Break-Fix: Diagnosing a Misbehaving Densitometer

The situation

Your densitometer has a certificate. It was calibrated on a good afternoon, its residuals were inside ±0.02 across most of the range, and you wrote a warm-up rule you have been obeying.

This morning it is lying to you, and it is lying quietly. Nothing has failed; there is no error message, no saturation flag, no blank screen. The numbers look like densities. They are simply wrong, and if you do not catch it, a fortnight of film characterisation goes into the notebook with a fault baked into every point of it.

Six instruments in six different states are set out below, each as a data set you can work on at a table. One idea sorts all six, and it is the only idea on the page:

An error that adds light bends the top of the scale and leaves the bottom alone. An error that multiplies light tilts or shifts the whole scale evenly. Re-zeroing removes the second and cannot touch the first.

By the end you should be able to read a residual plot’s signature in three seconds and say which of the two you are looking at, and then choose the one test that convicts the specific cause.

Before you start

You need the calibration experiment behind you, because every diagnosis here is made by comparing a run against a certificate you produced there. In particular you need your own stray-light ceiling, your repeatability figures at two densities, and your warm-up rule, because three of the six faults are diagnosed by showing that a reading has violated one of them.

You need the two equations that generate every signature on the page. They are derived once in The diagnosis and used six times.

And you need squared paper. Plot the residuals by hand at least once. A residual plot you have drawn yourself is one you will recognise across a room for the rest of your working life, and the whole method here is recognition.

📓How the data sets were produced, stated once and plainly

The course has no reference build and therefore no measured fault data to publish. Every table below is simulated from the arithmetic printed in this page, and the recipe for each is given with it so that you can regenerate every figure yourself in a spreadsheet.

Three of the six sets — A, B and E — carry no added scatter at all, so every digit is reproducible from the equation and a certified density. The other three — C, D and F — are about scatter, so scatter was added: normally distributed, with the standard deviation stated in each caption.

This is the honest form of a break/fix exercise the course cannot yet run on its own hardware, and it is labelled here rather than in a footnote. When you have induced these faults on your own instrument — which Further work tells you how to do, in about an hour — your data will be better than these tables, and you should replace them in your own copy.

Alternative route

You do not need an instrument to do this page, and that is deliberate rather than a concession. The six data sets are complete: they can be plotted, differenced and diagnosed at a kitchen table with a pencil, and the reasoning is the whole of the exercise.

A reader who has not built the densitometer — because the parts have not arrived, because they bought a second-hand commercial instrument, or because they are reading ahead — should work the page exactly as written and stop before The fix, which is the only section that needs hardware. The diagnostic method transfers to any instrument that reports a logarithm of a ratio, which is what the last section of the page is about.

The symptoms

Six sets. The letters are not in the order of the fault numbers, so do not try to match them by position. Each set was taken on the same instrument reading the same calibrated 21-step wedge, with the certified densities in the second column, except where the set is a time series or a set of repeats.

Set A — a linearity run

Every fifth step shown; the full set follows the same rule exactly. No scatter added.

Step	Certified D	Measured D	Residual
1	0.05	0.150	+0.100
4	0.50	0.600	+0.100
8	1.10	1.200	+0.100
11	1.55	1.650	+0.100
14	2.00	2.100	+0.100
17	2.45	2.550	+0.100
21	3.05	3.150	+0.100

Set B — a linearity run

No scatter added.

Step	Certified D	Measured D	Residual
1	0.05	0.050	−0.000
4	0.50	0.496	−0.004
8	1.10	1.080	−0.020
11	1.55	1.494	−0.056
14	2.00	1.856	−0.144
17	2.45	2.124	−0.326
21	3.05	2.312	−0.738

Set C — ten readings of step 11, sample lifted and replaced each time

Certified density of step 11 is 1.55. Scatter within each reading, 0.003 D.

Reading	1	2	3	4	5	6	7	8	9	10
D	1.554	1.599	1.555	1.602	1.551	1.603	1.547	1.601	1.554	1.604

Mean 1.577, standard deviation 0.026, spread 0.057. For comparison, ten readings of the same step without touching anything gave 1.551, 1.551, 1.553, 1.553, 1.551, 1.553, 1.552, 1.555, 1.551, 1.551: mean 1.552, standard deviation 0.0014, spread 0.004.

Set D — one step read every two minutes from switch-on

Step 11, certified 1.55. The zero was taken once at t = 0 and never retaken. Scatter 0.0012 D.

Minutes	0	2	4	6	8	10	12	14
D	1.5516	1.5508	1.5549	1.5569	1.5570	1.5603	1.5597	1.5587

Minutes	16	18	20	22	24	26	28	30
D	1.5620	1.5623	1.5610	1.5613	1.5601	1.5629	1.5621	1.5625

Set E — a linearity run

No scatter added.

Step	Certified D	Measured D	Residual
1	0.05	0.045	−0.005
4	0.50	0.450	−0.050
8	1.10	0.990	−0.110
11	1.55	1.395	−0.155
14	2.00	1.800	−0.200
17	2.45	2.205	−0.245
21	3.05	2.745	−0.305

Set F — the spread of a hundred readings, at seven densities

Nothing was moved between readings. The predicted column is the converter’s own input-referred noise divided by the square root of the forty-eight readings the firmware keeps, converted to density.

Density	Predicted spread, D	Measured spread, D
0.05	0.000003	0.0039
0.50	0.000009	0.0040
1.00	0.000004	0.0038
1.55	0.000013	0.0040
2.00	0.000036	0.0043
2.45	0.000101	0.0042
2.90	0.000284	0.0045

The three linearity runs on one pair of axes

• A perfect instrument
• Set A
• Set B
• Set E

Show the numbers behind this plot

Four traces on axes of certified density against reported density, both running from 0 to about 3.2. The dotted reference runs diagonally from corner to corner. Set A runs exactly parallel to it and one tenth of a density unit above it at every point, from 0.15 at a certified 0.05 to 3.15 at a certified 3.05, so its error is the same size everywhere. Set E runs as a straight line below the reference, starting almost on it at 0.045 and diverging steadily to 2.745 at a certified 3.05, so its error grows in proportion to the density and its line, extended backwards, passes through the origin. Set B begins indistinguishable from the reference, follows it closely to about 1.1, then bends away downwards ever more sharply, reading 1.494 at a certified 1.55, 1.856 at 2.00 and only 2.312 at 3.05, flattening towards a ceiling. Three completely different shapes from three different faults: a parallel offset, a proportional tilt, and a curve that only departs at the top.

Series	Certified density of the step	Density the instrument reported
A perfect instrument	0.05	0.05
A perfect instrument	3.05	3.05
Set A	0.05	0.15
Set A	0.50	0.60
Set A	1.10	1.20
Set A	1.55	1.65
Set A	2.00	2.10
Set A	2.45	2.55
Set A	3.05	3.15
Set B	0.05	0.05
Set B	0.50	0.50
Set B	1.10	1.08
Set B	1.55	1.49
Set B	2.00	1.86
Set B	2.45	2.12
Set B	3.05	2.31
Set E	0.05	0.04
Set E	0.50	0.45
Set E	1.10	0.99
Set E	1.55	1.40
Set E	2.00	1.80
Set E	2.45	2.21
Set E	3.05	2.75

Plotted directly from the tables above, which were computed from the equations in the next section rather than measured. The shapes are the diagnosis; the numbers are only how you get to them. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

Evidence to collect

Before you name anything, take these five pieces of evidence. Three of them cost under a minute and two of them will settle most faults on their own.

1. The residual, not the reading. Subtract the certified density from the measured one and plot the difference against the certified value. A plot of measured against certified is a beautiful straight line for every instrument that is not on fire, and it hides everything. The residual hides nothing.

2. A fresh zero, then the same reading again. This is the single most informative action on the page and it takes twenty seconds. If the error changes when you re-zero, it is multiplicative. If it does not, it is additive. A commercial instrument’s manual makes the same point from the other direction: the zero is the major factor of drift over a period of time, which is why that instrument has a one-button procedure for re-taking it alone.

3. The dark reading, on both ranges, now and twenty minutes from now. The dark is the only term you can measure in isolation, because switching the lamp off removes everything else. A dark that has moved is a fault with a name.

4. The opaque blank, with the lamp on and the room dark. This is the head page’s T2 and it gives the stray-light fraction and the ceiling as they are today, not as they were on the certificate.

5. The monitor column of the log. Every reading the firmware writes carries the monitor photodiode’s value beside it. A fault in which the monitor moves with the sample is in the lamp; a fault in which the monitor is flat while the sample moves is anywhere else. This one column separates half the possibilities without touching the instrument.

✦The five-minute check that would have caught three of these six

Before every session: take a dark, take a zero, read the calibration step, read the opaque blank.

Four readings, under five minutes. Compare each with the certificate. The step must be within 0.02 D of its certified value — the tolerance a commercial instrument specifies for exactly this check — and the blank must give the same stray-light fraction it gave on the certificate.

Faults A, B and E below would all have been caught by it before a single negative was read, and two of them would have been caught by the step alone.

Hints

Three levels, in increasing order. Stop at the first that lets you name a set.

Hint 1 — sort them into two piles before you name any of them. Three of the six are errors in the scale: they change what a given amount of light is reported as. Three are errors in the spread: they change how repeatable a reading is without necessarily biasing it. Do that sort first and you have halved the problem.

Hint 2 — for the scale faults, look at the residual’s shape and nothing else. There are exactly three shapes it can take and each has one cause-class. A horizontal line away from zero means everything is shifted by the same amount, which can only happen where the zero itself is wrong. A straight line through the origin means the error is proportional to the density, which can only happen where the scale factor is wrong. A curve that hugs the axis and then falls away means an error that is constant in light rather than in density, which is additive, and additive light is either stray or ambient.

Hint 3 — for the spread faults, compare the observed spread with what the arithmetic predicts. The electronics’ contribution to density noise grows with density, because the same voltage noise is a bigger fraction of a smaller signal. So a spread that is flat across the range did not come from the electronics. A spread that is bimodal did not come from noise at all, because noise does not have two preferred values. And a spread that grows faster than predicted at the dense end is the one case where the electronics really is the culprit.

The diagnosis

The two shapes, and the arithmetic behind them

Everything on this page follows from two equations, and it is worth deriving them once rather than appealing to them six times.

An additive error. Suppose a quantity of light I_s reaches the detector without passing through the sample — from inside the head, from the room, or manufactured by the electronics as a dark term that was not subtracted. It is present when the zero is taken and present when the sample is read, so:

D_measured = log₁₀ ( (I₀ + I_s) ÷ (I + I_s) )

An additive term in both readings

I₀ is the light through an empty stage, I the light through the sample and I_s the unwanted addition. Write f = I_s ÷ I₀ for the fraction and substitute I = I₀·10^−D:

D_measured = −log₁₀ ( (10^−D + f) ÷ (1 + f) )

The same thing in densities

At low density 10^−D is much larger than f and the error is invisible. At high density it is not, and as D grows without limit the reported density approaches a ceiling of −log₁₀ f. Additive errors bend the top of the scale and leave the bottom untouched.

A multiplicative error. Suppose the light is k times what it was when the zero was taken — the lamp has drifted, or the gain has changed. Then I becomes kI, but I₀ in the firmware is the recorded zero, so:

D_measured = log₁₀(I₀ ÷ kI) = D − log₁₀ k

A multiplicative change between zero and reading

That is a constant shift at every density: a horizontal line on a residual plot. And a wrong scale constant s — a mistyped certificate value, a mis-set slope — gives:

D_measured = s × D

A wrong scale constant

which is a straight line through the origin on a residual plot, of gradient s − 1.

Three shapes, three classes, and the discriminator between the first two is the twenty-second test: re-zero. A shift caused by k disappears the instant the zero is retaken with the lamp in its new state. An additive floor does not, because it contaminates the new zero exactly as it contaminated the old one.

From a residual plot to a named cause, in three questions

1. 1Is the residual a horizontal line away from zero?Then everything is shifted equally: the zero is wrong. Either it was taken through something (film base, a fingerprint, the wedge's clear leader), or the lamp changed level between the zero and the readings. Re-zero on a genuinely empty stage: if it goes, it was the lamp; if it stays, you were zeroing through something.
2. 2Is it a straight line through the origin?Then the error is proportional to density: the scale constant is wrong. A mistyped certificate value, the wrong step used for the calibration, or a slope left over from a previous session. Recalibrate against a step you have read the certificate for twice.
3. 3Does it hug zero and then fall away at the top?Then it is additive light. Run the opaque-blank test with the room dark, then again with the room lit. A difference between those two is ambient leakage through the shroud; no difference means it is stray light inside the head, or a stale dark reading.
4. 4Is the reading right but the spread wrong?Compare the measured spread with the arithmetic. Flat across the range means multiplicative noise, upstream of the film: the lamp or its supply. Bimodal means two mechanical states. Growing faster than predicted at the dense end means the electronics, and only then.

The order matters: the two scale questions are answered by one re-zero, so ask them before you take anything apart.

Set A — the zero was taken through the film base

Fault 4. A residual that is a horizontal line at exactly +0.100 at every step. Nothing else produces a shift that is identical at 0.05 and at 3.05.

By the arithmetic above there are two candidates, and they must be separated rather than assumed. Either the zero was taken through something of density 0.10 — the wedge’s clear leader, a piece of film base left on the stage, a fogged strip used as a spacer — or the lamp was 0.100 D brighter when the zero was taken than when the steps were read, which by D − log₁₀k means k = 0.794, a 21 per cent fall. Both give a horizontal residual and they are physically different faults.

The discriminating test. Re-zero on a genuinely empty stage and read step 11 again. If the residual vanishes, the zero was taken through something. If it stays at +0.100, the lamp is dropping 21 per cent between the zero and the reading — and the monitor column will have shown that all along, which is the second discriminator and the reason the column exists.

And a third possibility that is not a fault at all. If the sample is a negative rather than a wedge, part of that offset is the film’s own base density, and it is not an instrument error.

The course quotes no typical base density, because the datasheets it has read do not publish one. They describe the base by material, thickness and treatment instead — and FOMA’s description of Fomapan 100 is the reason a single figure could not be honest even if one were given: 120 roll film on a clear polyester base, 35 mm on a grey or grey-blue cellulose triacetate base, sheet film on clear polyester with an anti-halation layer. ILFORD describe FP4 Plus the same way, by material and thickness and backing. One emulsion, three bases, three different base densities, and the 0.10 in set A is a stipulation of the exercise rather than a published value.

So whether you zero on air or on film base is a decision, not a default, and the two answer different questions: air-zeroed gives the total density of everything in the light path, base-zeroed gives the density of the image alone. Write which you used on every data sheet, and measure your own base on an unexposed, processed strip of the same batch — because the difference is the size of the base, it depends on the format as well as the emulsion, and nobody can recover it afterwards.

Set B — stray light, and the ceiling arriving

Fault 2. The residual is indistinguishable from zero to about 1.0, then curves away, reaching −0.738 at the densest step. Fit the ceiling: the reported density flattens towards about 2.4, and −log₁₀ f = 2.4 gives f = 0.004, a stray-light fraction of 0.4 per cent. The whole set is reproduced by the additive equation with that one number.

The discriminating test, in two parts, because there are two additive sources. Put the opaque blank over the aperture with the lamp on and the room dark, and compute the fraction. Then repeat with the room lit. A fraction of 0.4 per cent in both cases is stray light inside the head — a seam, an unblackened surface, light around the aperture plate — and the cure is optical. A fraction that is small in the dark and 0.4 per cent with the lights on is ambient leakage through the shroud, and the cure is the foam seal, not the black paint. For scale, a commercial instrument in a sealed metal case specifies its ambient interference as a decrease in D of less than 0.25 per cent.

A third additive source that looks identical and is not optical at all. A stale dark reading. The detector’s pedestal is about 7.5 mV and its op-amp’s input bias current, 165 pA doubling every 10 °C through a megohm, adds a few hundred microvolts more that grows as the head warms. If the dark was taken cold and stored, the subtraction is too small and the residue is an additive term that behaves exactly like stray light. Take a fresh dark and repeat before you reach for the paint. This is the cheapest mistake on the page to avoid and the most expensive to make, because the cure for the wrong diagnosis is an evening of flocking.

Set C — the arm closes to two different places

Fault 6. Look at the ten replaced readings in order: 1.554, 1.599, 1.555, 1.602, 1.551, 1.603, 1.547, 1.601, 1.554, 1.604. They are not scattered. They are in two clusters, one near 1.552 and one near 1.601, about 0.05 apart, and the readings alternate between them.

Noise does not do this. Noise has one preferred value and a spread about it; a bimodal distribution means the instrument is in one of two states, and a state is mechanical: this is mechanical repeatability rather than noise. The undisturbed run confirms it from the other side: standard deviation 0.0014 with nothing touched, against 0.026 with the sample replaced. The electronics are eighteen times better than the mechanics.

The discriminating test. Read the step ten more times, deliberately closing the arm the same way each time — say, lowering it and letting it rest under its own weight, never pressing. If the bimodality disappears, the two states were you. Then do ten more, pressing firmly each time, and see whether that cluster is the higher or the lower one.

What produces two states, in the order to suspect them. An arm that can rest either on its spacers or slightly off them, so the gap is 3.0 mm or 3.3 mm depending on how it lands. Spacers of unequal height, so the arm tilts one way or the other. Film that can sit flat or slightly domed. A hinge with play, so the arm can be a millimetre forward or back and the aperture sees a different part of the step.

Why 0.05 D is a big number here. A 0.05 D change is a 10.9 per cent change in the light reaching the detector — a tenth of the signal, produced by a mechanism moving a fraction of a millimetre. That is the scale of the mechanical problem, and it is why the calibration page asks for two repeatability figures.

Set D — the lamp warming, seen through a stale zero

Fault 1, and it is zero drift. The reported density rises from 1.5516 at switch-on to about 1.562 at twelve minutes and then stops rising, creeping only a further thousandth over the remaining eighteen. Total excursion: +0.011 D, all of it in the first twelve minutes.

By the multiplicative equation, a shift of +0.011 D means k = 10^−0.011 = 0.975: the lamp is delivering 2.5 per cent less light than when the zero was taken. That is a lamp warming to its equilibrium junction temperature, and it is a documented behaviour of the class rather than a fault — LED datasheets plot relative flux against junction temperature, and both the output and the spectral mixture move with it.

The discriminating test, and it is one line. Re-zero and read the same step. The 0.011 vanishes completely, because a multiplicative change between zero and reading is exactly what re-zeroing removes. If it does not vanish, the drift is not in the lamp, and the monitor column will say so: a monitor that fell 2.5 per cent alongside the sample confirms the lamp; a monitor that stayed flat while the sample drifted means something in the optical path is moving — a diffuser lifting off its seat as its housing warms, an arm relaxing — which is a mechanical fault found by thermal means.

Which way does it go? Note the sign. The density reads high, because the lamp is dimmer than the zero assumed. A reader who expects “drift” to mean “reads low” will chase the wrong thing; the direction follows from the arithmetic and not from intuition.

Set D, and what the same instrument does when it re-zeroes

• Set D: zero taken once at t = 0
• The same step, re-zeroed before each reading

Show the numbers behind this plot

Two traces of reported density against minutes from switch-on, on a vertical scale from 1.545 to 1.567 with the certified value of 1.55 marked. The first, taken with the zero established once at switch-on, starts at 1.5516, climbs through 1.5549 at four minutes and 1.5603 at ten, and flattens after about twelve minutes at around 1.562, an excursion of 0.011 density in all. The second, of the same step with the zero retaken immediately before every reading, is flat at 1.551 within a thousandth across the whole half hour and sits essentially on the certified value. A vertical guide at twelve minutes marks where the first trace flattens. The pair is the diagnosis: the entire drift lives in the zero, so it is multiplicative, and it disappears when the zero is retaken.

Series	Minutes from switch-on	Reported density of step 11 (certified 1.55)
Set D: zero taken once at t = 0	0.00	1.552
Set D: zero taken once at t = 0	2.00	1.551
Set D: zero taken once at t = 0	4.00	1.555
Set D: zero taken once at t = 0	6.00	1.557
Set D: zero taken once at t = 0	8.00	1.557
Set D: zero taken once at t = 0	10.00	1.560
Set D: zero taken once at t = 0	12.00	1.560
Set D: zero taken once at t = 0	14.00	1.559
Set D: zero taken once at t = 0	16.00	1.562
Set D: zero taken once at t = 0	18.00	1.562
Set D: zero taken once at t = 0	20.00	1.561
Set D: zero taken once at t = 0	22.00	1.561
Set D: zero taken once at t = 0	24.00	1.560
Set D: zero taken once at t = 0	26.00	1.563
Set D: zero taken once at t = 0	28.00	1.562
Set D: zero taken once at t = 0	30.00	1.563
The same step, re-zeroed before each reading	0.00	1.551
The same step, re-zeroed before each reading	4.00	1.551
The same step, re-zeroed before each reading	8.00	1.552
The same step, re-zeroed before each reading	12.00	1.552
The same step, re-zeroed before each reading	16.00	1.551
The same step, re-zeroed before each reading	20.00	1.552
The same step, re-zeroed before each reading	24.00	1.551
The same step, re-zeroed before each reading	28.00	1.552

The set D column plotted against a re-zeroed comparison computed from the same arithmetic; neither is a measurement of an instrument. The shape of the pair is what matters, and it is the signature of every multiplicative fault on this page. The curve is drawn to show the shape, not measured from a real material. Your own materials will differ, and measuring them is what the sensitometry part of the course is for.

Set E — the wrong reference value

Fault 5. The residual is a straight line from −0.005 at 0.05 to −0.305 at 3.05. Fit it: the measured density is 0.900 times the certified one at every step, exactly, and the line through the residuals passes through the origin.

That is the signature of a wrong scale constant, and by the arithmetic there is only one place it can come from: the two-point calibration. Either the certificate value entered for the calibration step was wrong by 11 per cent, or the calibration was made against the wrong step, or a slope from a previous session was left in the firmware and never updated.

The distinction the covers of this page force you to make. The symptom is often described as “all densities scaled by 0.9”, and two causes are commonly offered: a wrong reference value, and the lamp level changing between the zero and the reading. Only the first fits this data. A lamp change is D − log₁₀k, an additive constant in density, which would give a residual that is horizontal — set A’s shape, not set E’s. A wrong reference value is s·D, proportional, which is set E’s shape and only set E’s. The two hypotheses make different predictions and the residual plot separates them in one glance. That is the whole reason for plotting residuals rather than readings.

The discriminating test. Re-read the calibration step and compare with its certificate before touching the slope. If the raw density of the calibration step, computed with a slope of exactly 1.000, disagrees with the certificate by 11 per cent, the fault is not in the calibration at all and you have a real instrument problem. If it agrees, the slope was simply wrong and recalibrating fixes it.

Why it is dangerous. Unlike stray light, this fault is invisible at the clear end, where most people sanity-check an instrument, and it is largest exactly where the important numbers live. A Dmax misreported by 11 per cent is a paper comparison silently reversed.

Set F — noise that does not grow with density

Fault 3. The measured spread is 0.004 D at every density from 0.05 to 2.90. The prediction, from the converter’s own input-referred noise divided by the square root of the readings kept, runs from 0.000003 D at the clear end to 0.00028 D at the dense end.

Two things are wrong and only one of them is obvious. The measured spread is sixteen times the prediction at the dense end and more than a thousand times it at the clear one, which says that something other than the converter is dominating. And it is flat, which says what.

Electronic noise is a voltage. Converted to density it is σ_V ÷ (V·ln 10), and V falls by a factor of ten for every unit of density, so the density noise from any voltage-like source must grow with density. A spread that is flat is a spread that is a constant fraction of the signal — which means it is multiplicative, which means it happened upstream of the film, in the lamp or in what drives it.

The discriminating test. Look at the monitor column’s own spread over the same hundred readings. If the monitor wobbles by the same 0.9 per cent — which is what 0.004 D corresponds to — the lamp is the source, and the causes are a supply that is not holding the drive current still, a constant-current element that is marginal, or a lamp sharing a rail with something that switches. If the monitor is quiet while the sample is not, the modulation is between the lamp and the detector, which means something is moving in the light path.

Why “low densities noisy, high densities stable” is how this fault is usually described. Nobody notices a flat 0.004 at the dense end, because at 2.9 D everyone expects a poor reading. What people notice is that the clear end is far worse than it ought to be, and that is the same observation. Comparing against the prediction rather than against expectation is what makes it diagnosable.

And the cure that does not work. Averaging more. This spread is not random about the true value from the film’s point of view; it is a real change in the light. More averaging buys the square root and the fault stays.

The fix

Each fix is stated with the retest that proves it, because a fix without a retest is a hope.

Set	Fault	The fix	The retest
A	Zero taken through film base, or a lamp change	Zero on a genuinely empty, clean stage; establish and record whether this instrument zeroes on air or on base, and put it on the certificate	Re-read all 21 steps. The residual must be flat at zero, not flat at +0.10
B	Additive light: stray, ambient, or a stale dark	In order of cost: fresh dark first; then the lights-on/lights-off blank to separate ambient from stray; then the shroud seal for ambient, or flocking and re-taping for stray	Opaque-blank test giving a fraction at or below the certificate’s, in both lighting conditions
C	Two mechanical states in the arm	Equal spacers measured with a caliper, a stiffer hinge, a positive stop the arm closes onto, and a closing action you perform the same way every time	Ten replaced readings that are unimodal, with a standard deviation inside your certificate’s replaced figure
D	Lamp warm-up seen through a stale zero	Obey the warm-up rule on the certificate, or re-zero before each reading, or both. Nothing needs repairing	The drift run repeated with re-zeroing: flat within repeatability across thirty minutes
E	Wrong scale constant	Re-read the certificate, re-run the two-point calibration against a step you have checked twice, and write the new slope and the step it came from into the firmware and the certificate together	The calibration step reading within 0.02 D of its certified value, and all 21 residuals back on a flat line
F	Multiplicative noise in the lamp or its supply	Decouple the driver, take the lamp off a shared rail, check the constant-current element is inside its compliance, and ratio every reading against the monitor if it persists	The spread at the clear end back near prediction, and the spread rising with density instead of flat

!The order in which to attempt fixes, and why it is not the order of the table

Cheapest and most reversible first, always. A fresh dark and a fresh zero cost forty seconds between them and resolve sets A, D and half of B. Reading the certificate again costs two minutes and resolves E. Only then do you start measuring spacers, and only after that do you open a tin of paint.

The reason is not laziness. Every irreversible fix destroys the evidence for the diagnosis you have not yet confirmed. Flock the arm’s underside and you can no longer measure what the old stray-light fraction was; adjust the spacers and set C’s two clusters are gone whether or not they were the fault. Take the cheap tests first because they leave the instrument in the state that produced the symptom.

What to record

For each fault, one entry, and it has six fields. This shape is the deliverable of the page and it is the shape Part IX’s fault diagnosis established for chemistry; nothing about it changes for an instrument.

1. Symptom, as observed, in numbers rather than adjectives. “Residual flat at +0.100 across all 21 steps” and not “reading high”.
2. Hypothesis, singular, and the shape it predicts. Naming the predicted shape is what makes the hypothesis falsifiable.
3. The single test, chosen because the two candidate causes make different predictions about its outcome. A test both hypotheses pass is not a test.
4. The result, written before you interpret it.
5. The fix applied, with the date and anything changed in the firmware or the hardware.
6. The retest against the reference wedge, with the numbers, and the certificate line updated or confirmed.

Then update the instrument certificate, or explicitly confirm that it still stands. A fault that was found and fixed without touching the certificate is a fault that will be rediscovered, because the certificate is the only thing any later page reads. These entries live with the certificate in the calibration records, whose reading-reference-deviation-action shape is what fields 1, 4 and 5 above are.

Transferring the method

The reasoning on this page is not about densitometers. It applies to any instrument that reports a logarithm of a ratio, and two of them are already in this course.

The sensitometer. Its axis is log exposure, and the same two classes apply with the same signatures. An additive error is light reaching the film that was not part of the intended exposure: a leak in the enclosure, a safelight, a lid that does not seal. It adds to every step equally in exposure, so it compresses the clear end of the strip — the opposite end from the densitometer’s, because the wedge is the other way round — and leaves the dense end alone. A multiplicative error is the lamp at the wrong level or the timer at the wrong interval, and it shifts the whole strip along the log exposure axis without changing its shape. Part XIV’s calibration page built its error budget on exactly that separation, and its rule that log-exposure and density terms must never be added is the same idea seen from the other side.

A lux meter. Its zero is additive: a meter that reads 3 lux with the cap on is reporting 3 lux of something that is not light, and the error is negligible in daylight and dominant in a darkroom. Its calibration constant is multiplicative: a meter reading 8 per cent low reads 8 per cent low at every level, and no amount of darkness reveals it. Cap the meter to find the additive term; compare with another meter to find the multiplicative one. Those are the same two tests as this page’s, in different clothes.

The general rule, which is worth carrying out of this part. Ask of any error: does it add to the signal, or scale it? An additive error is worst where the signal is smallest, and it can be measured by removing the signal entirely. A multiplicative error is the same fraction everywhere, and it can be measured only against something else. Two questions, two tests, and almost every instrument fault you will meet is one or the other.

Questions

1. An instrument reads a certified 0.05 step as 0.05 and a certified 2.45 step as 2.13. Give the stray-light fraction that fits, the ceiling it implies, and the one further reading you would take to confirm it. Then say what else, other than stray light, produces the same shape.
2. Your residual plot is a straight line of gradient +0.06 through the origin. Compute what the instrument reports for a true density of 2.00, name the fault, and say why re-zeroing will not change it by a thousandth.
3. Ten replaced readings give 1.881, 1.884, 1.930, 1.883, 1.931, 1.882, 1.929, 1.885, 1.932, 1.880. Compute the mean and the standard deviation, then say why quoting that standard deviation as the instrument’s repeatability would be misleading, and what you would quote instead.
4. A reader reports that their densities all read 0.07 too high and that a fresh zero fixes it completely. Give the value of k implied, say what physically changed, and state which log column would have shown it happening.
5. Your measured spread is 0.0009 D at 0.5 and 0.0011 D at 2.5, while the arithmetic predicts 0.00001 and 0.0003. Is the excess additive or multiplicative? Show the reasoning from the two shapes, and name one candidate cause for each answer you might have given.
6. A classmate blames a slope error of 0.94 on the analogue-to-digital converter, arguing that its gain must be out. Argue from the signature and from one datasheet figure why the converter is not the cause, and name the two places the error can actually be.

Further work

Induce all six on your own instrument, in about an hour. Every one of them can be created deliberately and none of them requires a modification beyond the design.

Fault A: take the zero with a piece of clear film base on the stage. Fault B: leave the shroud’s foam seal off and work with the room lit, or slide a strip of white card inside the head. Fault C: remove one spacer and read ten times. Fault D: switch on and start reading immediately, with no warm-up and no re-zero. Fault E: type a certificate value 10 per cent wrong into calibrate(). Fault F: run the lamp from the same rail as something that switches, or replace the constant-current element with a resistor fed from a sagging supply.

Then plot each residual, hand the six plots to somebody else with the labels removed, and see whether they can sort them. Data you generated yourself is worth more than the tables on this page, and this is the page that says so.

Build the five-minute check into the firmware. A preflight() routine that takes a dark, a zero, the calibration step and the opaque blank, compares each against constants from your certificate, and prints PASS or FAIL with the offending number. It is thirty lines, it runs in under two minutes, and it converts the discipline of this page into something that cannot be forgotten on a busy evening.

Find a fault this page does not list. Six is not all of them. A cracked solder joint that changes with temperature, a diffuser slowly yellowing, a wedge that has been fingerprinted, a firmware constant changed and not recorded: each has a signature, and each can be placed in the additive or multiplicative column before it is understood. Write the entry in the six-field shape above and it belongs in your notebook beside these.

Check your understanding

Question 1. You take a fresh zero and the error disappears entirely. What class of fault did you have?

• Additive, because zeroing subtracts a constant
• Multiplicative, because a change in the light between the zero and the reading is exactly what a new zero absorbs
• Mechanical, because you moved the sample to re-zero
• It is not diagnostic; re-zeroing changes every reading

Show the answer and why

Answer: Multiplicative, because a change in the light between the zero and the reading is exactly what a new zero absorbs

The zero records how much light an empty stage passes, and every density is a ratio against it. If the lamp is k times what it was when the zero was taken, every density is shifted by minus the logarithm of k, and taking a new zero with the lamp in its present state makes k equal to one again. An additive term contaminates the new zero exactly as it contaminated the old one, so it survives untouched. That asymmetry is what makes re-zeroing a twenty-second diagnostic rather than merely good practice, and it is why a commercial instrument provides a one-button procedure for retaking the zero alone.

Question 2. An instrument reads a certified 0.05 step correctly and a certified 3.05 step as 2.10. Estimate the stray-light fraction and the ceiling it implies.

• About 0.8 per cent, ceiling about 2.1 D
• About 3 per cent, ceiling about 1.5 D
• About 0.1 per cent, ceiling about 3.0 D
• It cannot be estimated without reading the opaque blank

Show the answer and why

Answer: About 0.8 per cent, ceiling about 2.1 D

At 3.05 D the light through the sample is about a thousandth of the light through air, so it is already small compared with any plausible additive term, and the reported density has nearly reached its ceiling. Since the ceiling is minus the logarithm of the fraction, a reading of 2.10 implies a fraction of about ten to the minus 2.1, which is 0.8 per cent. Option 4 is the cautious answer and is worth arguing with: the blank test measures the fraction directly and should always be run, but a wedge already contains the information, and being able to read it off a residual plot is what the page is teaching.

Question 3. Ten replaced readings of one step give 1.881, 1.884, 1.930, 1.883, 1.931, 1.882, 1.929, 1.885, 1.932, 1.880. Why is the standard deviation the wrong statistic to quote?

• Because ten readings are too few for a standard deviation
• Because the readings are bimodal, so a single spread describes two states rather than one distribution
• Because the mean is not the certified value
• Because a spread must be quoted as a range, never as a standard deviation

Show the answer and why

Answer: Because the readings are bimodal, so a single spread describes two states rather than one distribution

The readings fall into two clusters about 0.048 apart, five near 1.882 and five near 1.930, and alternate between them. A standard deviation is a description of scatter about one central value, and computing one here reports the distance between the clusters as though it were noise. The honest report names the bimodality and says what the two states are, because a fault with two states is mechanical and a fault with one is not. Option 3 is a different question, about trueness rather than precision, and option 1 mistakes a small sample for a wrong statistic.

Question 4. Your density noise is flat at 0.004 D from 0.05 to 2.9 D, while the arithmetic predicts a rise from 0.000003 to 0.00028. Where is the noise coming from?

• The converter, because 16 bits are not enough at the dense end
• Somewhere upstream of the film, because a noise that is a constant fraction of the signal is multiplicative
• The mechanics, because the sample was replaced between readings
• The dark reading, because it was not retaken

Show the answer and why

Answer: Somewhere upstream of the film, because a noise that is a constant fraction of the signal is multiplicative

Electronic noise is a voltage, and converting a voltage noise to a density divides by the signal, which falls tenfold per density unit. Any voltage-like source therefore produces a density noise that grows with density. A flat density noise is a constant fractional noise, which means the light itself is fluctuating by that fraction, which means it happened before the film. The monitor column settles it: if the monitor wobbles by the same 0.9 per cent, the lamp or its supply is the source. Note that averaging will not help, because the fluctuation is a real change in the light rather than scatter about a true reading.

Question 5. You want to characterise a lux meter with the same reasoning. Which pair of tests finds its additive and its multiplicative terms?

• Read it in bright light, then in dim light
• Cap the sensor to find the additive term; compare with a second meter to find the multiplicative one
• Compare it with a second meter twice
• Read it at two temperatures

Show the answer and why

Answer: Cap the sensor to find the additive term; compare with a second meter to find the multiplicative one

Capping the sensor removes the signal entirely, so whatever the meter still reports is additive: a dark offset, an internal leak, or an electrical zero error. It is negligible in daylight and dominant in a darkroom, which is exactly the density-dependence this page has been describing in another currency. The multiplicative term is a calibration constant and it cannot be found by removing the signal, because it scales whatever is there; it needs a second instrument or a known source. Option 4 is a useful third test that characterises how both terms move with temperature, but it identifies neither on its own.

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