Assignment: Interpreting Density Data

Four lessons have told you what a curve is, what its slope means and where its speed point sits. This one makes you do it: twelve tables of numbers, four families of curves drawn by hand, a slope and a speed for every one of them, and at the end a page of writing that recommends a film, a developer and a development time to somebody who has to go and use it.

Budget two hours and do it on paper. A spreadsheet will produce the same numbers in ten minutes and teach you almost nothing, because the errors this assignment exists to show you — a curve drawn wrong through scattered points, a speed point read off the toe by eye, a chord that is not quite the right length — are errors that only happen when a pencil is involved. Use the spreadsheet afterwards, to check.

!These tables are generated, not measured — read this before you plot anything

The data below is synthetic. It was produced from a stated mathematical model, printed in full at the foot of this page, and it describes no real film and no real developer. Nothing measured from it is a fact about any material you can buy.

The course would rather have supplied real data. It could not, honestly. Manufacturers publish characteristic curves as pictures — Kodak prints families of them for T-Max 100 and Tri-X, and ILFORD does the same — and turning a printed picture into a table means reading coordinates off a graph by eye. That is a legitimate technique when it is declared, and the course’s own rules would then require every number to be labelled a read-off approximation with a stated reading uncertainty. It is not legitimate to do it silently, and this course holds no published table of densities for two films in two developers at three development times, because no such table exists in its corpus.

So the tables are honestly generated instead. The model’s parameters were chosen to reproduce relationships the course can source — the way slope grows with development time, the size of the speed difference a solvent fine-grain developer costs, the shoulder a compensating developer manufactures — and the model is published so that you can regenerate every number and disagree with it.

What this exercise buys is the arithmetic, which is identical whatever the numbers came from. What it does not buy is a fact about the world. The real data is the next page: your own step wedge, your own developer, your own control strip. Everything you learn to do here, you do there on numbers nobody made up.

What you are given

Two films and two developers, none of them real, each named for what it is meant to represent.

Film M is a conventional cubic-grain medium-speed film with a long, gradual toe. Film T is a tabular-grain film with a shorter, more abrupt toe and a maximum density that keeps climbing further. The course refuses to classify any actual film as long-toe or short-toe, because it has verified no manufacturer statement doing so — that refusal is on the characteristic-curve lesson and stands. These two are model films, and the difference between their toes is a parameter the course chose so that the exercise has something to find.

Developer S is a solvent fine-grain developer of the ID-11 or PERCEPTOL kind. ILFORD describe PERCEPTOL as an extra fine grain developer designed for use when a decrease in film speed is not important, and the arithmetic behind that phrase is visible in their own development table: HP5 Plus is listed at EI 250/25 in PERCEPTOL stock and EI 400/27 in ID-11 stock, and DELTA 400 at EI 200/24 against EI 400/27. Part VIII’s solvent lesson gives the mechanism: sulfite dissolves the marginal latent-image speck before it can develop anything, which is the same reaction as the grain benefit and cannot be engineered away from it.

Developer H is a dilute high-acutance developer of the compensating kind. ADOX publish their FX-39 II times against a stated contrast figure and say plainly that the 1+9 dilution gives normal contrast with a speed increase while the 1+19 dilution acts to reduce contrast on high-contrast subjects, the speed utilisation falling as it does so — a published statement that speed and compensating ability are traded against one another. Part VIII’s compensation lesson gives that mechanism too, and one sentence of it is the whole signature you are about to find in the data: compensation acts where the density is high and leaves the toe alone, so it manufactures a shoulder.

Three development times for each combination, at 20 °C with the same agitation throughout. Every table carries its own control-strip reading: a strip of the same film, unexposed, processed alongside, as the speed criterion requires and as Part IX’s test-negative discipline established. Densities are quoted to 0.01 and you should treat them as carrying ±0.01 of reading uncertainty, which is about what a good instrument gives.

Log exposure is in millilux-seconds, in thirteen steps of 0.30 — one stop per step, which keeps the arithmetic in your head.

Film M in Developer S

Control strip (base plus fog): 9 min 0.15 · 14 min 0.18 · 22 min 0.23

log H	0.30	0.60	0.90	1.20	1.50	1.80	2.10	2.40	2.70	3.00	3.30	3.60	3.90
9 min	0.16	0.18	0.21	0.27	0.37	0.49	0.63	0.76	0.90	1.03	1.15	1.26	1.36
14 min	0.19	0.21	0.24	0.31	0.43	0.58	0.76	0.94	1.12	1.30	1.46	1.61	1.74
22 min	0.24	0.26	0.29	0.37	0.51	0.70	0.91	1.13	1.36	1.58	1.78	1.97	2.14

Film M in Developer H

Control strip: 5 min 0.13 · 8 min 0.15 · 13 min 0.18

log H	0.30	0.60	0.90	1.20	1.50	1.80	2.10	2.40	2.70	3.00	3.30	3.60	3.90
5 min	0.14	0.17	0.22	0.31	0.43	0.56	0.69	0.82	0.94	1.04	1.13	1.20	1.25
8 min	0.16	0.19	0.25	0.35	0.51	0.68	0.87	1.05	1.21	1.36	1.48	1.58	1.66
13 min	0.19	0.22	0.29	0.41	0.60	0.82	1.06	1.29	1.51	1.70	1.87	2.00	2.11

Film T in Developer S

Control strip: 9 min 0.15 · 14 min 0.17 · 22 min 0.20

log H	0.30	0.60	0.90	1.20	1.50	1.80	2.10	2.40	2.70	3.00	3.30	3.60	3.90
9 min	0.16	0.20	0.27	0.39	0.53	0.67	0.80	0.94	1.07	1.19	1.30	1.41	1.50
14 min	0.18	0.22	0.31	0.45	0.63	0.81	1.00	1.18	1.35	1.51	1.67	1.81	1.93
22 min	0.21	0.25	0.35	0.53	0.75	0.97	1.20	1.43	1.64	1.85	2.04	2.21	2.36

Film T in Developer H

Control strip: 5 min 0.14 · 8 min 0.15 · 13 min 0.17

log H	0.30	0.60	0.90	1.20	1.50	1.80	2.10	2.40	2.70	3.00	3.30	3.60	3.90
5 min	0.16	0.23	0.34	0.47	0.61	0.75	0.88	1.00	1.11	1.21	1.29	1.35	1.41
8 min	0.17	0.25	0.39	0.57	0.76	0.95	1.13	1.30	1.45	1.59	1.70	1.80	1.88
13 min	0.19	0.28	0.45	0.68	0.93	1.17	1.41	1.62	1.82	2.00	2.15	2.27	2.38

Task 1 — plot four families

One sheet of graph paper per film-and-developer pair; three curves on each sheet.

Get the scales right first, because one of them is not a free choice. The contrast-index construction measures distances along a chord, so it only works when 0.30 of density occupies the same distance on the paper as 0.30 of log exposure. Two centimetres per 0.30 on both axes fits an A4 sheet comfortably: 3.6 log units across is 24 cm, and 2.4 of density up is 16 cm. Draw the log exposure axis from 0.0 to 3.9 and the density axis from 0.0 to 2.4, and rule the grid at 0.30.

Plot base plus fog as a horizontal line before you plot a single point. It is the floor everything else is measured from, it is different for every one of the twelve conditions, and drawing it first stops you reaching for a number from the wrong strip. Label each line with its condition.

Then the points, then the curve — in that order, and not the other way round. Plot all thirteen points for one time. Look at them. Then draw a smooth curve, and draw it so it passes through the scatter rather than through every point: with ±0.01 on each reading, a curve that visits every dot exactly is a curve that has fitted your noise. A flexible rule, a French curve or a steady hand all work. What does not work is joining the dots with straight segments, because every slope you are about to measure is a tangent or a chord, and a polygon has neither.

One family, plotted: Film T in Developer H

• 5 min, contrast index 0.42
• 8 min, contrast index 0.58
• 13 min, contrast index 0.75

Show the numbers behind this plot

Three curves for the same model film at three development times, five, eight and thirteen minutes, all rising from a base plus fog near 0.15 at log exposure 0.30. The five-minute curve is the shallowest and flattens earliest, reaching about 1.41 at log exposure 3.90. The eight-minute curve is steeper and reaches 1.88. The thirteen-minute curve is the steepest and reaches 2.38. All three leave the base at almost the same place on the exposure axis, so the foot of the family is nearly common, and the separation between the curves grows steadily with exposure: the change made by development is almost all in the straight line and the shoulder and almost none of it in the toe.

Series	Log exposure (millilux-seconds)	Density
5 min, contrast index 0.42	0.30	0.16
5 min, contrast index 0.42	0.60	0.23
5 min, contrast index 0.42	0.90	0.34
5 min, contrast index 0.42	1.20	0.47
5 min, contrast index 0.42	1.50	0.61
5 min, contrast index 0.42	1.80	0.75
5 min, contrast index 0.42	2.10	0.88
5 min, contrast index 0.42	2.40	1.00
5 min, contrast index 0.42	2.70	1.11
5 min, contrast index 0.42	3.00	1.21
5 min, contrast index 0.42	3.30	1.29
5 min, contrast index 0.42	3.60	1.35
5 min, contrast index 0.42	3.90	1.41
8 min, contrast index 0.58	0.30	0.17
8 min, contrast index 0.58	0.60	0.25
8 min, contrast index 0.58	0.90	0.39
8 min, contrast index 0.58	1.20	0.57
8 min, contrast index 0.58	1.50	0.76
8 min, contrast index 0.58	1.80	0.95
8 min, contrast index 0.58	2.10	1.13
8 min, contrast index 0.58	2.40	1.30
8 min, contrast index 0.58	2.70	1.45
8 min, contrast index 0.58	3.00	1.59
8 min, contrast index 0.58	3.30	1.70
8 min, contrast index 0.58	3.60	1.80
8 min, contrast index 0.58	3.90	1.88
13 min, contrast index 0.75	0.30	0.19
13 min, contrast index 0.75	0.60	0.28
13 min, contrast index 0.75	0.90	0.45
13 min, contrast index 0.75	1.20	0.68
13 min, contrast index 0.75	1.50	0.93
13 min, contrast index 0.75	1.80	1.17
13 min, contrast index 0.75	2.10	1.41
13 min, contrast index 0.75	2.40	1.62
13 min, contrast index 0.75	2.70	1.82
13 min, contrast index 0.75	3.00	2.00
13 min, contrast index 0.75	3.30	2.15
13 min, contrast index 0.75	3.60	2.27
13 min, contrast index 0.75	3.90	2.38

Generated data, not measured: these are three rows of the table above, plotted so you can check the shape of your own drawing against it. Notice what Kodak's workbook says of a real family and this one reproduces — the toe stays put and the change is in the straight line and the shoulder. 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.

Task 2 — read base plus fog, the toe and the straight line

For each of the twelve curves, write down four things.

1. D₀, base plus fog, from the control strip and not from the first exposed step. On the shortest development of Film M in Developer S the first step reads 0.16 against a control of 0.15 — one hundredth apart, which is inside the reading uncertainty. Take the control’s figure anyway. The two agreeing is a check; treating them as interchangeable is how a fog difference gets lost.
2. Where the toe ends, judged by taking first differences down the row and finding where they stop growing. On Film M in Developer H at 8 minutes the differences run 0.03, 0.06, 0.10, 0.16, 0.17, 0.19, 0.18, 0.16, 0.15, 0.12, 0.10, 0.08. They peak around log H 2.1 and fall away after, so the straight run is roughly 1.5 to 2.4 and everything below is toe.
3. Where the shoulder begins, by the same test at the other end.
4. The highest density in the table, which on most of these conditions is not the film’s maximum density, because the strip has not reached it inside 3.9 log units. Say which it is.

📓First differences are the cheapest tool on this page

Subtracting each density from the next takes a minute a row and tells you three things a drawn curve tells you slowly and badly: where the toe ends, where the shoulder starts, and whether you have mis-transcribed a number. A difference that jumps out of sequence is nearly always a copying error, and finding it now saves re-drawing a curve later.

Task 3 — the speed point, and whether you may quote a speed

Use the course’s speed criterion exactly as written, and notice that it asks two questions rather than one.

Where is m? The exposure at which the density reaches D₀ + 0.10. Read it off the curve, then check it by interpolating in the table. For Film T in Developer H at 8 minutes: D₀ = 0.15, so m is at density 0.25, which the table gives at log H 0.60 exactly. Where it falls between two tabulated points, interpolate linearly.

Does the development qualify? Find the density 1.30 log units to the right of m and subtract. The criterion wants 0.80 ± 0.05. Continuing the same example, log H 0.60 + 1.30 = 1.90, where the curve reads about 1.01, and 1.01 − 0.25 = 0.76. Inside the window, so this development qualifies and a speed may be quoted for it.

S = 800 ÷ 10^{log H_m}

Speed from the criterion exposure, in millilux-seconds

Here S = 800 ÷ 10^0.60 = 800 ÷ 3.98 = 201, which rounds to the standard series value 200.

Now do it for all twelve. You will find that most of them do not qualify, and that is the exercise, not a flaw in the data. Where the rise over 1.30 falls short of 0.75, the strip is under-developed for the criterion; where it exceeds 0.85, it is over-developed. In both cases the exposure at m is still a perfectly good relative speed — it is the exposure needed to reach a stated density above a measured base — and you may compare it with another figure from the same table. What you may not do is call it a speed under the criterion, because the criterion includes the development condition and this development is not it.

✕The commonest wrong sentence a reader writes here

“Film T is ISO 200.” No page of this course may write that, and neither may your report.

Three separate reasons stack up. The course does not purchase the standards and quotes no part of them, so it cannot attribute a figure to one. The number came from a criterion the course states itself, modelled on ISO 6 and cited by number. And this particular number came out of a table generated from a model, so it is not a measurement of anything at all.

The sentence that is permissible: “Film T in Developer H at 8 minutes gives a speed of 200, measured under the criterion defined in Part XIII, which is modelled on ISO 6, from generated data.” Long, and every clause is load-bearing.

Task 4 — three slope measures on every curve

The slope lesson built three and warned that they disagree. Compute all three for at least four of the twelve conditions, and the contrast index for all twelve.

Gamma is the gradient of the straight run you identified in task 2. Take it over at least 0.90 of log exposure — three tabulated steps — and say which two points you used.

Average gradient over the whole strip, Ḡ, is the last density minus the first, divided by 3.60. It is the easiest of the three and the least comparable, and its value here is that it drops out of the table with no drawing at all.

Contrast index under the course convention needs the one-unknown solve. Let ΔD be the density difference between the two chord points; then

D₁ = D₀ + 0.1 ΔD and D₂ = D₀ + 1.1 ΔD

The two chord points

log H(D₂) − log H(D₁) = √(4 − ΔD²)

The condition that fixes ΔD

CI = ΔD ÷ √(4 − ΔD²)

And the answer

Three trials and an interpolation will get you to within 0.01. Worked on Film T in Developer H at 8 minutes, where D₀ = 0.15:

Trial ΔD	D₁	D₂	log H at D₁	log H at D₂	Actual run	Required √(4 − ΔD²)	Verdict
0.90	0.240	1.140	0.562	2.118	1.556	1.786	run too short, raise ΔD
1.10	0.260	1.360	0.621	2.520	1.899	1.670	run too long, lower ΔD
1.00	0.250	1.250	0.600	2.312	1.712	1.732	very slightly short
1.02	0.252	1.272	0.604	2.351	1.747	1.720	very slightly long

The answer lies between the last two, at ΔD ≈ 1.01, and

CI = 1.01 ÷ √(4 − 1.01²) = 1.01 ÷ 1.726 = 0.585

The contrast index of this condition

which is 0.58 to the two decimals worth quoting. Notice how little the last two trials moved the answer: two hundredths of ΔD moved CI by 0.016, so a trial-and-interpolate to within 0.02 of ΔD is plenty and there is nothing to be gained by a fourth trial.

🧮Why the arithmetic beats the straightedge here, and why you should still do both

The graphical construction — a marked paper edge slid until 0.20 sits on the base line and 2.20 touches the curve — is faster and it is what Kodak’s workbook prints. On a hand-drawn curve it will give you 0.57 or 0.59 where the table gives 0.58, and the spread between two people doing it is comfortably 0.02.

Do it graphically on four of the twelve to feel where the uncertainty lives, then do all twelve arithmetically. The reason the course’s convention is written as arithmetic rather than as an instruction is precisely that two people working from the same table then get the same number, and a convention that produces two numbers is two conventions.

Task 5 — tabulate, then plot slope and speed against time

Now the assignment turns from reading curves into reading a trend, which is what the whole apparatus was built for.

Make one table: twelve rows, and columns for film, developer, time, D₀, log H at m, relative speed, the rise over 1.30, γ, Ḡ and CI. Then two plots, both with development time along the bottom.

Time against contrast index, four curves, one per film-and-developer pair. This is the graph the whole part exists to let you draw. It answers a question no datasheet can answer for you: how long do I develop to get the contrast I want? Read it backwards — pick the contrast index, go across to the right curve, drop down to the time.

Time against relative speed, the same four pairs. Notice that this one is nearly flat, and notice how much flatter it is than the contrast plot. Between the shortest and longest development, Film M in Developer H moves from a relative speed of 93 to about 111 — about a quarter of a stop — while its contrast index moves from 0.39 to 0.67, which is nearly a doubling. Development is a contrast control that does a little to speed on the side, not a speed control. Hurter and Driffield said the same thing from the other end in the 1890s, when they came to define the inertia of a plate by the smallest exposure producing the slightest deposit, on the ground that very slight deposits are little altered by continued development.

Time against contrast index, all four combinations

• Film M in Developer S
• Film M in Developer H
• Film T in Developer S
• Film T in Developer H

Show the numbers behind this plot

Four rising lines of contrast index against development time. The two Developer H combinations run from about 0.40 at five minutes to about 0.67 for Film M and 0.75 for Film T at thirteen minutes, so they reach a given contrast in roughly half the time. The two Developer S combinations run from about 0.39 and 0.43 at nine minutes to 0.63 and 0.69 at twenty-two minutes. A horizontal line marks a target contrast index of 0.58, and it crosses the four curves at about eighteen minutes for Film M in Developer S, ten minutes for Film M in Developer H, fifteen minutes for Film T in Developer S and eight minutes for Film T in Developer H. Every curve is still climbing at its longest time, but the spacing between the plotted points narrows, which is the flattening the Sheppard and Mees relation predicts.

Series	Development time (minutes)	Contrast index, course criterion
Film M in Developer S	9.00	0.39
Film M in Developer S	14.00	0.52
Film M in Developer S	22.00	0.63
Film M in Developer H	5.00	0.39
Film M in Developer H	8.00	0.53
Film M in Developer H	13.00	0.67
Film T in Developer S	9.00	0.43
Film T in Developer S	14.00	0.56
Film T in Developer S	22.00	0.69
Film T in Developer H	5.00	0.42
Film T in Developer H	8.00	0.58
Film T in Developer H	13.00	0.75

Generated data. Draw this yourself from your own twelve contrast indices before you look at it; the point of the graph is the reading you take off it, and the reading is only worth anything if the curve is yours. 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.

Two readings to take off your own version, and to write down.

The development time for a contrast index of 0.58 in each of the four combinations. Interpolate between your plotted points; do not extrapolate beyond them, and say so where the answer would need you to.

The development time at which each combination first satisfies the criterion’s development condition, the rise of 0.80 over 1.30. For one of the four pairs that time lies beyond the longest development you were given, and the honest answer is to say that the data does not reach it and to name roughly how far short it falls.

Task 6 — compare the two developers, using only what you were given

This is the analytical heart of the assignment, and the rule is in the title: conclusions come from the tables in front of you, not from what you have read about developers elsewhere. Write down what the numbers show, then and only then compare it with the mechanisms Part VIII gave you.

Three findings are in the data. Find them before reading on.

Developer S costs speed. Compare like with like — the same film, the two developers, at developments giving about the same contrast index. Film M reaches CI 0.52 at 14 minutes in Developer S with a relative speed near 68, and CI 0.53 at 8 minutes in Developer H with a relative speed near 101. That ratio is 101 ÷ 68 = 1.49, which is 0.17 in log terms, or a little over half a stop. It is the same size as the difference ILFORD publish between the meter settings they list for the same film in PERCEPTOL and in ID-11, and it is the sensitometric shape of the mechanism Part VIII gave: the marginal speck dissolved before it could develop anything.

Developer H makes a shoulder. Look at the top end of each curve rather than the bottom. Take the last three tabulated differences of each row. In Developer S at 22 minutes, Film T runs 1.85, 2.04, 2.21, 2.36 — differences of 0.19, 0.17, 0.15, still nearly straight. In Developer H at 13 minutes, the same film runs 2.00, 2.15, 2.27, 2.38 — differences of 0.15, 0.12, 0.11, distinctly flattening. The high-acutance developer is running out of steam where the density is high, and only there, which is what compensation is: local exhaustion at highlight scale.

The toe is a property of the film, not of the developer. Take first differences at the bottom of each row this time. Film T at 8 minutes in Developer H begins 0.08, 0.14, 0.18 — it is at nearly full slope by the third step. Film M at the same development begins 0.03, 0.06, 0.10 and needs six steps to reach the same rate. Now change developer and check the ordering holds: in Developer S at 14 minutes, Film T begins 0.04, 0.09, 0.14 and Film M begins 0.02, 0.03, 0.07. The developer moved both films; it did not swap them over. Development sets the slope; the emulsion sets the shape of the foot; and exposure decides where on that foot your shadows land.

⚗Why the toe belongs to the emulsion

The toe is a picture of the distribution of latent-image centres across the grain population. At the lowest exposures only the few grains that happened to catch enough photons carry a developable centre, so density climbs slowly; as exposure rises, more of the population crosses the threshold and the climb accelerates until nearly every grain that will develop, develops. That distribution is set when the emulsion is made — by grain size, by the spread of grain sizes, by chemical sensitisation — and Part IV’s grain lesson and Part V’s precipitation lesson are where it comes from.

A developer can shift the threshold a little, which is what a solvent developer does by dissolving marginal specks and what a restrainer does by raising the barrier to development. What it cannot easily do is redistribute the population. That is why the toe moves less than anything else on the curve, and why the oldest quantitative reading in photography — the threshold step — is a reading about speed rather than about contrast.

Task 7 — predict the print

Take one subject: seven stops between the important shadow and the important highlight, which is 7 × 0.301 = 2.11 in log luminance. For each of your twelve conditions, compute the negative density range it would deliver.

Negative density range ≈ CI × subject log-luminance range

Negative density range

Then take that figure to ILFORD’s ISO Range table. Their instruction is explicit: multiply the effective negative density range by 100 and choose the nearest range figure. For MULTIGRADE RC DELUXE the published figures run 160, 130, 110, 90, 70, 60, 50 for filters 00 to 5, with 90 unfiltered.

Worked on Film T in Developer H at 8 minutes: 0.58 × 2.11 = 1.22, which is 122, nearest to the range figure of 130 — filter 0. At 13 minutes: 0.75 × 2.11 = 1.58, which is 158, nearest to 160 — filter 00, the softest filtration the paper has, and you have spent it on a subject of ordinary range. At 5 minutes: 0.42 × 2.11 = 0.89, which is 89, nearest to 90 — but 90 appears twice in that row, as filter 2 and as the unfiltered column, so read the filter column and say which you mean.

Tabulate all twelve, and then sit with the pattern, because it is not the one most readers expect. Every one of the twelve lands between filter 2 and filter 00 — the soft half of the paper’s range. The four that reach the makers’ own normal band of contrast index, 0.56 to 0.70, all land on filter 0 or 00, which is the softest filtration the paper has.

That is not an arithmetic slip. It is ILFORD’s own worked example seen from the other side: they take an effective density range of 1.32 and send you to the range figure of 130, and 1.32 is exactly what a seven-stop subject gives at a contrast index of 0.62. The middle grades are not for ordinary subjects developed normally. They are for longer subjects, or for lower contrast indices. Filter 2 has a range figure of 90, which a seven-stop subject reaches only at CI 0.43 — or which a ten-stop subject reaches at CI 0.30, or a five-stop subject at CI 0.60.

The practical consequence is worth writing on the front of the notebook. A grade is a match between two ranges, and the subject is one of them. A photographer who develops every roll to one contrast index and then complains that they always print on the same filtration has discovered nothing about their paper; they have discovered that they always photograph subjects of about the same range.

!Two things this arithmetic is not

It is not ILFORD’s advice, and it is not Kodak’s. Joining a film contrast index to a paper range figure is this course’s own inference from two manufacturers’ documents that were not written to be read together. The arithmetic is sound and the practice is ordinary, but the course says whose inference it is.

And ILFORD’s figure is not the number your densitometer reads. They specify the effective density range of the image as projected on the enlarger baseboard, which already contains your enlarger’s flare and its Callier effect. A condenser head projects a longer range than a diffusion head does from the same negative — about a grade’s worth, by ILFORD’s own account — so the same negative wants a different filter on the two machines. The paper lesson later in this part takes that up properly.

Task 8 — write the recommendation

One page. Not two, and not half. It is addressed to somebody who will act on it, which changes how it has to be written.

It must contain, in this order:

1. The intention. What kind of negative, for what kind of subject, printed how. “A negative for seven-stop daylight subjects, printed on variable-contrast RC paper under a diffusion head, aiming to land on filter 2 or 3 so that there is a grade in hand at both ends.” An intention is not a preference; it is the thing the numbers are measured against.
2. The recommendation. Film, developer, dilution as given, temperature, agitation and a time. One time, not a range.
3. The numbers that justify it, with their uncertainties. “Contrast index 0.58 ± 0.02, relative speed 200, negative density range 1.22 for a seven-stop subject.”
4. What you could not determine, named. Every honest technical report has this section and most bad ones do not.
5. The one measurement that would settle it, which for every recommendation on this page is the same one: your own step wedge, in your own developer, read on your own strip.

What to write it on

The course’s characteristic-curve record carries the fields this assignment asks for, and sheet SN-3 in particular is laid out for exactly the work of tasks 3 and 4: the criterion point, the rise over 1.30, the four trial rows for the contrast-index solve, and the three slope figures each with the construction that produced it written beside it. Print one per condition and you have twelve sheets that add up to the table task 5 wants.

Two of its sheets do not apply here and are deliberately left blank on a paper exercise: SN-1 records the light source and the wedge that made a strip, and SN-2 records twenty-one readings taken off one. Fill them in for the first time on the next page, where the strip is yours.

!There is no sensitometry calculator, and this page will not pretend otherwise

The course’s design lists a sensitometry calculator among its tools. It does not exist, and no page here links to one.

The gap is small on purpose. Every calculation in this assignment has been reduced to closed form so that a spreadsheet does it in one column and a pocket calculator does it in a minute: the speed is 800 ÷ 10^{log H}; the contrast index is ΔD ÷ √(4 − ΔD²) once ΔD is found by three trials; the negative density range is a contrast index times a subject range. A convention written as arithmetic does not need a tool, which is one of the reasons the course wrote it that way.

If the calculator is built later it will do this arithmetic and no more, and it will show its working.

📓What to keep

Keep the four plotted families, the twelve-row table, the two trend plots and the one-page recommendation. Date them, and write on the front of the table that the data is generated and where the model came from — because in six months you will find the table before you find this page, and a sheet of numbers with no provenance is exactly the thing this course exists to argue against.

You will want them again twice: on the step-wedge lab next, where you do all of this to numbers you measured, and in Part XV, when a real densitometer reads that strip again and you get to find out how good your interim reading was.

The model these tables came from

Published so that you can regenerate every number, check the arithmetic, and disagree with the modelling choices. This is not how film works; it is a function that has the right shape.

Density is the sum of a base term and two smooth ramps, the second subtracted from the first to make the shoulder:

D(x) = D₀ + γ · [ s(x − x_t, a) − s(x − x_s, b) ]

The model

s(z, a) = (1/a) · ln(1 + e^{a z})

The smooth ramp

x is log exposure. s is a ramp that sits at zero for negative z and rises with unit slope for positive z, bending over a width of about 1/a in between. So the first ramp makes the toe and sets the straight-line slope, the second bends the top over into a shoulder, and the four position and sharpness parameters do what their names say.

• D₀ is base plus fog. It is taken as a film-base constant plus a fog term proportional to development time, because fog grows with development.
• γ is the straight-line slope, taken from the relation Sheppard and Mees derived in 1907: γ = γ_∞(1 − e^−kt), with γ_∞ the ultimate development factor and k a velocity constant depending on the plate, the temperature and the developer.
• a is the toe sharpness and b the shoulder sharpness, both properties of the model film. Film M has the softer toe; Film T the harder one and the longer shoulder.
• x_t is set for each condition so that the criterion point lands where the model intends, and x_s = x_t + span, the span being a film property reduced for Developer H to model compensation.

Four modelling choices are worth naming because they are the ones that carry the teaching.

The speed point drifts by 0.05 in log exposure per doubling of development time, so that a longer development buys about a sixth of a stop. That is a modelling decision, chosen to be consistent with two sources that agree the toe is little altered by continued development: Hurter and Driffield’s account of very slight deposits, and Kodak’s observation of a real family of curves that most of the change is in the straight line and the shoulder.

Developer S is shifted 0.20 in log exposure to the right of Developer H, a speed loss of two-thirds of a stop. That figure was chosen to match the size of the difference ILFORD publish between the meter settings for one film in PERCEPTOL and in ID-11.

Developer H’s shoulder is brought forward by 0.55 log units relative to Developer S’s, which is how compensation appears on a curve. Part VIII’s compensation lesson is where the mechanism lives.

The two films differ only in toe sharpness, shoulder sharpness and speed. Everything else about them is identical, so that any difference you find between them is traceable to one of those three.

=What a real dataset would have looked like instead

Two things would replace this model, and both are within reach.

Published curves, digitised. Kodak’s T-Max 100 and Tri-X sheets and ILFORD’s film sheets print characteristic curves and contrast-index-against-time curves. Read off a printed graph, they would become a table of read-off approximations with a stated reading uncertainty — realistically ±0.02 in density and ±0.03 in log exposure, which is two to three times the uncertainty of the instrument that made them. That is a legitimate dataset if every number carries the label.

The course’s own measurements, which is what the next page starts. Twelve conditions is four sessions of darkroom work, and when this course has run them the tables here will be replaced and this section will say so.

Recap

• The data on this page is generated from a published model and describes no real material. The arithmetic is the transferable part; the numbers are not.
• Plot base plus fog first, from the control strip, and take it from the control strip every time.
• Draw a curve through the scatter, not through every point. With ±0.01 on each reading, a curve that visits every dot has fitted the noise.
• The speed criterion asks two questions: where the density reaches 0.10 above base, and whether the development qualifies. Most of these twelve conditions do not qualify, and the exposure at m is still a relative speed when they do not.
• Compute the contrast index arithmetically — three trials and an interpolation on ΔD — and graphically at least once, to feel where the 0.02 of spread lives.
• The time-against-contrast-index plot is the deliverable. Read it backwards to get a development time for a contrast you want.
• Development moves contrast a great deal and speed a little. Across these families, contrast index nearly doubles while relative speed moves by a third of a stop.
• A solvent developer costs speed; a compensating developer builds a shoulder; the toe belongs to the film. All three are visible in the tables, and all three have mechanisms in Part VIII.
• A negative density range is a contrast index times a subject range, and matching it to a paper’s published range figure is the course’s own inference from two manufacturers’ documents.

Check your understanding

Question 1. On one of the supplied conditions the control strip reads 0.17 and the curve reaches density 0.27 at log H 0.90, and at log H 2.20 the curve reads 1.15. May a speed be quoted for this development under the course criterion, and what is it?

• Yes; S = 800 / 10^0.90 = 101
• No; the rise from m over 1.30 log units is 0.88, which is above the 0.80 plus or minus 0.05 window, so the development is outside the criterion
• Yes; S = 800 / 10^2.20 = 5
• No; the criterion needs a density of 0.10 absolute, not 0.10 above base

Show the answer and why

Answer: No; the rise from m over 1.30 log units is 0.88, which is above the 0.80 plus or minus 0.05 window, so the development is outside the criterion

The criterion point m is correct: 0.17 plus 0.10 is 0.27, found at log H 0.90. But the criterion is a pair of conditions, and the second one fails. The density at m plus 1.30 is 1.15, and 1.15 minus 0.27 is 0.88, where the criterion allows 0.80 with a tolerance of 0.05, so anything above 0.85 is overdeveloped for the purpose. The exposure at m is still a perfectly usable relative speed for comparing this strip with another read the same way; what it is not is a speed under the criterion. Reporting it as one would quietly drop the development condition, which is the whole reason the criterion produces comparable numbers.

Question 2. You are solving for the contrast index on a curve whose base plus fog is 0.12. You try a density difference of 1.20 and find the two chord points fall 1.48 log units apart. What do you do next?

• Accept it; the contrast index is 1.20 divided by 1.48, which is 0.81
• Lower the trial density difference, because the required run at 1.20 is the square root of 4 minus 1.44, which is 1.60, and the actual run of 1.48 is shorter than that
• Raise the trial density difference, because a longer chord always gives a better fit
• Nothing can be decided without knowing the development time

Show the answer and why

Answer: Lower the trial density difference, because the required run at 1.20 is the square root of 4 minus 1.44, which is 1.60, and the actual run of 1.48 is shorter than that

The condition is that the chord must be exactly 2.00 units long, so its horizontal run has to be the square root of 4 minus the density difference squared: at a difference of 1.20 that is the square root of 2.56, which is 1.60. The run you measured is 1.48, shorter than required, which means the chord you drew is under 2.00 long and the trial difference is too large. Lower it and try again. The direction confuses people because both the numerator and the requirement move with the trial value, so it pays to write the required run beside the actual one in a column and let the sign of the difference tell you which way to go.

Question 3. Between its shortest and longest development in the same developer, a film moves from contrast index 0.39 to 0.67 while its relative speed moves from 93 to 111. What is the single most useful conclusion for a photographer?

• Longer development is a reliable way to gain film speed
• Development is a contrast control, and the speed change that comes with it is a third of a stop and not worth planning around
• The film is defective, because speed should not change with development at all
• The contrast index measurement must be wrong, since speed and contrast always move together

Show the answer and why

Answer: Development is a contrast control, and the speed change that comes with it is a third of a stop and not worth planning around

Contrast index rises by 72 per cent while speed rises by 19 per cent, which is 0.08 in log terms or about a quarter of a stop. Two sources ninety years apart explain why: Hurter and Driffield came to define the inertia of a plate by the smallest exposure producing the slightest deposit, precisely because very slight deposits are little altered by continued development, and Kodak observe of a real family of curves that most of the change is in the straight line and the shoulder while the toe remains basically the same. Practically this is the sentence behind all push processing: extending development raises midtone contrast and does very little for the shadows that were never exposed, which is why a pushed negative looks contrasty and still has empty shadows.

Question 4. Your subject measures six stops between the important shadow and the important highlight. You want to print it on a paper whose published ISO range figure at your intended filter is 110. What contrast index should you develop to?

• CI = 0.61
• CI = 1.10
• CI = 0.55
• The question cannot be answered without the film speed

Show the answer and why

Answer: CI = 0.61

Six stops is 6 times 0.301, which is 1.81 in log luminance. A range figure of 110 means the paper takes 1.10 log exposure units. Since the negative density range is about the contrast index times the subject log-luminance range, the contrast index wanted is 1.10 divided by 1.81, which is 0.61. Two caveats travel with the answer. Joining a film contrast index to a paper range figure is this course own inference from two manufacturers documents that were not written to be read together. And ILFORD specify the effective density range as projected on the enlarger baseboard, which already includes the flare and the Callier effect of your enlarger, so the figure your densitometer reads off the negative on a light box is not quite the same quantity.

Question 5. Why does this page insist the tables are synthetic rather than presenting them as approximate manufacturer data?

• Because synthetic data is more accurate than digitised data
• Because presenting generated numbers as measurements would attribute to real materials a behaviour nothing measured, and the course rule is that a gap with an honest note beats a confident error
• Because manufacturers forbid the reproduction of their curves
• Because the arithmetic only works on synthetic data

Show the answer and why

Answer: Because presenting generated numbers as measurements would attribute to real materials a behaviour nothing measured, and the course rule is that a gap with an honest note beats a confident error

The distinction is about provenance, not accuracy. A digitised manufacturer curve would be real data with a stated reading uncertainty, and would have been the better choice had the corpus held twelve such curves for two films in two developers at three times; it does not. What is never acceptable is generating numbers and letting a reader take them for measurements, because everything downstream then inherits a claim about a material that nobody made. The same rule is why this page also says what a real dataset would have looked like, and why the step-wedge lab that follows exists at all.

Question 6. Comparing the last three density differences of a strip developed in the high-acutance Developer H with the same film in the solvent Developer S at similar contrast, you find H flattening and S nearly straight. What has the developer done, and where would you look to confirm it is not simply a shorter development?

• H has fogged the film; look at the control strip
• H has built a shoulder by local exhaustion where density is high; confirm by checking that the toe and the mid-scale slope are comparable between the two, so only the top of the curve differs
• H is more dilute, so every part of the curve is lower; confirm from the base plus fog
• Nothing can be concluded from differences at the top of a strip

Show the answer and why

Answer: H has built a shoulder by local exhaustion where density is high; confirm by checking that the toe and the mid-scale slope are comparable between the two, so only the top of the curve differs

Compensation is local exhaustion at highlight scale: where the image is dense the developer in that small volume of solution is consumed and the reaction slows, while in the thin parts it never runs short. Its signature is therefore selective, appearing at the top of the curve and nowhere else. A shorter development would lower the gradient everywhere including the mid-scale, and a fog difference would lift the base, so the way to tell them apart is to check that the toe and the straight-line slope match while only the shoulder differs. That is also why compensation is worth having: it holds the highlights back without flattening the shadows, which no amount of cutting the time can do.

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