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Materials Science · Durability in real environments

Laboratory acceleration and real-service evidence

Useful background: Energy, stability and environmental context; Fatigue, creep and time-dependent damage; Joining, coatings and surface engineering. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

  • Use accelerated durability tests without assuming universal time conversion.
  • Define a service endpoint, observation clock and measured response before converting an exposure.
  • Derive a conditional temperature-acceleration factor and threshold time from a stated response model.
  • Propagate supplied rate bounds coherently and distinguish that model range from a confidence or lifetime guarantee.
  • Test a prediction against data excluded from fitting, then assess a changing temperature history and a changed mechanism.
01

Faster exposure can answer a narrower question

Raising temperature, concentration or cycling rate can accelerate damage. It can also change transport, phases or reaction pathways. A test is useful when its relation to service is physically justified, not merely because it produces failure quickly.

A film reaches a specified optical-loss endpoint after 250 h in a warm laboratory test. Is that 250 h of evidence about the intended environment, a predicted longer duration there, or neither? Answer by identifying the film state, endpoint, temperature history and model that connects the two environments. The word accelerated supplies none of those quantities by itself.

Service evidence includes what actually happened to identified material or components under recorded use: installation state, elapsed time, operating and idle conditions, maintenance, observed response and failure mode. A returned specimen may have survived to its removal without reaching the chosen endpoint. That is a bounded observation, not its complete life. Calendar hours, hours powered, cycles and hours above a specified temperature are different clocks.

Lifetime evidence, scatter and extrapolation supplies the distinction between an observed record, a population probability and a censored observation. This lesson instead develops one response-path transfer. It does not estimate a population failure percentile or repeat the separate short-time/creep property selection in Service conditions and reliable property data.

02

Connect the measured change to a named endpoint

Use an original fictional thin-film formulation aged unloaded in dry oxygen-containing gas, in the dark, at a uniform maintained temperature. All response readings use matched coupons from the same initial material/geometry, each removed once at its stated exposure age and measured at 23 °C with the same 650 nm optical procedure. Transfer is stipulated to add negligible change. These are supplied path indications, not a set of individual lifetimes, source data or a named product specification.

Let τ₀ be initial optical transmittance and τ(t) its value after age t under that common measurement procedure. Define d(t) = 1 − τ(t)/τ₀. It is dimensionless relative loss: d = 0.20 means 20% of the initial transmittance has been lost. Declare the endpoint to be the first d = 0.20 for this particular comparison. It is not complete breakage, the fraction of bonds cut, or a universal acceptable optical limit. Changing the criterion changes the time being predicted.

The laboratory record below gives nominal rates r₆₀ = 0.000400 h⁻¹ at 60 °C and r₈₀ = 0.000800 h⁻¹ at 80 °C. Within the modelled interval 0 ≤ d ≤ 0.20, stipulate d = rt from initial d = 0. The table is consistent with this linear path at its sampled ages. Those observations do not establish linearity at every intervening point or after the endpoint; a test of the assumption remains part of the evidence.

Chemical/structural observations in this fictional dossier report the same candidate oxidation-associated product family and no new resolved phase across the intended 40–80 °C comparison. That supports examining one temperature model; it does not prove a unique microscopic mechanism or exclude every unresolved change. Increased temperature can also change transport limitation, additive distribution, phase state or the response-to-chemistry relation. A short, hot test becomes useful for transfer only when those differences are addressed.

Original nominal path indications, not actual polymer-grade data. Matched coupons are measured by the common optical procedure after their stated exposure ages. Later rate bounds are supplied separately; these indications are not declared exact true rates.
Exposure temperatureAges (h)Indicated d at those agesNominal fitted rate (h⁻¹)
60 °C0 / 250 / 5000 / 0.100 / 0.2000.000400
80 °C0 / 125 / 2500 / 0.100 / 0.2000.000800
03

Derive a rate ratio before multiplying a duration

An Arrhenius-type relation represents a temperature-sensitive process over a justified range. Here write a normalised rate model r(T) = r₆₀ exp[B(1/T₆₀ − 1/T)]. T is absolute temperature in kelvins, T₆₀ = 333.15 K, and r₆₀ is the rate at that reference temperature. exp(x) is the exponential function; its inverse ln(x), the natural logarithm, recovers the exponent. B is a fitted temperature-sensitivity parameter in kelvins, so its product with a reciprocal-temperature difference is dimensionless. A fitted B is not proof of a single molecular activation barrier. Diffusion and phase transformations introduces the related thermal-activation form.

Use T in K = temperature in °C + 273.15. Thus 80 °C is 353.15 K and 40 °C is 313.15 K. Dividing the 80 °C rate by the 60 °C rate cancels the reference scale: 2 = exp[B(1/333.15 − 1/353.15)]. Taking ln gives B = ln(2)/(1/333.15 − 1/353.15) = 4077.50 K. Two nominal rate values determine the two parameters of this chosen model; they do not select the model family or independently validate an extrapolation.

Substituting 313.15 K gives r₄₀ ≈ 0.000183054 h⁻¹. This is a lower-temperature extrapolation beyond the two calibration temperatures, explicitly conditional on the same rate relation and response path. The model time from d = 0 to d* = 0.20 is t* = d*/r. At 60 °C it is 500 h, at 80 °C 250 h, and at 40 °C about 1092.58 h. The last time has not been observed in the lower-temperature record below.

Define AF₆₀→₄₀ = r₆₀/r₄₀ = 2.18515. Because the required increment d* is the same, t*₄₀/t*₆₀ = (d*/r₄₀)/(d*/r₆₀) = AF₆₀→₄₀. Thus 500 h × 2.18515 gives the modelled 1092.58 h at 40 °C. The rate ratio goes in the opposite direction from the time ratio. It does not turn the actual 500 h laboratory observation into 1092.58 h of observed service.

04

Keep parameter bounds inside the same fitted model

A separate supplied assessment permits r₆₀ between 0.000380 and 0.000420 h⁻¹ and r₈₀ between 0.000760 and 0.000840 h⁻¹. For this exercise every pair in that rectangle is allowed. These are stipulated input bounds, not confidence limits inferred from the two table rows. Refit B for each pair: r₆₀ and the fitted temperature sensitivity cannot be varied independently after using the same calibration rates.

The algebra can be written r₄₀ = r₆₀(r₆₀/r₈₀)^w, where w = (1/313.15 − 1/333.15)/(1/333.15 − 1/353.15) = 1.12773. The ratio inside the power is dimensionless. Since w is positive, r₄₀ increases with r₆₀ and decreases with r₈₀ when the other is fixed. The smallest possible rate uses the low 60 °C rate with the high 80 °C rate; the largest uses the opposite corner. Simply applying the same ±5% to the final projected time would miss this coupled extrapolation.

Those two rates are approximately 0.000155340 and 0.000215172 h⁻¹. Reversing them when dividing the fixed 0.20 increment gives modelled endpoint times from 929.49 to 1287.50 h. The central 1092.58 h is one member of this family. The range accounts only for the declared rate-input bounds inside the chosen linear/temperature model. It is not a population lifetime interval, a confidence band or a bound on an omitted mechanism.

A wrong functional relation can remain wrong for every parameter in a narrow range. A new phase, a different degradation pathway or a different optical response-to-state relation is a model issue, not another decimal-place adjustment. The next comparison therefore examines new lower-temperature evidence rather than interpreting the calculated range as its own validation.

05

Check the transfer

Compare damage morphology and mechanisms with service evidence. State measured test duration separately from predicted service life, including uncertainty and assumptions. Multiple environments may be needed for a component that experiences changing conditions.

Now use the separate 40 °C service-representative record in the table. Its values were withheld from the two-rate fit. All stated gas, geometry, initial state and probe conditions match the intended constant-temperature comparison. At 200, 400 and 600 h the nominal model predicts d ≈ 0.03661, 0.07322 and 0.10983; the observed indications are 0.039, 0.077 and 0.117. It is the observed history and chosen response that are compared, not just the final appearance.

For this invented comparison, declare in advance that the central prediction must be within 0.005 absolute d units of every reading. That is a comparison rule chosen for this exercise, not a standard, measurement-uncertainty statement or probability. The absolute residuals are 0.00239, 0.00378 and 0.00717. The last misses the rule, so the central prediction does not pass this declared check.

The input-bound family nevertheless spans d = 0.09320 to 0.12910 at 600 h, including 0.117. This does not erase the central-model miss. For example, an allowed pair r₆₀ = 0.000410 and r₈₀ ≈ 0.000792458 h⁻¹ gives r₄₀ = 0.000195 h⁻¹ and d = 0.039, 0.078 and 0.117 at those ages, close to all three readings. That shows compatibility with at least one common parameter pair rather than merely different favourable pairs at each time. If these service readings are used to choose that pair, they become calibration evidence for the updated model, not an independent test of the update.

The record ends at 600 h and d = 0.117, below the 0.20 endpoint. Neither the central time nor the input-range endpoint times were measured here. Additional independent lower-temperature/longer-duration observations, a measurement assessment and continued mechanism evidence would be needed to strengthen the transfer. Agreement at early ages would still not demonstrate every later shape or every future product’s behaviour.

Original lower-temperature comparison excluded from the nominal fit. “Model” is the central conditional prediction, not another observation. Values are rounded for reading; the declared absolute rule applies to the calculated values.
Age at 40 °C (h)Observed dCentral model dObserved − model
0000
2000.0390.03661+0.00239
4000.0770.07322+0.00378
6000.1170.10983+0.00717
Held-out response evidence and conditional endpoint timesOriginal fictional response-path comparison. The upper plot has linear age zero to 600 hours and relative optical-transmittance loss d zero to 0.22 at constant 40 degrees Celsius. Orange triangles are the held-out indications: zero at zero hours, 0.039 at 200, 0.077 at 400 and 0.117 at 600. The long-dashed green nominal prediction uses the two higher-temperature calibration rates and reaches 0.109832 at 600 hours. A violet diagonally hatched wedge, with short-dashed boundaries, is the family generated by the supplied rate bounds inside the same model; its range at 600 hours is 0.093204 to 0.129103. It is not a confidence or population prediction band. The declared endpoint d equals 0.20, above every lower-temperature indication. The central 600-hour residual is 0.007168, greater than the separately declared 0.005 comparison rule. The lower linear time axis runs from zero to 1400 hours. A circle marks the 80-degree laboratory endpoint indication at 250 hours and a square the 60-degree indication at 500 hours. At 40 degrees, a diamond marks the nominal projected endpoint time 1092.58 hours, with a bracket from 929.49 to 1287.50 hours for the supplied model/input family. Those projected endpoint times are beyond the actual 600-hour lower-temperature record. These are conditional response-path calculations, not observed service lifetimes, confidence bounds or a qualification.Check the prediction before calling it service evidenceOriginal fictional record · 40 °C data excluded from nominal fitRelative optical loss d0.000.050.100.150.200200400600Declared endpoint d = 0.20Age at 40 °C (h), linearTriangle: held-out indicationLong dash: nominal modelHatching: model / input familyAt 600 h: central residual 0.00717 exceeds the 0.005 rule.Endpoint times: indications versus projections80 °C60 °C40 °C250 h500 h1093 h9291288050010001400Time to the declared d = 0.20 endpoint (h)Circle / square: test indications. Diamond / bracket: model.The lower-temperature record stops at 600 h, below its endpoint.
Original fictional response-path comparison. The upper plot has linear age zero to 600 hours and relative optical-transmittance loss d zero to 0.22 at constant 40 degrees Celsius. Orange triangles are the held-out indications: zero at zero hours, 0.039 at 200, 0.077 at 400 and 0.117 at 600. The long-dashed green nominal prediction uses the two higher-temperature calibration rates and reaches 0.109832 at 600 hours. A violet diagonally hatched wedge, with short-dashed boundaries, is the family generated by the supplied rate bounds inside the same model; its range at 600 hours is 0.093204 to 0.129103. It is not a confidence or population prediction band. The declared endpoint d equals 0.20, above every lower-temperature indication. The central 600-hour residual is 0.007168, greater than the separately declared 0.005 comparison rule. The lower linear time axis runs from zero to 1400 hours. A circle marks the 80-degree laboratory endpoint indication at 250 hours and a square the 60-degree indication at 500 hours. At 40 degrees, a diamond marks the nominal projected endpoint time 1092.58 hours, with a bracket from 929.49 to 1287.50 hours for the supplied model/input family. Those projected endpoint times are beyond the actual 600-hour lower-temperature record. These are conditional response-path calculations, not observed service lifetimes, confidence bounds or a qualification.
06

Use the actual thermal clock, not just its average

For a separate explicitly conditional history calculation, retain the central rate model and additionally assume that increments add without state dependence: Δd = r(T)Δt in each uniform-temperature interval. Neglect transition durations and assume no induction period, recovery, changing transport, evolving geometry or new reaction mode. This is a teaching model for a response coordinate, not a general accumulation law for all damage.

Let the material spend 200 h at 40 °C and then 50.0 h at 60 °C. The modelled increments are 0.000183054 × 200 = 0.0366107 and 0.000400 × 50.0 = 0.0200, giving d = 0.0566107. On a 60 °C equivalent clock, the first interval is 200/2.18515 = 91.5268 h; add the actual 50.0 h to obtain 141.5268 equivalent hours. Multiplying that clock by r₆₀ gives the same d.

The elapsed time is 250 h and its time-weighted mean temperature is 44.0 °C. Substituting that mean into the nonlinear rate relation gives r(317.15 K) ≈ 0.000215725 h⁻¹ and d ≈ 0.0539312, a different result. Averaging temperature before calculating the rate loses information about how long the specimen spent at each temperature. The second figure shows the two assumed histories and their accumulated coordinates.

Reversing the two intervals leaves the same final d in this particular state-independent additive model, because the same two increments are added. It changes when the increments occur. A real sequence can additionally alter water distribution, chemistry, stress or transport and need not commute. The actual 40 °C comparison has already exposed a central-model miss; these arithmetic projections therefore remain conditional, rather than becoming established service histories.

Temperature history and an equivalent model clockOriginal conditional model calculations, with no measured service points. The first history spends 200 hours at 40 degrees Celsius and then 50 hours at 60 degrees; transitions have zero duration in this idealisation and temperatures are uniform. The comparison history remains at the same time-weighted mean, 44 degrees Celsius, for all 250 hours. Both temperature and elapsed-time axes are linear. Blue solid paths identify the two-interval temperature and loss histories. Circular markers identify calculated corners on the lower loss path; a square marks the lower orange long-dashed constant-mean-temperature endpoint. The lower plot uses linear time zero to 250 hours and loss d zero to 0.065. The two-interval path passes through d zero at zero hours, 0.0366107 at 200 hours and 0.0566107 at 250 hours. The constant-44-degree model ends at 0.0539312. Under the stated state-independent accumulation rule the two-interval history equals 141.5268 hours at 60 degrees, even though 250 calendar hours elapsed. Averaging temperature before evaluating the nonlinear rate does not preserve the accumulated coordinate. Reversing the same two intervals preserves only the final total in this model; real state feedback or different mechanisms can invalidate that order independence. Neither calculated history validates the nominal model, which missed a separate declared comparison rule.Two histories, two conditional calculationsUniform temperature · transition times neglected · original modelExposure temperature (°C)4050600100200250Elapsed time (h)Accumulated relative optical loss d0.000.020.040.060100200250Elapsed time (h), linearTwo intervals: final d = 0.0566107Constant 44 °C: final d = 0.0539312Two intervals: 250 elapsed hours; 141.5268 equivalent h at 60 °C.Modelled corners and joins; no observed service data are shown.
Original conditional model calculations, with no measured service points. The first history spends 200 hours at 40 degrees Celsius and then 50 hours at 60 degrees; transitions have zero duration in this idealisation and temperatures are uniform. The comparison history remains at the same time-weighted mean, 44 degrees Celsius, for all 250 hours. Both temperature and elapsed-time axes are linear. Blue solid paths identify the two-interval temperature and loss histories. Circular markers identify calculated corners on the lower loss path; a square marks the lower orange long-dashed constant-mean-temperature endpoint. The lower plot uses linear time zero to 250 hours and loss d zero to 0.065. The two-interval path passes through d zero at zero hours, 0.0366107 at 200 hours and 0.0566107 at 250 hours. The constant-44-degree model ends at 0.0539312. Under the stated state-independent accumulation rule the two-interval history equals 141.5268 hours at 60 degrees, even though 250 calendar hours elapsed. Averaging temperature before evaluating the nonlinear rate does not preserve the accumulated coordinate. Reversing the same two intervals preserves only the final total in this model; real state feedback or different mechanisms can invalidate that order independence. Neither calculated history validates the nominal model, which missed a separate declared comparison rule.
07

Use field evidence to test the link, not decorate a fit

Consider a further fictional 90 °C test that reaches d = 0.20 at 120 h but produces a reaction layer absent from the lower-temperature specimens. Dividing 500 by 120 would produce 4.17 as a time ratio. It does not establish an acceleration factor for the original mode: the changed mechanism breaks the transfer assumption even if another curve can fit the extra number. Compare chemistry, morphology, spatial distribution and the response definition as well as rates.

A concrete model-epoxy weathering study developed predictions from controlled laboratory exposures and compared them with outdoor specimens while recording changing ultraviolet spectrum/intensity, temperature and humidity. It also examined versions updated using early outdoor readings. Those updated predictions answer a different question from predictions made without those specimen readings. The outdoor specimens were in a covered chamber, leaving other real exposure factors outside the demonstrated scope. This illustrates both a useful transfer method and its boundary.

A returned component is informative when its material, geometry, initial processing state, installation and environmental/load history can be connected to the model. A film coupon in dry unloaded gas does not establish the performance of a stressed joint in changing humidity. Unknown repairs or selective retrieval can further change what the returned set represents. Combined mechanical and environmental damage develops such interacting histories; Protection, inspection and condition monitoring addresses detection capability and maintained protection.

A defensible statement keeps its layers explicit: what was observed and for how long, the chosen endpoint and response, the assumed model and domain, the input/model limitations, and which new observations tested the prediction. A concise report for the worked dossier would say that the nominal two-temperature model projects about 1093 h, with a 929–1288 h input-bound range to the stated optical endpoint at constant 40 °C, but the independently held-out 600 h comparison misses the declared central-model rule. It would not call that a measured service lifetime or an approved operating duration.

Worked examples

Work through the case

A hot test produces a reaction layer absent in service. Can one constant acceleration factor be assumed?

  1. Identify the changed mechanism.
  2. Examine whether the proposed rate model still represents service.
  3. Seek relevant evidence before extrapolation.

No. A different mechanism undermines an unqualified conversion.

A conditional transfer produces a prediction with a named endpoint

Use the two nominal laboratory rates to find B, the rate at 40 °C, and the time from d = 0 to d = 0.20. State the direction of the acceleration factor and which duration was actually observed.

  1. Convert 60, 80 and 40 °C to 333.15, 353.15 and 313.15 K. Fit B = ln(0.000800/0.000400)/(1/333.15 − 1/353.15) ≈ 4077.50 K.
  2. The chosen rate relation gives r₄₀ ≈ 0.000183054 h⁻¹. The central ratio r₆₀/r₄₀ ≈ 2.18515 is dimensionless.
  3. At a fixed increment, time is increment/rate: t*₄₀ = 0.20/r₄₀ ≈ 1092.58 h = 2.18515 × 500 h. The slower rate gives the longer model time.
  4. The laboratory 500 h and 250 h endpoint indications are in the supplied record. The 40 °C endpoint is extrapolated, not observed; its lower-temperature comparison only extends to 600 h.

B ≈ 4077.50 K, r₄₀ ≈ 0.000183054 h⁻¹ and t*₄₀ ≈ 1092.58 h under the chosen model. The 2.18515 factor multiplies 60 °C endpoint time into a conditional 40 °C time, not additional observation.

A model range and a held-out check answer different questions

Propagate the supplied two-rate bounds to the 40 °C endpoint, then assess the observed 600 h loss of 0.117 using the declared central-prediction rule. Can including the reading inside a parameter family be called a pass of that rule?

  1. For w > 0 in r₄₀ = r₆₀(r₆₀/r₈₀)^w, use (0.000380,0.000840) for the smallest projected rate and (0.000420,0.000760) for the largest. Refit B for each pair.
  2. The rates are about 0.000155340 and 0.000215172 h⁻¹. Divide 0.20 by the largest rate for the earliest model endpoint, 929.49 h, and by the smallest for the latest, 1287.50 h.
  3. At 600 h the central prediction is 0.109832. The absolute residual is 0.117 − 0.109832 ≈ 0.007168, greater than the declared 0.005 limit. It misses the central-prediction check.
  4. At the same time, the allowed model family gives 0.093204–0.129103. The reading lies inside that family; a suitable common allowed pair is shown in the section. Selecting it using these readings is an update, requiring new independent testing.

The stipulated model/input family gives about 929–1288 h. The central prediction misses the 600 h comparison rule; family compatibility does not change that result or create a confidence/lifetime guarantee.

Add modelled exposure increments before compressing the clock

Under the additional state-independent assumptions, find d after 200 h at 40 °C and 50.0 h at 60 °C. Compare with treating the same 250 h as a constant 44.0 °C exposure.

  1. At 40 °C, the central rate gives 0.0366107 loss in 200 h. At 60 °C, 50.0 h contributes 0.0200. Sum to d = 0.0566107.
  2. Alternatively, 200 h at 40 °C is 91.5268 h on the 60 °C clock; add 50.0 h to get 141.5268 h. Multiplying by 0.000400 h⁻¹ recovers the same loss.
  3. The mean-temperature substitution gives 250 × r(317.15 K) ≈ 0.0539312. The nonlinear rate makes this smaller than the two-interval value.
  4. The two clocks and all results belong to the declared conditional model. They do not validate it against the failed central-model comparison or establish how a changing real environment will act.

The two-interval model gives d ≈ 0.0566107, or 141.5268 equivalent hours at 60 °C. A constant-mean-temperature model gives 0.0539312; the two histories are not equivalent.

Try the reasoning

  1. Why compare field observations?

    Show the answer and reasoning

    They test the accelerated model’s relevance.

    A laboratory fit alone may not transfer.

  2. Does passing one exposure establish unlimited durability?

    Show the answer and reasoning

    No.

    It establishes survival under the investigated conditions and duration.

  3. The initial transmittance is τ₀ = 0.900. What absolute transmittance corresponds to d = 0.20? A report instead stops at τ = 0.800. Which relative loss did it actually use?

    Show the answer and reasoning

    The declared endpoint is τ = 0.720. Stopping at τ = 0.800 uses d = 1 − 0.800/0.900 = 0.1111, about 11.11%.

    A 20% relative loss means retaining 80% of the initial value, not retaining an absolute transmittance of 0.80. The denominator and endpoint must match before transferring time.

  4. Under the same nominal linear/temperature model from a fresh initial state, change only the endpoint to d = 0.10. Find the 40 °C time. Does this new number need a new material mechanism?

    Show the answer and reasoning

    About 546.29 h. The arithmetic changes because the endpoint increment halves; it need not imply a new mechanism.

    t = 0.10/0.00018305354 ≈ 546.29 h. This lies in the stipulated 0–0.20 response range, but it is still a conditional prediction rather than an observed crossing or a different safety rating.

  5. A separately assessed same-state response starts at d = 0.060. Retain the nominal model and fixed 40 °C. How much additional time reaches d = 0.20? Can its clock be reset to a fresh 1092.58 h?

    Show the answer and reasoning

    About 764.80 additional hours; the clock cannot be reset by relabelling the age.

    The remaining increment is 0.20 − 0.060 = 0.140, so Δt = 0.140/r₄₀ ≈ 764.80 h. The model assumes the same unchanged state-independent rate; evidence of a different mechanism or recovery would require a different calculation.

  6. A new service specimen has d = 0.19 at 400 h and d = 0.22 at 500 h, with no intervening observations. If d is continuous and nondecreasing, what crossing-time statement is justified for d = 0.20?

    Show the answer and reasoning

    The crossing is later than 400 h and no later than 500 h. An exact 450 h is not established.

    A chosen interpolation law could supply a modelled crossing, but the two observations alone provide an interval. Calendar times also need to refer to the same start and actual exposure history.

  7. Choose the allowed pair r₆₀ = 0.000410 and r₈₀ ≈ 0.000792458 h⁻¹ because it matches the 40 °C readings. Can those same readings then independently validate that updated pair?

    Show the answer and reasoning

    No. They were used to select the update, although the pair is compatible with the stated rate bounds.

    The fitted pair gives r₄₀ ≈ 0.000195 h⁻¹ and predicts 0.039/0.078/0.117 at 200/400/600 h. Future or otherwise held-out observations are needed to test the updated prediction without reusing its calibration evidence.

  8. The 80 °C nominal line would reach d = 0.30 after 375 h. Is this a supported prediction from the stated model and record?

    Show the answer and reasoning

    No. The arithmetic extends beyond the modelled 0–0.20 response interval and beyond the 250 h endpoint record.

    0.30/0.000800 = 375 h is an extrapolated line value. The dossier does not state that the same path persists after the declared endpoint; a changed response shape or mechanism remains possible.

  9. The further 90 °C test reaches d = 0.20 at 120 h and produces a new reaction layer. A learner reports AF = 500/120 ≈ 4.17 for all field conditions. What remains valid and what fails?

    Show the answer and reasoning

    4.17 is the observed endpoint-time ratio for those two test records. A transferable same-mechanism acceleration factor is not established.

    The new layer is evidence of a changed process. Fitting more temperatures cannot substitute for checking whether the same physical relation and endpoint represent service.

  10. Reverse the two stipulated intervals: 50.0 h at 60 °C followed by 200 h at 40 °C. What final d does the additive model give, and does that prove order never matters in service?

    Show the answer and reasoning

    The model still gives d ≈ 0.0566107. It does not prove general order independence.

    The same two state-independent increments add in either order. Their timing differs, and real moisture, chemistry, stress or transport can evolve with history. Such feedback was excluded explicitly rather than disproved.

What to carry forward

  • Accelerate the relevant mechanism.
  • Separate observed survival from extrapolated life.
  • A response coordinate needs a defined starting state, endpoint, observation procedure and clock.
  • Derive a rate ratio and invert it correctly for time; name lower-temperature extrapolation as a prediction.
  • Propagate the supplied parameter bounds coherently, and keep that range distinct from omitted-model error or population uncertainty.
  • A held-out comparison can challenge a central model even when an allowed parameter family contains the readings.
  • Changing exposure histories and new mechanisms require physical evidence beyond a good fit.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. National Institute of Standards and Technology (NIST)4 sources
    1. e-Handbook of Statistical Methods · 8.3.1.4 Accelerated life testsNIST/SEMATECHSelected reference teachingitl.nist.govRead source
    2. e-Handbook of Statistical Methods · 8.1.4 Physical accelerationNIST/SEMATECHSelected reference teachingitl.nist.govRead source
    3. e-Handbook of Statistical Methods · 8.1.5.1 ArrheniusNIST/SEMATECHSelected reference teachingitl.nist.govRead source
    4. e-Handbook of Statistical Methods · 8.4.2.3 Degradation data instead of failuresNIST/SEMATECHSelected reference teachingitl.nist.govRead source
  2. Development of an Accelerated Test Methodology to the Predict Service Life of Polymeric Materials Subject to Outdoor WeatheringYuanyuan Duan, Yili Hong, William Q. Meeker, Deborah L. Stanley and Xiaohong Gu · 2017 author preprintSelected primary author preprintarxiv.orgRead source