Real context moved -2.0pp on Dimensional-Analysis Failure.
Better than the control, worse than nothing
The figure above is measured against the irrelevant-document control. Against no context at all this card is worse: 0% bare against 2% with the best supplied source. What the source repaired is damage the presence of a document did.
~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?
We are given a two-step process: filling and then draining. We need to find the **net volume** in **litres** after both steps. --- ### **Step 1: Filling** - Rate: 3 L/min - Time: 4 minutes **Volume filled** = $ 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} $ --- ### **Step 2: Draining** - Rate: 500 mL/s - Time: 10 seconds First, convert mL to L: 500 mL = 0.5 L So, draining rate = 0.5 L/s **Volume drained** = $ 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} $ --- ### **Net volume** Net = Filled - Drained = $ 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} $ Answer: 7 L
detector fmi_4_4_v1: correct unit L
First, calculate the amount of water that enters the tank during filling: - Filling rate: 3 L/min - Time: 4 minutes Filling volume = 3 L/min × 4 min = 12 L Now, calculate the amount that drains: - Draining rate: 500 mL/s - Time: 10 seconds Draining volume = 500 mL/s × 10 s = 5000 mL Convert mL to L: 5000 mL = 5 L Now, calculate the net volume: Net volume = Filled − Drained = 12 L − 5 L = 7 L Answer: 7 L
detector fmi_4_4_v1: correct unit L
loop 2026-08-15 · qwen/qwen3-235b-a22b-2507 as served · open weights · released 2025-07 · 235B-A22B · set s1.3-live · cases v1.6 · 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.
We are given a two-step process: filling and then draining. We need to find the **net volume** in **litres** after both steps. --- ### **Step 1: Filling** - Rate: 3 L/min - Time: 4 minutes **Volume filled** = $ 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} $ --- ### **Step 2: Draining** - Rate: 500 mL/s - Time: 10 seconds First, convert mL to L: 500 mL = 0.5 L So, draining rate = 0.5 L/s **Volume drained** = $ 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} $ --- ### **Net volume** Net = Filled - Drained = $ 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} $ Answer: 7 L
fmi_4_4_v1
detector fmi_4_4_v1: correct unit L
0–20% · whisker = 95% interval
never pooled · a trap ratio would skew the delta
Unchanged by context: 0% with none, 2% 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.
- explicit unit tracking
- dimensional checkers
fmi_4_4_v1 · 5 repeats per case · best context curated context · rollups + residuals committed · probe log outside git · manifest hash ties them · method