A radar feature can appear to move even when none of its stipulated components changes location. A centroid is an average position. Change the contributions to that average and its position can change. That is the short answer; deciding whether it explains an observed track requires a precise definition of the feature, its weights and the observation window.
This guide develops a small mathematical model to separate those possibilities. Every coordinate, weight and time interval below is invented for teaching. The components are labelled pieces of a paper feature, not identified air parcels. Their weights are not asserted to be conserved water mass, and the calculation is not a description of an operational storm-tracking algorithm.
Begin with the Weather Radar Convective Cell Tracking subject guide if you need the wider distinction between recognising an echo, associating features across scans and interpreting a track. Here we isolate one question: what information does a change in an average position actually preserve?
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Choose your closest question, or follow the worked argument in order. The chapter index groups the reading while the explanations remain open.
ROUTE 1 · CHAPTERS 1–6
Define the displayed centre
Work fixed locations, alternative histories, rates and coordinate checks.
ROUTE 2 · CHAPTERS 7–12
Check weighting and membership
Separate geometric centres, nonlinear weights, thresholds, clipping and background.
ROUTE 3 · CHAPTERS 13–18
Derive the rate identity
Extend to two dimensions and separate component motion from weight changes.
ROUTE 4 · CHAPTERS 19–25
Audit information and processing
Test hidden movement, finite attribution, uncertainty, grouping, shape and smoothing.
ROUTE 5 · CHAPTERS 26–31
Choose evidence and transfer
Match extra observations to alternatives, practise and follow verified sources.
Full chapter contents · Weather Radar Convective Cell Tracking subject guide
Chapter contents
Define the displayed centre · Chapters 1–6
1. Start with two fixed locations
2. Define the average before interpreting the track
3. Reproduce the first calculation
4. Build a second history with the same displayed positions
Check weighting and membership · Chapters 7–12
7. Recognise that a geometric centre is a different definition
8. Check what a nonlinear weight transformation changes
9. See how a threshold can move a centre
10. Preserve the rule at the threshold boundary
Derive the rate identity · Chapters 13–18
13. Extend the calculation to two dimensions
14. Use nonnegative weights to check the result
15. Derive a rate equation for a fixed set of components
16. Follow the quotient rule one line at a time
Audit information and processing · Chapters 19–25
19. Distinguish instantaneous reasoning from two-frame attribution
20. Bound the centroid when the weights are uncertain
21. Combine groups using their total weights
22. Test whether two different shapes share one centre
23. Keep a smoothing decision visible
Choose evidence and transfer · Chapters 26–31
26. Build a practical comparison record
27. Repair an explanation that overstates the result
28. Try a changed-context problem independently
29. Check the transfer and choose a next step
Put two labelled components on a horizontal line, at 0 kilometres and 10 kilometres. Give them equal weights of 2 units. Their weighted average position is (2 × 0 + 2 × 10)/(2 + 2), which is 5 kilometres. The midpoint is expected because equal contributions balance at the middle.
Keep both locations fixed. Now reduce the left weight to 1 unit and increase the right weight to 3 units. The new average is (1 × 0 + 3 × 10)/(1 + 3), or 7.5 kilometres. The average has moved 2.5 kilometres to the right, while each labelled location has moved zero kilometres.
Total weight is 4 units in both frames. Consequently, even the extra statement that total feature weight stayed constant does not repair the inference. The missing information concerns its distribution. This is a counterexample to treating centroid displacement as sufficient proof that all parts of a feature translated together.
CHAPTER 2 OF 31 · Define the displayed centre
2. Define the average before interpreting the track
Write X = Σ(wᵢxᵢ)/W, where W = Σwᵢ, each wᵢ is nonnegative and W is positive. The symbol xᵢ denotes the coordinate of component i. The summation includes exactly the components admitted by the chosen feature definition and observation window.
A weight of zero contributes nothing. If every weight is zero, W is zero and this centroid is undefined. A software display might use a placeholder, retain an earlier position or suppress a feature; none of those choices follows from the formula. In particular, undefined does not mean that the feature is located at coordinate zero.
Units deserve attention. If weights have arbitrary common units, numerator units are weight times kilometres and division returns kilometres. Multiplying all weights by the same positive constant changes numerator and denominator equally, leaving X unchanged. Relative contributions determine this position; their overall scale does not.
Use the following two frames as a complete paper record. Positions remain at 0 and 10 kilometres throughout. There is no measurement noise, hidden component or changing threshold in this first model.
| Frame | Left weight | Right weight | Total weight | Centroid in kilometres |
|---|---|---|---|---|
| Initial | 2 | 2 | 4 | 5 |
| Later | 1 | 3 | 4 | 7.5 |
The numerator rises from 20 to 30 weight-kilometres. The denominator stays at 4. Dividing gives the shift directly. No velocity observation has been supplied, and the change in weights has not been assigned a physical cause.
A useful first check is to ask which line of the computation would record translation. Translation changes coordinates xᵢ. Here only wᵢ changes. The same final average can therefore emerge from different kinds of input change. Keeping a table of the inputs makes that distinction visible before the answer becomes a line on a map.
CHAPTER 4 OF 31 · Define the displayed centre
4. Build a second history with the same displayed positions
Return to equal weights of 2 and initial positions 0 and 10 kilometres. This time keep the weights fixed and move both components 2.5 kilometres right. Their later positions are 2.5 and 12.5 kilometres. The later centroid is (2 × 2.5 + 2 × 12.5)/4, again 7.5 kilometres.
Both histories begin with centroid 5, end with centroid 7.5 and have total weight 4 in each frame. In one history neither component translates. In the other, both translate together. The pair of centroid values and total weights cannot distinguish these histories.
Their full spatial arrangements differ. Observing the outer locations would distinguish these two particular constructions. That does not establish that a particular extra statistic resolves every possible real feature. A counterexample identifies an insufficiency; selecting additional evidence requires checking the alternatives relevant to the actual measurement.
CHAPTER 5 OF 31 · Define the displayed centre
5. Turn displacement into a rate carefully
Suppose the invented frames are five minutes apart. Centroid displacement divided by elapsed time is 2.5/5 = 0.5 kilometres per minute, or 30 kilometres per hour. That calculation describes the rate of change of this defined average over the interval.
The fixed-location history still gives that rate. Calling it a component’s travel speed would add information that the record does not contain. Calling it an air-parcel speed, a surface-wind observation or a future hazard speed would require further physical and observational justification.
An interval-average displacement rate also does not specify the path within the interval. The centroid could change steadily, change late or temporarily move back before reaching its endpoint. Adding a time unit makes the arithmetic dimensional; it does not recover the missing trajectory or explain the process that produced it.
CHAPTER 6 OF 31 · Define the displayed centre
6. Separate translation from a change of origin
If every coordinate increases by the same amount a while weights stay fixed, the new centroid is Σwᵢ(xᵢ + a)/W = X + a. This translation property is an important consistency check. A rigid translation of the entire stipulated feature produces the same translation of its centroid.
Changing the coordinate origin can produce the same numerical transformation without physical motion. If the later frame uses coordinates measured from a different origin, subtracting its displayed position from an earlier coordinate mixes two reference systems. The frames must first be expressed in a common system.
Similarly, converting kilometres to metres multiplies every coordinate and the centroid by 1,000. A large number after unit conversion is not a large displacement. Record the coordinate system, units and time reference with the centroid, especially when comparing tracks assembled from different data products.
CHAPTER 7 OF 31 · Check weighting and membership
7. Recognise that a geometric centre is a different definition
A geometric centroid of admitted equal-area pixels gives each included pixel equal weight. An intensity-weighted centroid gives pixels different contributions according to an explicitly chosen quantity. These definitions answer different averaging questions even when computed from the same image.
For two admitted equal-area pixels at 0 and 10 kilometres, the geometric centroid is 5 kilometres. If assigned positive weights are 1 and 3, the weighted centroid is 7.5 kilometres. Neither number is automatically an error. The disagreement can arise entirely from which definition was chosen.
The comparison becomes misleading when a report simply says “centre” without specifying the weighting. Before analysing changes, write down whether the weights describe equal pixels, a transformed signal quantity or some other feature measure. An interpretation attached to one definition cannot be transferred to another merely because both results are positions.
CHAPTER 8 OF 31 · Check weighting and membership
8. Check what a nonlinear weight transformation changes
Consider two positions at 0 and 10 with positive weights 1 and 9. Their centroid is 9 kilometres. If the chosen weighting rule instead uses the square roots of those values, the contributions become 1 and 3 and the centroid becomes 7.5 kilometres.
The spatial inputs have not changed. A nonlinear transformation of the relative contributions has changed the average. Multiplication by a common constant would leave it unchanged, but square roots alter their ratio. This is why the exact measurement scale and transformation belong in the centroid definition.
Do not infer a particular radar processing convention from this example. It uses arbitrary positive teaching weights. Some signal encodings can include zero or negative numbers, which would violate this guide’s nonnegative-weight assumptions if inserted directly. An explicit, defensible transformation is required before applying a formula whose interpretation depends on those assumptions.
CHAPTER 9 OF 31 · Check weighting and membership
9. See how a threshold can move a centre
Keep two equal-area pixels fixed at 0 and 10 kilometres. Let their reported values initially be 2 and 1 units. Define a feature by retaining pixels whose values are at least 1.5. Initially only the left pixel is retained, so the geometric centroid of the retained set is 0.
In the later frame, let the reported values be 2 and 2. Both pixels now pass the unchanged threshold. Their geometric centroid is 5 kilometres. No pixel location has moved. Membership in the feature has changed because the right pixel crossed the inclusion rule.
For a related threshold-defined observable, the Weather Radar Echo Top Height Estimation guide explains why a reported top also depends on a declared inclusion rule.
This mechanism differs from reweighting a fixed set, although both change an average. A useful audit separates the admitted set from the weights applied within it. Otherwise a jump caused by a membership boundary may be discussed as if the same collection of components simply travelled to a new location.
CHAPTER 10 OF 31 · Check weighting and membership
10. Preserve the rule at the threshold boundary
“At least 1.5” and “greater than 1.5” admit different pixels whose value is exactly 1.5. That difference may be irrelevant for one dataset and decisive for another. State the rule explicitly rather than assuming every implementation interprets a threshold identically.
A small perturbation around the boundary can cause a discrete membership change. With one retained pixel and one pixel just below the threshold, a small value adjustment may double the admitted pixel count in this two-pixel model. The centroid change need not be small relative to the spatial separation.
This does not imply that any particular threshold is poor. It explains why robustness to threshold choice is an empirical question. Recomputing with nearby, justified thresholds can expose sensitivity, but such a comparison is not a confidence interval and does not by itself identify the physically correct feature boundary.
CHAPTER 11 OF 31 · Check weighting and membership
11. Treat the observation window as part of the definition
Suppose equal-weight components lie at 0 and 10 kilometres. A window containing both gives centroid 5. A later window that contains only the right component gives centroid 10. The components can remain stationary throughout; the observation window has changed what enters the average.
Clipping is particularly easy to overlook when a visible feature approaches an image edge or when a region is cropped for a display. A centroid of the visible portion is a well-defined statistic of that portion, but it is not automatically the centroid of the complete feature outside the window.
A practical comparison should record whether feature membership changed through observed signal evolution, a changed inclusion rule or an observation boundary. These explanations can coexist. The numerical shift alone does not select among them, and retaining an earlier label does not restore the missing outside-window information.
CHAPTER 12 OF 31 · Check weighting and membership
12. Work the effect of an added background contribution
Suppose the feature has total weight W and centroid X. An added nonnegative contribution has total weight B and centroid Xᵦ. Combining them gives Xnew = (WX + BXᵦ)/(W + B), provided the combined total is positive.
For an invented feature with W = 4 and X = 7.5 kilometres, add B = 4 centred at 2.5 kilometres. The combined centroid is (4 × 7.5 + 4 × 2.5)/8 = 5 kilometres. The added contribution pulls the result toward its own centroid without relocating the original components.
This is an accounting identity, not a claim that a real background is known, spatially uniform or harmless. If a background correction is estimated inaccurately, the retained weights and their validity need review. Subtraction can produce negative values, which require a different interpretation or an explicitly stated handling rule outside the nonnegative model used here.
CHAPTER 13 OF 31 · Derive the rate identity
13. Extend the calculation to two dimensions
Let each location be a pair, (xᵢ, yᵢ). Compute the weighted average of each coordinate using the same weights and total. The result is the vector X = Σwᵢxᵢ/W, with the boldface-vector convention replaced here by words to keep the notation readable.
For equal-weight components at (0, 0) and (10, 4) kilometres, the centroid is (5, 2). Changing weights from 2 and 2 to 1 and 3 moves it to (7.5, 3), while both component locations stay fixed. The centroid displacement is (2.5, 1) kilometres.
Its displacement magnitude is the square root of 2.5² + 1², approximately 2.69 kilometres. That geometric length is still a displacement of the defined average. Computing both direction and distance does not convert the statistic into evidence that every component followed that vector.
CHAPTER 14 OF 31 · Derive the rate identity
14. Use nonnegative weights to check the result
A centroid with nonnegative weights and positive total lies within the convex hull of the included locations. On a line, it lies between the smallest and largest included coordinates. For the two positions 0 and 10, a computed centroid of 12 cannot satisfy those assumptions.
Such an outcome prompts an input or definition check: perhaps negative weights were used, a coordinate was omitted from the stated range, units were mixed or a calculation was transcribed incorrectly. It should not immediately be explained as exceptional motion. The check concerns the arithmetic of a single frame.
In two dimensions the centroid lies within the convex hull, not necessarily on an occupied pixel or inside every irregular part of the feature. An average can fall in a gap between components. Therefore absence of a bright pixel exactly at the centroid does not alone invalidate the calculation.
CHAPTER 15 OF 31 · Derive the rate identity
15. Derive a rate equation for a fixed set of components
Now assume a fixed set of labelled components whose positions and weights vary differentiably with time. Let each component’s velocity be vᵢ = dxᵢ/dt. Differentiate X = Σwᵢxᵢ/W using the product and quotient rules.
The result is dX/dt = Σwᵢvᵢ/W + Σ(dwᵢ/dt)(xᵢ − X)/W. The first term is a weighted average of component velocities. The second records how changes in weights shift the average toward or away from locations relative to its current centroid.
The derivation assumes W remains positive and the labelled set stays meaningful over the interval considered. A threshold-induced appearance or disappearance may violate the differentiable fixed-membership setup. In that case compute finite-frame centroids directly or specify a different model; do not silently apply a smooth equation across a discrete change.
CHAPTER 16 OF 31 · Derive the rate identity
16. Follow the quotient rule one line at a time
Define N = Σwᵢxᵢ so X = N/W. Its derivative is N′/W − NW′/W². Differentiate N to obtain Σ(wᵢvᵢ + wᵢ′xᵢ), while W′ = Σwᵢ′. Substitute N/W = X into the second part.
The weight-change pieces become Σwᵢ′xᵢ/W − XΣwᵢ′/W. Combine their numerators to get Σwᵢ′(xᵢ − X)/W. This step explains why a component’s position relative to the current centre matters, rather than its absolute coordinate alone.
For a component on the right of the centre, increasing weight contributes positively to this term. Increasing weight on the left contributes negatively. A decrease reverses the sign. These are statements about the stipulated average, not independent explanations of why an echo’s measured contribution strengthened or weakened.
CHAPTER 17 OF 31 · Derive the rate identity
17. Identify conditions that remove the weight-change term
If every weight is constant, each wᵢ′ is zero and the centroid rate equals the weighted mean component velocity. This is a sufficient condition within the model. It is stronger than observing an unchanged total W, since weights can redistribute while their sum remains fixed.
Another sufficient condition is identical proportional growth: wᵢ′ = gwᵢ for one common g at the instant considered. Then the extra numerator is gΣwᵢ(xᵢ − X), which is zero by the definition of X. Relative contributions remain unchanged even as their overall scale grows or decays.
Neither condition should be assumed from a smooth-looking track. A track may be smooth while relative weights change smoothly. The rate identity identifies the additional conditions needed for a particular interpretation; it does not certify that those conditions hold in an observed feature.
CHAPTER 18 OF 31 · Derive the rate identity
18. Show that a stationary centroid can hide movement
At one instant, put components at 0 and 10 kilometres with weights 2 and 2. Let both stipulated velocities be 1 kilometre per minute to the right. Their weighted mean velocity is therefore 1 kilometre per minute.
At that same instant let the left weight increase at 0.4 weight units per minute and the right weight decrease at 0.4. With X = 5 and W = 4, the weight-change term is [0.4 × (0 − 5) − 0.4 × (10 − 5)]/4 = −1 kilometre per minute. The two terms cancel, giving zero instantaneous centroid velocity.
This is a local construction. The weights remain nonnegative only over a suitably limited interval if those rates continue. It demonstrates that zero centroid rate does not force zero component velocities. A stable displayed centre and a changing spatial field can be mathematically compatible.
CHAPTER 19 OF 31 · Audit information and processing
19. Distinguish instantaneous reasoning from two-frame attribution
The derivative identity separates contributions at an instant under a specified smooth model. Two observed frames usually do not provide every intermediate position, weight or rate needed to evaluate it continuously. Substituting endpoint differences into its terms without checking assumptions can introduce an additional approximation.
A finite-frame decomposition can be constructed by deciding which weights to hold fixed while changing positions and which positions to hold fixed while changing weights. Changing that order generally allocates interaction effects differently. The total centroid difference is fixed, but its partition into named contributions may depend on the chosen convention.
For the first fixed-location example no such ambiguity is needed: component positions are stipulated unchanged, so weight redistribution alone produces the shift. For a real changing feature, state the attribution method and the required observations before treating the separate contributions as uniquely measured quantities.
CHAPTER 20 OF 31 · Audit information and processing
20. Bound the centroid when the weights are uncertain
For fixed locations 0 and 10, the centroid is 10w₂/(w₁ + w₂). Suppose w₁ can range from 1 to 2 and w₂ from 2 to 3, and every pair in that rectangle is admissible. Increasing w₂ moves the average right; increasing w₁ moves it left.
The smallest centroid is therefore 10 × 2/(2 + 2) = 5 kilometres. The largest is 10 × 3/(1 + 3) = 7.5 kilometres. These are exact conditional bounds for the specified admissible set, not a statistical confidence interval or a radar accuracy specification.
If the weights are linked, the corner combinations may be impossible. For example, a known fixed total restricts the admissible pairs to a line rather than the full rectangle. Preserve those constraints before computing extremes. Independent-looking error bars do not justify independent variation unless that is part of the uncertainty model.
CHAPTER 21 OF 31 · Audit information and processing
21. Combine groups using their total weights
Suppose one group has centroid 2 kilometres and total weight 1, while another has centroid 8 and total weight 3. Their combined centroid is (1 × 2 + 3 × 8)/4 = 6.5 kilometres. Simply averaging the two group centres gives 5 and discards their unequal contributions.
This identity lets a larger collection be summarised consistently when each group’s centroid and total use the same definitions. It does not let incompatible feature measures be combined. A group total defined as pixel count and another defined as a signal sum cannot be treated as interchangeable weights without a justified conversion.
When a tracked feature is split or merged in an analysis, preserve the totals used to combine its pieces. A new centre after regrouping may reflect a changed aggregation rule. The displayed label alone does not establish continuity of the same underlying collection or a unique physical identity through that event.
CHAPTER 22 OF 31 · Audit information and processing
22. Test whether two different shapes share one centre
Place equal weights at 0 and 10 kilometres. Their centroid is 5. Now place the same equal weights at 4 and 6 kilometres. The centroid remains 5 even though the separation falls from 10 to 2 kilometres. A centre preserves one first-moment summary, not the full spatial arrangement.
A spread measure can distinguish these particular shapes. With normalised equal weights, the mean squared distance from the centre is 25 square kilometres in the first arrangement and 1 in the second. That additional statistic answers a different question about dispersion. It does not turn the centroid into a complete description.
Many more complicated arrangements can share both centre and spread. Therefore adding a second statistic should be justified by the alternatives under consideration rather than treated as a universal cure for information loss. If the question concerns deformation, retain evidence about the spatial field and shape changes instead of inferring them from a centre alone.
CHAPTER 23 OF 31 · Audit information and processing
23. Keep a smoothing decision visible
Suppose three successive observed centroid positions are 0, 6 and 6 kilometres at equal time intervals. A trailing two-position mean at the last two times gives positions 3 and 6. The displayed change is then 3 kilometres, although the corresponding last raw change was zero.
No arithmetic error is required. The displayed series answers a different averaging question and includes an earlier position in its later interpretation. The first smoothed value is influenced by the initial zero. A rate calculated from the smoothed positions therefore cannot be labelled as the raw last-interval displacement rate without qualification.
This example supplies no recommended smoothing window. It shows why a track’s processing history matters alongside the centroid definition within each frame. A temporal filter and a spatial weighting rule operate at different stages. Preserve both definitions when comparing a plotted track with the underlying observations, and evaluate any filter against the intended use rather than assuming a smoother line contains more direct evidence of motion.
CHAPTER 24 OF 31 · Audit information and processing
24. Ask what a feature association actually establishes
Associating a feature in one scan with a feature in another provides a proposed correspondence under an association rule. That rule may use positions and additional attributes. A centroid calculation is one possible summary within a much larger observation and interpretation problem.
Even a correct numerical association does not imply that every part of the later echo consists of the same material as the earlier echo. A labelled weather feature can evolve. Conversely, the paper counterexamples do not prove that an observed association is wrong; they show why a particular inference requires more than two average positions.
The NWS description of the Warning Decision Support System provides historical context for severe-storm analysis and identifies a storm-cell identification and tracking component. It does not establish that the illustrative weighting rule in this guide is that algorithm’s implementation, and this guide makes no such claim.
CHAPTER 25 OF 31 · Audit information and processing
25. Decide which extra evidence addresses the alternative
For the two constructed histories, observing both component locations distinguishes fixed locations with changed weights from common translation. For a threshold explanation, retaining values around the inclusion boundary and the exact threshold rule is more informative than saving only the centroid.
For a window-clipping explanation, preserve coverage and the feature boundary relative to the image edge. For a weighting explanation, preserve both the underlying field and its transformation. Each extra observation should be selected because it can discriminate a named alternative, rather than because a longer list of statistics appears reassuring.
The NOAA MRMS information page describes a broader multi-source radar and precipitation setting. It is useful context for understanding that products have defined inputs and processing histories. It supplies no accuracy guarantee for this paper centroid model and does not make its weights a physical mass measurement.
CHAPTER 26 OF 31 · Choose evidence and transfer
26. Build a practical comparison record
A useful record contains both timestamps, the coordinate reference, spatial units, the admitted pixels or components, their weights, the weighting transformation and the total W in each frame. Record thresholds and coverage before reducing the field to a single position.
Then compute the two centroids independently. Check their units, positive denominators and containment within the admitted locations’ convex hull. Only after those checks calculate the displacement and interval-average rate. Label that rate according to what was actually averaged.
Retain the numerical precision used to calculate the centre as well as the rounded display. Two displayed coordinates can be equal after rounding while their unrounded difference is nonzero. Conversely, a small displayed change may be comparable to the coordinate quantisation used by the product. Neither observation supplies an uncertainty distribution. It identifies another part of the measurement definition that must accompany an interpretation of small movements. For a paper check, compare rounding each position before subtraction with subtracting first and rounding the result; these operations need not agree.
Finally list plausible changes in membership, relative weights and positions. Some may be excluded by retained evidence; others may remain unresolved. Keeping unresolved alternatives explicit makes the result more useful. It tells a later reader what additional observation could change the interpretation instead of hiding missing information behind a precise decimal.
CHAPTER 27 OF 31 · Choose evidence and transfer
27. Repair an explanation that overstates the result
Suppose a learner writes, “The centre moved 2.5 kilometres, so every part moved 2.5 kilometres.” Repair the earliest unstable step: the jump from an average’s displacement to a shared displacement of its components. Recalculate the two stationary-component frames to show exactly where that conclusion fails.
Once that distinction is stable, extend the explanation: common translation with unchanged relative weights is compatible with the observed centre shift, but so is the specified redistribution of weights. Compatibility is weaker than identification. The sentence should reflect which conditions are observed and which are assumptions.
A second repair is needed for “The total did not change, so growth cannot matter.” Return to weights 2 and 2 becoming 1 and 3. The total remains 4 while the balance changes. Keeping the sum fixed constrains one aggregate; it does not freeze the distribution that determines the centroid.
CHAPTER 28 OF 31 · Choose evidence and transfer
28. Try a changed-context problem independently
Place two fixed components at 2 and 14 kilometres. Initially their weights are 3 and 1. Later the weights are 1 and 3. Compute both centroids, their displacement and the apparent interval-average rate if the frames are six minutes apart.
Before calculating, predict the direction. The later distribution gives more weight to the right-hand component, so the centre should move right. Both answers must remain between 2 and 14 kilometres. These checks can catch an inverted weighting or a denominator mistake before the final arithmetic.
Then construct a different history with unchanged initial weights that produces the same centroid displacement through common translation. Keep the time interval unchanged. Explain which retained observations would distinguish your two histories. Do not answer only with a speed; the purpose is to preserve the difference between a computed summary and a claim about its components.
CHAPTER 29 OF 31 · Choose evidence and transfer
29. Check the transfer and choose a next step
Initially X = (3 × 2 + 1 × 14)/4 = 5 kilometres. Later X = (1 × 2 + 3 × 14)/4 = 11 kilometres. The centroid displacement is 6 kilometres and its six-minute average rate is 1 kilometre per minute, or 60 kilometres per hour.
For the translation alternative, keep weights 3 and 1 and move positions 2 and 14 to 8 and 20 kilometres. The centre again moves from 5 to 11. Observing the component locations separates these specific histories; observing just the two centroids and totals does not.
If the arithmetic was difficult, return to the numerator and denominator in the first example. If the arithmetic was secure but the interpretation slipped, practise writing two compatible histories. If both are secure, extend to two dimensions or impose linked weight constraints and derive the resulting admissible centroid range.
Does this mean centroid tracks are useless? No. A clearly defined summary can be useful while preserving less information than the full field. Its usefulness depends on the question and the evidence supporting the interpretation attached to it.
Does a moving centroid prove a feature is strengthening on one side? No. That is one compatible mechanism in a specified model. Translation, membership changes, clipping and combinations of changes can also affect the average. The observed shift alone does not establish the cause.
Can an unchanged centroid prove nothing changed? No. Balanced changes can leave the average fixed, and the local rate example shows cancellation even while stipulated components move. Can a source describing an operational system validate these invented numbers? No. The numbers test the mathematical distinction under declared assumptions; they are not performance measurements or forecasts.
CHAPTER 31 OF 31 · Choose evidence and transfer
31. Follow the reading routes and evidence limits
For feature definitions, association and wider interpretation, return to the Weather Radar Convective Cell Tracking guide. For a related measurement question, read the Weather Radar Rainfall Estimation guide to separate a reported radar quantity from the precipitation estimate built from it.
For institutional context, the NWS WDSS page describes a historical severe-storm analysis system, while NOAA's MRMS information page introduces a wider radar-product setting. Both pages were checked on 24 October 2026. They support context, not a claim that the teaching centroid reproduces their operational algorithms. No current warning, instrument calibration or measured tracking accuracy is supplied here.
To continue across subjects, use How X Works and The Gold Standard of Education: Learning Transfers into Life. The transferable skill is to identify what an average preserves, construct a compatible alternative and ask which additional observation can actually distinguish it.
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