CVP reproduce
Independent re-run pack: fixed input vectors, O3 expected values, oracle version, commands, and the last result summary. This is how a reader moves from “CalculatorX says it passed” to “I can reproduce why it passed.”
Calculation version, protocol version, and evidence revision are distinct.
O3 PASS is a tabulated-vector claim, not a whole-domain proof.
O2: delta vs a separate-module identity (≤2 ULP). Not midspan under the same triangle, and not a station curve. ≤2 ULP vs O3 applies only to the published tabulated triangular-load-maximum vectors.
Regenerate the table from the public generator. Independent REST check needs only this page's reproduce.json plus stdlib Python — no repo clone and no mpmath. The CVP runner remains a maintainer command.
Re-running the generator without changing seed or inputs should reproduce table SHA-256 d45d14c2d2706f33abe423b77f5636feb13e52cb3f371ec0a9e930c9770f941a. Compare each expected_f64 to the production result within the declared ULP threshold.
Every O3 vector used for the ≤2 ULP claim. Expected values come from the published mpmath table, not from the implementation under test.
| ID | Kind | Inputs | Expected (f64) | Actual | ULP | Status |
|---|---|---|---|---|---|---|
o3-check | check | {"w0":"24","L":"2","E":"1","I":"1"} | {"oracle_layer":"analytic","delta":2.504518745057035,"x":1.0386592447184564,"w0":24,"L":2,"E":1,"I":1} | {"model":"triangular_load_max_deflection","w0":24,"L":2,"E":1,"I":1,"x":1.0386592447184564,"delta":2.5045187450570356,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 1 | PASS |
o3-SI | SI | {"w0":"1000","L":"2","E":"200000000000","I":"0.00001"} | {"oracle_layer":"analytic","delta":0.0000521774738553549,"x":1.0386592447184564,"w0":1000,"L":2,"E":200000000000,"I":0.00001} | {"model":"triangular_load_max_deflection","w0":1000,"L":2,"E":200000000000,"I":0.00001,"x":1.0386592447184564,"delta":0.000052177473855354896,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 1 | PASS |
o3-alias | alias | {"w":"48","span":"2","modulus":"1","Ix":"1"} | {"oracle_layer":"analytic","delta":5.00903749011407,"x":1.0386592447184564,"w0":48,"L":2,"E":1,"I":1} | {"model":"triangular_load_max_deflection","w0":48,"L":2,"E":1,"I":1,"x":1.0386592447184564,"delta":5.009037490114071,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 1 | PASS |
o3-awkward | awkward | {"w0":"13.7","L":"2","E":"210000","I":"0.83"} | {"oracle_layer":"analytic","delta":0.000008202310864238235,"x":1.0386592447184564,"w0":13.7,"L":2,"E":210000,"I":0.83} | {"model":"triangular_load_max_deflection","w0":13.7,"L":2,"E":210000,"I":0.83,"x":1.0386592447184564,"delta":0.000008202310864238235,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 0 | PASS |
o3-neg | signed | {"w0":"-24","L":"2","E":"1","I":"1"} | {"oracle_layer":"analytic","delta":-2.504518745057035,"x":1.0386592447184564,"w0":-24,"L":2,"E":1,"I":1} | {"model":"triangular_load_max_deflection","w0":-24,"L":2,"E":1,"I":1,"x":1.0386592447184564,"delta":-2.5045187450570356,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 1 | PASS |
o3-small | small | {"w0":"10","L":"2","E":"1000","I":"0.01"} | {"oracle_layer":"analytic","delta":0.1043549477107098,"x":1.0386592447184564,"w0":10,"L":2,"E":1000,"I":0.01} | {"model":"triangular_load_max_deflection","w0":10,"L":2,"E":1000,"I":0.01,"x":1.0386592447184564,"delta":0.1043549477107098,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 0 | PASS |
o3-stiff | stiff | {"w0":"500","L":"2","E":"210e9","I":"2e-4"} | {"oracle_layer":"analytic","delta":0.0000012423208060798786,"x":1.0386592447184564,"w0":500,"L":2,"E":210000000000,"I":0.0002} | {"model":"triangular_load_max_deflection","w0":500,"L":2,"E":210000000000,"I":0.0002,"x":1.0386592447184564,"delta":0.0000012423208060798786,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 0 | PASS |
o3-long | long | {"w0":"200","L":"4","E":"70e9","I":"5e-5"} | {"oracle_layer":"analytic","delta":0.00009541023790693468,"x":2.0773184894369128,"w0":200,"L":4,"E":70000000000,"I":0.00005} | {"model":"triangular_load_max_deflection","w0":200,"L":4,"E":70000000000,"I":0.00005,"x":2.0773184894369128,"delta":0.00009541023790693468,"solver":"simply supported triangular load maximum","convention":"Simply supported span. Intensity is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the unloaded end. I is an input. The midspan value is its own page."} | 0 | PASS |
