Parallax (S. B. Rowbotham): Mathematician's review (Step 2)

Reviewer: PhD mathematician. Scope: derivations, notation, units, hidden assumptions. Physics is the physicist's job; where a mathematical check needs a physical input (refraction, the Coriolis term) it is stated and tagged.

Tags: [read: p.N] = text layer of the 1865 book (PDF page; book page = PDF page - 5); [read: page image p.N] = I rendered the page; [calc] = I computed it (Python, numpy; scripts not kept, every input is stated); [web] = URL given; [memory] = recalled, with confidence.

Source note. The book is read through an OCR text layer with number errors; the author's own words are quoted from it. Only the pendulum section had a deep read. I read myself: pp.12-22 (Bedford Level, Irish Sea, Brighton), 33-39 (perspective), 45-46 (Polaris), 55-57 (southern circuit), 68-79 (air-gun, tubes, Sun distance), 105-107 (moonlight), 112-113 (tides), 137-139 (Neptune), 162-174, 179-180 (ships, lighthouses, Southsea-Ryde), and the Gleason map image. The tides, eclipse accounts, burning interior and scripture sections I read only in digest (facts.json summary) or in part.

Corrections to Step 1

1. "43-mile Southsea to Ryde sight" (claims.md claim 2 and facts.json) is an OCR error. The text layer reads "43 statute miles" [read: p.173], but the same paragraph says the centre of the arc would be "40 inches higher" and that the steamer would climb for "2 1/4 miles". Those follow from 4 1/2 miles: (2.25)^2 x 8 in = 40.5 in [calc]. A 43-mile sight would give 21.5^2 x 8 in = 308 ft. The diagram caption reads "ACROSS 4 1/2 MILES" in the OCR. The Southsea-Ryde argument is a 4.5-mile test, not a 43-mile one. Step 1's own text calls it "43 miles" in two places; both need fixing.
2. "Eye 24 feet" at Southsea Common is also an OCR error. The page image says "the eye was 2 1/2 feet", and the Cowes passage says "2 1/2 feet" [read: page image pp.164-165]. With 24 ft the horizon would be 6 miles away and the 7-mile ship's copper would be visible by geometry; with 2.5 ft the author's "16 feet" claim is the right order (see claim 2).
3. "1,944 feet" (Irish Sea, p.18) is in the book, not an OCR slip [read: page image p.18], but it matches neither the arc's mid-height (30^2 x 8 in = 600 ft) nor the full-chord figure (60^2 x 8 in = 2,400 ft). It equals 54^2 x 8 in; I cannot say why 54 miles. It does not affect the argument, which the author then redoes with 47 miles (1,472 ft, correct [calc]).
4. The southern-circuit comparison (p.56): the book's "45 degrees" parallel gives 17,600 miles, not 14,282. 14,282 is 24,901 x cos 55 [calc]. The book's text says the average is 45 degrees; the figure fits 55. Also 22,657 nautical miles is 26,073 statute miles (x1.1508), not the 26,433 printed [calc]. Whether these are OCR errors or the author's I cannot tell without a page image of p.56-57.
5. claims.md claim 4 says "Sun under 4,000 miles". The book never states a value; it says "considerably less than 4,000" [read: p.78]. From the book's own two data sets I get 2,508 miles (London/Ackworth) and 2,858 miles (the 800-mile pair) [calc]. Two stations give two numbers that disagree by 14%, and the book does not say they should agree or on what dates.
6. Pendulum deep read: confirmed. I re-did its figures: Manchester rate 12.09 deg/h (one degree per 4.97 min) [calc]; half-period of a 7.62 m pendulum 2.77 s [calc]. Its "11 m = 30 ft" note is also right (11 m = 36 ft).
7. Step 1's "no source gap" holds, but the central arithmetic was digest-level. I checked it (below); nearly all the book's curvature arithmetic is correct. The errors are in the hypotheses, not the multiplications.

Mathematical frame (what a learner needs)

Curvature drop. For a sphere of radius R the surface falls below a tangent line by d^2/(2R) at distance d (valid while d << R). With R = 3,959 mi, 1/(2R) = 1/7,918 mile, which is 8.0 inches [calc]. So "8 inches times the square of the distance in miles" is correct, and the book quotes it correctly [read: pp.12-13]. Horizon distance for eye height h: d_h = sqrt(2Rh), which is about sqrt(1.5 h) miles for h in feet. Anything beyond the horizon is hidden up to a height (d - d_h)^2 x 8 in. The author uses exactly this method on pp.164 (Nab light-ship) and gets right answers [calc]. The step he omits throughout is refraction: light bends toward the denser air below, which effectively enlarges R to R/(1-k), with k about 0.07 to 0.2 typical over land and sea [memory: medium; see claim 2 for the check I could do]. He mentions it once (1/12 of the height, p.179-180), i.e. k about 0.08.

Claim 1. The Earth is a stationary plane (north pole at the centre, the south a ring of ice)

a. Internal consistency. The book's own plane-trigonometry method (Section 3, pp.77-79: heights from angles and a measured base) can be applied to any celestial object seen from several stations. Applied to Polaris it gives inconsistent heights. On the plane, Polaris hangs over the pole at height H, and its altitude angle from a station at distance x from the pole is atan(H/x). Observed altitude is the latitude (a measurement the book accepts: it uses latitudes throughout). With x = (90 - latitude) x 69.17 miles (the Gleason map's own scale: an equal-distance map where each degree of latitude is 69.17 miles):

[calc] One object cannot have heights from 0 to 3,600 miles. A sphere with a distant Polaris gives altitude = latitude for every station with no free parameter. The same test on the Sun at the equinox (noon altitude 90 - latitude, so cot(altitude) must grow linearly with ground distance on a plane): latitudes 0, 30, 60 give a Sun height of 3,593 miles from the first pair and 1,797 miles from the second [calc]. The equal 69-mile degrees are a surveyed fact [memory: high]; the book also relies on them (map, tables). The plane therefore needs a different "Sun height" at every pair of stations. The book's own London-Ackworth figure supports this: the observed altitude difference, 2 deg 16', is the latitude difference (about 2.15 deg to 2.19 deg for 151 to 156 miles [calc]) to within about 6 minutes of arc. On a plane with a nearby Sun the 800-mile pair (7 deg difference) and the 151-mile pair (2.27 deg) would have to give the same height on the same day. They give 2,858 and 2,508 miles, on unstated dates.

Second internal check, the southern ring. The book argues that sailing distances round the southern ocean (22,000 to 26,000 miles) exceed what a globe allows (pp.55-57). On the Gleason map (an equal-distance projection centred on the north pole, which is what the map image shows [read: page image flat-map.png]) the circuit of the 45 S parallel has radius 135 x 69.17 = 9,338 mi and length 58,700 miles; the equator is 39,100 miles; the rim (90 S) is 78,200 miles [calc]. So the plane the book draws needs a southern circuit more than double the book's own 26,000-mile figure and over three times the globe's 17,600. The one measured figure fits the globe better than the plane. [The book's route sum also adds three ship routes that are not along one parallel; I treat the comparison as the author's own argument, not as a clean measurement.]

Third, the map's distances. Taking city coordinates [memory: high] and r = (90 - latitude) x 69.17 mi: Sydney to Cape Town 15,707 mi on the map against 6,843 mi on a globe; Sydney to Santiago 15,975 against 7,051; Perth to Johannesburg 11,432 against 5,169; Buenos Aires to Sydney 16,618 against 7,331; London to New York 3,699 against 3,463 [calc]. In the northern hemisphere the map agrees with a globe within 7%; in the south it is out by a factor of 2.2 to 2.3. This is the mathematics of mapping: a sphere cannot be flattened without distorting lengths (Gauss's Theorema Egregium: a surface's curvature cannot be changed by bending). The Gleason patent itself calls its departure from a globe "extorsion" (distortion) [web: https://patents.google.com/patent/US497917A/en, via the Investigator]. The patent holder describes the map as a time-calculating chart with globe-map geography, and that description is a party's statement beside its "As It Is" slogan and the author's book.

b. Experimental consistency. Against the plane: the 69 miles per degree is uniform north and south of the equator [memory: high]; stars' altitudes (Polaris, southern constellations) and the southern cross seen from the south need a single set of heights, shown above to be impossible; southern distances. For the plane (what the book offers): see claim 2.

c. Tested? The model has not been tested as a whole by the author; its tests are the Bedford-type level sights (claim 2) and the 1864 Plymouth committee. The Investigator found no controlled test of the plane model as a map or in navigation.

d. If true. Surveying, navigation by great circle, latitude from Polaris altitude and the whole of southern-hemisphere distance measurement would need a different geometry. Not a small change: the equal-degree survey data alone are incompatible.

Verdict. Strong form (a plane with the Sun and Polaris within 4,000 to 6,000 miles, consistent with the book's own method): Refuted by calculation and Internally inconsistent (Polaris heights; southern ring against the map). Weak form (that a chart can be drawn centred on the north pole with time-zone dial): Established; this is the azimuthal equidistant projection, a valid map with large southern distortion.

Claim 2. Level water proves it (Bedford Level, Southsea-Ryde, lighthouses, horizon at eye level)

a. Internal consistency. The curvature arithmetic is correct [calc]: - Bedford, eye 8 in, flag 3 ft, 6 miles. Horizon 1.0 mile away (sqrt(2 x 3,959 x 8/12/5,280) = 1.0), so the sight line at the boat is (5)^2 x 8 in = 16.7 ft above the water and the flag top (3 ft) would be 13.7 ft below it. The book says "16 feet 8 inches" and 13 ft [read: pp.16-17]. Correct. The book's 6 ft figure for the middle of the 6-mile arc (3^2 x 8 in = 72 in) is also correct [read: p.16]. Not the 6-mile result, which is the tangent calculation. - Nab light-ship: 6^2 x 8 in = 24 ft; minus 10 ft bulwarks gives 14 ft hidden [read: page image p.164]. Correct, with the author's one misprint "6 x 8 in." for "6^2 x 8 in.". - Southsea Common to the quarantine ship: eye 2.5 ft gives horizon 1.94 miles; remaining 5.06 miles; 5.06^2 x 8 in = 17.1 ft; the book's "16 feet" is right in size [read: page image p.165; calc]. - Southsea-Ryde (4.5 miles): 40.5 in [calc]. - Irish Sea 47 miles: 1,472 ft [calc]. But the geometry is wrong in one respect: the book puts Orm's Head (600 ft) 872 ft below the sight line from a 100-ft observer. Two-ended visibility is d_1 + d_2 = 12.2 + 30.6 = 42.8 miles, and the real distance is about 60. The book's visibility claim ("from an altitude not exceeding 100 feet the Great Orm's Head is visible") is stated without date, observer or source [read: p.19]. - Brighton line: the ship at 20 miles would be 266 ft "below" (20^2 x 8 in = 266.7 ft [calc]). But what a taut string between poles can measure is an angle: the horizon seen from 100 ft is depressed by 0.177 deg (10.6 minutes of arc) in every direction; over a 40-mile track the rectilinear view of that circle differs from a straight line by about 0.17 deg [calc]. Neither a globe nor a plane predicts a visibly bent horizon at this resolution. The test cannot discriminate. - The 43-mile figure is an OCR error (see Corrections).

The inference is where the mathematics fails. "No curvature seen along 6 miles" gives "the Earth is a plane" only if light travels in straight lines. The book's own computation shows how large a departure it needs. For the Bedford flag-top to be just visible I need k = 0.73, and for the waterline k above 0.95 (an effective radius 3.7 times larger or more) [calc]. Values that large are physically possible only in unusual temperature gradients just above water [memory: medium; not checked on the web]. At 13 ft or more above the water, as in Wallace's 1870 test, the line of sight is above the strongest gradient. The Investigator's check for standard refraction (k = 0.13): flag hidden by 11.1 ft; I reproduce 14.09 ft of sight-line height minus 3 ft of flag, equal to 11.1 ft [calc]. The Bedford reading therefore rests on a hypothesis (k about 0.7 to 1) the book neither states nor measures.

Three Bedford descriptions differ, per the Investigator: the 1849 pamphlet (theodolite, "a fall of 24 feet", a boat on the New Bedford), the 1865 book (bather's eye 8 in, flag 3 ft on the Old Bedford) and a summer 1839 date against Wikipedia's 1838 [web: Schadewald Ch.1, via sites.md]. 24 feet is 6^2 x 8 in x 1.0 = 24 ft, the curvature at 6 miles from the eye's tangent; the book's two different figures (24 ft versus 16 ft 8 in with a hidden flag) are the same arithmetic with different eye heights, which fits the hypothesis that the descriptions are two readings of one observation, but I cannot show this.

Horizon at eye level (balloons, mirror). The dip below eye level is 0.044 deg (2.6') at 6 ft, 0.177 deg (10.6') at 100 ft, 1.29 deg (77') at 1 mile, 2.17 deg at 15,000 ft [calc]. In a balloon the horizon sits 1 to 2 degrees below eye level, which looks "high" against the nadir (90 deg below); that is how a horizon "rises with the balloon" and why a balloonist such as Le Fevre (Investigator: Balloon Society president, 1882, saw the surface as concave, more so with height [web: Denbighshire Free Press 4 Feb 1882, via sites.md]) reports a bowl. The book's mirror test: at 10 in viewing distance and h = 100 ft the horizon image is lowered 0.031 in [calc; matches the Investigator], below what an eye sees. A test that cannot show a difference cannot support either side. The 110-ft Camera-house observation is consistent with both.

b. Experimental consistency. Tests by others in the Investigator's file: Wallace, 5 Mar 1870: three signals 13 ft 4 in high, 3 miles apart, the centre one more than 5 ft above the line; the umpire J. H. Walsh of The Field found for curvature [web: Nature 1, 581 (1870)]. A 6-mile chord gives 6.00 ft of bulge, or 5.22 ft with k = 0.13 [calc], so "more than 5 ft" is what refraction-corrected curvature predicts. Hampden (the wager's other party) and Carpenter disputed the result in print; they are interested parties and their reports are in the record. H. Yule Oldham 1901 (middle pole about 6 ft), Stretton 1905 (Ashby Canal, about 8 in at 1 mile, which is the expected 8 in [calc]). Lady Blount and Clifton, 11 May 1904: telephoto photograph of a white sheet at 6 miles with the camera 2 ft above the Old Bedford. [web: Wikipedia via the Investigator; primary not read.] With the camera at 2 ft the horizon is 1.7 miles away and the sheet is 4.3 miles beyond it, hidden up to 4.3^2 x 8 in = 12.3 ft [calc]; a sheet of unknown height would need k near 0.7 as well, or to be taller than the 12 ft. The Koreshan Unity (Morrow, 1896, Illinois Drainage Canal) used the same method and concluded the Earth is concave (Investigator): the same sights read as plane, convex or concave according to the refraction assumed, which is the hidden-assumption point.

c. Tested? Yes, repeatedly, by both camps. Results reported for curvature: Wallace, Oldham, Stretton. Results reported against: Rowbotham (1838/39 or 1849 or 1865 accounts, not reproducible from the text), Blount/Clifton 1904. No test in either camp measured refraction at the same time, which is what a clean test needs.

d. If true. If light followed the surface over 6 miles with k = 0.73 to 1, surveyors' levelling would fail by hundreds of feet, and geodetic distances would not close. If the Earth were flat and refraction were 0.13, the flag would be visible: it is not at standard refraction.

Verdict. The curvature arithmetic: Established (correct). The claim that the Bedford sight "proves a plane": Tested, not supported by the 1870 and later controlled sights on the same canal, with the original observation resting on an unmeasured refraction value of 0.7 to 1. Not dismissed without testing: Wallace, Oldham and Stretton ran tests, though some testers were parties.

Claim 3. Perspective, not curvature, explains ships sinking hull first (vanishing point at about 3,000 times the object's height; telescope restores hulls)

a. Internal consistency. The rule comes from a quotation from Mayhew's Wonders of Science: the smallest visible angle is one minute of arc, so an object is "just distinguishable" at 3,000 times its own diameter, and a shilling (1 in) at 250 ft [read: p.37]. 3,000 in = 250 ft is correct [calc]. The author then writes "a line passing over the hull ... must converge to the vanishing point at the distance of about 3000 times its own elevation": a 10-ft hull vanishes at 3,000 x 10 ft = nearly 6 miles; a mast-head at 30 ft is visible to 17 miles [read: pp.37-38]. This swaps size for height above the water. The rule is about an object's angular size: a thin mast 1 ft across, by the author's rule, vanishes at 3,438 ft (0.65 mile), long before a 10-ft-high, 300-ft-long hull [calc]. The mast-head is "visible for 17 miles" only if the rule is applied to its height, and the hull "vanishes at 6 miles" only if applied to its height, not its length. By the rule as quoted, a ship's hull never vanishes at a distance that depends on the height of anything above the water. So the central step is a misapplication of its own quoted law: internally inconsistent.

Second, what the rule predicts is wrong in kind. If perspective shrinks things, a ship receding on a plane shrinks uniformly: both hull and masts become smaller and merge into a dot. What is observed (and what the author's own Plymouth and Nab accounts record: "the hull ... entirely out of sight", "half the hull", "only the upper part of the bulwarks" [read: pp.163, 168]) is a ship cut off at the waterline by a sharp horizon, with the visible part at full proportion. On a plane the horizon is at eye level at infinity (zero dip) and the hull-top is always below it; a mast just above eye level is always visible and a hull is never "cut". The hull-down shape needs a sharp horizon edge at finite distance, which is what a sphere's d_h = sqrt(2Rh) gives. The perspective model has no such edge. And the data the author quotes (hull-down distances) fit the horizon formula with k about 0.08 to 0.13 to a few minutes: the Nab example fits in the author's own arithmetic.

Third, the telescope. A telescope of magnification M multiplies angles by M, so a perspective-limited object returns at 3,000 M times its size (the author's argument). A hull hidden by a bulge of water is not returned by magnification, since the water stays in the way. What a telescope does show, on a calm day, is the part of the ship that is just above the sight line with more detail; and under changing refraction hull-down boundaries move by several feet (the same Nab light-ship was fully visible and fully hidden on different days from the same 2 ft 8 in [read: p.163], which fits refraction and waves, not a fixed perspective law). The claim is stated by the author as "distinctly restored" without a magnification, a distance or a record of what was seen.

Fourth, Polaris (pp.45-46). The book says the globe would make Polaris invisible "for a single degree beyond the equator" because the line of sight becomes a tangent. That is a straw man: from a mast 100 ft up the horizon dip is 0.18 deg and refraction at the horizon is about 0.5 deg [memory: high], so Polaris is visible to roughly 0.7 deg south of the equator, about 50 miles. The case cited (Captain Wilkins, The Times 13 May 1862, "23 deg 58 minutes of latitude") gives no hemisphere in the OCR; if it is north, it is not an example at all. The author's own method gives Polaris altitude as a perspective angle atan(H/x), which I showed in claim 1 requires a different H at every station.

Fifth, circumnavigation and "loss of a day" (pp.39-43). On a circular plane with the pole at the centre, a ship can circle the pole sailing "due east", as the table-and-thread example shows [read: p.42]. This is true as geometry. It says nothing against the sphere, since both permit it. Where it has a cost is the ring's size: the plane's east-west circumference must grow with distance south, so a circle of latitude 45 S is 58,700 miles on the map, against 17,600 on a globe and 26,000 in the book's own sailing sum (claim 1).

b. Experimental consistency. Hull-down distances reported by sailors fit the geometric horizon plus ordinary refraction. The Eddystone case the author quotes: his own source (Findlay, Lighthouses of the World, 1862, p.28) says the range tables assume an eye 10 ft above sea level [read: p.166]. The Admiralty's 13 nautical miles (15 statute) is the range for a 10-ft eye. For a 5-ft eye (the Plymouth committee's, Oct 1864) the geometric limit is sqrt(1.5 x 5) + sqrt(1.5 x H); for the lantern at 82 to 89 ft that is 13.8 to 14.3 statute miles [calc]. The committee was at 14 statute miles, at the predicted limit. Seeing nothing at 14 miles with a 5-ft eye is what a globe with the Admiralty range predicts; the author uses the 10-ft figure for a 5-ft observer. (The Eddystone height is the book's: 89 ft above the foundation [read: p.167]; tide state unknown.)

c. Tested? Yes: Plymouth committee (Rowbotham ("Parallax") and others, 25 Oct 1864, per the Western Daily Mercury as quoted [read: pp.167-168]). Proctor's contrary account of the Plymouth tests is in the Investigator's lead file [web: Schadewald Ch.1]; the Investigator also lists the book's account beside it, and both stay in the record. No controlled refraction measurement.

d. If true. Optics would need a new law under which angular size depends on height above a plane, contradicting the standard theory of projection (a lens maps lines through the centre of projection; parallel lines meet at one point on the eye's level; the author's Figures 17 and 18 deny that, p.35-36 [read]). That is a geometric statement, not an experiment, and a camera's pictures of parallel rails contradict it daily.

Verdict. The vanishing-point rule as used: Internally inconsistent (size and height confused) and Refuted by calculation (a perspective cause gives no sharp waterline cut, and no dip). The telescope-restoration observation: Tested, not supported as proof of perspective; variations are consistent with refraction. Circumnavigation and the lost day are compatible with a plane: Plausible, untested as a plane-versus-globe discriminator (it discriminates nothing; the plane fails on the southern circuit).

Claim 4. The Sun, Moon and stars are small and near (Sun under 4,000 miles; stars within 6,000 miles; the Moon self-luminous)

a. Internal consistency. 1. Sun distance. Given altitudes a_S (south) and a_N (north) at one moment and a base b: H = b / (cot a_N - cot a_S). London 55 deg 18', Ackworth 53 deg 2', b = 151 mi: H = 2,508 mi [calc]. Southern 45 deg, northern 38 deg, b = 800 mi: H = 2,858 mi [calc]. Both are "under 4,000", as the book says. But a globe predicts that the altitude difference equals the latitude difference whatever the Sun's distance. For 151 mi that is 2.19 deg against the observed 2.27 deg (close); for 800 mi it is 11.6 deg, against the 7 deg the book quotes for the Southampton officer's data [calc]. That second data set, unsourced ("an officer ... at Southampton"), disagrees with a globe and with any consistent plane (it needs a nearer Sun on the day for the pair 800 mi apart than the one that matches the 151-mile pair). The author does not give dates, so the two sets may not be simultaneous; the method needs them to be. As a mathematical method the formula is correct; its hidden assumption is that the ground between the stations is flat. 2. The Sun's size. On a plane the Sun at height H over a disc moves along a circle at distance x from the centre. Its distance changes from H to sqrt(H^2 + x^2). With H = 3,000 mi and x from the Tropic (6,200 mi) to the south ring (20,000 mi) the Sun's distance changes by a factor 2.3 to 6.7 [calc], so its apparent diameter would change by the same factor between a station's noon and its sunset. Measured angular diameter changes by about 3% over a year and not at all between noon and evening [memory: high]. The book explains sunrise and sunset by perspective (a "vanishing point"), where the Sun gets lower as it recedes; a Sun at 3,000 miles height never goes below 8.5 deg altitude from the farthest point of a 20,000-mile disc [calc], so on the book's numbers the Sun does not set. 3. The tubes (p.74-75). The experiment: two parallel iron tubes one yard apart aimed at a star; a delay between the two observers' signals is said to show that the same star is not seen at the same moment in two parallel lines, so the Earth would show a similar shift if it moved a yard in six months. Two parallel lines of sight to a very distant star both hit it at once; a delay exists only if the lines differ. For the book's own star distance (6,000 miles) the shift between tubes is 0.9144 m / 9.66 x 10^6 m = 9.5 x 10^-8 rad = 0.0195 arcsec, a delay of 0.0195/15.04 = 1.3 ms at the sky's apparent speed of 15 arcsec per second [calc]. A knock signal cannot show 1.3 ms. The six-month version has a worse sensitivity problem: a globe moves 186 million miles across, not a yard, and the largest parallax (nearest star, 0.77 arcsec [memory: high]) shifts the end of a 2-yard tube sideways by 6.8 micrometres [calc]. The test as described is millions of times too coarse to see the effect the book says it excludes, and it does not touch aberration (about 20 arcsec [memory: high]), a known effect of orbital motion. 4. Moonlight (pp.105-107). Focused moonlight raising no thermometer is said to prove the Moon self-luminous. Moonlight is about 1/400,000 of sunlight [memory: high]; a 6,000x concentrator (Forbes) would give 0.0025 W/m^2 x 6,000 = 15 W/m^2 at best [calc], about 1.4% of full sunlight. That does not fuse gold, which is the book's contrast. Maximum possible irradiance from a focused image of the Moon cannot exceed the Moon's surface brightness times the lens's solid angle: for a perfect f/0.5 lens with Moon albedo 0.12, at most about 12% of full sunlight (120 W/m^2) [calc], and for Forbes's f/1.4 lens about 14 W/m^2. So "a brilliant focus, no heat" is a statement about size, not source. And the claim "no effect" was superseded: Piazzi Smyth made what is described as the first positive detection of the Moon's heat in 1856, and Lord Rosse's thermopile work (1869 onward) measured it [web: https://en.wikipedia.org/wiki/Charles_Piazzi_Smyth; web: https://www.scientificamerican.com/article/heat-from-the-moon/ (Scientific American's account of Rosse: the full Moon's surface raised more than 380 deg F above boiling water)]. The book's own Lancet report (a drop of 8 degrees from moonlight, p.107) is an observation with no mechanism, probably radiative cooling.

b. Experimental consistency. Star parallax (Bessel's 61 Cygni, 1838, about 0.3 arcsec [memory: high]) and aberration (Bradley 1727) were measured before the book. Both need Earth's orbital motion and stars at light-years. The book's tube test is not sensitive enough to bear on either.

c. Tested? The Sun triangulation is a computation with unsourced data, not a test. The author's "officers of the Ordnance Survey" data was not found independently by the Investigator. The moon-heat claim was tested (1856 onward) by others with a positive result.

d. If true. Parallax, aberration, the Sun's constant angular size, planetary orbits (Kepler's third law: 36.15^1.5 = 217 years, 30.07^1.5 = 165 years [calc]) and the whole scale of the solar system would need replacing; the Sun's output at 3,000 miles height would be astronomically different from the measured solar constant.

Verdict. Near Sun and stars: Refuted by calculation (Polaris and Sun heights inconsistent between stations; Sun's size change; Sun never sets). Moon self-luminous from no-heat: Tested, not supported (heat detected in 1856 and after) and a non sequitur (Internally inconsistent reasoning: low focal heat follows from low brightness, not source). Tubes: Not testable as stated at the sensitivity described.

Claim 5. The Foucault pendulum does not prove the Earth rotates

I checked this one most closely, since the deep read had its figures in hand.

a. Internal consistency. The mathematics of the standard result. In the rotating frame, with x east, y north, small swings, pendulum frequency w0 = sqrt(g/L) and Omega_z = Omega sin(latitude) the vertical component of the Earth's spin, the equations are

x'' = -w0^2 x + 2 Omega_z y', y'' = -w0^2 y - 2 Omega_z x'.

(The extra terms are the Coriolis term, the sideways apparent force on a moving body in a rotating frame.) Writing z = x + iy gives z'' + 2i Omega_z z' + w0^2 z = 0, whose solution is z = e^(-i Omega_z t)(A e^(i w t) + B e^(-i w t)) with w = sqrt(w0^2 + Omega_z^2). A swing along a line is the case |A| = |B|; the factor e^(-i Omega_z t) rotates the whole plane at rate Omega_z, i.e. 15 deg/h x sin(latitude). I integrated these equations numerically for the Manchester pendulum (L = 7.62 m, 53.48 N, Omega_z = 12.09 deg/h) over 7 minutes: the swing plane turned 1.34 deg, against Omega_z x 390 s = 1.31 deg [calc]. The book's figures for Paris, London, Edinburgh, Cairo and Calcutta agree with the rule (deep read, re-checked above).

"B. A. C.'s" argument (the deviation on the way out is cancelled on the way back, so only the pole works) [read: pp.150-152] is wrong, and the numerical integration shows it. The Coriolis force reverses with the velocity, but the displacement it makes on the outward leg is still there when the bob turns, so the path becomes an open loop and each swing's plane is rotated slightly: the cumulative effect is the rotation I found, with no cancellation. "T. T.'s" argument (swings are artificial, so the experiment is humbug) is philosophical, and the author reports that he does not dispute the facts [read: p.152].

The author's three points: 1. The plane varies, so it proves nothing. The premise is partly right: pendulum planes do vary. The main cause is elliptical motion. If the release has a sideways error b (and a is the swing amplitude), the ellipse itself precesses at about (3/8) w0 a b / L^2 radians per second [memory: medium; the formula is Airy's of 1851, and I could not confirm it on the web: Wikipedia's Foucault article names Kamerlingh Onnes's analysis of the elliptical drift but not the formula (https://en.wikipedia.org/wiki/Foucault_pendulum)]. A direct numerical check: for Jones's pendulum (a = 1.46 m from the 360-inch chalk circle's radius), a sideways release error of 6.2 mm (a quarter of an inch) gives 13.7 deg/h of extra precession, which is exactly the excess Jones saw (observed about 25.7 deg/h against the predicted 12.1 deg/h, ratio 2.1) [calc; the formula's inputs are mine, the ratio is the book's]. A 6-mm release error is far below what the chalk lines (one inch per degree) could show. And the sign of the error sets whether the extra precession adds or subtracts. That is how a well-run experiment gives "the wrong direction" and "the wrong rate" with a careless release, and Jones had warned of "some slight lateral bias at starting" beforehand [read: page image p.155]. The scatter is real; it is explained by an error term, which makes the experiment careful, not worthless. I checked by integration that the pure Coriolis case at the same length gives the clean 1.31 deg per 390 s; the elliptical term is a separate additive effect. 2. The observation cannot say whether the Earth moves under the pendulum or the pendulum over the Earth. This is a logical point. In pure kinematics the two descriptions are equivalent, and the author says so correctly. What decides is that in Newtonian mechanics (the same mechanics he uses to compute a falling ball in his Fig.22) the plane of a free pendulum is fixed relative to the distant stars, and the sine-of-latitude rate then follows from the Earth's spin with no other parameter. The book has no model that gives sin(latitude) in the other description. A plane that rotates at exactly 15 sin(latitude) deg/h, zero at the equator and a full turn per day at the pole, for any pendulum, is a nontrivial quantitative fact the plane-Earth view must explain; the book does not try. The author's charge of circular reasoning is fair against the 1851 popular accounts (rotation assumed, then used as the explanation); it is not fair against Foucault's calculation, which predicts the rate before measuring. 3. A ball-and-socket suspension twists and causes the turning. The deep read found that the only "globular piece in a bell-shaped nut" described is Jones's [read: p.155]; the book says "as was M. Foucault's" without a source [read: p.159]. Foucault is reported to have used a steel wire hung from a steel plate [read: p.145]. A twisting joint would add a random, not a systematic, turn; it cannot produce a rate that follows 15 sin(latitude) at five different latitudes, a point the author's own table supplies (his table is predicted, not measured, per the deep read). The author reports no experiment of his own.

The magnetic-meridian reports (an unnamed writer; "a scientific gentleman in Dundee", Liverpool Journal 17 May 1851) are observations without numbers. Both bobs described (Foucault's and Jones's) were brass [read: pp.144, 154]. This remains in the record: untested as stated, though testable (turn the pendulum's release azimuth and compare with the magnetic meridian; the sine law predicts latitude dependence and no magnetism).

b. Experimental consistency. The rule and the cities' periods match the book's own Foucault figures (deep read). Wikipedia states the standard result, Omega_p = Omega sin(phi) [web: https://en.wikipedia.org/wiki/Foucault_pendulum]. I have no source here for the number of replications or the Pantheon details (deep read: Pantheon pendulum about 67 m, 28 kg [memory: high]).

c. Tested? By Foucault (Comptes Rendus 3 Feb 1851), Arago (Observatory, 11 m), the Pantheon, Jones (Manchester), and unnamed others. Jones's case reported the right direction and 2.1 times the rate, in one interval under 7 minutes; the passage does not say if it continued or how he explained it [read, via deep read].

d. If true. If the plane's turning had nothing to do with the Earth's spin, the sine-of-latitude rate would be a coincidence at every latitude. Mechanics would need an unexplained force with Coriolis's exact form; the gyroscopic compass [memory: high] and the weather-pattern rotation, which share it, would also fail.

Verdict. Strong form (the pendulum plane's turn is erratic and has no relation to the Earth's rotation): Tested, not supported, and the argument for it is Internally inconsistent in the B. A. C. step (Refuted by calculation, shown above). Weak form (short, carelessly released pendulums give erratic results, and the popular 1851 reasoning assumed what it set out to show): Established. Magnetic meridian: Never properly tested (reported without data; not dismissed by testing in these pages).

Claim 6. Orthodox astronomy rests on assumed premises and has failed its tests (Neptune, air-gun and mast-drop, tides)

a. Internal consistency. - Neptune [read: pp.137-139]. The quoted figures are right as figures: Le Verrier's planet at 36 times the Earth's distance with a 217-year period, against 30 and 165 years for Neptune. Kepler's third law gives 36.15^1.5 = 217.4 and 30.07^1.5 = 164.9 [calc], so the book's numbers are mutually consistent. The inference ("all his theory falls to the ground") does not follow: Le Verrier named a position, and Galle found the planet within about 1 degree of it on 23 Sep 1846 [web: https://en.wikipedia.org/wiki/Neptune]. Mathematically, over the short arc observed in 1846 the position depends on a combination of distance and mass that cannot be separated from a few decades of perturbations; the predicted orbit later diverged, as the book's quotation says. This is a statement about the precision of an inverse problem, not about heliocentrism, and the same Kepler's law holds on both sides. The book's own source says Le Verrier "to a great extent admitted his own error" about the orbit [read: p.138]. - Mast-drop and air-gun [read: pp.68-74]. Internally inconsistent. Dropping a ball from a moving mast (the author grants this, "the explanation is granted", p.69) works because the ball keeps the ship's horizontal motion while falling. Throwing it upward the book says the horizontal motion and the upward motion both "expend themselves" at the top and the ball then falls only by gravity, "far behind the ship" [read: p.71]. No force acts horizontally in either case. The same retained momentum that carries the dropped ball along must carry the thrown ball along; the horizontal motion cannot vanish at the top of the path. The railway-carriage ball, thrown up and caught in the hand, is the everyday case, and the author's description of it is contradicted by the same principle he granted. The numbers: 25,000 mi x cos 51 deg is 15,700 miles in 24 h, i.e. 961 ft/s, not 700 ft/s; the book says "more than 700 feet per second", and then multiplies by 8 seconds to get 5,600 ft [read: pp.72-74; calc]. At 961 ft/s the figure would be 7,700 ft. It does not matter: both are the wrong quantity. The effect expected on the rotating Earth for a 16-second flight (apex about 1,030 ft in vacuum) is the Coriolis deflection (4/3) Omega cos(latitude) g t^3 with t = 8 s: about 0.31 m (12 in) westward [calc]. So the globe predicts about a foot, not 5,600 ft; the book's reported result ("within a few inches of the gun; twice on the very mouth of the barrel") is well under that foot, but with air drag the real flight is far shorter than 16 s from a pellet and wind drift dominates a foot-scale test, so the result is consistent with a null and with a small effect alike. The experiment is anonymous and gives no date, wind or muzzle speed. - Tides (pp.112-113, digest and read in part). The cause is a floating plane rocking under atmospheric pressure. A steady pressure moves nothing; an oscillating one would have to follow the lunar day (12 h 25 min) with metre-scale amplitude. A pressure change of 1 hPa raises the equilibrium water level by about 1.0 cm (hydrostatics, delta-h = delta-P / (rho g) = 100 / (1000 x 9.81) = 0.0102 m [calc]); storms of 20 to 30 hPa move it by 20 to 30 cm [calc], and the semi-diurnal atmospheric pressure oscillation is about a hectopascal [memory: medium] against tidal ranges of metres. The book offers no equations, amplitude or timing, so there is nothing to check but the order of magnitude, which fails by two orders.

b. Experimental consistency. Neptune, observed 1846; Coriolis effects as in claim 5. Tides keep lunar timing [memory: high].

c. Tested? The book's cited experiments are anonymous, single and unreplicated. No evidence the air-gun test was repeated.

d. If true. Galilean relativity (a ball thrown up in a uniformly moving frame comes down in the same frame) would fail, with consequences for all mechanics, including the book's own ship experiments.

Verdict. Neptune: Established as fact (the orbit differed) but the conclusion drawn is Internally inconsistent with the planet being found where predicted; the claim that it shows astronomy "failed" is Tested, not supported. Air-gun and mast-drop: Internally inconsistent and Refuted by calculation (retained momentum; Coriolis deflection about 0.3 m). Tides: Refuted by calculation (order-of-magnitude, pressure-to-height).

Claim 7. Scripture is literally true on the form of the Earth

a-d. Not a mathematical or physical claim in the form given (book Section 14, pp.178-229, digest only). The only numerical items in the facts are a roughly 6,000-year chronology and a Herschel 1800 figure of 1,900,000 years of light travel; I did not read these pages in full. Interpretation of the text is outside my scope. Verdict: Not testable as stated. It stays in the record as the author's stated motive: the "zetetic" method (facts first, theory rejected) is applied to the shape of the Earth and then used to answer "atheism" and the "plurality of worlds" (facts.json).

Investigator's findings I keep, with my assessment

• Bedford geometry (sites.md): reproduced, 13.7 ft without refraction and 11.1 ft at k = 0.13 [calc].
• Wallace 1870: reproduced, 6.00 ft and 5.22 ft [calc]; the umpire's finding is documented in one published source (Nature 1, 581, 1870) and in the Cambrian of 15 Apr 1870 (via the Investigator); the movement's own printed reports (Hampden, Carpenter) are by interested parties.
• Oldham 1901, Stretton 1905: curvature found, 8 in per mile at Stretton matches the formula [calc].
• Blount/Clifton 1904: a photograph of a sheet at 6 miles at 2 ft; unread primary; needs the sheet's size. For a camera at 2 ft the hidden height is 12.3 ft at 6 miles, so a standard sheet should not be visible without strong refraction [calc]. The photo's existence is reported by Wikipedia from The Earth vol.5, which I have not read; the record remains open.
• Le Fevre (balloon horizon concave): first-hand, one source (Denbighshire Free Press, 4 Feb 1882). The geometry is explained by dip against the nadir (above).
• Mirror test: reproduced (0.031 in) [calc].
• Gleason patent US 497,917: the patent describes geography agreeing with globe maps and the departure as distortion; this is the patent holder's description. It is consistent with my map-distance check.
• The Anti-Newtonian (1819) as the likely source of the north-centred disc, Koreshan concave Earth (1896), followers' works: kept as related work. The Koreshan result matters mathematically: the same sight lines are read as concave.
• RAS, 8 Dec 1848 and the lecture trail: historical record. De Morgan's remark that no account is in the Notices is a report about the Society's records; it does not by itself test the claim. Deciding record: the RAS minute book.
• Record going quiet: the Investigator found no disappearance; the record continues through the Zetetic Society (1883) and its successor (1893). Nothing for me to add.

Summary of verdicts

Limits of this review: web checks done: Foucault pendulum (Wikipedia: rate formula confirmed, Airy formula not shown), Neptune (Wikipedia: 1-degree miss; no orbital elements on the page read), lunar heat (search results: Smyth 1856, Rosse). Not checked: the refraction range k = 0.07 to 0.2 [memory: medium], Findlay's range tables, the semi-diurnal pressure figure [memory: medium], the 1,000-ft/16-s air-gun flight. No page image of pp.56-57 for the southern-circuit figures.