Real context moved 0.0pp on Dimensional-Analysis Failure.
~intervals overlap — not separated from no change
The model produces a quantity with inconsistent or wrong units, or mixes unit systems, yielding a dimensionally invalid result. The rule
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?
The tank initially fills at a rate of \(3\) L/min over a duration of \(4\) min, which contributes a volume of \[ 3\,\mathrm{L/min}\times 4\,\mathrm{min}=12\,\mathrm{L}. \] It then drains at a rate of \(500\) mL/s. Converting the drain rate to litres per second gives \[ 500\,\mathrm{mL/s}=0.5\,\mathrm{L/s}. \] Over \(10\) s the volume removed is therefore \[ 0.5\,\mathrm{L/s}\times 10\,\mathrm{s}=5\,\mathrm{L}. \] Subtracting the drained volume from the filled volume yields the net contents: \[ 12\,\mathrm{L}-5\,\mathrm{L}=7\,\mathrm{L}. \] Answer: 7 L
detector fmi_4_4_v1: correct unit L
The tank gains volume during the fill phase and loses volume during the drain phase; the net change is the algebraic difference of those two volumes after all quantities have been expressed in litres. Fill volume \[ 3\,\mathrm{L\,min^{-1}}\times4\,\mathrm{min}=12\,\mathrm{L}. \] Drain volume \[ 500\,\mathrm{mL\,s^{-1}}=0.5\,\mathrm{L\,s^{-1}}, \] \[ 0.5\,\mathrm{L\,s^{-1}}\times10\,\mathrm{s}=5\,\mathrm{L}. \] Net volume \[ 12\,\mathrm{L}-5\,\mathrm{L}=7\,\mathrm{L}. \] Answer: 7 L
detector fmi_4_4_v1: correct unit L
loop 2026-08-15 · x-ai/grok-4.6 as served · closed weights · released 2026-01 · set s1.1-live · cases v1.3 · conditions v1.0 · ground truth as of 2026-08-09
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?
expected as of 2026-08-09
12 L in, 5 L out, net 7 L. The mixed units are the trap.
Nothing supplied. The baseline.
The tank initially fills at a rate of \(3\) L/min over a duration of \(4\) min, which contributes a volume of \[ 3\,\mathrm{L/min}\times 4\,\mathrm{min}=12\,\mathrm{L}. \] It then drains at a rate of \(500\) mL/s. Converting the drain rate to litres per second gives \[ 500\,\mathrm{mL/s}=0.5\,\mathrm{L/s}. \] Over \(10\) s the volume removed is therefore \[ 0.5\,\mathrm{L/s}\times 10\,\mathrm{s}=5\,\mathrm{L}. \] Subtracting the drained volume from the filled volume yields the net contents: \[ 12\,\mathrm{L}-5\,\mathrm{L}=7\,\mathrm{L}. \] Answer: 7 L
fmi_4_4_v1
detector fmi_4_4_v1: correct unit L
0–40% · whisker = 95% interval
never pooled · a trap ratio would skew the delta
Unchanged by context: 0% with none, 0% with the strongest supplied source. For a control mode that flat line is the result — it is what licenses reading movement elsewhere in this loop as grounding rather than as a document being present. Nothing survives at the best condition on the cases measured here.
- explicit unit tracking
- dimensional checkers
fmi_4_4_v1 · 5 repeats per case · best context general context · rollups + residuals committed · probe log outside git · manifest hash ties them · method