"Parallax" and the level water of the Bedford canal: a primer

Sources. No source gap. The project folder holds one book and one map. The book is Zetetic Astronomy. Earth Not a Globe! by "Parallax", the pen name of Samuel Birley Rowbotham (1816-1884) [memory: high; web: https://en.wikipedia.org/wiki/Samuel_Rowbotham]. It is the 1865 edition, printed at Bath, as a 229-page PDF; the book's page number is the PDF page minus 5 [read]. The text layer is OCR with number errors, so figures were checked against page images where they matter. The map is Alexander Gleason's 1892 "New Standard Map of the World", a pole-centred flat map (two copies of one image). The Investigator found the 1881 third edition online, the Gleason patent, newspaper reports of the 1870 wager on the same canal, and a critic's history (Schadewald, The Plane Truth). Not read by anyone in this review: the 1849 pamphlet, the 1881 edition in depth, Hampden's and Carpenter's 1870 pamphlets, Lady Blount's 1904 report and Wallace's own account. Where this primer relies on them it says so and names the go-between source. After the quality check (qe.md), six more parts of the book were read closely (the eclipse and moonlight records, pp.94-110; the long sightings and balloon quotations, pp.18-25; the geodesy arguments, pp.49-67; the Sun's circle and the New Zealand seasons, pp.80-90; the lighthouse list, pp.172-178; southern navigation, pp.26-30 and 55-58), and the arbiter added a "QE revision" with new sub-verdicts. This primer carries them.

Tags. [read: p.N] = the book, PDF page N. [read: page image p.N] = read from a rendered image of that page. [web] = Internet source with URL (most were found by the Investigator or the reviewers). [memory] = recalled from training, with confidence. [calc] = computed in example.py, which you can run yourself:

.venv/bin/python review/Parallax/example.py

This primer adds no verdicts. Verdicts are the arbiter's (arbiter.md).

Corrections carried from the reviews. Earlier versions of facts.json and claims.md said "the 43-mile Southsea to Ryde sight"; this is an OCR error, and the distance is 4.5 miles [read: PDF pp.172-173, via arbiter]. The eye height on Southsea Common, and again at West Cowes, is 2 1/2 ft, not 24 ft [read: page images pp.164-165, via QE]. The air-gun ball is said to have fallen "within a few inches" of the gun, not "feet" [read: pp.73-74, via QE]. Both files are now corrected.

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1. The idea in one paragraph

In the summer of 1838 or 1839, a young man living in an Owenite (co-operative) colony at Manea Fen in the Cambridgeshire fens sat in the Old Bedford River, a straight, still drainage canal six miles long, with his eye a few inches above the water and a telescope in his hand. A boat with a flag three feet high rowed away from him. He reported that he could see the flag, and the boat down to the waterline, the whole six miles. On a round Earth, he calculated, the boat should have sunk out of sight behind a hump of water more than sixteen feet high. He concluded that standing water is flat and therefore that the Earth is a flat, motionless plane, with the north pole at its centre and the south a ring of ice. Over the next forty years, as "Parallax", he lectured and wrote on this "zetetic" (from the Greek for "inquiring") method: start from what you can see and measure yourself, and distrust theories built on assumptions. His book applies it to everything: ships that seem to sink hull first, the distance of the Sun, the Foucault pendulum, tides and the Bible. His arithmetic for the curvature of water is correct, and his insistence on testing it directly is sound. What he left out is that light does not travel in perfectly straight lines through air. This primer shows how much bending his observation needs, why that amount is unusual but not impossible over cold water in summer, and why it means the canal could not decide the question on its own. (Sources for this paragraph: the book [read: pp.16-18]; the colony and dates from Schadewald and Wikipedia [web, via Investigator]. The 1865 book itself does not name the observer or give a year: it says only "the observer" in "the summer season" [read: p.16]. That the observer was Rowbotham, in 1838 or 1839, comes from those later sources.)

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2. What you need to know first

Circles, tangents and Pythagoras. A tangent line touches a circle at one point and is at right angles to the radius there. If you walk a distance d along a tangent from a circle of radius R, the right triangle formed by the centre, the touching point and where you stand gives (R + h)² = R² + d², where h is your height above the circle. This one triangle gives the curvature rule, the distance to the horizon and the height hidden behind it. Read: any trigonometry text; Stewart, Calculus, for the series expansions used below.

Small-angle and series approximations. For x much smaller than 1, √(1 + x) ≈ 1 + x/2. This turns the exact Pythagoras answer into the simple rule h ≈ d²/(2R). You will also meet the idea of an angle in radians: an angle θ (theta) in radians is arc length divided by radius, and for small θ, tan θ ≈ θ. Read: Stewart, Calculus, on Taylor series.

Refraction and the refractive index. Light slows down in a material by a factor n, the refractive index (n = 1 in vacuum; n ≈ 1.0003 in air at sea level). When n changes from place to place, a ray bends toward the region of higher n (Snell's law, applied layer by layer). Air density, and so n, falls with height, so a near-horizontal ray curves gently downward, following the Earth a little. How much depends almost entirely on how temperature changes with height. Read: Hecht, Optics (refraction and Fermat's principle); Minnaert, Light and Color in the Outdoors (mirages and looming) [memory: high].

Temperature inversions and mirages. Normally air cools with height. When air near the ground is colder than the air above it, as over cold water on a warm day, the gradient is "inverted", density falls unusually fast with height, and rays bend downward more strongly. Distant objects then "loom": things below the geometric horizon come into view, and in extreme cases a superior mirage (an image above the object) appears. Over a hot road the gradient goes the other way and you get the familiar inferior mirage ("water" on the road). Read: Minnaert; Andrew T. Young's web pages on mirages and green flashes (San Diego State University) [memory: high].

Plane triangulation. If two observers a known distance apart measure the angle up to the same object at the same moment, plane trigonometry gives the object's distance. Surveyors do this every day for objects near enough that the Earth's curvature and the direction to the object both change appreciably between the stations. It fails for very distant objects, because the two sight lines are almost parallel. Read: any trigonometry text.

Rotating frames and the Coriolis effect. Newton's laws hold in a frame that does not rotate. On a rotating Earth we see two extra apparent forces: the centrifugal force and the Coriolis force, which pushes a moving body sideways, to the right in the northern hemisphere. Its size is 2mΩ × v, where m is the mass, Ω (omega) is the Earth's rotation rate (7.29 × 10⁻⁵ rad/s, one turn per sidereal day) and v the velocity. It explains why a Foucault pendulum's swing plane turns and why a dropped stone lands slightly east of the plumb line. Read: Taylor, Classical Mechanics, chapter 9 ("Mechanics in Noninertial Frames") [memory: high]; Kleppner and Kolenkow, An Introduction to Mechanics.

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3. The textbook view

3.1 How far does water fall below a level sight line?

Take a sphere of radius R. Stand at a point on it and look along the tangent (a perfectly level sight line). At a distance d along that line, the surface has fallen away by

h = √(R² + d²) − R ≈ d²/(2R) (1)

where h is the drop, d is the distance and R the radius. The second term of the series, −d⁴/(8R³), is far too small to matter at these distances [calc, Part A]. With the mean Earth radius R = 3,959 statute miles (6,371 km) [memory: high], 1/(2R) in inches per square mile is 5280 × 12 / 7918 = 8.00. This is the famous "8 inches times the square of the miles" rule. Rowbotham quotes it, with the same derivation from Euclid, from the Encyclopaedia Britannica article "Levelling", which uses 2R = 7,912 miles and gets 8.008 inches [read: page image p.14; calc]. At 6 miles: 36 × 8 in = 288 in = 24.0 ft [calc].

3.2 The horizon and the hidden height

An eye at height e above the water sees a sight line that grazes the water at the horizon, a distance

d_h = √(2Re) (2)

away (equation 1 run backwards). Beyond the horizon, the sight line rises above the water again by equation (1) measured from the grazing point. So an object at distance d > d_h is hidden from the water up to a height

H_hidden = (d − d_h)² / (2R) (3)

An object whose top is at height f just shows when d = √(2Re) + √(2Rf).

3.3 Refraction as a bigger Earth

A near-horizontal ray in air curves downward. Call the ray's curvature (1 over its radius of curvature) k times the Earth's curvature 1/R. The number k is the refraction coefficient. Geometrically, a curved ray over a sphere of radius R behaves exactly like a straight ray over a larger sphere of effective radius

R_e = R / (1 − k) (4)

So every formula above holds with R replaced by R_e. Three landmarks: k = 0 means no bending; k = 1 means the ray curves exactly as the water does, so a sphere looks flat; k > 1 means the surface looks concave (bowl-shaped). Surveyors use a standard value of about k = 0.13 [memory: high]. Rowbotham's own Britannica quotation says the same: refraction "may at a mean compensate for about a seventh of the curvature of the earth, it sometimes exceeds a fifth, and at other times does not amount to a fifteenth" [read: page image p.15]. A seventh is k = 0.14.

k comes from the vertical temperature gradient. For a horizontal ray near the ground

k = 503 × (P / T²) × (0.0343 + dT/dh) (5)

where P is air pressure in millibars, T the air temperature in kelvins, and dT/dh the rate at which temperature changes with height h, in kelvins per metre (positive means warmer higher up) [web: https://en.wikipedia.org/wiki/Atmospheric_refraction, as used in arbiter.md]. The 0.0343 K/m is the gradient at which air density stops changing with height. In normal air (cooling by about 0.0065 K/m) at 1013 mbar and 290 K, k = 0.17 [calc, Part D]. Make the air warmer with height and k grows quickly.

3.4 The dip of the horizon

Because the horizon is below the level line, you look slightly down to see it. The dip angle is about √(2e/R) radians: 2.6 minutes of arc at 6 ft, 10.6 minutes at 100 ft, about 1.3 degrees at a mile [calc, mathematician]. A minute of arc is 1/60 degree. Navigators correct sextant readings for it routinely [memory: high]. It is far too small to see by eye, so the horizon always seems to be "at eye level" and to "rise with the observer". The ground directly below drops away by 90 degrees, so from a balloon the scene looks like a bowl [physicist; mathematician].

3.5 The Sun's distance and the two-station method

Two stations on a north-south line measure the noon altitude of the Sun (its angle above the horizon). If the Sun were a nearby lamp over a flat plane, plane trigonometry would give one height for it from any pair of stations. If the Sun is very far away over a sphere, the difference in altitude equals the difference in latitude, the angle between the two stations' verticals at the Earth's centre. The test is whether one Sun height fits all stations. Today's distance, about 150 million km, comes from radar and spacecraft [memory: high].

3.6 The Foucault pendulum

A long pendulum free to swing in any vertical plane keeps that plane fixed relative to the distant stars (strictly, against non-rotating axes). The Earth turns under it. At latitude φ (phi) the plane appears to turn at

ω_p = Ω sin φ (6)

where ω_p is the turning rate of the plane and Ω = 15 degrees per hour (15.04 against the stars). At the pole it turns once a day, at the equator not at all, and in the southern hemisphere it turns the other way. In the rotating frame this is the Coriolis force at work [memory: high; web: https://en.wikipedia.org/wiki/Foucault_pendulum]. Short pendulums have a known error: if the bob moves in a thin ellipse instead of a line, the ellipse itself turns ("Airy precession") at a rate (3/8)(ab/L²) ω₀, where a and b are the semi-axes of the ellipse, L the length and ω₀ = √(g/L) the swing frequency [web: https://arxiv.org/html/2502.12230v1, via arbiter]. A few millimetres of sideways motion at release can swamp the Earth's effect in a short pendulum.

3.7 Four more results the evidence table uses

These come in with the book's geodesy, eclipse and seasons arguments (see the QE revision in arbiter.md).

• The seconds pendulum. A simple pendulum of length L in gravity g has period T = 2π√(L/g). A "seconds pendulum" takes one second for each swing one way, so T = 2 s and L = g/π². The length it needs is therefore a direct measure of local gravity: about 39.0 inches at the equator and 39.2 at the pole, because gravity is about 0.5 percent stronger at the poles on a flattened, spinning Earth [memory: high; arbiter, from WGS84 constants].
• Spherical excess. On a flat plane a triangle's angles add to 180°. On a sphere of radius R they add to 180° plus E = A/R² radians, where A is the triangle's area (Legendre's and Girard's theorems) [web: https://en.wikipedia.org/wiki/Spherical_trigonometry, via arbiter]. E is a property of the angles between the stations, so it does not depend on the instrument. It is tiny: about 0.009 arcseconds for a triangle 6,600 ft on a side, 23 arcseconds for one 64 miles on a side [calc, arbiter]. An arcsecond is 1/3,600 of a degree.
• Refraction at the horizon and the Earth's red shadow. At the horizon the air lifts the image of the Sun or Moon by about 34 minutes of arc, a little more than the width of either disc [memory: high]. So in a lunar eclipse near sunrise or sunset, both the Sun and the eclipsed Moon can appear just above opposite horizons. This has a name, a selenelion [web: https://en.wikipedia.org/wiki/Selenelion, via arbiter]. In totality the Moon does not vanish: sunlight passing through the Earth's atmosphere is bent into the shadow, and the blue is scattered out, so the Moon is lit a dim copper red that varies from eclipse to eclipse [memory: high].
• Day length and latitude. On a globe, the Sun rises when its hour angle H₀ (how far, at 15° per hour, the sky has to turn from noon) satisfies cos H₀ = (sin a − sin φ sin δ)/(cos φ cos δ), with φ the latitude, δ the Sun's declination (its angle north or south of the celestial equator, ±23.44° at the solstices) and a = −0.833°, the Sun's centre at sunrise once refraction and the disc's half-width are counted [memory: high]. Day length is 2H₀/15 hours. The longest day gets longer, and twilight gets longer, the farther you are from the equator.

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4. Where the inventor departs

Rowbotham accepts the textbook geometry and uses it correctly. He departs at the step from the geometry to the conclusion. The table puts each point side by side.

The pivot of the whole book is the second row. Rowbotham printed, on PDF page 15, an authority telling him that refraction is "extremely variable and uncertain". Ten lines later he concludes that a sight line skimming eight inches above water for six miles shows the water's true shape [read: page image pp.15-16]. (The reviewers wrote that the book mentions refraction only once, at p.179; the passage on p.15 is a second place, inside the Britannica quotation. This does not change any verdict.)

To be fair to him: the textbook gloss "the Earth is round, so it must have been refraction" is also an inference, not a measurement, unless someone measures the refraction. No one on either side did, on any of the canal trials in this record [arbiter; contrarian].

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5. Worked example: how much must light bend for the Bedford view?

All numbers are reproduced by example.py, Parts A to G. (Part H checks two further arguments, the pendulum lengths and the New Zealand day lengths; they are worked in section 7.) Try each step with a calculator first. Units are statute miles and feet; 1 mile = 5,280 ft.

Step 1. Check the rule (Part A)

1/(2R) = 1/(2 × 3,959) = 1.263 × 10⁻⁴ per mile. Times 5,280 × 12 = 63,360 inches per mile gives 8.00 inches per square mile [calc]. The exact Pythagoras answer at 6 miles is 24.0060 ft and the rule gives 24.0061 ft [calc]: the approximation is perfect at canal distances.

Step 2. Rowbotham's own numbers (Part B)

The setup, from the book [read: pp.16-17]: eye "not exceeding eight inches" above the water (e = 0.667 ft); flag "three feet above the water" (f = 3 ft); "six statute miles" from Welney Bridge to Welche's Dam (d = 6 mi).

• Horizon: d_h = √(2 × 3,959 × 0.667/5,280) = √1.000 = 1.00 mile. The book says the tangent point "is 1 mile from the eye". Correct.
• Beyond the horizon: 6 − 1 = 5 miles, so the hidden height is 5² × 8 in = 200 in = 16.67 ft. The book: "16 feet 8 inches". Correct.
• The flag top is 16.67 − 3 = 13.67 ft below the sight line. The book: "13 feet below the line-of-sight". Correct (rounded down).
• Mid-arc bulge of the 6-mile stretch: 3² × 8 in = 72 in = 6.0 ft. The book: "6 feet higher in the centre". Correct.

So on straight light, Rowbotham should have seen nothing of the boat, flag included. His arithmetic is not the problem.

Step 3. Add refraction (Part C)

Replace R by R_e = R/(1 − k) and redo Step 2.

[calc, Part C]

Solve exactly for the threshold. The flag top just shows when d = √(2R_e e) + √(2R_e f), so R_e = d² / (2(√e + √f)²) with e and f in miles. That gives R_e = 14,614 miles and k = 1 − 3,959/14,614 = 0.729 [calc]. For the boat "down to the water's edge" the hidden height must shrink to near zero: within 6 inches needs k = 0.903, within 1 inch needs k = 0.949, and fully visible needs k = 1 [calc]. These reproduce the arbiter's and reviewers' figures.

So the report needs k between about 0.73 (flag only) and 1 (the whole boat), against a typical 0.13 and the top of Rowbotham's own source's range, 0.2.

Step 4. What air would do that? (Part D)

Rearrange equation (5): dT/dh = k / (503 P/T²) − 0.0343. At P = 1013 mbar and T = 290 K (17 °C, a summer day), 503 × 1013 / 290² = 6.06 [calc].

[calc, Part D]

The air must get warmer with height by about 0.09 to 0.13 °C per metre along the whole six miles, just above the water. Wikipedia's article on the experiment gives 0.11 °C per metre for the flat-looking case [web: https://en.wikipedia.org/wiki/Bedford_Level_experiment]. Over a canal fed by cold fen water under warm summer air, the bottom metre can be in exactly such an inversion. Rowbotham says it was "the summer season" and he was bathing [read: p.16]. Nothing in his account rules it out; nothing in it shows it either, because he gives no date, time, weather or temperatures.

Step 5. Same canal, higher sight line: Wallace, 1870 (Part E)

On 5 March 1870 Alfred Russel Wallace accepted a £500 wager from John Hampden, a follower of Rowbotham. Three signals, each 13 ft 4 in above the water, stood 3 miles apart along the same six miles of the Old Bedford. Because all three are at the same height, the eye height cancels: looking from one end signal to the other, the middle one should stand above the straight line by the mid-arc bulge, 6.0 ft, times (1 − k) [web: Nature 1, 581 (1870); Cambrian 15 Apr 1870, https://newspapers.library.wales/view/3330809/3330814/29/; via Investigator].

• k = 0: 6.00 ft; k = 0.13: 5.22 ft; k = 0.73: 1.62 ft; k = 1: 0 [calc].
• Reported: the middle signal stood "more than 5 ft" above the line; the umpire, J. H. Walsh, editor of The Field, found for Wallace [web, same sources].
• "More than 5 ft" means 6.0 (1 − k) > 5, so k < 0.17 along Wallace's sight line that day [calc].

Both results can be true at once. Rowbotham's line ran 8 inches above the water in summer; Wallace's ran 13 ft up in early March. The refraction coefficient can differ that much between the bottom metre of a summer canal and a line four metres up on a cool day. That is the lesson: a single sight line measures curvature plus refraction together, and only measuring the air (or sighting at several heights at once) separates them. Hampden and his referee, Carpenter, disputed the result in print, and Hampden was later convicted of libelling Wallace; a court later held the wager void [web: Wikipedia, via Investigator]. They and The Field each had a stake. Their pamphlets have not been read in this review.

Step 6. Lady Blount's photograph, 1904 (Part F)

On 11 May 1904 Edgar Clifton, a photographer from the Dallmeyer firm working for Lady Blount (head of the Universal Zetetic Society), photographed a white sheet 6 miles away along the Old Bedford with his camera 2 ft above the water. He was reported to be "surprised" to obtain the image [web: Wikipedia, citing The Earth vol.5 nos.49-50; primary not read]. With a 2 ft camera the horizon is 1.73 miles away and the water hides 12.1 ft at 6 miles on straight light, 10.0 ft at k = 0.13 [calc]. The sheet's size is not in the record we have. If its top was 3 ft up it needs k ≈ 0.59 to show; 6 ft, k ≈ 0.38; 10 ft, k ≈ 0.13 [calc]. (The arbiter quoted the mathematician's rougher "k above 0.7 or a sheet taller than 12 ft"; for the sheet's bottom edge at the water, which is what "flat" requires, k must approach 1 as before.) So the photograph is a real datum that fits either strong refraction or a tall target; the original article and photograph would decide which.

Step 7. Side check: one Sun for the plane? (Part G)

At an equinox the noon Sun is overhead at the equator, and at latitude φ it stands 90° − φ above the horizon. On the flat plane the observer is φ × 69.1 miles from the point under the Sun, and the Sun's height would be that distance times tan(90° − φ).

[calc, Part G; the arbiter's figures, 3,865 to 2,392, differ only by the miles-per-degree used]

No single height fits. On a sphere with a distant Sun every row follows at once. This calculation uses nothing but the book's own geometry and ordinary noon sights, which is why the arbiter calls it decisive for the near-Sun claim.

What the example teaches

The canal is the most honest experiment in the book and the one a reader can most easily repeat. Its geometry is right. What it measures is the product of two unknowns, the Earth's curvature and the bending of light near the water. Rowbotham's conclusion needs the second to be zero; the reported view needs it to be very large. A clean answer needs the temperature profile, or several eye heights on one morning.

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6. Evidence table

Verdicts are the arbiter's final verdicts (arbiter.md, "Verdict table").

Where the record goes quiet

The overall record does not go dark: Rowbotham lectured until at least 1882, became president of the Zetetic Society in 1883 and died in 1884, and the movement continued (Universal Zetetic Society 1893; Carpenter, Hampden, Gleason, Blount) [web, via Investigator]. The arbiter lists eight places where the record is silent where it should continue (six in its first pass, two added in its QE revision), each with competing explanations. In short:

1. Jones's pendulum run (Manchester, 1851). Known only from the Manchester Examiner of 24 May 1851 as reprinted in the book. Possible: a casual local demonstration nobody followed up; a transcription error; more data in the original. Deciding record: that newspaper and the Manchester Philosophical Society minutes.
2. The founding Bedford observation. Two differing versions and no independent witness. Possible: one observation retold; two trials; embellishment; a real inversion that day. Deciding records: the 1849 pamphlet; Manea Fen records 1838-41 (Cambridgeshire Archives).
3. The Royal Astronomical Society, 8 December 1848. The 1849 pamphlet says its argument was "a paper read before the Royal Astronomical Society". Augustus De Morgan, the Society's secretary, wrote that no account appears in its Notices and that he read some of the arguments aloud at the close of the meeting "for your amusement" [web: Schadewald Ch.1, quoting A Budget of Paradoxes]. This is one printed account by an officer of the institution, a party with a stake. Possible: rejection for the Notices; an informal reading; the pamphlet overstating its credit. Deciding record: the RAS minute book and MNRAS vol. 9.
4. The Southampton officer's data. One printed report, no name or date. Deciding record: Ordnance Survey correspondence of 1864-65.
5. The unnamed Nab, Southsea and air-gun observers, and the Plymouth committee of October 1864 (named in the book [read: p.167]). R. A. Proctor, a critic, wrote that from the beach half the Eddystone lantern was visible and blamed weaker-than-usual refraction [web: Schadewald Ch.1]; the book says it was entirely out of sight. Deciding record: the Western Daily Mercury report.
6. Blount/Clifton 1904. The primary article was not reached (the movement's online library was blocked by a security error). Deciding record: The Earth vol.5 nos.49-50 and the photograph.
7. The Lancet cooling report of 14 March 1856. Known only from one sentence in the book. Possible: a poorly controlled thermometer exposed to the clear sky; an overstated citation by the book; a genuine measurement of something else. Deciding record: the Lancet item.
8. The Wilkes letter ("I made 70 degrees of it"). A single quotation in the book, with no date or archive. Possible: an accurate letter, a paraphrase, or a misattribution. Deciding record: the Wilkes and Ross papers.

How Rowbotham handled tests: a critic's accounts. Robert Schadewald, a critic who wrote against the flat-Earth movement (his stake), reports from newspapers that the Investigator found cited but did not read: at Burnley in December 1849 Rowbotham did not appear for his second lecture after a question about ship hulls (Blackburn Standard); he reportedly walked out of an agreed repeat of the marker test on the river Teign because a committee had placed the markers without him; and his 1849 Trowbridge lectures were given under the name "S. Goulden" (Wiltshire Independent) [read: research/2026-10-06-cantab-planetruth.md, via arbiter QE revision]. These count against the inventor on how he handled testing. They sit beside the one documented institutional episode on the other side, De Morgan at the RAS (item 3). Deciding records: the Blackburn Standard of December 1849, the Wiltshire Independent of January 1849, and the Teign committee's own report.

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7. What a learner should take away

What is solid. - The curvature rule (8 in per mile squared, horizon at √(2Re)) and every Bedford figure the book derives from it. Rowbotham's arithmetic is right. - The zetetic demand to test curvature directly over standing water is a good experimental idea. A still canal is a fine instrument. - Refraction near water really is "extremely variable and uncertain", as his own source says, and it can be large enough to flatten the view. - Some 1851 popular accounts of the Foucault pendulum did assume the rotation they set out to prove, and short pendulums released by hand really do scatter. Le Verrier's orbit for Neptune really was wrong in its elements, because of an assumption (Bode's law). - The horizon really does seem to stay at eye level, and from a balloon the land really can look like a bowl. - Many of the observations the book quotes are real: the eclipsed Moon seen with the Sun still up, the copper-red Moon in totality, the Wellington and England day lengths, Ross's southward fall in pressure, the seconds-pendulum table. What fails is the inference drawn from them, not the observations.

What is wrong, and why it is instructive. - An experiment measures everything that affects it. A sight line over water measures curvature and refraction together. Treating one of them as zero, without measuring it, is the central mistake. The same rule cuts the other way: "it must have been refraction" is not a measurement either. - A rescue that explains everything explains nothing. "Perspective" and "waves" are invoked when a hull is hidden; level water when it is seen. A model that predicts both outcomes is not being tested. - Check a model against all the data, not one point at a time. Plane triangulation is real trigonometry; it fails only when you ask for one Sun height from four latitudes. The southern-hemisphere distances fail the same way. - Scatter is not absence. A noisy measurement of a real effect still has a mean. The Foucault rates Rowbotham quotes follow sin φ to a few minutes of arc. - A quoted law must be applied to the quantity it is about. The eye's one-arcminute limit is about an object's angular size, not its height above the water. - Put the proposed cause and the effect side by side in numbers. The book puts the different seconds-pendulum lengths at the equator and the pole down to temperature, quoting Noad: 30 °F changes a pendulum's length by 1/5000. Over its own table's 84.2 °F that is (84.2/30) × (1/5000) = 5.6 × 10⁻⁴ of the length. The table's own lengths change by (39.197 − 39.027)/39.027 = 4.36 × 10⁻³, 7.8 times more. Add Derham's air effect (2 seconds in an hour, another 5.6 × 10⁻⁴) and the two causes still cover only about a quarter. The book never did this division. Worse, its own standard length is quoted "in vacuo … at the temperature of 62°": the table is already corrected for air and temperature, so what is left is the effect to explain, gravity [read: deep read pp.51-54; calc, Part H]. - Check the premise before the conclusion. The book quotes real almanack day lengths for Wellington and England and finds them "altogether incompatible" with a globe, because New Zealand's "latitude is about the same as in England". It is not: Wellington is at 41.3° S, London at 51.5° N. Put the real latitudes into the globe's sunrise formula (section 3.7) and you get Wellington 15 h 10 m and 9 h 11 m, London 16 h 38 m and 7 h 50 m, within about 12 minutes of the almanack. The data the book offers against the globe are the globe's own prediction [calc, Part H]. The brief southern twilight it quotes is what a lower latitude gives too. - Read the quotation to the end. Glaisher wrote that the Sun was "apparently above the horizon" during horizontal eclipses; the book keeps "apparently" in the quotation and drops it in its conclusion. Findlay wrote that refraction can lengthen his lighthouse ranges; the book quotes this on p.166 and then treats the ranges as straight-line sightings. The balloonist Wise called the bowl-shaped view "an optical illusion".

What is genuinely open. - What the air was doing on the Old Bedford on the day of the 1838/39 view, on Lady Blount's day in 1904, and on the Nab days in 1864. A modern repeat with temperature measured at several heights would tell how often a summer canal reaches k ≈ 1. This is a good amateur project. - What Captain Wilkins saw in 1862, and where; what the Southampton figures were. Both need the original records. The book's own framing puts Wilkins south, where his report would be a real anomaly; a northern reading is only marginal. - What The Lancet of 14 March 1856 actually measured; how bright Walkey's 1848 Moon really was (the Phil. Mag. original); what Findlay's lighthouse "ranges" mean. Each needs one original document. - The 4.4 mm southward deviation in Reich's 1831 drops remains unexplained [web, via arbiter]. It is unrelated to a plane but is a real residual. - Whether The Anti-Newtonian (1819) was Rowbotham's unacknowledged source (Schadewald). It bears on history, not on physics.

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8. Check yourself

1. Using 8 inches per mile squared, how much is hidden at 10 miles for an observer whose eye is 6 ft above the water, with no refraction?
2. With k = 0.13, what is the effective radius of the Earth, and what does the 8-inch rule become?
3. In Wallace's three-signal test, why does the observer's eye height drop out? What would a reading of "3 ft" at the middle signal have told you about k?
4. A Foucault pendulum in Edinburgh (latitude 55.95° N) turns at 12.4° per hour. What would it do in Sydney (33.9° S), and why does that matter for the "pendulum over the Earth" objection?
5. Rowbotham says the Moon must shine by its own light because focused moonlight does not warm a thermometer. Why does the null result fail to show that, and why is it also too weak to confirm a reflecting Moon?
6. The book says New Zealand and England are at "about the same" latitude, yet England's longest day is 1 h 36 m longer. Taking Wellington at 41.3° S and London at 51.5° N, use the sunrise formula in section 3.7 (δ = 23.44°, a = −0.833°) to find Wellington's longest day. What does the result say about the book's argument?

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9. Further reading

The inventor's own works. - "Parallax" (S. B. Rowbotham), Zetetic Astronomy. Earth Not a Globe! (Bath and London, 1865), the project's book. Third edition, 1881, 430 pp., online at https://sacred-texts.com/earth/za/index.htm [web, via Investigator]. - Zetetic Astronomy (pamphlet, 1849, 16 pp.; Library of Congress QB638 .R87) [web: https://guides.loc.gov/flat-earth/books]. - The Inconsistency of Modern Astronomy and its Opposition to the Scriptures (date unknown).

The 1870 wager and other canal tests. - "The Rotundity of the Earth", Nature 1, 581 (7 April 1870) [web: https://www.nature.com/articles/001581a0]. - A. R. Wallace, My Life (1905), vol. 2, pp. 381-393 [web: LoC guide]. - J. Hampden, Is Water Level or Convex After All? (Swindon, 1870), at Wallace Online S179aa [web]. An interested party's account. - H. Yule Oldham, British Association Report 1901, pp. 725-726 [web, via Wikipedia].

History. - Christine Garwood, Flat Earth: The History of an Infamous Idea (2007) [memory: high]. - Robert J. Schadewald, The Plane Truth, online at https://www.cantab.net/users/michael.behrend/ebooks/PlaneTruth/ (a critic's history that quotes the primaries at length). - Augustus De Morgan, A Budget of Paradoxes (1872), on the RAS episode [memory: high].

Original records the book quotes and this review did not read (each named by the arbiter as deciding a point). - A. G. Findlay, Lighthouses of the World (1862): what its "ranges" mean. - Philosophical Magazine no. 220 (August 1848): Walkey's account of the 19 March 1848 eclipse. - The Lancet, 14 March 1856: the moonlight cooling report. - The Times, 13 May 1862: Captain Wilkins and the pole star.

Mainstream physics. - M. Minnaert, Light and Color in the Outdoors: refraction, looming and mirages [memory: high]. - E. Hecht, Optics: refraction and Fermat's principle. - J. R. Taylor, Classical Mechanics, chapter 9: rotating frames, Coriolis force, the Foucault pendulum [memory: high]. - L. Foucault, "Démonstration physique du mouvement de rotation de la Terre au moyen du pendule", Comptes Rendus 32 (1851) [memory: high]. - Wikipedia, "Atmospheric refraction" (refraction coefficient formula) and "Bedford Level experiment" (a map of the primary records). - Wikipedia, "Selenelion" (Sun and eclipsed Moon both seen above the horizon) and "Spherical trigonometry" (spherical excess).

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Answers

1. Horizon for a 6 ft eye: √(2 × 3,959 × 6/5,280) = 3.0 miles. Beyond it: 10 − 3 = 7 miles, so 7² × 8 in = 392 in ≈ 32.7 ft hidden. (A ship with a 30-ft hull would be entirely hull-down.)
2. R_e = 3,959/(1 − 0.13) = 4,551 miles. The rule becomes 8 × (1 − 0.13) ≈ 7.0 inches per mile squared.
3. All three signals are at the same height above the water, so the line from end to end and the middle signal are equally raised; only the bulge between them counts, 6.0 ft × (1 − k). A 3 ft reading would mean 6.0(1 − k) = 3, so k = 0.5: strong refraction, but still a curved surface.
4. In Sydney it turns at 15 × sin 33.9° = 8.4° per hour, and the other way round (anticlockwise seen from above). A pendulum "moving over a fixed Earth" would need a separate reason to turn at a rate set by latitude and to reverse at the equator. A rotating Earth gives both from one number, Ω.
5. Full moonlight is about 2.5 × 10⁻⁶ of full sunlight. Concentrated ×3,000 to ×6,000, as in Forbes's test, it gives roughly 1 percent of full sun on the receiver before losses, and no lens can make an image brighter than the Moon's own surface. A blackened bulb would warm by a few tenths of a kelvin at most, at the edge of what a good 1840s instrument could see and easily cancelled by the bulb radiating heat to the clear night sky [calc, arbiter QE revision]. So "no heat" does not imply "self-luminous": a sunlit Moon also gives only a faint, hard-to-measure signal. But the nulls are equally too insensitive to confirm a reflecting Moon; with no stated sensitivity, losses or control they decide little either way. Piazzi Smyth detected the Moon's heat in 1856 with better instruments [web]. The Lancet's reported cooling of more than 8 degrees is a different matter, an observation with no details, which the original article would decide.
6. Wellington, 21 December: sin a = sin(−0.833°) = −0.01454; sin φ sin δ = sin(−41.3°) sin(−23.44°) = (−0.6600)(−0.3978) = 0.2626; cos φ cos δ = 0.7513 × 0.9175 = 0.6893. So cos H₀ = (−0.01454 − 0.2626)/0.6893 = −0.4020, H₀ = 113.7°, and the day is 2 × 113.7/15 = 15.16 h = 15 h 10 m. The almanack says 14 h 58 m. London's longest day on the same formula is 16 h 38 m (almanack 16 h 34 m). The difference the book finds is the difference between 41° and 51.5° of latitude, which is what a globe predicts; the premise "about the same latitude" is the mistake [calc, Part H].