Classical-Mechanics · Unit 13 · Video 3 · Interactive Practice

Feynman, the O-Rings, and Challenger: When a Mathematical Model Hides Its Uncertainty

IKey Formulas

Formula	Name	What it says
$E = C\,Q^{0.58}$	Empirical erosion model	Erosion depth $E$ from the stagnation heat $Q$ — the exponent came from a fit, not from a law
$\log E = \log C + 0.58\log Q$	Log-log form	A power law is a straight line on log axes, and the exponent is its slope
$\tfrac{1}{2}E_{\text{fit}} \le E \le 2E_{\text{fit}}$	Factor-of-two scatter	Points sit twice above and twice below the line, so twice the predicted erosion is ordinary
$3 \times \tfrac{1}{3}r = r$	The margin arithmetic	The deepest observed erosion was $\tfrac{1}{3}$ of the ring radius $r$ ; three times that is the whole radius

Key Insight: The physics is sound as far as the stagnation point; past it the model is a curve fit. A fit interpolates the cases it has already seen — and its scatter is the honest measure of what it does not know.

IIThe Chain from a Cold Seal to a Lost Vehicle

Every link in the loss of Challenger is ordinary physics, and each one needs the link before it.

Propagation 7/7

IIIWhere the Exponent 0.58 Came From

The exponent is not a law: it is the slope that misses the scattered points by the least.

💡 The physics is sound only as far as the stagnation point: the step from heat to erosion is a formula suggested by data on a similar material, not on the O-ring rubber itself.

IVWhat a Factor-of-Two Scatter Allows

A fit whose points scatter by a factor of two predicts a range of erosion depths, not a number.

Heat (log Q) 1.95

💡 Erosion had been seen on earlier flights and read as evidence of margin. Feynman's verdict: "Erosion was a clue that something was wrong. Erosion was not something from which safety can be inferred."

VQuiz Questions

Problem 1 · Reading the Power Law

Given: the fitted model $E = C\,Q^{0.58}$ . If the stagnation heat $Q$ is multiplied by $10$ , by what factor does the predicted erosion depth $E$ grow?

✅ Correct!

10^{0.58} = 3.8

: a tenfold heat gives less than a fourfold bite, because the exponent is well below

1

❌ That is the answer for an exponent of $1$ . The fitted exponent is

0.58

, so the factor is

10^{0.58}

, not

10^{1}

❌ Not quite. The exponent is a power, not a multiplier:

E

grows by

10^{0.58}

Show solution

Take the ratio of the model at the two heats — the constant $C$ cancels:

\frac{E_2}{E_1} = \frac{C(10Q)^{0.58}}{C\,Q^{0.58}} = 10^{0.58} = 3.80

The same statement on log axes: one decade of $\log Q$ raises $\log E$ by $0.58$ , and $10^{0.58} \approx 3.8$ .

Problem 2 · What the Scatter Permits

Given: at one heat the fitted curve predicts $E_{\text{fit}} = 0.10\,r$ , and the twelve measured cases scatter by a factor of two about the line — some twice above it, some twice below. Which erosion depths are consistent with the data at that heat?

✅ Correct! A factor of two either way runs from

E_{\text{fit}}/2 = 0.05\,r

2E_{\text{fit}} = 0.20\,r

— an erosion twice as deep as predicted is not an outlier.

❌ A factor of two is not a $20\%$ band. Twice above means

2 \times 0.10\,r

; twice below means

0.10\,r \div 2

❌ The band runs both ways. Half the cloud lies below the fitted line, so the interval starts at

0.05\,r

❌ Not quite. "Twice above and twice below" means multiply and divide the prediction by

2

Show solution

The scatter is multiplicative, so it is symmetric on log axes, not on a linear scale:

\tfrac{1}{2}E_{\text{fit}} \le E \le 2E_{\text{fit}} \quad\Longrightarrow\quad 0.05\,r \le E \le 0.20\,r

On the log-log plot this is a band of constant width $\log 2 = 0.30$ above and below the fitted line — and several of the twelve points sit right on its edges.

Problem 3 · Getting the Exponent Off the Plot

Given: two points on the fitted straight line, $(\log Q,\ \log E) = (1.0,\ -1.27)$ and $(2.0,\ -0.69)$ , with $E$ measured in units of the ring radius $r$ .

What exponent does this line give?

What erosion depth does the line give at $\log Q = 2.0$ ?

✅ Correct! Slope

0.58

on log axes is the exponent, and

10^{-0.69} = 0.20

❌ Check the run. It is

2.0 - 1.0 = 1.0

, not

2.0

❌ Check the direction.

-0.69

is greater than

-1.27

, so the line rises and the slope is positive.

❌ Check the slope. Use rise over run on the logarithms, not the ratio of the two

\log E

values.

❌ Undo the logarithm.

\log E = -0.69

means

E = 10^{-0.69}

, a depth smaller than

r

, not a negative one.

Show solution

Step 1 — slope of the log-log line:

0.58 = \frac{-0.69 - (-1.27)}{2.0 - 1.0} = \frac{0.58}{1.0}

Because $\log E = \log C + n\log Q$ , that slope is the exponent $n$ : $E \propto Q^{0.58}$ .

Step 2 — back out the depth:

E = 10^{-0.69}\,r = 0.20\,r

Nothing physical fixed the value $0.58$ — it is the slope that lands nearest the scattered cloud.

Problem 4 · The Next Flight (Transfer)

Given: the deepest erosion ever measured was $\tfrac{1}{3}r$ , at stagnation heat $Q_0$ . The next flight runs hotter, at $Q = 2Q_0$ , and the fit still carries its factor-of-two scatter.

By what factor does the fit predict the erosion to grow?

Allowing for the scatter, what is the deepest erosion consistent with the data at $2Q_0$ ?

✅ Correct!

2^{0.58} = 1.5

, so the fit predicts

\tfrac{1}{3}r \times 1.5 = 0.50\,r

— and the scatter alone allows twice that, a bite as deep as the whole ring radius.

❌ That assumes an exponent of $1$ . Doubling

Q

multiplies

E

2^{0.58}

, not by

2

❌ $0.58$ is an exponent, not a multiplier. The factor is

2^{0.58}

, not

2 \times 0.58

❌ Check the power. The growth factor for a doubling of

Q

2^{0.58}

❌ The scatter reaches a factor of two above the line too. Double the predicted

0.50\,r

Show solution

Step 1 — scale the deepest measured case with the fitted power law:

\frac{E(2Q_0)}{E(Q_0)} = 2^{0.58} = 1.49 \quad\Longrightarrow\quad E \approx \tfrac{1}{3}r \times 1.49 = 0.50\,r

Step 2 — add the uncertainty the fit already carries:

E_{\max} = 2 \times 0.50\,r = 1.00\,r

The prediction is $0.50\,r$ , but the honest report is " $0.25\,r$ to $1.00\,r$ " — an upper edge equal to the ring's full radius. Feynman made the same arithmetic with his own factor: erosion three times more severe than the deepest case observed, $3 \times \tfrac{1}{3}r$ , is again the whole radius.

The precise-looking exponent never narrowed that spread; only physical understanding of how the rubber erodes could.

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