Separate a changed input from a delayed environmental stock

Connect the Earth-system mechanism

Environmental stocks retain material from earlier inputs. A reduction in present input can therefore coexist with continued loss or accumulation while the system approaches a new balance. Comparing only this year’s input with this year’s stock can confuse a transient response with the absence of an intervention effect.

Build the declared budget

The original one-pool model begins at 1000 kg and updates yearly: first remove fraction k of the current stock, then add input I. Thus Snext=(1−k)S+I. Its conditional equilibrium is I/k, and the deviation from that equilibrium is multiplied by 1−k each year. The ordering matters; newly added material is not removed until the following update.

Separate the competing quantities

After n updates, S=I/k+(S 0−I/k)(1−k)^n. The formula separates the eventual equilibrium from the decay of the initial deviation. The displayed fractional half-decay time ln(0.5)/ln(1−k) interpolates that geometric envelope; the actual model observations occur at integer-year steps.

Challenge the inference

Real soils and sediments contain pools with different accessibility and turnover, plus transport, fire, erosion, microbial and mineral interactions. A single short record may fit several input/removal combinations yet imply different long-run stocks. Use independent flux and pool evidence, inspect residuals and test alternative models. A good fit to one trajectory is not a measured carbon age or proof of permanent storage.

Worked original packet

At I 40 kg/year and k 0.1, equilibrium is 400 kg. First-year stock is 940; after five years 400+600×0.9⁵=754.294 kg. The initial deviation halves after about 6.58 years under the supplied geometric model, not after one year because input changed.

Investigation

Compare inputs 40 and 80 at k 0.1, then compare removal fractions 0.1 and 0.2 at input 40. Separate the equilibrium shift from the approach rate and identify a missing flux that could mimic a changed coefficient.

Changed-case transfer

At I 100 and k 0.1 with initial 1000, what happens? At I 40 and k 0.2, calculate equilibrium and first-year stock.

Feedback

The first case is already at equilibrium 1000, although removal 100 and input 100 continue. The second has equilibrium 200 and first-year stock 840. Zero net change at equilibrium does not imply zero material turnover.

Sources and evidence boundary

Original prose, numerical packets and exercises. Mechanism/method context checked 30 October 2026: primary source. These are conditional computational models, not measured field records, carbon credits, safe yields, treatment standards or independent mastery. No physical extraction, chemical addition or environmental release is requested.