Curved Film Plane and Multiple-Pinhole Experiments

Bend the paper into a cylinder centred on the hole and two of the three cosines in the falloff law disappear — but only in one direction, and the price is paid in the shape of every straight line that is not vertical. This session predicts all of that on paper first, then photographs the same doorway twice to find out whether the prediction was right.

Purpose

Two experiments in one session, sharing a camera, a scene and a processing run.

Experiment 1, flat against curved. The hypothesis: a cylindrical back of radius equal to the focal distance, with the pinhole on the cylinder’s axis, removes the inverse-square term and the film-obliquity term from the illumination law across the wrap only, leaving a single cosine there and the full fourth power in the axial direction; and that on the same cylinder, the only world lines rendered straight are those whose plane through the pinhole contains the cylinder’s axis, plus the one horizontal at the height of the pinhole. The control is the flat-back negative: same scene, same viewpoint, same plate, same exposure time, processed in the same solutions in the same session. The variable is the shape of the recording surface and nothing else.

Experiment 2, more than one hole. The hypothesis: two pinholes a distance s apart put two complete images on the film, displaced by s(1 + f/u), with a doubled-exposure band where the two image circles overlap and single-exposure strips at the ends. The control here is a single-hole negative of the same scene at the same exposure; the variable is the number and separation of the holes.

Every prediction is written before the sheet goes in the camera. That is not a style preference. A prediction written after the negative is dry is not a prediction, and the reconciliation paragraph at the end of the session — where the numbers you wrote meet the numbers you read — is the actual output of this page.

Learning objectives

By the end you will be able to:

• Derive which factors of the cos-fourth law survive on a cylindrical film surface concentric with the pinhole, and in which direction each survives.
• Predict, before exposure, which straight lines in a scene will render straight on a curved film plane and which will bow, from the geometry rather than from experience.
• State the angular scale of a cylindrical projection and say how it differs from the flat plane’s tangent scale.
• Compute the image displacement from two pinholes of known separation and the width of the overlap band on a sheet of known width.
• Predict the falloff asymmetry produced by an off-centre pinhole used as shift.
• Estimate a density difference in stops by eye against a step ladder, and state the limits of that method.
• Write a reconciliation that says which predictions held, which failed and by how much.

Prerequisites

Commissioning the pinhole camera, which must have passed: a camera that leaks will put a gradient across a sheet and you will read it as falloff. The geometry comes from pinhole geometry and field of view and the curved insert from designing the modular camera; the exposure arithmetic from pinhole exposure and reciprocity correction. You will also need the curved insert built in stage 9 of the build, and at least one two-hole plate made by the method of making and measuring pinholes.

Safety classification

Level A, and identical in kind to the commissioning session, which is why this section is short.

• Substances. Working-strength paper developer, citric acid stop at 1+19 and ammonium thiosulfate rapid fixer at 1+4, at their makers’ dilutions, in volumes of a litre or less.
• Energies. None indoors beyond a safelight; outdoors, the sun.
• Procedures. Trays and tongs; a tripod outdoors.
• Waste. Used fixer is silver-bearing and is collected.
• New this session. Drilling one or two extra pinhole plates, which carries the needle and shim precautions of the pinhole lab: a plate held in a vice or against a block rather than in the hand, and swarf collected rather than brushed off with a palm.

What is not a hazard here, and why. Nothing generates a vapour, a gas or a dust: the solutions are cold, dilute and aqueous, the stop is citric rather than acetic and its maker describes it as low odour, and the only solid handled is a piece of shim brass that is cut, not abraded. The silver stays locked in hardened gelatin until the fixer dissolves it, which is the whole reason the fixer is the one bath that is collected. Bending paper is not a hazard either, but it is worth naming what it does do: a sheet under tension in a curved back can spring out when the back is opened, and it will take a fingernail across the emulsion with it if your hand is in the way.

Hazards

Paper developer. Hydroquinone’s aggregated GHS classification includes skin sensitisation and serious eye damage; the working solution is alkaline. Tongs and gloves, as ILFORD instruct.

Stop and fixer. Mild irritants at working strength, and eye irritants in particular.

Cross-contamination. A trace of fixer in the developer gives inconsistent results or blank prints. On this page the whole method is comparison between sheets, so a drifting bath does not spoil one negative, it spoils the experiment.

Outdoors. Sun exposure over a long session; a tripod in a line of travel; and, if the scene includes the sun near the frame edge, the instruction from the commissioning page stands — aim by the camera’s shadow and never sight along the axis towards the sun.

Sharp shim. Cut edges on 0.05 mm brass are sharp in the way paper is sharp. Deburr them, and collect the offcuts in a closed tin rather than loose in a bin.

Required PPE

• Nitrile gloves for wet work; HSE’s COSHH essentials sheet for manual film and plate development takes single-use nitrile as splash protection where the SDS gives nothing more specific.
• Eye protection from before the first bottle is opened, and again when a needle is being pushed through metal.
• Print tongs, one pair per tray, marked and never swapped.
• Outdoors, sun protection is the PPE: a hat, sleeves and sun cream, for a session that is mostly standing still in the open.

Ventilation

An openable window or an extractor in the processing room, and not a sealed cupboard, which is ILFORD’s own general instruction for darkroom areas. Extraction is not among the controls here because none of the three baths generates a vapour that needs it. Air movement is present for the ordinary reason: a small dark room with three trays becomes stuffy, and this session’s whole method depends on a person who is still paying attention at sheet fourteen.

Materials

Item	Quantity	Notes
Variable-contrast RC paper	14 sheets	Twelve for the plan, two spare. Fibre paper is easier on the tight curve; see the note in the preparation.
Shim brass, 0.05 mm	2 small pieces	For the two-hole and three-hole plates.
Tracing paper or drafting film, larger than your sheet	2 sheets	For the line-tracing overlay.
Fine permanent pens, two colours	2	The overlay must distinguish the two negatives without relying on colour alone, so use a solid line for one and a dashed line for the other.
Black card	1 sheet	Masks for opening the holes in sequence.
Black photographic tape	1 roll
A printed step ladder on your own paper	1	Made as in the commissioning session, in half-stop or one-stop steps.
Pencil and a printed prediction sheet	—	Predictions on the left, observations on the right, with a fold so you cannot see one while writing the other.

Chemicals

Chemical	Quantity	Form
Paper developer concentrate	100 mL	Diluted 1+9 to make 1 L. ILFORD’s MULTIGRADE developer is used at 1+9 for 1 minute at 20 °C.
Stop bath concentrate	50 mL	Diluted 1+19 to make 1 L; 10 seconds at 18–24 °C. ILFOSTOP is a citric acid stop.
Rapid fixer concentrate	200 mL	Diluted 1+4 to make 1 L; 30 seconds for RC paper. An ammonium thiosulfate fixer.
Water	about 15 L	Dilution and washing.

Those are ILFORD’s published figures for their own paper in their own developer. Use your own maker’s sheet if you are using another material, and write down which sheet you took the numbers from.

Equipment

The camera with its flat back and its curved insert, both frames, and the pinhole register; a tripod; a timer; three trays and a wash tray; a 1 L graduate; a thermometer; three pairs of tongs; a tested safelight; a loupe; a light box or bright window; a steel rule; and a needle, a block and a fine abrasive for making the multi-hole plates.

Estimated cost

Cost band £. Fourteen sheets of paper, two small pieces of shim and some tracing paper, plus chemistry you already have. Dated prices live in the laboratory planner.

Estimated consumables cost

Fourteen sheets of paper, two small pieces of shim, some tracing paper, and chemistry you already have. This is one of the cheapest experiments in the course to run a second time, which matters because the curved-plane comparison is worth repeating.

Consumed	This session	Sourced price	Cost this session
Variable-contrast RC paper	14 sheets: 12 for the plan, 2 spare	£16.06–£44.71 per 25 to 100 sheets, 5 x 7 in, variable contrast RC (£0.45–£0.64 a sheet)	£6.26–£8.99
Paper developer concentrate	100 mL, diluted 1+9 to make 1 L	£10.52–£20.03 per 500 ml to 1 L of concentrate, diluted 1+9	£2.00–£2.10
Stop bath concentrate	50 mL, diluted 1+19 to make 1 L	£10.66–£12.18 per 500 ml of citric acid concentrate, diluted 1+19	£1.07–£1.22
Rapid fixer concentrate	200 mL, diluted 1+4 to make 1 L	£21.05–£25.98 per 1 L of ammonium thiosulfate concentrate, diluted 1+4 for film	£4.21–£5.20
Shim brass, 0.05 mm	two small pieces, for the two- and three-hole plates	None. A named price gap: pinhole-making stock — shim, wet-and-dry abrasive paper, a pin vice and a bought laser-drilled pinhole	—
Tracing paper or drafting film	2 sheets, larger than your negative	None. card-and-paper-stock carries a cost band and no dated figure	—
Black card and black photographic tape	one sheet, a few strips	None. tape-and-adhesives carries a cost band and no dated figure	—
Water	about 15 L	Metered supply; the planner prices no water	—

The priced rows come to £13.54 to £17.51 for one run of this session, at the retail ranges read on 5 September 2026 and recorded in the laboratory planner. That is a floor, not a total: 3 of the 8 rows carry no dated price, so they are counted as nothing here and are certainly not free. A priced entry is a dated range to plan against, never a quotation.

The fine permanent pens, the steel rule, the needle, the block and the abrasive are equipment. Shim is a named price gap and is bought once for the whole of Part VI.

Waste streams

Used fixer is silver-bearing and is collected, with the first change of wash water, into the labelled container Part II established. Used developer and stop carry no silver. Brass offcuts and swarf go into a closed tin for scrap metal, not loose into a recycling bag. Fixed and washed paper is ordinary household waste. Local regulations govern, and they differ between authorities even within the United Kingdom.

Alternative route

Most of this experiment is done with a pencil. The predictions — the falloff derivation, the mapping of straight lines onto a cylinder, the scale along the curve, and where the second image from a second hole must land — are worked and written down before any paper is exposed, and they are where the physics is. That half needs nothing but the folded prediction sheet.

No curved insert. A strip of thin card sprung into the back against two glued stops makes the same cylinder as a moulded insert. Measure the chord and the sagitta and write both in the log, because the radius you compute from them is what every prediction on this page is keyed to; a curve you did not measure is a curve you cannot reconcile against.

No darkroom. Loading and unloading can be done in a changing bag, but the tray sequence cannot: a 5 × 7 inch three-tray line needs a space you can stand in. A windowless room at night with a red LED at the distance the commissioning session established is the usual answer. If you have no such space at all, do the whole prediction stage, expose the twelve sheets over as many days as you like — exposed paper keeps in a light-tight box — and process the batch in one borrowed or hired session. Stage 4 was written as a single batch for exactly this reason.

No two-hole plate. Two separate single-hole plates in two carriers, opened in sequence with the black card mask, produce the same double image and make the sequencing easier to see. You lose the simultaneous exposure and gain a cleaner test of which image belongs to which hole.

What cannot be substituted is the flat-against-curved pair at identical exposure. Everything the experiment concludes rests on the two negatives differing in one thing only.

Preparation

1. Derive the falloff before you build anything

This is the piece of arithmetic the whole page turns on, and the design lesson gave only its conclusion. Here is the derivation.

Put the pinhole at the origin, on the axis of a cylinder of radius R = f. Take a point P on the cylinder at wrap angle φ around the axis and at axial distance z along it, and define ψ by tan ψ = z/R. The three factors that made cos⁴ on a flat plane are still the same three, and each is now evaluated on this surface.

Distance. |OP| = √(R² + z²) = R/cos ψ. By the inverse-square law this costs cos² ψ, and notice what it does not contain: φ. Going round the cylinder does not move you further from the hole.

Aperture foreshortening. The hole is seen from P at the true field angle θ between the ray and the plate’s normal, and a short piece of vector algebra gives cos θ = cos φ cos ψ. So this factor is cos φ cos ψ, and it is the one that survives the wrap.

Film obliquity. The surface normal at P is radial, in the plane perpendicular to the cylinder axis. The ray direction dotted with that normal is R/√(R² + z²) = cos ψ, so this costs cos ψ — again with no φ in it, because on a cylinder concentric with the hole the light arrives along the surface’s own normal all the way round.

Multiply the three:

E(φ, ψ) = E(0) · cos φ · cos⁴ ψ

Illumination on a cylinder concentric with the pinhole

Read that carefully, because it is not what most accounts say. Across the wrap, at ψ = 0, it is plain cos φ: one cosine where a flat plane has four. Along the axis, at φ = 0, it is cos⁴ ψ: exactly the flat law, unimproved. And that is right, because a cylinder cut by a plane containing its axis is flat in that direction — there is nothing for the curve to fix.

🧮What that predicts for a 4 × 5 sheet at f = 50 mm

Wrap the 127 mm dimension on a 50 mm radius and leave the 101.6 mm dimension along the axis.

Point on the sheet	Flat back, cos⁴ θ	Curved back, cos φ · cos⁴ ψ	Curved advantage
Centre	0	0	—
Middle of the long (wrapped) edge	θ = 51.8°: 2.77 stops	φ = 72.8°, ψ = 0: 1.75 stops	1.02 stops
Middle of the short (axial) edge	θ = 45.5°: 2.05 stops	φ = 0, ψ = 45.5°: 2.05 stops	none
Corner	θ = 58.4°: 3.73 stops	φ = 72.8°, ψ = 45.5°: 3.80 stops	none — slightly worse

The wrapped edge is a stop better and the axial edge is not better at all. The corner is a fraction worse, and there is no contradiction in that: the wrapped sheet is covering 145.5° of horizontal angle against the flat back’s 103.6°, so its corner is looking much further off-axis for the same place on the paper.

The honest comparison is at equal field angle, not at equal position on the sheet, and that is what the plot below shows. The single most testable prediction on this page is the third row: the cylinder buys nothing vertically. If your curved negative is brighter at the top and bottom edges as well as at the sides, something other than the geometry is doing it.

What the cylinder is worth, in stops, against field angle

• Flat back: cos⁴ θ
• Curved back, axial direction: cos⁴ ψ — identical to flat
• Curved back, across the wrap: cos φ

Show the numbers behind this plot

Three curves of illumination loss in stops against field angle from zero to seventy-five degrees. The uppermost is the flat film plane's cos-fourth law: zero on the axis, 0.20 stop at 15 degrees, 0.83 at 30, exactly 2.00 at 45, 2.77 at 51.8 degrees which is the middle of the long edge of a 4 by 5 inch sheet at 50 millimetres, exactly 4.00 at 60, and 6.19 at 70. Lying exactly on top of it is a second curve, the curved back measured in its axial direction, which is identical at every angle because a cylinder is flat along its own axis and the full fourth power survives there. The third and much lower curve is the curved back measured across the wrap, a plain cosine: zero on the axis, 0.05 stop at 15 degrees, 0.21 at 30, exactly 0.50 at 45, 0.69 at 51.8, exactly 1.00 at 60, 1.55 at 70 and 1.75 at 72.8 degrees, the edge of a 127-millimetre sheet wrapped on a 50-millimetre radius. The vertical gap between the top pair and the bottom curve is what the cylinder is worth in the wrap direction: 0.62 stop at 30 degrees, 1.50 stops at 45, 2.08 at 51.8, 3.00 at 60 and 4.64 at 70. The point of drawing the axial curve on top of the flat curve rather than omitting it is that the cylinder's benefit is entirely one-dimensional.

Series	Field angle from the axis, degrees	Illumination loss, stops
Flat back: cos⁴ θ	0.00	0.00
Flat back: cos⁴ θ	10.00	0.09
Flat back: cos⁴ θ	15.00	0.20
Flat back: cos⁴ θ	20.00	0.36
Flat back: cos⁴ θ	30.00	0.83
Flat back: cos⁴ θ	40.00	1.54
Flat back: cos⁴ θ	45.00	2.00
Flat back: cos⁴ θ	51.80	2.77
Flat back: cos⁴ θ	55.00	3.21
Flat back: cos⁴ θ	60.00	4.00
Flat back: cos⁴ θ	65.00	4.97
Flat back: cos⁴ θ	70.00	6.19
Flat back: cos⁴ θ	72.80	7.02
Curved back, axial direction: cos⁴ ψ — identical to flat	0.00	0.00
Curved back, axial direction: cos⁴ ψ — identical to flat	10.00	0.09
Curved back, axial direction: cos⁴ ψ — identical to flat	15.00	0.20
Curved back, axial direction: cos⁴ ψ — identical to flat	20.00	0.36
Curved back, axial direction: cos⁴ ψ — identical to flat	30.00	0.83
Curved back, axial direction: cos⁴ ψ — identical to flat	40.00	1.54
Curved back, axial direction: cos⁴ ψ — identical to flat	45.00	2.00
Curved back, axial direction: cos⁴ ψ — identical to flat	51.80	2.77
Curved back, axial direction: cos⁴ ψ — identical to flat	55.00	3.21
Curved back, axial direction: cos⁴ ψ — identical to flat	60.00	4.00
Curved back, axial direction: cos⁴ ψ — identical to flat	65.00	4.97
Curved back, axial direction: cos⁴ ψ — identical to flat	70.00	6.19
Curved back, axial direction: cos⁴ ψ — identical to flat	72.80	7.02
Curved back, across the wrap: cos φ	0.00	0.00
Curved back, across the wrap: cos φ	10.00	0.02
Curved back, across the wrap: cos φ	15.00	0.05
Curved back, across the wrap: cos φ	20.00	0.09
Curved back, across the wrap: cos φ	30.00	0.21
Curved back, across the wrap: cos φ	40.00	0.38
Curved back, across the wrap: cos φ	45.00	0.50
Curved back, across the wrap: cos φ	51.80	0.69
Curved back, across the wrap: cos φ	55.00	0.80
Curved back, across the wrap: cos φ	60.00	1.00
Curved back, across the wrap: cos φ	65.00	1.24
Curved back, across the wrap: cos φ	70.00	1.55
Curved back, across the wrap: cos φ	72.80	1.75

Computed from the derivation above; nothing here was measured by this course, and the flat and axial curves are drawn as two series on purpose, because they coincide. That coincidence is the prediction this experiment is most likely to falsify if the insert's radius is not really equal to the focal distance. 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.

2. Predict the mapping of straight lines

A pinhole is a central projection. Every straight line in the world, together with the pinhole, defines a plane; the image of that line is wherever that plane cuts the recording surface.

On a flat plane, a plane cuts a plane in a straight line. That is the whole reason a pinhole on a flat back is rectilinear, and it is why the geometry lesson could say so without qualification.

On a cylinder, a plane cuts the surface in a conic — an ellipse in general — which is a curve, and stays a curve when the cylinder is unrolled. There are exactly two exceptions, and both come out of the same argument:

1. A plane containing the cylinder’s axis cuts it in two straight generators. The pinhole is on the axis, so this is the case for any world line whose plane through the hole contains the axis — in practice, with the camera level and the wrap horizontal, the verticals. They stay straight and they stay parallel.
2. The plane perpendicular to the axis through the hole cuts the cylinder in a circle, and a circle unrolls to a straight line. That is the one horizontal at the height of the pinhole, and it maps to the centre line of the sheet.

Everything else bows. A horizontal above the pinhole’s height bows one way, one below it bows the other, and the further from centre height, the more. This is the behaviour of every rotating panoramic camera ever made, and it follows from three sentences of geometry rather than from a rule of thumb.

A rectangular grid, projected onto the cylinder and then unrolled

1. 1Flat back — every straight line stays straight; horizontal spacing widens towards the edges as f tan θ
2. 2Cylindrical back, unrolled — verticals straight and parallel; the horizontal at pinhole height straight; all other horizontals bow
3. 3The centre line — the one horizontal that survives — the circle at the pinhole’s own height, which unrolls straight

The flat plane keeps the lines straight and stretches the edges by sec θ; the cylinder keeps the angular scale even and bends the lines instead. Neither is a distortion in the sense of a fault: both are exact central projections onto different surfaces.

📓Where this argument comes from, and where it does not

The derivations in this section are the course’s own, worked from the definition of a central projection and the intersection of a plane with a cylinder. They are elementary, and you should check them rather than take them.

This course has not obtained Eric Renner’s Pinhole Photography or the Pinhole Journal, which are the standard references for the history and practice of curved-plane and multiple-pinhole cameras, and it therefore makes no historical attribution on this page: no names, no dates, no claims about who built the first curved-back pinhole camera. Matt Young’s 1971 Pinhole optics is cited elsewhere in this part, and its abstract states that a pinhole camera’s angular field can be made to exceed 90°, but the course has read only that abstract and it says nothing about cylinders.

What the course does have from the period is Wall’s 1912 account of the plain box camera, which notes only that “the larger the plate the wider the angle and the greater the distance the larger the image”. That is the flat-plane trade, and it is exactly the constraint a cylinder is built to escape.

3. Predict the scale, and where the flat plane stretches

On a flat plane, a point at field angle θ lands at f tan θ from the centre; on the cylinder it lands at arc length f φ. So the cylinder has a uniform angular scale — equal angles occupy equal widths of paper, everywhere — and the flat plane does not.

The consequence people notice is the wide-angle stretch, and it has a number. On a flat plane the horizontal scale at field angle θ is f sec²θ per radian and the vertical scale is f sec θ per radian, so the horizontal is stretched relative to the vertical by a factor of sec θ: 1.41 at 45°, 2.00 at 60°. A face at the edge of a very wide flat-plane frame is widened by that factor, which is why it looks wrong. The cylinder removes that stretch entirely across the wrap, and pays for it by bending lines instead. That is the trade, stated as two numbers rather than as a taste.

4. Make the plates and check the curved insert

Pierce and mount the multi-hole plates by the pinhole lab’s method, and measure the separation as carefully as you measured the diameters — the separation is the whole experiment. A steel rule under a loupe against the scan is enough; record the uncertainty.

Then check the insert. Its radius must equal the focal distance you are going to use, because the derivation assumed R = f, and it must be seated so that the pinhole really is on the axis. Measure from the pinhole to the paper at the centre and at both wrapped edges: on a correct insert those three distances are equal. If the edges are further away than the centre, the radius is too large and the cosine advantage will be smaller than predicted.

🛠Which paper, and why the resin-coated sheet fights you

Resin-coated paper has a polythene layer on both faces and is noticeably stiffer than fibre paper. On a 50 mm radius it resists, and it will spring flat at the ends unless the insert holds it — the build page suggests a card tongue at each end of the arc. A sheet that has partly sprung flat is no longer at the focal distance across its whole width, so the blur and the local f-number both change, and the falloff you measure is not the falloff you derived.

Fibre-based paper takes the curve more willingly. If you have some, use it for the curved sheets, and then use it for the flat control sheets too, so the only variable is still the shape. Note that ILFORD warn against wet times over 15 minutes for RC papers because of edge penetration and curl; fibre paper has its own, longer, wash requirement, which Part XII owns.

5. Write the predictions down, then fold the sheet

Before anything is exposed, fill in the left column: the falloff in stops you expect at centre, mid and edge on both backs; which lines you expect to stay straight; the displacement you expect from the two-hole plate; the exposure difference you expect in the overlap band; and the falloff asymmetry you expect from the shift. Then fold the sheet so you cannot see them while you are reading the negatives. Confirmation bias is not a moral failing, it is a perceptual one, and the fold is the control for it.

Procedure

Stage 1 — Flat against curved, on one scene (60 minutes)

Choose an architectural subject with a strong grid in it: brickwork, a panelled door, railings, a tiled wall. It must contain true verticals, a horizontal at about the height you will put the camera, and horizontals well above and below that height.

1. Set the camera to f = 50 mm with your best plate, on a tripod, levelled. Levelling matters more here than anywhere else in the part: the whole prediction about which lines stay straight assumes the cylinder’s axis is vertical and the horizon passes through the hole.
2. Expose two sheets on the flat back at the computed time, one as the working negative and one as a spare.
3. Without moving the tripod, fit the curved insert and expose two sheets at the identical time.
4. Mark every sheet in pencil on the back as it comes out, with the back used and the time.

Why identical exposure and not “correct” exposure for each. The axial illumination is the same on both backs — E(0) does not care what shape the surface is — so an identical exposure should give an identical centre density, and any difference at the centre is a fault rather than a finding. It is the edges that are supposed to differ, and holding the exposure constant is what makes that difference readable.

Flat and cylindrical film surfaces from above, with the rays that decide the falloff

1. 1Flat plane: all three factors — distance f/cos θ costs cos²θ; oblique arrival costs cos θ; the foreshortened hole costs cos θ — total cos⁴ θ
2. 2Arc of radius f, centred on the hole — every point is exactly f away, so the distance factor goes; every ray arrives along the normal, so the obliquity factor goes
3. 3What is left — the hole is still seen at an angle from every off-axis point: cos φ, and nothing else

The cancellation is a property of the plane of this drawing only. Turn the drawing through ninety degrees and the cylinder is a flat surface again, which is why the axial direction keeps the full fourth power.

Stage 2 — Multiple pinholes (45 minutes)

1. Fit the two-hole plate, with the holes separated horizontally by a measured s. Photograph a scene with one strong isolated feature — a window, a signpost, a tree against sky — because two overlaid copies of a busy scene are unreadable.
2. Expose one sheet with both holes open, and, before that, one sheet each with one hole masked, so that you have the two single images separately as well as their sum. Black card and tape make the masks; open them in sequence without moving the camera.
3. Fit the three-hole plate and repeat with all three open.
4. Fit the off-centre single hole, or move the carrier to its shift stop, and photograph a tall building from ground level with the camera level, so the verticals stay parallel. Expose one sheet centred and one shifted, for comparison.

Stage 3 — The aesthetic reference set (20 minutes)

Two more sheets of one scene at f = 120 mm, with the same framing, changing only the plate:

Plate	N	Predicted total blur at f = 120 mm	Exposure relative to the 0.30 mm plate
0.15 mm, badly undersized	f/800	1.03 mm, 42 % above the best	×4
0.30 mm, near optimum	f/400	0.74 mm	×1
0.60 mm, badly oversized	f/200	0.82 mm, 13 % above the best	×0.25

Keep these. They are not a test of anything — the numbers are already known — they are a reference set of what too small and too large actually look like, and Part VII will ask you to choose between them for a picture rather than for a measurement. The undersized hole should glow; the oversized one should be plainly, evenly soft. That difference in character at almost the same blur width is the whole reason both are kept.

Stage 4 — Process everything together (40 minutes)

Twelve to fourteen sheets, one session, same solutions, same times: developer at 1+9 for 1 minute at 20 °C, stop at 1+19 for 10 seconds, rapid fixer at 1+4 for 30 seconds, wash 2 minutes in running water above 5 °C, all as ILFORD publish for MULTIGRADE RC. Include one unexposed control sheet, as always. Develop by the clock, never by inspection.

Expected observations

The two flat sheets should match each other, and so should the two curved sheets. If the pairs do not match, the session has a repeatability problem and the flat-against-curved comparison cannot carry any weight.

The centres of the flat and curved sheets should match in density. They received the same illumination for the same time.

The wrapped edges of the curved sheet should be about a stop brighter than the corresponding edges of the flat sheet — brighter on the negative means denser, because a negative is inverted — while the axial edges should match. That asymmetry is the signature of the derivation.

The curved sheet should show more scene, 145.5° against 103.6° at f = 50 mm, so it will include things the flat sheet does not. That makes the two images different pictures of the same place, which is exactly why the tracing overlay is needed to compare them.

Verticals should be straight on both. Horizontals should be straight on the flat sheet and bowed on the curved one, except the one at the height of the pinhole.

The two-hole sheet should show two complete images displaced by about s, with a brighter — that is, denser — band where they overlap.

What is happening chemically

Nothing new: the chemistry of this session is the chemistry of the commissioning session, and Part IV owns all of it. The developer amplifies whatever developable centres exist, whether they were made by image light, by a leak or by nothing at all; the stop halts it at a known instant; the fixer dissolves the halide that was never exposed.

The one point worth making here is about where the geometry and the chemistry meet. An illumination difference of one stop at the edge of a sheet is not one stop of density difference on the negative. The paper’s own characteristic curve decides how an exposure difference becomes a density difference, and near the ends of that curve — in the toe and in the shoulder — a whole stop of exposure can produce very little density change. So a curved back that really is delivering a stop more light at its edges may show much less than a stop of visible difference there, if the flat back’s edge had already fallen into the toe.

That is not an excuse for a failed prediction. It is a real, predictable effect, it works in one direction only, and it is the reason the reconciliation below asks for the difference in estimated stops rather than for a verdict. Part XIII turns this into a curve and makes it exact.

Data to record

The prediction sheet, both columns, and the fold intact until stage 4 is over.

Falloff table, six rows, filled in by eye against the step ladder:

Sheet	Centre	Mid wrapped edge	Mid axial edge	Corner
Flat, predicted	0	2.77	2.05	3.73
Flat, estimated
Curved, predicted	0	1.75	2.05	3.80
Curved, estimated
Difference, predicted	0	1.02	0	−0.07
Difference, estimated

Multiple-pinhole table: measured separation s with its uncertainty; subject distance u; predicted displacement s(1 + f/u); measured displacement on the dry negative; predicted overlap-band width; measured overlap-band width; the density difference between the overlap band and the single-image ends, in steps of the ladder.

Shift table: the offset used; the predicted field angle and falloff at each of the four corners; the estimated falloff at each corner; whether the verticals stayed parallel.

An exposure-log row per sheet, as always, the first ten fields written before the shutter opens.

Analysis

1. Trace the two negatives onto one overlay

This is the measurement that reads the distortion, and it costs nothing.

Tape the flat negative to a light box, lay tracing paper over it, and trace the principal lines: the verticals, the horizontal at camera height, and two horizontals well above and below it. Use a solid line. Then, without moving the tracing paper’s registration marks, replace the negative with the curved one — aligned on the same central vertical and the same centre point — and trace the same features with a dashed line.

Now read the overlay. The verticals should coincide in direction and diverge in spacing, because the two projections have different scales away from the centre. The centre horizontal should coincide. The upper and lower horizontals should be straight on the solid tracing and bowed on the dashed one, and the bow should grow with distance from the centre line. Write a paragraph describing the mapping in your own words; if you can describe it accurately, you understand the projection.

2. Reconcile the falloff, honestly

Fill in the estimated rows and compare with the predicted ones. Three outcomes and what each means:

• The wrapped edge is about a stop better and the axial edge is not: the derivation held. Say by how much it held — “predicted 1.02 stops, estimated one and a half steps of the ladder, which is 0.75 stop, within the resolution of the method” is a real result.
• Both edges improved. Something other than the geometry is at work. The likeliest candidates are that the flat sheet was not flat, or that the two exposures were not really identical, or that the light moved between them.
• Neither edge improved. Check the insert’s radius against the focal distance, and check that the paper actually held the arc rather than springing flat.

3. Reduce the multiple-pinhole result to one number

Measure the displacement between two identifiable features that are the same feature in both images. Compare with s(1 + f/u). Then propagate the uncertainty: if you measured s to ±0.5 mm and u to ±10 per cent, what is the uncertainty in the prediction, and does the measured value lie inside it? That is the calculation measurement and uncertainty exists for, and it is the difference between “it looked about right” and a result.

🧮Two holes, worked all the way through

Given. Two holes, s = 20.0 mm apart, at f = 60 mm, on a sheet 100 mm wide.

A distant subject, u = 30 m. f/u = 60/30000 = 0.002, so Δ = 20.0 × 1.002 = 20.04 mm, which for any practical purpose is 20 mm: at infinity the displacement is simply the hole separation.

The overlap band. Each hole throws a complete image; the two are offset by 20 mm. On a 100 mm sheet the region receiving both images is 100 − 20 = 80 mm wide, and there is a 10 mm strip at each end receiving only one. In the band, two images add, so the exposure there is twice the exposure at the ends: one stop.

A near subject, u = 600 mm. Now f/u = 0.1, so Δ = 20.0 × 1.1 = 22 mm, and the overlap narrows to 78 mm. The displacement grew by a tenth because the subject came ten focal distances close.

Three holes at the same 20 mm spacing. Images at 0, 20 and 40 mm. A central band 100 − 40 = 60 mm receives all three, and is three times the single exposure: 1.58 stops above one image. Two bands 20 mm wide, one at each end, receive two of the three, at one stop. Add those up — 60 + 20 + 20 — and they already fill the sheet: with three holes at this spacing on this width, no part of the negative receives only one image. The single-exposure region has been squeezed out of existence, and noticing that is the point of doing the arithmetic before the exposure rather than after it.

So what exposure do you give? Whichever band you want correct — but notice how much easier this is than it first looked. The step you are actually composing with is the 0.58 stop between the triple band and the two doubles, not the 1.58 stops you would face if a single-image strip existed. Expose for the triple band and the ends are 0.58 stop under; expose for the ends and the middle is 0.58 stop over. Either way it is well inside what a paper negative will carry, which is why three holes at a close spacing is a friendlier plate than two holes far apart.

The unifying observation. Δ = s(1 + f/u) is the blur formula b = d(1 + f/u) with s in place of d, and that is not a coincidence: a single hole of diameter d is a continuous set of pinholes spanning d, and its blur circle is the displacement between the two extreme ones. Drilling two holes 20 mm apart is making that displacement large enough to read as two pictures rather than as softness.

Two holes, three holes, and one hole moved off centre

1. 1Two holes, s = 20 mm, f = 60 mm — displacement 20 mm at infinity; 80 mm doubled band, 10 mm single strip each end
2. 2Three holes at 20 mm — 60 mm triple band (+1.58 stops), two 20 mm double bands (+1 stop), single ends
3. 3One hole, offset 20 mm — shift — image circle moves by 20 mm; verticals stay vertical; corner falloff becomes asymmetric, 1.47 against 0.78 stops

All three panels are the same geometry: the image circle belongs to the hole and goes where the hole goes. Δ = s(1 + f/u) is the blur formula with the hole separation in place of the hole diameter.

4. Read the shift result

With the hole 20 mm off centre at f = 120 mm on a 4 × 5 sheet, the four corners are no longer at one field angle. The near pair sit at arctan(97.7/120) = 39.2°, losing 1.47 stops; the far pair at arctan(66.9/120) = 29.1°, losing 0.78. A difference of 0.69 stop across the frame, which you should be able to see. Check that against the negative, and check the thing shift is actually for: with the camera level, the verticals of the building should be parallel, not converging.

5. Say what the method cannot do

Every density on this page was estimated by eye against a ladder calibrated in stops of exposure. That resolves about half a stop at best. The predicted difference at the wrapped edge is 1.02 stops, which is comfortably resolvable; the predicted difference at the axial edge is zero, which is a prediction you can only confirm to within the method’s half-stop. Measured densities wait for Part XV, and saying so is part of the result rather than an apology for it.

Troubleshooting

What you find	Likely cause	What to do
Curved and flat centres differ in density	The exposures were not identical, or the light moved between them	Re-shoot the pair back to back and re-meter between them
No falloff advantage anywhere on the curved sheet	The insert’s radius does not equal the focal distance, or the hole is not on the axis	Measure hole-to-paper at centre and both wrapped edges; they must be equal
Advantage at the top and bottom edges as well	Not the geometry: a leak, a flare gradient, or a flat control sheet that bowed	Re-run with a fresh control and check the flat back’s pressure plate
The curved sheet is sharp in the middle and soft at the ends	The paper sprang partly flat, so the ends are not at the focal distance	Card tongues at each end of the arc, or fibre paper
Verticals bow on the curved sheet	The camera was not level, so the cylinder axis was not vertical	Level it; this is the one prediction that a tripod head can falsify by itself
Two-hole images are displaced by more than s	The subject is nearer than you assumed	Compute s(1 + f/u) with the real distance before calling it an error
Two-hole images overlap almost completely	s is too small relative to the frame	A larger separation, or a nearer subject, which multiplies the displacement
The overlap band is not a stop denser	The paper’s toe or shoulder is compressing the difference	Read a mid-tone rather than a shadow or a highlight, and note the effect
The shifted sheet loses a corner entirely	The shift pushed the worst corner past the useful image circle	Reduce the shift, or use the longer focal setting where the corner angles are smaller

Clean-up

Pour used developer and stop away according to local rules; collect the used fixer and the first change of wash water into the labelled silver waste container. Rinse trays, graduate and tongs. Return the paper to its bag inside its box before the room light goes on. Take the curved insert out of the camera, wipe it, and check that no paper fibres are caught in the end tongues.

Storage

Dry sheets go into sleeves numbered to match the log rows. Keep the aesthetic reference set together and label it as such; Part VII asks for it by name, and a set that has been filed among ordinary negatives is a set you will not find. Keep the tracing overlay flat with the two negatives it was made from.

Working solutions: date every bottle you keep and write the number of sheets it has processed on the label.

Disposal considerations

Used fixer is silver-bearing, because thiosulfate has taken the unexposed halide into solution as a soluble complex, and silver is toxic to aquatic organisms. That is the reason for the labelled container, and Part XII teaches recovering the silver rather than merely handing it over. Developer and stop carry no silver; the developer is alkaline and contains a sensitiser, and the stop is a weak organic acid. Brass swarf is scrap metal.

Local regulation governs disposal and it differs between authorities, even within the United Kingdom. ILFORD’s guidance for UK domestic users is that local authorities usually accept small quantities of chemical waste at Household Waste and Recycling Centres, bottled separately and labelled, and that different wastes are never mixed for disposal.

Questions

1. Derive, in your own words, why the film-obliquity factor vanishes on a cylinder concentric with the pinhole but the aperture-foreshortening factor does not.
2. Your curved negative is brighter at all four edges than the flat one. Give two explanations that are consistent with the derivation being correct, and one that is not.
3. Two pinholes 20 mm apart at f = 60 mm photograph a subject 1.2 m away on a 100 mm sheet. Compute the displacement and the width of the doubled band, and state the exposure difference between the band and the ends in stops.
4. Sketch how a doorway’s verticals and a kerb running across the frame at ground level will render on a curved back with the camera level at 1.5 m. Which stays straight, and why?
5. The flat plane stretches horizontally relative to vertically by sec θ. What is that factor at the middle of the long edge of a 4 × 5 sheet at f = 50 mm, and what does the cylinder do with it?
6. Why does this page insist that the flat and curved sheets be given the same exposure rather than the correct one for each?

Further experiments

Make an insert of the wrong radius on purpose — say 80 mm when the focal distance is 50 — and repeat the falloff comparison. The advantage should be smaller and the edges softer, because the paper is no longer at the focal distance. Predict both before you shoot, and put a number on how much smaller.

Photograph a plumb line and a spirit level in the same frame on the curved back, one vertical and one horizontal, both at the height of the pinhole. Both should be straight. Then raise the camera by 300 mm without re-levelling the subject and shoot again: the horizontal should now bow, and the amount of bow is a direct measurement of the projection.

Wrap the short dimension instead of the long one. The wrap angle falls and the axial angle rises, so the advantage moves from the long edges to the short ones. It is the same derivation with φ and ψ exchanged, and it is a good test of whether you believe the derivation or have merely memorised its conclusion.

Open the three holes one at a time onto one sheet, giving a third of the total exposure to each, and compare with all three open for the full time. The reciprocity behaviour of paper is unpublished, so the two are not obviously equivalent, and the comparison is a small unpublished measurement of your own.

Shift in two axes at once, if your carrier allows it, and predict the falloff at all four corners before processing. The worst corner is the one whose displacement adds to the half-diagonal in both directions, and it runs out long before the mechanism does.

Check your understanding

Question 1. On a cylindrical film surface of radius R = f with the pinhole on the axis, which factors of the cos-fourth law survive, and in which direction?

• All three vanish, so the illumination is uniform
• Only the aperture-foreshortening factor survives, and it survives in every direction equally
• Across the wrap, only the aperture foreshortening survives, giving cos φ; along the cylinder axis all three survive, giving the full cos⁴ ψ — so the illumination is E(0) cos φ cos⁴ ψ
• The inverse-square factor survives and the other two vanish

Show the answer and why

Answer: Across the wrap, only the aperture foreshortening survives, giving cos φ; along the cylinder axis all three survive, giving the full cos⁴ ψ — so the illumination is E(0) cos φ cos⁴ ψ

Take the three factors one at a time on the cylinder. The distance from the hole to a point at wrap angle φ and axial coordinate z is the square root of R² plus z², which contains no φ at all — going round the cylinder does not move you further from an axis you are concentric with — so the inverse-square factor becomes cos² ψ, a purely axial term. The film-obliquity factor is the cosine of the angle between the ray and the surface normal, and on a concentric cylinder the ray arrives along the radius, which is the normal in the wrap plane; that leaves cos ψ, again purely axial. The aperture foreshortening depends on the true field angle from the plate's normal, and cos θ = cos φ cos ψ, so it carries both. Multiply: cos²ψ × cos φ cos ψ × cos ψ = cos φ · cos⁴ ψ. The practical consequence is the one that surprises people: at the middle of the wrapped edge of a 4 × 5 sheet at f = 50 mm the cylinder is worth about a stop, and at the middle of the axial edge it is worth exactly nothing.

Question 2. Two pinholes 20 mm apart at a focal distance of 60 mm photograph a subject 1.2 m away, on a sheet 100 mm wide. What is the image displacement and how wide is the doubled band?

• Displacement 20 mm; doubled band 80 mm
• Displacement 21 mm; doubled band 79 mm
• Displacement 20 mm; doubled band 60 mm
• Displacement 24 mm; doubled band 76 mm

Show the answer and why

Answer: Displacement 21 mm; doubled band 79 mm

Δ = s(1 + f/u), so with s = 20 mm, f = 60 mm and u = 1200 mm the bracket is 1 + 60/1200 = 1.05 and Δ = 21 mm. The doubled band is the sheet width less the displacement: 100 − 21 = 79 mm, with a 10.5 mm single-image strip at each end. Option 1 is the infinity answer and is wrong here by a millimetre — small, but the point of computing it is that the bracket is exactly the one in the blur formula b = d(1 + f/u), because a hole of diameter d is a continuous set of pinholes spanning d and its blur circle is the displacement between the two extreme ones. Option 3 subtracts twice the separation, which is the three-hole answer. Note the exposure consequence too: the band receives both images and is therefore one stop denser than the ends, so no single exposure is correct across the whole sheet, and choosing which part to be right about is the decision a multi-hole plate forces on you.

Question 3. With the camera level and the wrap horizontal, which lines in the world render straight on the curved back?

• All of them, because a pinhole is rectilinear
• Only the verticals, plus the single horizontal at the height of the pinhole
• Only the horizontals
• None of them; every line curves

Show the answer and why

Answer: Only the verticals, plus the single horizontal at the height of the pinhole

A pinhole is a central projection, so the image of a world line is where the plane containing that line and the hole cuts the recording surface. A plane cuts a plane in a straight line, which is why a pinhole on a flat back is rectilinear and Option 1 is right about the flat case and wrong about this one. A plane cuts a cylinder in a conic, generally an ellipse, which is a curve and stays a curve when the cylinder is unrolled — with two exceptions. A plane containing the cylinder's axis cuts it in two straight generators, and the pinhole sits on that axis, so any line whose plane through the hole contains the axis stays straight: with the camera level and the wrap horizontal, those are the verticals. And the plane perpendicular to the axis through the hole cuts the cylinder in a circle, which unrolls to a straight line: that is the one horizontal at the height of the pinhole, and it lands on the sheet's centre line. Every other horizontal bows, one way above centre height and the other way below, which is the behaviour of every rotating panoramic camera ever made.

Question 4. You give the flat and curved sheets identical exposures and the two centres come out at visibly different densities. What does that tell you?

• That the cylinder gains light at the centre as well as the edges
• That something other than the geometry changed between the two sheets, because the axial illumination does not depend on the shape of the surface at all
• That the curved back needs more exposure
• That the paper behaved differently on a curve

Show the answer and why

Answer: That something other than the geometry changed between the two sheets, because the axial illumination does not depend on the shape of the surface at all

On the axis, φ = 0 and ψ = 0, so the cylinder formula gives E(0) cos 0 cos⁴ 0 = E(0): exactly the same as the flat plane. The point on the film directly behind the hole is at distance f, square on to the ray, seeing the hole face-on, whatever shape the rest of the surface is. So the centre densities must match if the exposures matched, and a difference there is evidence about the session rather than about the optics. The likely causes, in order: the light changed between the two exposures — check the times you re-metered — the timing was not actually identical, the curved insert is not at the focal distance so the axial distance changed too, or the two sheets came from different boxes or different parts of a processing run. This is why the procedure insists on identical exposure rather than a correct exposure for each: it turns the centre of the negative into a built-in control on everything except the shape of the surface.

Question 5. A single pinhole is moved 20 mm off the format centre at f = 120 mm on a 4 × 5 inch sheet. What happens, and what limits how far you can go?

• The image rotates, and the limit is the shutter
• The whole image circle moves by 20 mm and verticals stay vertical; the limit is falloff, because the corner nearest the shift moves to a field angle of 39.2° and loses 1.47 stops against the opposite corner's 0.78
• The image is magnified on one side, and the limit is the tripod
• Nothing changes except the framing, and there is no limit

Show the answer and why

Answer: The whole image circle moves by 20 mm and verticals stay vertical; the limit is falloff, because the corner nearest the shift moves to a field angle of 39.2° and loses 1.47 stops against the opposite corner's 0.78

Sliding the hole sideways slides the whole circle of illumination with it, because the circle belongs to the hole. Nothing rotates and nothing tilts, so a levelled camera keeps its verticals parallel — which is precisely the point of shift, and the reason a pinhole can photograph a tall building from ground level without the converging verticals that tipping the camera up would give. What runs out is not the mechanism but the light. The corner nearest the shift is now √((63.5 + 20)² + 50.8²) = 97.7 mm from the axis, so its field angle is arctan(97.7/120) = 39.2° and its cos-fourth loss is 1.47 stops; the opposite corner is 66.9 mm out at 29.1° and loses 0.78. That 0.69 stop of asymmetry across the frame is visible, and it grows quickly: double the shift and the near corner passes 45°, where the loss is two stops by itself. The longer focal setting tolerates more shift than the wide one, for the same reason it has darker corners to begin with — smaller half-diagonal-to-focal-distance ratio.

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