Challenge an exact fit with a withheld trajectory point
Interpolation can hide a wrong functional model.
Learning outcome: Compare zero and nonzero jerk with the same offset, then change only the offset. Freeze the first three fitting points before revealing the fourth and retain separate deterministic observation-bound versus omitted-model limits.
Starting recap
Read derivatives as local rates and integrals as signed accumulation. Review power and chain rules, vector components, units and initial conditions where used.
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Try
At j=.6 and offset−.1, a_fit=3.6 m/s², v_fit=1.8 m/s and intercept .9 m. The withheld residual is .6 m. Correcting the common offset changes only the intercept.
Open the experimentPredict
A zero residual on three points used to fit three quadratic coefficients establishes:
Experiment
Compare zero and nonzero jerk with the same offset, then change only the offset. Freeze the first three fitting points before revealing the fourth and retain separate deterministic observation-bound versus omitted-model limits.
Test a fitted acceleration against a withheld point
Worked teaching model · synthetic values · Synthetic x=1+2t+1.5t²+j t³/6 plus common readout offset. Fit quadratic exactly to t0,1,2 s; t3 is withheld. Per-point independent error bounds .01 m are considered separately from the disclosed systematic cubic deviation; no probability model or unique force inference.
Reset keeps your prediction, captures and practice history.
Fitting points and a genuinely withheld time
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: supplied positions including withheld t3 (solid).2: quadratic fitted only to t0,1,2 (dashed). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Fitting points and a genuinely withheld time
supplied positions including withheld t3
| time (s) | position (m) |
|---|---|
| 0 | 1.1 |
| 1 | 4.666667 |
| 2 | 11.63333 |
| 3 | 22.4 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
quadratic fitted only to t0,1,2
| time (s) | position (m) |
|---|---|
| 0 | 1.1 |
| 0.05 | 1.197583 |
| 0.1 | 1.303667 |
| 0.15 | 1.41825 |
| 0.2 | 1.541333 |
| 0.25 | 1.672917 |
| 0.3 | 1.813 |
| 0.35 | 1.961583 |
| 0.4 | 2.118667 |
| 0.45 | 2.28425 |
| 0.5 | 2.458333 |
| 0.55 | 2.640917 |
| 0.6 | 2.832 |
| 0.65 | 3.031583 |
| 0.7 | 3.239667 |
| 0.75 | 3.45625 |
| 0.8 | 3.681333 |
| 0.85 | 3.914917 |
| 0.9 | 4.157 |
| 0.95 | 4.407583 |
| 1 | 4.666667 |
| 1.05 | 4.93425 |
| 1.1 | 5.210333 |
| 1.15 | 5.494917 |
| 1.2 | 5.788 |
| 1.25 | 6.089583 |
| 1.3 | 6.399667 |
| 1.35 | 6.71825 |
| 1.4 | 7.045333 |
| 1.45 | 7.380917 |
| 1.5 | 7.725 |
| 1.55 | 8.077583 |
| 1.6 | 8.438667 |
| 1.65 | 8.80825 |
| 1.7 | 9.186333 |
| 1.75 | 9.572917 |
| 1.8 | 9.968 |
| 1.85 | 10.37158 |
| 1.9 | 10.78367 |
| 1.95 | 11.20425 |
| 2 | 11.63333 |
| 2.05 | 12.07092 |
| 2.1 | 12.517 |
| 2.15 | 12.97158 |
| 2.2 | 13.43467 |
| 2.25 | 13.90625 |
| 2.3 | 14.38633 |
| 2.35 | 14.87492 |
| 2.4 | 15.372 |
| 2.45 | 15.87758 |
| 2.5 | 16.39167 |
| 2.55 | 16.91425 |
| 2.6 | 17.44533 |
| 2.65 | 17.98492 |
| 2.7 | 18.533 |
| 2.75 | 19.08958 |
| 2.8 | 19.65467 |
| 2.85 | 20.22825 |
| 2.9 | 20.81033 |
| 2.95 | 21.40092 |
| 3 | 22 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Fitted initial position including offset | 1.1 | m |
| Fitted initial velocity | 1.866667 | m/s |
| Fitted constant acceleration | 3.4 | m/s² |
| Withheld position at3 s | 22.4 | m |
| Predicted position at3 s | 22 | m |
| Withheld observed minus predicted | 0.4 | m |
| Independent .01 m error bound on fitted acceleration | 0.04 | m/s² |
An exact fit to three points cannot validate the assumed constant-acceleration model.
The second difference x2−2x1+x0 gives3+j, and solving the first increment gives initial velocity2−j/3. The withheld residual at3 s is j metres with these numerical time units. A common position offset shifts the intercept but cancels from derivatives/residuals. Independent bounded errors ±.01 m give acceleration bound .01+2(.01)+.01=.04 m/s²; that does not cover an omitted cubic term or prove a confidence level.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
An exact fit to three points cannot validate the assumed constant-acceleration model.
The second difference x2−2x1+x0 gives3+j, and solving the first increment gives initial velocity2−j/3. The withheld residual at3 s is j metres with these numerical time units. A common position offset shifts the intercept but cancels from derivatives/residuals. Independent bounded errors ±.01 m give acceleration bound .01+2(.01)+.01=.04 m/s²; that does not cover an omitted cubic term or prove a confidence level.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
Understand
Apply
Public practice checks a specific claim. It does not establish independent mastery.
Why is a withheld point no longer independent if you use it to retune the model?
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