A weathering front advances quickly at first, then more slowly. Must the transport coefficient have fallen? Not necessarily. In a simple retained-layer model, the reactant must cross a progressively thicker altered layer to reach fresh material. The same coefficient can then produce a smaller flux because the concentration drop is spread over a longer distance.
The short answer is that constant conditions need not produce constant front speed. Under the particular planar, quasi-steady assumptions developed here, front thickness squared grows linearly with elapsed time. The result follows from reactant conservation and a changing transport distance. It is not a universal clock for real weathering rinds.
Use the Weathering subject guide for physical breakdown, chemical alteration, exposed area and field interpretation. This branch works through one original moving-front teaching model, then changes its reaction assumption and tests what a measured thickness can actually identify. All numerical parameters are fictional. The aim is to understand a mechanism and its limits, not to date a rock or assess a site from its appearance.
eduKateSG · Weathering
Find your next learning step
Start with the question closest to yours, or read in order. Worked chapters remain open; the grouped index helps you return to a check.
ROUTE 1 · CHAPTERS 1–5
Define the transport model
Draw the layer and name the supply, demand and quasi-steady assumptions.
ROUTE 2 · CHAPTERS 6–10
Derive and test slowing
Balance reactant, integrate thickness and distinguish interval from instantaneous rates.
ROUTE 3 · CHAPTERS 11–15
Interpret parameters and reactions
Examine identification, initial state and the changed front-reaction assumption.
ROUTE 4 · CHAPTERS 16–21
Compare limits and geometry
Work finite reaction, check resistance limits and keep amount and shape distinct.
ROUTE 5 · CHAPTERS 22–29
Test evidence and transfer
Challenge mechanism claims, check a new case and separate model consistency from support.
Chapter contents
Define the transport model · Chapters 1–5
1. Draw a planar layer and a moving boundary
2. Name the reactant supply and the material demand
3. Define the transport coefficient carefully
Derive and test slowing · Chapters 6–10
6. Balance delivered reactant against front advance
7. Integrate from a positive initial thickness
8. Work a complete thickness history
Interpret parameters and reactions · Chapters 11–15
11. Interpret K as a composite parameter
12. Explain why one thickness is not an age
13. Examine the scale behind quasi-steady transport
14. Replace instantaneous front reaction with finite reaction
Compare limits and geometry · Chapters 16–21
16. Work the changed reaction case
17. Check the transport and reaction limits
18. Use the resistance ratio before choosing a simplification
19. Keep total reacted amount distinct from thickness
Test evidence and transfer · Chapters 22–29
22. Avoid treating a square-root pattern as unique proof
23. Transfer to an independent parameter set
24. Audit a rate claim and plan a learning route
25. Reconstruct the flux from a proposed thickness history
26. Examine which measurements could separate explanations
27. Compare parameter changes at a fixed thickness
CHAPTER 1 OF 29 · Define the transport model
1. Draw a planar layer and a moving boundary
Imagine a flat exposed face with an altered layer behind it and fresh parent material beyond the layer. Let ℓ be the altered-layer thickness measured inward from the fixed face. Reactant enters at the face, travels through the layer and is consumed at a sharp front where parent material becomes altered.
We assume the altered material remains in place. There is no erosion of the outer face, no swelling that changes the distance convention and no fracture shortcut. The exposed area A stays fixed. The model is one dimensional, so concentration varies with inward distance alone.
These assumptions create a clean question: how does an increasing transport distance affect the rate of advance? A real front may be diffuse, mineral-specific or irregular. Begin with the drawing so the mathematical thickness has a defined meaning. A visible colour boundary in a field specimen is not automatically the sharp chemical front represented by this model.
CHAPTER 2 OF 29 · Define the transport model
2. Name the reactant supply and the material demand
Let c₀ be the fixed reactant concentration at the outer face. It has units of amount of substance per volume of fluid. Let q be the amount of reactant required to alter one unit bulk volume of parent material, with units of amount per volume of parent material.
The two concentrations refer to different physical volumes even though their dimensional units can be written alike. Their ratio is dimensionless when expressed consistently, but its meaning depends on the definitions. q incorporates the stipulated reaction demand for the transformation represented by our front. It is not simply the measured dissolved concentration in the pore water.
If the front advances a small distance dℓ across area A, it alters bulk volume A dℓ and requires reactant amount qA dℓ. Write that demand before deciding how quickly the material can change. The front cannot consume an arbitrary amount merely because transport is fast; conservation connects the delivered reactant to the transformed volume.
CHAPTER 3 OF 29 · Define the transport model
3. Define the transport coefficient carefully
Use an effective diffusion coefficient D with units of length squared per time. Here its convention is chosen so that multiplying D by the concentration gradient gives reactant flux per unit gross exposed area. Effects such as accessible pathways and tortuosity belong in that stipulated effective coefficient where appropriate.
Do not substitute a free-solution diffusivity without examining whether it matches this convention. Some formulations write porosity or other factors separately. Either convention can be used consistently, but combining them carelessly can count a factor twice or omit it.
For the first model, D, c₀ and q are positive constants. Constant D means the coefficient is fixed, not that the amount crossing the layer each day is fixed. The flux also depends on the concentration gradient. Holding one part of a transport law constant does not hold the entire product constant when the geometry changes.
CHAPTER 4 OF 29 · Define the transport model
4. State the quasi-steady approximation
Assume that transport through the layer adjusts sufficiently rapidly that, at each front position, the concentration profile can be approximated by a steady linear profile. Concentration is c₀ at the outer face and zero at the reaction front. The front moves slowly enough for that approximation to be useful.
This is a quasi-steady moving-boundary model, not a full transient diffusion solution. We neglect reactant storage changes within the altered layer in the leading front balance. The approximation requires support in any physical application; writing a linear profile does not establish that support.
A useful scale comparison comes later, but no single numerical threshold in this article certifies a real material’s regime. The model’s value is that its assumptions are visible. If transport adjustment and front motion occur on comparable time scales, a more complete treatment can be needed. The following derivation should not be presented as an exact solution of every diffusion-reaction weathering problem.
CHAPTER 5 OF 29 · Define the transport model
5. Turn the concentration drop into a flux
At thickness ℓ greater than zero, the linear profile drops by c₀ across distance ℓ. Its inward gradient magnitude is c₀/ℓ. The inward flux magnitude is therefore J = Dc₀/ℓ under our convention.
Check the units. D contributes length squared per time; concentration contributes amount per length cubed; division by thickness contributes inverse length. Their product is amount per area per time, exactly the units needed for a flux across the exposed face.
Now hold D and c₀ fixed and double ℓ. The same concentration drop is spread over twice the distance, so the gradient and flux each halve. No change in the coefficient is required. This is the first connection to stabilise before introducing a time law. A learner who can explain the doubled-distance case already understands why constant transport properties can coexist with slowing front advance.
CHAPTER 6 OF 29 · Derive and test slowing
6. Balance delivered reactant against front advance
Over a short interval dt, supplied amount is AJ dt. Required amount for advance dℓ is qA dℓ. Equating the two and cancelling the fixed area gives q dℓ/dt = J.
Insert the transport expression to obtain dℓ/dt = Dc₀/(qℓ). Define K = Dc₀/q. K has units of length squared per time, and the equation becomes dℓ/dt = K/ℓ.
The cancellation of A is informative. In this ideal planar model, enlarging the face supplies more reactant and exposes proportionally more parent material to be altered. The thickness speed stays unchanged. Total reacted amount still scales with area, as a later check will show. Area cancellation is a consequence of the shared geometry and uniform conditions, not a universal claim that specimen size never matters in weathering.
CHAPTER 7 OF 29 · Derive and test slowing
7. Integrate from a positive initial thickness
Let the initial thickness at time zero be ℓ₀ greater than zero. Multiply the speed equation by ℓ to get ℓ dℓ/dt = K. The left side is one half of the time derivative of ℓ squared. Integrating gives ℓ² = ℓ₀² + 2Kt.
Therefore ℓ(t) is the positive square root of ℓ₀² + 2Kt. The positive root matches the physical thickness convention. A useful inverse expression is t = (ℓ² − ℓ₀²)/(2K), but it is conditional on the model, its parameters and the initial state.
Starting at positive ℓ₀ avoids the zero-thickness singularity in the speed expression. We have not modelled the earliest formation of an altered layer. Extending the formula mathematically towards zero does not resolve the physical assumptions there. Keep the initial thickness and the meaning of time zero attached to every worked answer.
Take fictional K = 0.5 square millimetres per day and initial thickness one millimetre. Then ℓ² = 1 + t when thickness is expressed in millimetres and time in days. After three days, thickness is two millimetres; after eight days, three; after fifteen days, four.
| Elapsed time in days | Thickness in millimetres | Instantaneous speed in millimetres per day |
|---|---|---|
| 0 | 1 | 0.5 |
| 3 | 2 | 0.25 |
| 8 | 3 | 1/6 |
| 15 | 4 | 0.125 |
The first additional millimetre takes three days, the next takes five, and the next takes seven. D, c₀ and q remain constant throughout the example. The slowing follows from the growing transport distance.
These are exact values within a fictional calculation, not observed weathering rates. The table also separates instantaneous speed at a stated thickness from average speed over an interval. Confusing those two rates can conceal the same slowing that the example is designed to explain.
CHAPTER 9 OF 29 · Derive and test slowing
9. Calculate interval times without assuming constant speed
For advance from thickness a to thickness b, with b greater than a, elapsed time is (b² − a²)/(2K). The formula uses both endpoints. Dividing the thickness increase by the initial speed assumes that speed stays at its initial value, which our model explicitly rejects.
From two to three millimetres with K = 0.5, the exact interval is five days. The initial speed is one quarter millimetre per day, which would suggest four days if incorrectly held constant. The final speed is one sixth, which would suggest six days if incorrectly held constant. The true five-day interval lies between those two endpoint estimates.
This bracket is a useful reasonableness check because speed decreases continuously over the interval. It is not a general numerical rule for all front models. Here it follows directly from K/ℓ. Begin with the mechanism, then use the inequality to catch a mistaken constant-speed calculation.
CHAPTER 10 OF 29 · Derive and test slowing
10. Keep average speed and instantaneous speed separate
Between two and three millimetres, average speed is one millimetre divided by five days, or 0.2 millimetres per day. Instantaneous speed starts at 0.25 and ends at approximately 0.1667. A single interval average cannot be substituted for the instantaneous law at every position.
For any interval a to b in this model, average speed is 2K/(a + b). It equals the instantaneous speed at thickness (a + b)/2. That is an algebraic property of the quadratic time relation, not evidence that the front spends equal amounts of time at every thickness.
A useful independent check is to compare equal-duration intervals instead of equal-thickness intervals. With the same initial conditions, the advance during the first three days is one millimetre. During the next three days, thickness rises from two to √7 millimetres, an increase of about 0.646. Equal elapsed times therefore produce shrinking advances even though the external concentration remains fixed.
CHAPTER 11 OF 29 · Interpret parameters and reactions
11. Interpret K as a composite parameter
K combines transport, supplied concentration and material demand. A fitted value of K does not separately identify D, c₀ and q. Several different parameter combinations can produce the same thickness history within this model.
For example, fictional D = fifty square millimetres per day and c₀/q = 0.01 give K = 0.5. D = one hundred with c₀/q = 0.005 gives the same K. Every thickness prediction from our diffusion-limited equation is then identical when the same initial thickness is used.
This is a structural information limit, not an arithmetic inconvenience. To distinguish those combinations, an investigation would need additional information appropriate to the parameters. Printing more time points does not automatically separate parameters that enter the equation only through their product or ratio. The learner should identify which combination the observation constrains before attaching a physical interpretation to a fitted number.
CHAPTER 12 OF 29 · Interpret parameters and reactions
12. Explain why one thickness is not an age
A specimen has a three-millimetre altered layer. Our fictional history reached that thickness after eight days, but that does not establish an age for the specimen. The calculation also used K = 0.5, initial thickness one millimetre, a fixed face and all the other model assumptions.
If K were 0.25 with the same initial state, reaching three millimetres would take sixteen days. If the initial thickness were two millimetres and K remained 0.5, it would take five. A final thickness alone cannot choose between those histories.
The existing rind thickness and altered-volume article explains why geometry, altered fraction, mass and time are different targets. Our transport model adds a possible mechanism under explicit assumptions. It does not remove that earlier warning or turn a visible rind into a universal clock.
CHAPTER 13 OF 29 · Interpret parameters and reactions
13. Examine the scale behind quasi-steady transport
A diffusion adjustment time across thickness ℓ has scale ℓ²/D. A front-advance time based on the current thickness and speed has scale ℓ divided by K/ℓ, or ℓ²/K. Their ratio is K/D = c₀/q in this formulation.
A small supplied reactant amount relative to the parent material’s reaction demand can therefore support the intuition that the concentration field adjusts faster than the front advances. The same ratio also points to why neglected dissolved-reactant storage may be relatively small in a suitable regime.
This is a scale argument within our chosen conventions, not a sufficient checklist for real rocks. Effective transport, accessible volume, geometry and reaction behaviour still need examination. A small ratio cannot repair an unmodelled fracture or an eroding boundary. Use it to ask whether the approximation has a plausible separation of time scales, then seek the appropriate physical evidence rather than treating a classroom number as certification.
CHAPTER 14 OF 29 · Interpret parameters and reactions
14. Replace instantaneous front reaction with finite reaction
The first model set reactant concentration at the front to zero, representing consumption fast enough that front reaction does not add a limiting resistance. Now change that one assumption. Let front concentration be c_f and let reaction flux obey J = k c_f, where k is a positive constant with units of length per time.
The layer transport flux is now D(c₀ − c_f)/ℓ. Equate it to the reaction flux. This links delivery and consumption at the same front instead of choosing either expression alone.
Solving gives J = c₀/(ℓ/D + 1/k). The denominator contains a transport contribution that grows with thickness and a fixed front-reaction contribution. This is another deliberately simple model. The linear reaction law is stipulated, not asserted for every mineral or aqueous reaction. Its purpose is to show how changing an assumption changes the thickness-time relationship.
CHAPTER 15 OF 29 · Interpret parameters and reactions
15. Derive the finite-reaction front law
Continue using q dℓ/dt = J. Define α = D/k, which has units of length. With the same K = Dc₀/q, front speed becomes dℓ/dt = K/(ℓ + α).
Integrating from ℓ₀ gives (ℓ² − ℓ₀²)/2 + α(ℓ − ℓ₀) = Kt. Equivalently, ℓ(t) = √((ℓ₀ + α)² + 2Kt) − α. Substitute t = zero to confirm that thickness begins at ℓ₀.
When α is positive, the denominator is larger than in the instantaneous-reaction model at the same thickness, D, c₀ and q. Speed is smaller. The additional term is not an arbitrary correction factor pasted onto an earlier answer; it follows from retaining a nonzero front concentration and satisfying both transport and reaction equations. That derivation makes the condition attached to the new formula visible.
Keep K = 0.5 square millimetres per day and initial thickness one millimetre. Now set α = two millimetres. The time to reach two millimetres is [(4 − 1)/2 + 2(2 − 1)]/0.5, which is seven days.
To reach three millimetres, time is [(9 − 1)/2 + 2(3 − 1)]/0.5, or sixteen days. The earlier diffusion-limited model gave three and eight days respectively. The difference results from changing the front-reaction assumption while retaining the other stated quantities.
At initial thickness one millimetre, speed is 0.5/(1 + 2), or one sixth millimetre per day. At thickness two, it is one eighth. The front still slows as the layer grows, but its early behaviour differs. A thickness observation should therefore not be interpreted through a chosen time law without examining the reaction assumption that made that law possible.
CHAPTER 17 OF 29 · Compare limits and geometry
17. Check the transport and reaction limits
If k grows without bound while D, c₀ and q stay fixed, α approaches zero. The finite-reaction law approaches the original diffusion-limited model. This is a useful consistency check on both the algebra and the physical interpretation of a very fast front reaction.
If D grows without bound while k, c₀ and q stay fixed, the layer resistance ℓ/D becomes negligible. Flux approaches k c₀, and speed approaches k c₀/q. Thickness then grows linearly with time in that limiting model.
Do not increase D while artificially holding K fixed and claim to have tested the same physical limit. K contains D. A limiting argument must state which independent quantities remain fixed and which derived quantities change. This habit matters beyond weathering: a parameter defined from other parameters cannot always be varied independently without changing the meaning of the experiment.
CHAPTER 18 OF 29 · Compare limits and geometry
18. Use the resistance ratio before choosing a simplification
The transport resistance ℓ/D divided by reaction resistance 1/k is ℓk/D, or ℓ/α. When this ratio is large, transport contributes most of the resistance in the stipulated model. When it is small, front reaction contributes most. Near comparable values, both terms matter.
These are qualitative limiting statements. No universal numerical cutoff is supplied here for neglecting a term. The required accuracy and supported parameter ranges determine whether a simplification is adequate for a particular calculation.
As the layer thickens while D and k remain fixed, ℓ/α increases. A system that initially has an important reaction resistance can become increasingly affected by transport distance. That changing balance is another reason not to infer a changing material coefficient from slowing alone. The geometry can change the relative importance of fixed process properties.
CHAPTER 19 OF 29 · Compare limits and geometry
19. Keep total reacted amount distinct from thickness
The total reactant amount consumed in advancing from ℓ₀ to ℓ is qA(ℓ − ℓ₀). Under our assumptions, doubling A doubles that amount and doubles the instantaneous total reactant delivery AJ. It does not change the thickness history because the demand per unit advance doubles as well.
For a fictional one-square-metre face advancing one millimetre, altered bulk volume is 0.001 cubic metres. If q is stipulated as two hundred moles per cubic metre of parent material, consumption is 0.2 moles. A two-square-metre face requires 0.4 moles for the same thickness change.
This calculation is about the reactant represented by q. It is not automatically rock mass lost, dissolved product mass exported or carbon dioxide permanently removed. Reaction stoichiometry, density, retained solids and chemical pathways would be needed to connect those targets. A front-position model should not quietly acquire claims about quantities it has not defined.
CHAPTER 20 OF 29 · Compare limits and geometry
20. Change geometry only by rebuilding the balance
A planar layer has the same gross area at every depth in this idealisation. A sphere or irregular fragment does not. Transport area, reaction-front area and path geometry can change together. Replacing ℓ by a spherical rind thickness inside our planar law does not derive a spherical model.
The existing rind-geometry branch already works through the relationship between shell thickness and altered volume. Use it for that geometric question. To model transport through a curved retained shell, the concentration field and front balance must be formulated for the curved geometry and its boundary conditions.
This is a practical transfer boundary. The useful habit that travels is conservation: identify the delivered reactant, the area through which it travels and the volume transformed by front movement. The final planar formula travels only when its assumptions remain appropriate. Recognising where to restart a derivation is stronger understanding than applying the same expression to every shape.
CHAPTER 21 OF 29 · Compare limits and geometry
21. Ask whether the altered layer remains intact
Our growing-distance mechanism assumes the altered layer stays between the supply and the fresh material. Erosion can remove part of that layer. Cracks can create shorter pathways. Changes in porosity, mineralogy or reaction products can alter effective transport. Each change can invalidate the fixed-coefficient retained-layer representation.
That does not make the simple model useless. It tells a reader which observations would matter when judging whether its mechanism is relevant. An intact-looking surface is not proof of uniform internal transport, and a complex field setting is not evidence that every process must be changing simultaneously.
Keep a record of what was observed and what was assumed. If a proposed explanation relies on transport through a retained layer, evidence about that layer and its pathways matters. If the boundary itself moves through erosion, the thickness coordinate and material history need a revised model. Do not repair an incompatible geometry merely by fitting a different K.
CHAPTER 22 OF 29 · Test evidence and transfer
22. Avoid treating a square-root pattern as unique proof
A model can predict a pattern without being the only mechanism capable of producing it. A thickness history resembling a square-root law is therefore not, by itself, proof that the particular diffusion-limited front model here governs the system.
Hoke and Turcotte's 2002 study of weathering-rind growth develops a damage-diffusion model with its own assumptions and erosion context. It is related specialist background, not the source of our exact sharp reactant-front construction. Braun and colleagues' 2016 regolith study illustrates that square-root growth can also arise in a model involving water flow and solute removal under specified conditions.
Those studies motivate a careful evidence question: which additional measurements distinguish candidate explanations? Our derivation supplies one conditional prediction. It does not establish uniqueness or replace the mineralogical, hydrological and geomorphic evidence needed for field interpretation.
CHAPTER 23 OF 29 · Test evidence and transfer
23. Transfer to an independent parameter set
Take fictional K = two square millimetres per day and initial thickness two millimetres. Under the instantaneous-front-reaction model, calculate the time to reach six millimetres. The squared-thickness increase is thirty-two square millimetres, so time is thirty-two divided by four, or eight days.
Now keep K and the initial thickness but add α = one millimetre. The finite-reaction time is [(36 − 4)/2 + 1(6 − 2)]/2, or ten days. The additional resistance adds two days for this particular interval.
Finally, double the exposed area without changing any other model input. Both predicted times remain unchanged, while total reacted amount doubles. Explain each answer through the balance rather than through memorised formula substitutions. Then remove the known value of K and ask whether the final thickness alone determines time. The correct response is to identify missing information, not to reuse a previous case’s parameter without support.
CHAPTER 24 OF 29 · Test evidence and transfer
24. Audit a rate claim and plan a learning route
A report says, “The front slowed, therefore diffusion became weaker.” Ask whether it measured a transport coefficient or only a front displacement over time. In our first model, slowing occurs with constant D because the gradient decreases as the layer thickens.
Next inspect the coordinate, initial state, supply concentration, material demand and reaction assumptions. Check whether the layer is retained and whether the reported front is chemically comparable across observations. A fitted curve can be useful evidence, but its parameters need interpretation within the model that produced it.
To repair understanding, double a drawn layer’s thickness and explain the flux change. To stabilise, balance delivered amount against transformed volume and work one interval. To extend, change the reaction assumption or geometry and identify which derivation must be rebuilt. The learner’s strongest result is a prediction accompanied by the conditions that make it meaningful.
CHAPTER 25 OF 29 · Test evidence and transfer
25. Reconstruct the flux from a proposed thickness history
A derived formula should satisfy the balance that produced it. For the diffusion-limited solution, differentiate ℓ² = ℓ₀² + 2Kt. This gives 2ℓ dℓ/dt = 2K, so q dℓ/dt = qK/ℓ = Dc₀/ℓ. That is exactly the supplied flux per gross area.
For the finite-reaction solution, differentiation gives (ℓ + α)dℓ/dt = K. Multiplying by q and substituting K and α recovers c₀/(ℓ/D + 1/k). The proposed time law therefore reproduces the simultaneous transport and reaction relationship at each positive thickness.
This forward check is more useful than checking one numerical endpoint alone. A mistyped square, omitted factor of two or misplaced α can sometimes pass a convenient endpoint but fail the governing balance. Check the initial state, derivative, units and positive-root choice together.
Algebraic consistency still does not establish that the model describes a real specimen. It confirms that the solution answers the equations and assumptions we wrote. Physical support remains a separate task. That distinction lets a learner be exact about the mathematics while remaining appropriately cautious about a field interpretation.
CHAPTER 26 OF 29 · Test evidence and transfer
26. Examine which measurements could separate explanations
Imagine two candidate accounts of a slowing front. One uses increasing diffusion distance with fixed properties. Another uses a changing supply concentration. A thickness record might be compatible with both over a limited interval. Calling the first account plausible does not eliminate the second.
Relevant additional evidence depends on the candidate models. Measurements that constrain boundary concentration, transport properties, reaction demand or the internal concentration profile could address assumptions that thickness alone leaves unresolved. Observations of fractures, erosion or mineral changes could challenge the retained planar layer directly. This is a guide to questions, not a field sampling protocol or a promise that any one measurement resolves every ambiguity.
Within the fictional diffusion-limited model, a supported K and a known initial thickness would allow a predicted thickness history to be compared with later observations. Systematic disagreement would motivate examining assumptions and uncertainties. Agreement would support consistency over that range, while remaining insufficient to prove a unique mechanism.
The practical learner route is to write two columns: what the thickness data constrain, and what the proposed interpretation additionally assumes. A good explanation connects the columns with evidence. When the connection is missing, state the remaining possibilities rather than turning a fitted curve into a complete causal account.
CHAPTER 27 OF 29 · Test evidence and transfer
27. Compare parameter changes at a fixed thickness
At a fixed positive thickness in the first model, doubling D doubles speed if c₀ and q stay fixed. Doubling c₀ has the same proportional effect. Doubling q halves speed. These comparisons hold thickness fixed so that they isolate parameter dependence at that position.
Comparing thicknesses at a fixed elapsed time is a different question. With the same initial thickness, doubling K gives ℓ² = ℓ₀² + 4Kt rather than ℓ₀² + 2Kt. It does not generally double ℓ because the initial squared thickness and the square root remain in the expression.
For example, with initial thickness one millimetre after three days, K = 0.5 gives thickness two. Doubling K to one gives thickness √7, about 2.646 millimetres, rather than four. The speed at a specified thickness doubled, but the thickness reached at a specified time did not.
In the finite-reaction model, changing D also changes both K and α when k, c₀ and q remain fixed. Use J = c₀/(ℓ/D + 1/k) to see the consequence directly: increasing D reduces transport resistance, while the fixed reaction resistance limits how far the flux can rise. This comparison protects the meaning of a parameter experiment and prevents derived quantities from being varied as though they were independent controls.
Does a slowing front prove that D decreased? No. The growing distance can reduce the gradient with constant D. Does constant boundary concentration mean constant flux? Not when the transport distance or front concentration changes.
Can the square-root law be used from zero thickness without qualification? The speed expression is singular there, and our physical treatment starts from a positive retained layer. Is K the diffusion coefficient? It combines D with supplied concentration and reaction demand. Does a measured rind thickness determine age? It needs a supported model, parameters and initial history.
Does this article give a field weathering-rate calibration? Its numbers are fictional teaching inputs. The linked research provides specialist context, and each study has its own model and evidence. The practical result is a set of questions and conservation checks a reader can apply to a supplied explanation, not a site-specific dating or engineering assessment.
CHAPTER 29 OF 29 · Test evidence and transfer
29. Continue with weathering, geometry and geochemistry
For a measurement comparison, continue to Comparing Mass Loss Across Unequal Specimens and Exposed Areas. It asks how to normalise an observed change, while the present model asks how a stipulated supply and demand move a front. Those targets can inform each other only after their physical quantities are connected.
As a final evidence check, imagine a specimen with an uneven rind, a crack reaching fresh material and an unknown exposure history. Identify the planar, retained-path and initial-state assumptions that are now unsupported. Do not calculate an age by choosing whichever example produces a familiar number. A useful result can instead be a clear account of the information required before a transport model could be applied. That is practical transfer: using the model to recognise both a possible mechanism and the boundary of the evidence.
Return to Weathering for the wider processes. Use rind thickness and altered volume to separate shell geometry from mass and time. Continue to Geochemistry for chemical inventories, fluxes and the distinction between thermodynamic possibilities and rates.
The two primary research links were checked on 24 October 2026. Their authors’ models should be read on their own terms. The planar reactant balance, fictional parameter sets, finite-reaction comparison and transfer exercises here are original teaching constructions with stated assumptions.
For the educational route, use The Gold Standard of Education: learning transfers into life and How X Works. Keep the central insight precise: a process rate can change because the distance and gradient change, even when the coefficient remains fixed. Then keep the equally useful question beside it: which observations would show that this is the right model for the material being discussed?