1Demonstration 1 of 4
Inside the data, a lookup and a fit agree
Where the evidence exists, does it matter whether the bend is held as a table or a formula?
The lookup interpolates between the two neighbouring observations. A degree 5 polynomial passes through all six points, so between them it stays near the same curve. A low degree cannot pass through every point and drifts a little further from the table.
Load is the input, from 0 to 5. Effectiveness is a fraction. The six observed points are (0, 0), (1, 0.35), (2, 0.6), (3, 0.78), (4, 0.88) and (5, 0.93). The lookup draws straight lines between them; the fit is a least-squares polynomial of the chosen degree.
Predict first. At a load of 2.5, will the degree 5 fit and the lookup differ by more than 0.05?
Choose an example
Constructed example: the chapter's six saturation points, read with the pack's Lookup and fit_polynomial.
Calculated values
- Lookup reading
- 0.690
- Fit reading
- 0.699
- Fit minus lookup
- 0.009
At a load of 2.5 the lookup reads 0.60 + (0.78 - 0.60) x (2.5 - 2.0)/(3.0 - 2.0) = 0.690. The degree 5 fit reads 0.699. Fit minus lookup = 0.699 - 0.690 = 0.009. Inside the six observations the two are close, so on this evidence alone there is no reason to prefer one.
Use the idea
When two representations agree inside the evidence, choose between them on what happens outside it.
Where the conclusion applies
Six constructed observations treated as exact. Agreement here says nothing about loads above 5, and a fit of low degree can disagree with the table even inside the data.
Check your understanding: At a load of 1.5, what does the lookup read, using the points at loads 1 and 2?
Chapter 15 source: section "Two ways to hold a bend". Demonstration C15-D01.
2Demonstration 2 of 4
Outside the data, the fit answers and the lookup refuses
What do the two representations say about a load nobody observed?
Past the last observation the highest order term of the polynomial takes over, so the answer is set by the functional form and not by the data. The lookup has nothing to interpolate from, so it raises OutsideDomain.
Load asked is the input. The lookup's observed domain is 0 to 5. The effectiveness is a fraction, so any honest value lies between 0 and 1.
Predict first. At a load of 12 with the degree 5 fit, is the fit's answer inside 0 to 1?
Choose an example
Constructed example: the chapter's saturation points; the load 12 value 47.76 is the chapter's own printed result.
Calculated values
- Load asked
- 12.0
- Lookup
- refuses (OutsideDomain)
- Fit
- 47.76
- Fit within 0 to 1
- no
The observed domain is 0 to 5 and 12.0 - 5.0 = 7.0 lies beyond its end, so the lookup refuses and names the missing evidence. The degree 5 fit answers anyway: 47.76 - 1.00 = 46.76 above the largest value a fraction can take. The fit carries no flag that it has left the data.
Use the idea
Treat the refusal as information: it turns an invisible extrapolation into a decision to get an observation, state an assumption, or restrict the operating range.
Where the conclusion applies
Effectiveness is a fraction bounded by 0 and 1. A lower degree can look plausible at some loads and still be unsupported there; the failure is that it is silent, not that it is always large.
Check your understanding: If a degree 5 fit returns 47.76 for a fraction, how far above its ceiling is that?
Chapter 15 source: section "Two ways to hold a bend". Demonstration C15-D02.
3Demonstration 3 of 4
Monotonic and bounded are claims, so test them
How does a lookup fail the two checks the chapter asks it to pass?
Interpolation does not guarantee either property. A noisy point can introduce a wiggle, and a value above 1.0 breaks the ceiling of a fraction. Writing both as tests makes a later edit that introduces them fail loudly.
The lookup holds six points. Monotonic means no step between neighbouring points is negative. Bounded means every value lies from 0.0 to 1.0. The two controls change the observation at load 2 and at load 5.
Predict first. If the value at load 5 falls to 0.85 while load 4 stays at 0.88, does the monotonic check pass?
Choose an example
Constructed example: the chapter's six points with two values changed for this reader to trigger each failure.
Calculated values
- Monotonic
- True
- Bounded between 0 and 1
- True
- Smallest step
- 0.05
- Largest value
- 0.93
Step changes between neighbouring points are 0.35, 0.25, 0.18, 0.10, 0.05. Monotonic = True. Largest value 0.93 against the ceiling 1.00: 0.93 - 1.00 = (-0.07), so bounded between 0 and 1 = True.
Use the idea
Keep these assertions beside the lookup so a later edit cannot add a wiggle unnoticed.
Where the conclusion applies
A flat step counts as monotonic because the relationship never falls. Whether it should be strictly rising is a domain decision the check does not make.
Check your understanding: If the value at load 5 is 0.88 and load 4 is 0.88, is the lookup still monotonic?
Chapter 15 source: section "What the lookup asserts". Demonstration C15-D03.
4Demonstration 4 of 4
Does the conclusion survive a change of shape?
If a line, a curve and a threshold pass through the same ends, do they lead to the same conclusion?
Away from the shared ends the shapes differ most. Near load 3 a threshold is still low and the curve is already high, so the conclusion can flip. Near load 4 all three have risen enough to agree.
Three relationships share the points (0, 0) and (5, 0.93). The requirement is an effectiveness of 0.60. The control picks the shape and the load at which the conclusion is read.
Predict first. At a load of 3, do all three shapes meet the 0.60 requirement?
Choose an example
Constructed example: the chapter's curve plus two alternative shapes defined for this reader, all read with the pack's Lookup.
Calculated values
- Shape chosen
- Gentle curve
- Reading at this load
- 0.780
- Meets 0.60
- yes
- Shapes that meet 0.60
- 1 of 3
- Do the three shapes agree
- no
Chosen shape, gentle curve, at load 3.0: 3.0 is an observed point, so the reading is its own value 0.78. Against the requirement, 0.780 - 0.60 = 0.180, so it meets 0.60. All three readings: straight line 0.558, gentle curve 0.780, sharp threshold 0.100. 1 of 3 meet it. The conclusion flips between shapes, so it is a finding about the shape, and the shape needs evidence.
Use the idea
Before trusting a finding, redraw the lookup as a line, a gentle curve and a threshold and rerun; if the finding flips, collect evidence for the shape.
Where the conclusion applies
Three hand-drawn shapes, one operating load at a time and one requirement. A conclusion can also depend on the endpoints, which this test holds fixed.
Check your understanding: At load 4, the curve reads 0.88. Does the straight line from (0, 0) to (5, 0.93) meet 0.60 there?
Chapter 15 source: section "How much the shape matters". Demonstration C15-D04.