chord is shorter than the circular arc connecting its endpoints. Adding precisely measured chords therefore still omits curved distance. A finer ruler can reduce chord-measurement error but does not replace chords by arcs. Increasing NN, or applying a justified model-based correction, addresses this separate geometric discrepancy.

(c) More samples do not supply a universal convergence guarantee.

The circle itself is a counterexample to the claim of infinity: its chord sums increase yet approach the finite value 2πR2\pi R. For an unknown boundary, unresolved bends may add further length. A few increasing sums do not bound that missing contribution.

A defensible accuracy claim needs additional information, such as a validated shape or curvature bound, an appropriate convergence/error model, and a clear definition of the spatial scale at which the boundary is to be measured. The latest chord sum alone establishes neither infinite length nor a 1%1\% error limit.