Two wetland water-level records can have the same mean and still imply different mean stored volumes. The missing step is the basin’s geometry. If a rising water surface spreads across a widening basin, each extra centimetre can cover more area than the previous centimetre. Averaging the levels before converting them can then discard information the volume calculation needs.
The short answer is to convert each supported level to storage with the appropriate level–volume relation, then average those volumes using the durations they represent. In the invented basin used here, V(h) = 50h² cubic metres, with h measured in metres above the chosen bottom datum. A constant level of one metre gives mean storage fifty cubic metres. Equal periods at zero and two metres have the same mean level, but mean storage one hundred cubic metres.
This is a mathematical teaching basin, not a surveyed wetland or a flood design. Its fixed geometry and complete plateau durations are stipulated. We use it to distinguish mean level, mean area, mean volume, end-to-end storage change and flow. None of those quantities should silently replace another.
For ecological context and the variety of wetland settings, begin with the Wetlands subject guide. This article addresses one narrower interpretation problem: what a mean level can and cannot tell you once the storage conversion is nonlinear.
eduKateSG · Wetlands
Find your next learning step
Convert supported level histories with a specified basin geometry before calculating duration-weighted storage.
ROUTE 1 · CHAPTERS 1–5
Define the conversion
Establish the datum, build the basin model and compare equal-mean histories.
ROUTE 2 · CHAPTERS 6–10
Preserve timing and geometry
Use variance, duration weights, interpolation limits and the constant-area exception.
ROUTE 3 · CHAPTERS 11–15
Separate neighbouring questions
Test shelves, changing land, storage change, flow and bounds.
ROUTE 4 · CHAPTERS 16–20
Check the evidence
Preserve uncertainty, datum, connectivity and ecological distinctions.
ROUTE 5 · CHAPTERS 21–24
Practise and continue
Solve a fresh record, target repairs and choose related guides.
Chapter contents
Define the conversion · Chapters 1–5
1. Define the level and its reference
2. Build volume from horizontal layers
3. Distinguish storage at the mean from mean storage
Preserve timing and geometry · Chapters 6–10
6. Connect the discrepancy with variance
7. Give unequal durations their proper weights
8. Separate complete histories from isolated readings
Separate neighbouring questions · Chapters 11–15
11. Work with a shelf instead of a smooth widening basin
12. Preserve the geometry when the land changes
13. Keep mean storage apart from storage change
Check the evidence · Chapters 16–20
16. Translate level intervals into volume intervals
17. Quantify a shared datum shift
18. Ask whether the basin is hydraulically connected
19. Keep surface-water storage separate from ecological condition
Practise and continue · Chapters 21–24
21. Solve a fresh history independently
22. Repair, stabilise and extend the method
Water level is a vertical coordinate relative to a chosen reference, often called a datum. It is not automatically water depth at every point. In a basin with a sloping bottom, the same water surface can stand above a deep central region while leaving a high shelf dry.
Our h is measured upward from a model bottom reference. We allow h from zero to two metres unless another range is explicitly supplied. Zero means no above-bottom surface-water volume in this particular model. It does not mean that a real wetland’s soil contains no water, or that a sensor reading zero at another datum means a dry basin.
The reference must stay fixed across the history. Combining a reading measured above one gauge zero with another measured above a different gauge zero can create artificial variation. Converting units will not fix a shifted reference; it requires a datum adjustment based on the record.
Before calculating, write one sentence identifying the quantity, reference and domain. For example: “These are surface levels above the fixed basin-bottom datum, in metres, within the surveyed range.” That sentence establishes what the conversion is allowed to interpret and prevents an attractive equation from being applied to the wrong vertical coordinate.
Imagine filling the basin with many thin horizontal layers of water. A thin layer’s volume is approximately its horizontal area multiplied by its thickness. Add the layers from the bottom up to the water surface and you obtain the stored volume.
For our invented geometry, the flooded horizontal area at level h is A(h) = 100h square metres. At one metre it is one hundred square metres; at two metres it is two hundred. The basin widens as it fills. Because area changes with height, multiplying the final surface area by the full depth would overcount lower layers.
Adding the layers gives V(h) = 50h² cubic metres. Readers using calculus can write V(h) as the integral of A(z) from zero to h. Readers working geometrically can recognise a triangle under the straight area-versus-height graph: one-half times height h times top area 100h.
The coefficient has units. In A(h), one hundred has units of metres, so multiplying by h metres gives square metres. In V(h), fifty likewise has units of metres, so h squared gives cubic metres. Keeping units visible is a useful check when the same geometry is recorded in centimetres.
CHAPTER 3 OF 24 · Define the conversion
3. Distinguish storage at the mean from mean storage
Suppose a level record has duration-weighted mean one metre. One possible calculation inserts that number into the conversion: V(1) = 50 cubic metres. This is the storage associated with the mean level. It describes one hypothetical state.
Mean storage is a different calculation. Convert each actual state to its volume, multiply by the duration of that state, add the products and divide by total duration. This describes an average of the history’s volumes. For a curved conversion, the two operations need not give the same result.
The distinction is easiest to see before any complicated monitoring issues appear. We will assume exact levels, exact durations and exact geometry. If the answers differ even there, the difference cannot be blamed on rounding, sensor error or missing visits. It is already built into the order of the mathematical operations.
Name both outputs if you report them. “Volume at the mean level” and “duration-weighted mean volume” are inspectable descriptions. The phrase “average water” does not identify a calculation and can hide a substantial change of meaning.
History A stays at one metre for ten hours. Every state has volume fifty cubic metres, so its mean level is one metre and its mean volume fifty cubic metres.
History B stays at zero metres for five hours and two metres for five hours. Its mean level is (5 × 0 + 5 × 2)/10 = one metre. The corresponding volumes are zero and two hundred cubic metres. Its mean volume is (5 × 0 + 5 × 200)/10 = one hundred cubic metres.
| Supplied complete history | Mean level | Volume at mean level | Mean stored volume |
|---|---|---|---|
| Ten hours at 1 m | 1 m | 50 m³ | 50 m³ |
| Five hours at 0 m, five at 2 m | 1 m | 50 m³ | 100 m³ |
The transitions are idealised as taking negligible time in this comparison. We are not claiming physically instantaneous filling of a real site. The records are complete plateau histories chosen to isolate averaging. If transition durations mattered, they would need their own supported volume history.
The equal mean level therefore leaves an ambiguity of fifty cubic metres between these two examples. It does not uniquely determine mean storage, even when the basin geometry itself is perfectly known.
CHAPTER 5 OF 24 · Define the conversion
5. Explain the difference without memorising a rule
At the low state in History B, a one-metre decrease from the mean removes fifty cubic metres. At the high state, a one-metre increase from the mean adds one hundred and fifty cubic metres. The volume changes on the two sides are unequal because the basin gets wider upward.
The equal positive and negative level departures cancel when levels are averaged. Their volume departures do not cancel. That is why the same mean level can conceal a larger mean volume. The explanation depends on this geometry; it is not a claim that every nonlinear function behaves in this direction.
You can check the pattern with levels half a metre and one and a half metres for equal durations. Their volumes are 12.5 and 112.5 cubic metres, and the mean is 62.5. The mean level remains one metre, but the smaller spread creates a smaller discrepancy than the zero-and-two history.
A useful prediction comes before the arithmetic: for this upward-curving volume relation, a spread around a fixed mean should raise mean volume above the volume at that mean. The next chapter quantifies that prediction exactly for our stipulated quadratic relation.
CHAPTER 6 OF 24 · Preserve timing and geometry
6. Connect the discrepancy with variance
Let μ be the duration-weighted mean level. Define the duration-weighted variance as the mean of (h − μ)² across the same history. This is a descriptive population variance of the supplied time record, using total duration in the denominator. It is not the sample variance with a correction for estimating a wider population.
The identity mean(h²) = μ² + variance gives mean(V) = 50μ² + 50variance for our basin. The first term is V(μ); the second is the discrepancy. Thus mean volume minus volume at mean level equals fifty times the level variance, with the coefficient’s appropriate units retained.
For History B, μ is one and variance is one square metre. The correction is fifty cubic metres. For the half-and-one-and-a-half history, variance is 0.25 square metres and the correction 12.5 cubic metres. Both agree with the direct calculations.
This identity is exact for this quadratic model and these time weights. It is not a universal wetland correction factor. Another basin’s level–volume relation may require more than mean and variance. If only the mean was archived, the missing variation cannot be reconstructed merely by naming a statistical identity.
CHAPTER 7 OF 24 · Preserve timing and geometry
7. Give unequal durations their proper weights
Consider a third complete history: level half a metre for three hours and one and a half metres for one hour. Its mean level is (3 × 0.5 + 1 × 1.5)/4 = 0.75 metres. Its mean volume is (3 × 12.5 + 1 × 112.5)/4 = 37.5 cubic metres.
The volume at the mean level is 50 × 0.75² = 28.125 cubic metres. The discrepancy is 9.375 cubic metres. The duration-weighted variance is 0.1875 square metres, so the variance identity gives the same difference.
A simple average of the two listed levels would give one metre. That would treat a three-hour plateau and a one-hour plateau as equally important. The error would enter before the level-to-volume conversion, producing a second and separate problem.
Write duration beside each row even when the durations happen to be equal. It makes the weighting rule visible and supports later changes to the record. A spreadsheet that averages rows is appropriate only when each row represents equal time or when the question really asks for an average per observation.
CHAPTER 8 OF 24 · Preserve timing and geometry
8. Separate complete histories from isolated readings
Our plateau examples specify what happened throughout each interval. A field notebook containing a level at 09:00 and another at 13:00 does not automatically establish a constant, linear or smoothly varying level between them. The two readings are endpoints, not a complete four-hour history.
If you assume linear interpolation, state it. The inferred volume history will still be curved when V is quadratic. Averaging only the endpoint volumes amounts to a trapezoidal approximation of that curve, rather than an exact duration average under the linear-level assumption.
For example, let level rise linearly from zero to two metres over an interval. Mean level is one metre. Mean h squared is four-thirds square metres, so mean volume is 200/3, about 66.67 cubic metres. Averaging endpoint volumes would give one hundred, while using the mean level would give fifty. The three results describe different operations.
This example does not prescribe interpolation for a real wetland. It shows why the between-reading assumption matters. Keep the observed timestamps, any gap flags and the reconstruction method. Additional calculations cannot turn unsupported interpolation into an observation.
CHAPTER 9 OF 24 · Preserve timing and geometry
9. Average area and volume as separate quantities
In our model, flooded area is a linear function of level: A(h) = 100h. Therefore mean area is exactly one hundred times mean level for every supported history. Histories A and B both have mean flooded area one hundred square metres, despite having different mean stored volumes.
This is a useful surprise. An area summary may agree while the volume summary differs. Agreement in one quantity does not validate the conversion of another quantity. The extra power of h in the volume relation changes what information averaging must preserve.
Nor does a mean area tell us how long a particular patch was wet. A high shelf can be flooded only during the high state, while a low region may remain wet much longer. Patch-level inundation duration requires the level history and the patch elevation, together with the appropriate connection assumptions.
For that ecological timing question, use the existing guide to comparing hydroperiods when timing and duration differ. Here we keep the calculation focused on surface-water storage while recognising that habitat interpretation may need a different summary of the same record.
CHAPTER 10 OF 24 · Preserve timing and geometry
10. Find the constant-area exception
A straight-sided basin with constant horizontal area A has volume V(h) = Ah above its bottom. In that case mean volume equals A times mean level. Converting the mean and averaging converted levels agree exactly, provided the same time weights and fixed geometry are used.
This exception explains the general issue more clearly than a blanket warning about averages. The problem is the curvature of the conversion, not averaging itself. A linear conversion commutes with averaging; a curved conversion generally does not.
A constant added storage offset also passes through a mean correctly. If V(h) = V₀ + Ah over a stated range, mean V is V₀ + A mean h. The reference volume V₀ must belong to that range and datum; it is not a licence to ignore water below an arbitrary gauge zero.
Before choosing a complicated method, inspect the actual conversion. Is it linear over the levels in question? Is the relation valid over the entire observed range? If the record crosses a shelf or another geometric transition, a single linear segment may no longer apply. State the segment and its limits rather than assuming a small-looking level range is automatically safe.
CHAPTER 11 OF 24 · Separate neighbouring questions
11. Work with a shelf instead of a smooth widening basin
Consider a second invented basin. Its horizontal area is one hundred square metres from level zero to one metre, then three hundred square metres above one metre. The shelf becomes inundated at the one-metre level. For this teaching model, volume is 100h below the shelf and 100 + 300(h − 1) above it.
At half a metre, volume is fifty cubic metres. At one and a half metres, volume is two hundred and fifty. Equal durations at those levels give mean volume one hundred and fifty. Their mean level is one metre, whose volume is one hundred cubic metres.
Within either straight segment alone, averaging and conversion agree. Crossing the shelf introduces the bend and the discrepancy. This shows why a locally linear approximation must be checked against the history’s full range.
The shelf model is not a claim that real basins have perfectly abrupt area steps. It is a manageable representation of a change in flooded extent. If the geometry comes from a survey table, use that table’s interpolation rules and uncertainty. Do not choose a convenient quadratic correction merely because it worked for our first basin.
CHAPTER 12 OF 24 · Separate neighbouring questions
12. Preserve the geometry when the land changes
A fixed level–volume relation assumes a fixed basin shape over the period being interpreted. Sediment accumulation, erosion or changes in connectivity can alter the relation. A water-level comparison using one old conversion may then combine actual level change with an outdated geometry.
The existing guide to separating surface accretion from land-elevation change helps distinguish vertical land measurements. A surface accretion measurement is not itself a complete new basin-volume survey, and a changed water level does not establish which land process occurred.
For a repeated storage analysis, identify the geometry version, the survey date and the datum used. If separate periods use separate level–volume relations, document that choice. If a common relation is retained for comparability, explain its limits.
A numerical example makes the issue visible. If a revised fictional survey gives V(h) = 40h² instead of 50h² at the same datum, level one corresponds to forty rather than fifty cubic metres. The level did not change, but the conversion did. That is a changed geometry assumption, not a change generated by averaging.
CHAPTER 13 OF 24 · Separate neighbouring questions
13. Keep mean storage apart from storage change
Mean storage describes the average amount present over a period. Storage change compares the ending amount with the starting amount. A basin can begin and end at the same level while experiencing large fluctuations and a high mean volume between those endpoints.
Under fixed geometry, storage change is V(h_end) − V(h_start). If both endpoints are one metre, that change is zero regardless of whether the intervening record stayed constant or rose and fell. Zero net change does not mean no water moved through the basin.
Conversely, mean storage alone does not identify either endpoint. A record averaging one hundred cubic metres might start high and end low, start low and end high, or begin and end together. The order matters for a change question even when it does not matter for the average of a fully specified multiset of states and durations.
Use an explicit target statement before calculating: “mean amount present,” “end-to-end change,” or “largest amount present.” These are all storage questions, but they require different operations on the history. A correct mean cannot substitute for the maximum when the concern is a threshold exceeded during a short high-water period.
CHAPTER 14 OF 24 · Separate neighbouring questions
14. Do not infer flow from a mean volume
Flow has units of volume per time. Storage has units of volume. A mean storage is still a volume, even though time weights were used to calculate it. Dividing it by the monitoring period does not automatically produce a meaningful inflow, outflow or residence time.
A water balance needs relevant inflows, outflows and storage change, with processes and boundaries defined. The existing article on flood drainage, rainfall intensity and storage explains the separate accounting question. Its actual published route should be followed for that broader problem rather than replacing flow evidence with a level average.
In our plateau example, a high mean volume could coexist with small exchange or large exchange. The stored amount alone cannot distinguish those situations. A basin containing one hundred cubic metres while one cubic metre per hour enters and leaves has the same storage as one with ten cubic metres per hour entering and leaving, under a compatible steady account.
Those invented alternatives show nonuniqueness. They do not describe a real wetland’s mixing or ecology. To estimate a turnover time, define the intended model and obtain the required flow evidence. The conversion from level to volume supplies one input, not the whole inference.
CHAPTER 15 OF 24 · Separate neighbouring questions
15. Use range information to bound an unknown mean
Suppose only the mean level μ = one metre and the range zero to two metres are known. In our quadratic basin, mean volume cannot be below fifty cubic metres. A constant level at the mean attains that lower bound.
The range also gives an upper bound. For h between zero and two, h² is at most 2h. Averaging yields mean h² at most two times mean h, or two square metres here. Mean volume is therefore at most one hundred cubic metres. Equal durations at the endpoints attain that upper bound.
The two complete histories from the beginning are thus the extremes for this particular mean and range. We can report a supported interval of fifty to one hundred cubic metres even without knowing the full distribution. We still cannot choose one value inside it as though it were measured.
More generally, with levels between zero and H, mean h squared lies between μ² and Hμ. Both the mean and the range must refer to the same duration-weighted history. A few observed extremes do not establish hard bounds on unobserved intervals, so do not silently treat a sampling range as a guaranteed physical range.
CHAPTER 16 OF 24 · Check the evidence
16. Translate level intervals into volume intervals
If a supported level lies between 0.9 and 1.1 metres, the monotone quadratic relation gives volume between 40.5 and 60.5 cubic metres. The volume at one metre is fifty. The interval is asymmetric around fifty because the conversion is curved.
For a history with separately bounded plateau levels and known durations, convert each allowed lower and upper level and apply the same duration weights. This gives conservative bounds when the extremes can occur together. If the uncertainties share a common datum offset, their dependence should be retained.
A common offset is especially different from independent noise. If every level is shifted upward by 0.1 metres, every conversion changes in a related way. Averaging more readings does not remove an unknown fixed datum error. Likewise, applying the variance identity to a biased level record does not repair its mean.
Keep geometric uncertainty separate from level uncertainty. A perfectly measured h passed through an uncertain volume curve still gives uncertain storage. Reporting both sources helps another reader see whether a better gauge, a datum check or a better basin survey would address the main limitation.
Let every supported level in the quadratic model be shifted by the same amount δ while the conversion is incorrectly left tied to its original bottom reference. The calculated volume becomes 50(h + δ)². Expanding and averaging gives an error of 100δμ + 50δ² cubic metres, with the coefficient units understood and μ the original mean level.
For μ = one metre and δ = 0.1 metres, the error is 10.5 cubic metres. This is different from the variance correction: a constant shift does not change the variance, but it does change the squared mean. Keeping only a variance statistic would therefore miss this systematic reference problem.
If the datum is intentionally changed, the conversion must change with it. With new coordinate h′ = h + δ, the same physical volume is V = 50(h′ − δ)², not 50h′². That subtraction restores the old bottom reference inside the formula. The water has not moved because its coordinate label changed.
This is a useful transfer check for any measurement-to-quantity table: coordinates and calibration must travel together. A new zero printed on a gauge is not a newly surveyed basin, and a consistent numerical shift is not necessarily a physical change.
CHAPTER 18 OF 24 · Check the evidence
18. Ask whether the basin is hydraulically connected
Our single-level model assumes that the stored water represented by the basin shares the stated water-surface level in the relevant equilibrium approximation. A disconnected pool behind a barrier may not follow the same gauge. Adding its mapped area to the conversion without checking connection can invent water that the gauge does not describe.
This is a modelling boundary, rather than a universal claim about how every wetland behaves. A wetland may include multiple compartments, channels, depressions and exchanges. Decide which storage the gauge is intended to represent and whether one relation is adequate.
In a two-compartment desk model with independent levels h₁ and h₂, total volume is V₁(h₁) + V₂(h₂). A mean of the two levels is not generally an adequate replacement. Even if the compartments have the same shape, unequal levels contribute their own squared terms in a quadratic example.
The practical check is to draw the compartments and connections before adding areas or volumes. Identify which level observations belong to each component. A good diagram can reveal missing evidence before a long calculation gives a false impression that the whole wetland has been quantified.
CHAPTER 19 OF 24 · Check the evidence
19. Keep surface-water storage separate from ecological condition
A larger mean surface-water volume is not automatically a healthier wetland. The appropriate water regime depends on the organisms, setting and management question. Our examples compare a physical quantity, without assigning one history a higher ecological score.
The histories also differ in dry periods, high-water exposure and variation. Those differences may matter to a separate ecological question, but the volume formula cannot determine their biological effect. Bring species and habitat evidence to that interpretation rather than asking a storage average to carry it.
Likewise, a larger volume does not automatically mean greater available flood storage. Available storage depends on the relevant upper limit and the initial state, while downstream effects depend on connections and flows. A basin that is already high may hold more water while having less remaining room below a specified limit.
An informative report keeps the columns separate: level history, converted surface-water volume, hydroperiod measures, ecological observations and the question each supports. Connecting them is useful. Collapsing them into a single unexplained “wetland performance” number loses the distinctions that make the evidence interpretable.
Use one row per supported interval, with start time, end time, duration, level, conversion version, computed volume and duration-times-volume. Add durations and weighted volumes separately. Divide only at the end to obtain the mean.
Include the duration-weighted level mean as a diagnostic column, but label V(mean h) separately from mean V. If they happen to agree, test whether that follows from a constant level or linear geometry. An accidental match on one dataset does not establish a general interchange rule.
Retain raw gauge data and flags outside the derived table. Missing intervals should remain visible, along with the rule used to handle them. A mean over the observed intervals is a statement about those intervals; extending it to the whole season needs a justified treatment of the gaps.
For an independent check, calculate one interval by hand, verify units and test zero and the upper model level. Then use the variance identity for the quadratic example. Two independently organised calculations are more likely to reveal a weighting or unit error than repeating the same spreadsheet formula in another column.
Use the same quadratic basin. Level is 0.4 metres for two hours and 1.2 metres for six hours. Find the mean level, mean volume and volume at mean level. Decide which quantity answers “the average amount of surface water present during the eight hours.”
The mean level is (2 × 0.4 + 6 × 1.2)/8 = one metre. The two volumes are eight and seventy-two cubic metres. Their duration-weighted mean is (2 × 8 + 6 × 72)/8 = fifty-six cubic metres. The volume at the mean level remains fifty.
The level variance is [2 × (0.4 − 1)² + 6 × (1.2 − 1)²]/8 = 0.12 square metres. Multiplying by fifty gives the six-cubic-metre discrepancy. The average amount present is fifty-six cubic metres under the supplied history.
Now change only the durations so that the two levels each last four hours. The mean level becomes 0.8 metres and mean volume forty cubic metres, while V(0.8) is thirty-two. Explain why the old answer cannot survive a change of weights, even though the two listed level values have not changed.
CHAPTER 22 OF 24 · Practise and continue
22. Repair, stabilise and extend the method
If your mean level is wrong, repair the duration weights first. If the level mean is correct but mean storage is wrong, inspect the conversion order. If both arithmetic steps are correct but the answer is still unsupported, inspect the datum, geometry, gaps and connection assumptions.
To stabilise understanding, alternate constant-area, widening and shelf basins. Ask which histories cross a bend in the conversion. Predict whether averaging first should work before computing. A learner who chooses the method across changed geometry has transferred the reasoning beyond the first formula.
For extension, start from only a mean and hard range and derive bounds rather than inventing a full history. Then give two histories with the same mean and different variance. Explain which additional summary would recover mean volume in the quadratic model and why that may fail for another curve.
A concise mastery check combines a worked calculation with an evidence statement: “I converted supported states before averaging, and my result depends on this fixed geometry and these durations.” That sentence makes the method portable while keeping the supported claim appropriately narrow.
Can the mean level ever be sufficient?
Yes. A constant level has no variation to lose, and an affine level–volume relation commutes with averaging. The quadratic model also permits recovery from mean and variance together, using the same time weights.
Does a higher mean level guarantee higher mean storage?
Not when variation differs and only the means are compared. Under the quadratic relation, both the squared mean and the variance contribute. Compare complete supported histories, or derive bounds from the information available.
Can I square the level after changing to centimetres?
Only if the coefficient is converted consistently. The expression fifty times h squared assumes h in metres and gives cubic metres. Inserting centimetres without changing the coefficient multiplies the squared input by ten thousand.
Is a gauge record a complete wetland water balance?
No. A level record needs geometry to describe storage and additional evidence to describe relevant flows. Soil water, disconnected pools and changes in geometry may require separate representations rather than an enlarged interpretation of one gauge.
CHAPTER 24 OF 24 · Practise and continue
24. Continue with the right neighbouring question
For water regimes and ecological setting, return to Wetlands. For timing, use hydroperiod comparisons. For changing ground references, use surface accretion and land-elevation change. Each route answers a related question with a different required record.
For a real example of geometry-specific conversion, the USGS report Elevation-area-capacity relationships of Lake Powell in 2018, published 21 March 2022, describes survey-based elevation, area and capacity relationships. It is reservoir evidence, not a calibration for our invented wetland. The numerical histories, quadratic relation and shelf comparison above are original teaching constructions.
Use the Gold Standard of Education for the wider principle of independent transfer, and the How X Works hub for other subjects where converting a summary in the wrong order changes its meaning. The practical habit is simple to state and worth practising: preserve the history and its weights until the required quantity has been calculated.