Sources page biographical material

6.BF.2. SLIDING SPEAR = LEANING REED
A spear (or ladder) of length H stands against a wall. Its base moves out B from the wall, causing the top to slide down D. Hence B² + (H - D)² = H².

The leaning reed has height H. It reaches D out of the water when it is straight up. When it leans over, it is just submerged when it is B away from its upright position. This is identical to the sliding spear turned upside down.

In these problems, two of H, B, D are given and one wants the remaining value. I will denote them, by e.g. H, D = 30, 6.

See Tropfke, p. 619 & 621.

BM 85196. Late Old Babylonian tablet in the British Museum, c 1800. Transcribed, translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte II; Springer, Berlin, 1935, pp. 43++. Prob. 9 -- translation on pp. 47-48, commentary on p. 53. Quoted in B. L. van der Waerden; Science Awakening; OUP, 1961, p. 76. See also: J. Friberg; HM 8 (1981) 307-308. Sliding beam(?) with H, D = 30, 6 and with H, B = 30, 18.

BM 34568. Seleucid period tablet in the British Museum, c 300. Transcribed, translated and commented on by O. Neugebauer; Mathematische Keilschrift-texte III, Springer, Berlin, 1937, pp. 14-22 & plate 1. Prob. 12 -- translation on p. 18, commentary on p. 22. Quoted in van der Waerden, pp. 76 77. See also: J. Friberg, HM 8 (1981) 307-308. Sliding reed or cane, B, D = 9, 3.

Papyri Cairo J. E. 89127 30 & 89137 43. c 260. Shown and translated in: Richard A. Parker; Demotic Mathematical Papyri; Brown Univ. Press, Providence, 1972; pp. 1, 3 4, 35 40 & Plates 9 10.

Prob. 24 26: H, B = 10, 6; 14½, 10; 10, 8.

Prob. 27 29: H, D = 10, 2; 14½, 4; 10, 4.

Prob. 30 31: B, D = 6, 2; 10, 4.

Chiu Chang Suan Ching (Jiu Zhang Suan Shu). c 150? Chap. IX.

Prob. 6, p. 92. [English in Mikami, p. 22.] Leaning reed, B, D = 5, 1.

Prob. 7, p. 93. Version with a rope hanging and then stretched giving B, D = 8, 3.

Prob. 8, p. 93. [English in Mikami, p. 22 and in Swetz & Kao, pp. 30 32 and in HM 4 (1977) 274.] Ladder, but with vertical and horizontal reversed. B, D = 10, 1. Mikami misprints the answer as 55 rather than 50.5.

Bhaskara I. 629. Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97 103, with reproductions of original diagrams on pp. 101-102; English version of the examples is on pp. 296-300. The material of interest is examples 6 and 7. In the setup of 6.BF.3, the lotus is OBA and LBM is the water level.

Ex. 6: B, D = 24, 8. Shukla notes this is used by Chaturveda.

Ex. 7: B, D = 48, 6.

Chaturveda. 860. Commentary to the Brahma sphuta siddhanta, chap. XII, section IV, v. 41, example 3. In Colebrooke, pp. 309 310. Leaning lotus: B, D = 24, 8.

Tabari. Miftāh al-mu‘āmalāt. c1075. P. 124, no. 42. ??NYS - cited by Tropfke 621.

Bhaskara II. Lilavati. 1150. Chap. VI, v. 152 153. In Colebrooke, pp. 66. Leaning lotus: B, D = 2, ½.

Bhaskara II. Bijaganita. 1150. Chap. IV, v. 125. In Colebrooke, p. 204. Same as Lilavati.

Fibonacci. 1202. P. 397 (S: 543). Sliding ladder: H, B = 20, 12.

Leonardo Fibonacci. La Practica di Geometria. Volgarizzata da Cristofano di Gherardo di Dino, cittadino pisano. Dal Codice 2186 della Biblioteca Riccardiana di Firenze, 1448. Ed. by Gino Arrighi, Domus Galilaeana, Pisa, 1966. P. 37 and fig. 18. Same as Fibonacci 1202.

Zhu Shijie. Siyuan Yujian (Precious Mirror of the Four Elements). 1303. ??NYS -- English given in Li & Du, p. 179. Questions in verse, no. 1. Two reeds, 14 apart, which reach 2½ and 1 out of the water. When they lean together, they just touch at the water surface. The water is assumed to have the same depth at both reeds, which reduces the problem from two variables to one variable.

Gherardi?. Liber habaci. c1310. Pp. 139 140. H, B = 20, 12

Gherardi. Libro di ragioni. 1328. Pp. 77 78. Ship's mast with H, D = 131, 4. B&W picture on p. 77, from f. 46v.

Columbia Algorism. c1350.

No. 135, pp. 138 139. Sliding ladder: H, B = 10, 6.

No. 140, pp. 149 150. Leaning tree: H, B = 20, 10.

(I have colour slides of the illustrations to these problems.)

Pseudo-dell'Abbaco. c1440. No. 167, pp. 138-140, with B&W picture on p. 139. Sliding spear: H, D = 30, 4. I have a colour slide of this.

della Francesca. Trattato. c1480. F. 44r (108). B, D = 6, 2. English in Jayawardene.

Pacioli. Summa. 1494. Part II, ff. 54v-55v.

Prob. 25 (misprinted 52). B, D = 6, 2. = della Francesca.

Prob. 26. H, H - D = 10, 8.

Prob. 27. H, D = 10, 4.

Prob. 28. H, D = 10, B/3.

Prob. 29. D + H = 12, DH = 12.

Prob. 30. H = D + 4, DH = 12.

Prob. 41. Tree of height 40 with a rope of length 50 tied to the top which reaches to the ground at a point 30 away. Length 10 of the rope is pulled, causing the tree to lean. How high is the top of the tree now? We now have a triangle of sides 40, 40, 30 and want the altitude to the side of length 30.

PART II.

F. 68r, prob. 98. Tree of height 30 has a rope of length 50 tied to the top, so it reaches the ground 40 away. How much rope has to be pulled to move the top of the tree to being 8 away from the vertical? He neglects to give the value 50 in the problem statement and the ground distances of 8 and 32 are interchanged in the diagram.

van Etten. 1624. Prob. 89 (86), part V (4), p. 135 (214). Sliding ladder: H, B = 10, 6.

Ozanam. 1694. Prob. 42 & fig. 48, plate 10, 1696: 123-124; 1708: 129; 1725: 320-321 & plate 10 (11). Ladder 25 long with foot 7 from wall. Foot is pulled out 8 more -- how much does the top come down?

Vyse. Tutor's Guide. 1771? Prob. 16 (in verse), 1793: p. 179, 1799: p. 190 & Key p. 228. H, B = 100, 10.

Mittenzwey. 1880. Prob. 294, pp. 53-54 & 104; 1895?: 324, pp. 57 & 106; 1917: 324, pp. 52 & 100. Leaning reed, B, D = 5, 1. [I have included this and the next entry as this problem is not so common in the 19C and 20C as in earlier times.]

Clark. Mental Nuts. 1904, no. 69; 1916, no. 95. The boatman's puzzle. Leaning pole, B, D = 12, 6. Find H - D.
6.BF.3. WELL BETWEEN TWO TOWERS
The towers have heights A, B and are D apart. A well or fountain is between them and equidistant from the tops of the towers. I denote this by (A, B, D). Vogel, in his DSB article on Fibonacci, says the problem is Indian, and Dold pointed me to Mahavira. Pseudo dell'Abbaco introduces the question of a sliding weight or pulley -- see Pseudo dell'Abbaco, Muscarello, Ozanam-Montucla, Tate, Palmaccio, Singmaster.

I have just found that Bhaskara I gives several unusual variations on this.

See Tropfke, p. 622. See also 10.U.
INDEX of A, B, D problems, with A  B.
0 4 8 Chaturveda

0 9 27 Bhaskara II

0 12 24 Bhaskara I

0 18 81 Bhaskara I

5 6 12 Bhaskara I

10 10 12 Bhaskara I

13 15 14 Mahavira

18 22 20 Mahavira

20 24 22 Mahavira

20 30 30 Gherardi?

20 30 50 Perelman

30 40 50 Fibonacci, Muscarello, Cardan

30 50 100 Bartoli

30 70 100 Tate

40 50 30 Muscarello

40 50 70 Pseudo-dell'Abbaco

40 50 100 della Francesca

40 60 50 Lucca 1754

60 80 100 Calandri c1485

70 100 150 Columbia Algorism, Pacioli

80 90 100 Calandri 1491
Bhaskara I. 629. Commentary to Aryabhata, chap. II, v. 17, part 2. Sanskrit is on pp. 97 103; English version of the examples is on pp. 296-300. The material of interest is examples 2 and 3.

These are 'hawk and rat problems'. A hawk is sitting on a wall of height A and a rat is distance D from the base of the wall. The rat tries to get to its hole, in the wall directly under the hawk. The hawk swoops, at the same speed as the rat runs, and catches the rat when it hits the ground. Hence this is the same as our two tower problem, but with B = 0, so I will denote this version by (A, 0, D). Bhaskara I attributes this type of problem to unspecified previous writers. Shukla adds that later writers have it, including Chaturveda and Bhaskara II, qqv.

Ex. 2: (12, 0, 24).

Ex. 3: (18, 0, 81). Bhaskara I explains the solution in detail and Shukla gives an English precis of it. Let ABOC be the horizontal diameter of a circle and let LBM be a vertical chord. LB is our pole, with the hawk at L, and the rat is at C and wants to get to B. The point of capture is O, because LO = OC. From LB² = AB x BC, we can determine AB and hence the other values.

Looking at Chaturveda (below), I now see that turning this sideways gives the same diagram as the broken bamboo problem -- the tree was BC and breaks at O to touch the ground at L. So the broken bamboo problem (H, D) is the same as the two towers or hawk and rat problem (D, 0, H).

Bhaskara I. 629. Ibid. Examples 8 and 9 are 'crane and fish problems'. A fish is at the NE corner of a rectangular pool and a crane is at the NW corner and they move at the same speeds. The fish swims obliquely to the south side, but the crane has to walk along the edge of the pool. The fish unfortunately gets to the side just as the crane reaches the same point and gets eaten. This again like our two tower problem, but with one pigeon unable to fly, so it has to walk down the tower and across. Because the pool is rectangular, the two values A and B are equal.

Ex. 8: (6, 6, 12).

Ex. 9: (10, 10, 12). The meeting point is 3 3/11 from the SW corner.

Chaturveda. 860. Commentary to the Brahma sphuta siddhanta, chap. XII, section IV, v. 41, example 4. In Colebrooke, p. 310. Cat and rat, where the cat behaves like the hawk of Bhaskara I: (4, 0, 8).

Mahavira. 850. Chap. VII, v. 201-208, pp. 249-251.

He gives several problems, but he usually also asks for the equal distance from the top

of each tower to the fountain.

v. 204. Two pillars, with a rope between them which touches the ground but with equal lengths to the tops. (13, 15, 14).

v. 206. Two hills with mendicants who are able to fly along the hypotenuses. (22, 18, 20)

v. 208. Same context. (20, 24, 22).

Bhaskara II. Lilavati. 1150. Chap. VI, v. 149 150. In Colebrooke, pp. 65 66. Peacock and snake version of the hawk and rat problem: (9, 0, 27).

Fibonacci. 1202. De duabus avibis [On two birds], pp. 331 332 (S: 462-463). (40, 30, 50). He does the same problem differently on pp. 398 399 (S: 544-545).

Gherardi?. Liber habaci. c1310. P. 139. (20, 30, 30).

Lucca 1754. c1330. F. 54v, pp. 120-121. (60, 40, 50).

Columbia Algorism. c1350. No. 136, pp. 139 140. (70, 100, 150).

Bartoli. Memoriale. c1420. Prob. 10, f. 76r (= Sesiano 138-139 & 148-149, with reproduction of the relevant part of f. 76r on p. 139). (50, 30, 100).

Pseudo-dell'Abbaco. c1440.

Prob. 80, p. 72, with picture on p. 71. (50, 40, 70).

Prob. 158 159, pp. 129 133, with illustrations on pp. 130 & 132, deal with the related problem where a rope with a sliding weight hangs between two towers, and the diagram clearly shows the weight in the air, not reaching the ground, so that the resulting triangles are similar. [I found it an interesting question to determine when the rope was long enough to reach the ground, and if not, how much above the ground the weight was -- see Muscarello, Ozanam-Montucla, Singmaster below.]

Prob. 158 has A, B, D = 40, 60, 40 and a rope of length L = 110, so the rope is more than long enough for the weight to reach the ground, but all he does is show that the two parts of the rope are 66 and 44, which is a bit dubious as there is slack in the rope. The diagram clearly shows the weight in the air. I have a colour slide of this.

Prob. 159 has A, B = 40, 60 with L = 120 such that the weight just touches the ground -- find the distances of the weight to the towers.

Prob. 160, p. 133. (40, 30, 50).

Muscarello. 1478.

Ff. 95r-95v, pp. 222-223. A, B, D = 50, 40, 30. Place a rope between the towers just long enough to touch the ground.

Ff. 95v-96r, pp. 223-224. A, B, D = 30, 20, 40. A rope of length L = 60 with a sliding weight is stretched between them -- where does the weight settle?

F. 99r, pp. 227-228. Fountain between towers for doves: (40, 30, 50).

della Francesca. Trattato. c1480. F. 22r (72). (40, 50, 100). English in Jayawardene.

Calandri. Aritmetica. c1485. Ff. 89r-89v, pp. 178 179. (60, 80, 100). (Tropfke, p. 599, shows the illustration in B&W.)

Calandri. Arimethrica. 1491. F. 100v. Well between two towers. (80, 90, 100). Nice double size woodcut picture.

Pacioli. Summa. 1494. Part II.

F. 59v, prob. 62. (70, 100, 150) = Columbia Algorism.

Ff. 59v-60r, prob. 63. A, B, D = 30, 40, 20 with rope of length 25 between the towers with a sliding lead weight on it. How high is the weight from the ground.

Ff. 61r-61v, prob. 66. Three towers of heights A, B, C = 125, 135, 125, with distances AB, AC, BC = 150, 130, 140. Find the point on the ground equidistant from the tops of the towers.

Cardan. Practica Arithmetice. 1539. Chap. 67.

Section 9, f. NN.vi.r (p. 197). (40, 30, 50).

Section 10, ff. NN.vi.r - NN.vii.v (pp. 197-198). Three towers of heights A, B, C = 40, 30, 70, with distances AB, AC, BC = 50, 60, 20. Find the point on the ground equidistant from the tops of the towers. Same idea as Pacioli, prob. 66.

Ozanam Montucla. 1778. Vol. II, prob. 7 & fig. 5, plate 1. 1778: 11; 1803: 11-12; 1814: 9 10; 1840: 199. Rope between two towers with a pulley on it. Locate the equilibrium position. Uses reflection.

Carlile. Collection. 1793. Prob. XLV, p. 25. Find the position for a ladder on the ground between two towers so that leaning it each way reaches the top of each tower. (80, 91, 100). He simply states how to do the calculation, x = (D² + A² - B²)/2D for the distance from the base of tower B.

T. Tate. Algebra Made Easy. Chiefly Intended for the Use of Schools. New edition. Longman, Brown, Green, and Longman, London, 1848. P. 111.

No. 36. A, B, D = 30, 70, 100. Locate P such that the towers subtend the same angle, i.e. the two triangles are similar. Clearly P divides D as 30 to 70.

No. 37. Same data. Locate P so the distance to the tops is the same. This gives 70 to 30 easily because A + B = D.

No. 38. Same data. Locate P so the difference of the squares of the distances is 400. Answer is 68 from the base of the shorter tower.

Perelman. MCBF. 1937. Two birds by the riverside. Prob. 136, pp. 224-225. (30, 20, 50). "A problem of an Arabic mathematician of the 11th century."

Richard J. Palmaccio. Problems in Calculus and Analytic Geometry. J. Weston Walch, Portland, Maine, 1977. Maximum-Minimum Problems, No. 3, pp. 9 & 70-71. Cable supported pulley device over a factory. A, B, D = 24, 27, 108 with cable of length L = 117. Find the lowest point. He sets up the algebraic equations corresponding to the the two right triangles, assumes the distance from one post and the height above ground are implicit functions of the length of the cable from the pulley to the same post and differentiates both equations and sets equal to zero, but it takes half a page to get to the answer and he doesn't notice that the two triangles are similar at the lowest point.

David Singmaster, proposer; Dag Jonsson & Hayo Ahlburg, solvers. Problem 1748: [The two towers]. CM 18:5 (1992) 140 & 19:4 (1993) 125-127. Based on Pseudo-dell'Abbaco 158-159, but the solution by reflection was later discovered to be essentially Ozanam Montucla.

David Singmaster. Symmetry saves the solution. IN: Alfred S. Posamentier & Wolfgang Schulz, eds.; The Art of Problem Solving: A Resource for the Mathematics Teacher; Corwin Press, NY, 1996, pp. 273-286. Gives the reflection solution.

Yvonne Dold-Samplonius. Problem of the two towers. IN: Itinera mathematica; ed. by R. Franci, P. Pagli & L. Toti Rigatelli. Siena, 1996. Pp. 45-69. ??NYR. The earliest example she has found is Mahavira and it was an email from her in about 1995 that directed me to Mahavira.

6.BF.4. RAIL BUCKLING.
A railway rail of length L and ends fixed expands to length L + ΔL. Assuming the rail makes two hypotenuses, the middle rises by a height, H, satisfying H² = {(L+ΔL)/2}²   (L/2)², hence H  (LΔL/2).

However, one might assume the rail buckled into an arc of a circle of radius r. If we let the angle of the arc be 2θ, then we have to solve rθ = (L + ΔL)/2; r sin θ = L/2. Taking sin θ  θ - θ³/6, we get r²  (L + ΔL)³/ 24 ΔL. We have H = r (1 - cos θ)  rθ²/2 and combining this with earlier equations leads to H  {3(L+ΔL)ΔL/8} which is about 3 / 2 = .866... as big as the estimate in the linear case.

The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. P. 149, prob. 12. L = 1 mile, ΔL = 1 ft or 2 ft -- text is not clear. "Answer: More than 54 ft." However, in the linear case, ΔL = 1 ft gives H = 51.38 ft and ΔL = 2 ft gives H = 72.67 ft, while the exact answers in the circular case are 44.50 ft and 62.95 ft.

Sullivan. Unusual. 1943. Prob. 15: Workin' on the railroad. L = 1 mile, ΔL = 2 ft. Answer: about 73 ft.

Robert Ripley. Mammoth Believe It or Not. Stanley Paul, London, 1956. If a railroad rail a mile long is raised 200 feet in the centre, how much closer would it bring the two ends? I.e. L = 1 mile, H = 200 ft. Answer is: "less than 6 inches". I am unable to figure out what Ripley intended.

Jonathan Always. More Puzzles to Puzzle You. Tandem, London, 1967. Gives the same question as Ripley with answer "approximately 15 feet". The exact answer is 15.1733.. feet or 15 feet 2.08 inches.

David Singmaster, submitter. Gleaning: Diverging lines. MG 69 (No. 448) (Jun 1985) 126. Quotes from Ripley and Always.

David Singmaster. Off the rails. The Weekend Telegraph (18 Feb 1989) xxiii & (25 Feb 1989) xxiii. Gives the Ripley and Always results and asks which is correct and whether the wrong one can be corrected -- cf Ripley above.

Phiip Cheung. Bowed rail problem. M500 161 (?? 1998) 9. ??NYS. Paul Terry, Martin S. Evans, Peter Fletcher, solvers and commentators. M500 163 (Aug 1998) 10-11. L = 1 mile, ΔL = 1 ft. Terry treats the bowed rail as circular and gets H = 44.49845 ft. Evans takes L = 1 nautical mile of 6000 ft and gets almost exactly H = 50 ft. Fletcher says it took 15 people to lift a 60ft length of rail, so if someone lifted the 1 mile rail to insert the extra foot, it would need about 1320 people to do the lifting.
6.BF.5. TRAVELLING ON SIDES OF A RIGHT TRIANGLE.
New section. See also the Mittenzwey example in 10.A.6.
Brahmagupta. Brahma sphuta siddhanta. 628. Chap. XII, sect. IV, v. 39. In Colebrooke, p. 308. Rule for the problem illustrated by Chaturveda.

Mahavira. 850. Chap. VII, v. 210-211, pp. 251-252. A slower traveller goes due east at rate v. A faster traveller goes at rate V and starts going north. After time t, he decides to meet the other traveller and turns so as to go directly to their meeting point. How long, T, do they travel? This gives us a right triangle with sides vT, Vt, V(T-t) leading to a quadratic in T whose constant term drops out, yielding T = 2t V²/(V²-v²). If we set r = v/V, then T = 2t/(1-r²), so we can determine T from t and r without actually knowing V or v. Indeed, if we let ρ = d/D, we get 2ρ = 1 - r². v, V, t = 2, 3, 5.

Chaturveda. 860. Commentary to the Brahma sphuta siddhanta, chap. XII, section IV, v. 39. In Colebrooke, p. 308. Two ascetics are at the top of a (vertical!) mountain of height A. One, being a wizard, ascends a distance X and then flies directly to a town which is distance D from the foot of the mountain. The other walks straight down the mountain and to the town. They travel at the same speeds and reach the town at the same time. Example with A, D = 12, 48.

Bhaskara II. Lilavati. 1150. Chap. VI, v. 154-155. In Colebrooke, pp. 66-67. Similar to Chaturveda. Two apes on top of a tower of height A and they move to a point D away. A, D = 100, 200.

Bhaskara II. Bijaganita. 1150. Chap. IV, v. 126. In Colebrooke, pp. 204-205. Same as Lilavati.
6.BG. QUADRISECT A PAPER SQUARE WITH ONE CUT
This involves careful folding. One can also make mn rectangles with a single cut.
Mittenzwey. 1880. Prob. 193 & 194, pp. 36 & 89; 1895?: 218 & 219, pp. 41 & 91; 1917: 218 & 219, pp. 37 & 87. Make four squares. Make four isosceles right triangles.

Walter Gibson. Big Book of Magic for All Ages. Kaye & Ward, Kingswood, Surrey, 1982. Tick-tack-toe, pp. 68-69. Take a 4 x 4 array and mark alternate squares with Xs and Os. By careful folding and cutting, one produces eight free squares and a connected lattice of the other eight squares, with the free squares being either all the Os or all the Xs, depending on how the final part of the cut is made.

David Singmaster. Square cutting. Used in my puzzle columns.

Weekend Telegraph (18 & 25 Mar 1989) both p. xxiii.

G&P, No. 16 (Jul 1995) 26. (Publication ceased with No. 16.)
6.BH. MOIRÉ PATTERNS
I have not yet found any real history of this topic. One popular book says moirés were first made in 15C China. The OED has several entries for Moire and Moiré. It originally refers to a type of cloth and may be a French adaptation of the English word mohair -- Pepys refers to 'greene-waterd moyre' and this is the earliest citation. In the early 19C, the term began to be used for the 'watered' effect on cloth and metal. At some point, the term was transferred to the optical phenomena, but the OED does not have this meaning.
Journal of Science and Arts 5 (1818) 368. On the Moiré Metallique, or Fer blanc moiré. ??NYS -- cited in OED as their first citation for the noun use of the term.

John Badcock. Domestic Amusements, or Philosophical Recreations, ... Being a Sequel Volume to Philosophical Recreations, or Winter Amusements. T. Hughes, London, nd [Preface dated Feb 1823]. [BCB 16-17; OCB, pp. 180 & 196. Heyl 21. Toole Stott 78 80. Wallis 34 BAD. HPL [Badcock]. These give dates of 1823, 1825, 1828.] Pp. 139-141, no. 169: Moiré Metal, or Crystallised Tin & no. 170: Moiré Watering, by other Methods. "Quite new and splendid as this art is, .... M. Baget, a Frenchman, however, claims the honour of a discovery of this process, attributing the same to accident, ...." Cited in the OED as the first adjectival use of the term, though the previous entry seems to also have an adjectival usage.

Rational Recreations. 1824. Experiment 16, p. 15: Metallic watering, or, fer blanc moire. Says it is of Parisian invention and gives the method of applying sulphuric acid to tin.

Endless Amusement II. 1826? Pp. 24-25: Application of the moiré métallique to tin-foil. This deals with obtaining a moiré effect in tin-foil and is quite different than Badcock.

Young Man's Book. 1839. Pp. 312-314. Identical to Endless Amusement II.

Tom Tit, vol. 2. 1892. Le papier-canevas et les figures changeantes, pp. 137-138. Uses perforated card.

Hans Giger. Moirés. Comp. & Maths. with Appls. 12B:1/2 (1986) [= I. Hargittai, ed., Symmetry -- Unifying Human Understanding, as noted in 6.G.] 329 361. Giger says the technique of moiré fabrics derives from China and was first introduced into France in 1754 by the English manufacturer Badger (or Badjer). He also says Lord Rayleigh was the first to study the phenomenon, but gives no references.
6.BI. VENN DIAGRAMS FOR N SETS
New topic. I think I have seen more papers on this and Anthony Edwards has recently sent several more papers.
Martin Gardner. Logic diagrams. IN: Logic Machines and Diagrams; McGraw Hill, NY, 1958, pp. 28-59. Slightly amended in the 2nd ed., Univ. of Chicago Press, 1982, and Harvester Press, Brighton, 1983, pp. 28-59. This surveys the history of all types of diagrams. John Venn [Symbolic Logic, 2nd ed., ??NYS] already gave Venn diagrams with 4 ovals and with 4 ovals and a disconnected set. Gardner describes various binary diagrams from 1881 onward, but generalised Venn diagrams seem to first occur in 1909 and then in 1938-1939, before a surge of interest from 1959. His references are much expanded in the 2nd ed. and he cites most of the following items.

John Venn. On the diagrammatic and mechanical representation of propositions and reasonings. London, Edinburgh and Dublin Philos. Mag. 10 (1880) 1-18. ??NYS -- cited by Henderson.

John Venn. Symbolic Logic. 2nd ed., Macmillan, 1894. ??NYS. Gardner, p. 105, reproduces a four ellipse diagram.

Lewis Carroll. Symbolic Logic, Part I. 4th ed., Macmillan, 1897; reprinted by Dover, 1958. Appendix -- Addressed to Teachers, sections 5 - 7: Euler's method of diagrams; Venn's method of diagrams; My method of diagrams, pp. 173-179. Describes Euler's simple approach and Venn's thorough approach. Reproduces Venn's four-ellipse diagram and his diagram for five sets using four ellipses and a disconnected region. He notes that Venn suggests using two five-set diagrams to deal with six sets and does not go further. He then describes his own method, which easily does up to eight sets. The diagram for four sets is the same as the common Karnaugh diagram used by electrical engineers. For more than four sets, the regions become disconnected with the cells of the four-set case being subdivided, using a simple diagonal, then his 2-set, 3-set and 4-set diagrams within each cell of the 4-set case. ?? -- is this in the 1st ed. -- ??NYS date??

Carroll-Gardner, p. 61, says this is in the first ed. of 1896.

William E. Hocking. Two extensions of the use of graphs in elementary logic. University of California Publications in Philosophy 2:2 (1909) 31(-??). ??NYS -- cited by Gardner who says Hocking uses nonconvex regions to get a solution for any n.

Edmund C. Berkeley. Boolean algebra and applications to insurance. Record of the Amer. Inst. of Actuaries 26:2 (Oct 1937) & 27:1 (Jun 1938). Reprinted as a booklet by Berkeley and Associates, 1952. ??NYS -- cited by Gardner. Uses nonconvex sets.

Trenchard More Jr. On the construction of Venn diagrams. J. Symbolic Logic 24 (Dec 1959) 303-304. ??NYS -- cited by Gardner. Uses nonconvex sets.

David W. Henderson. Venn diagrams for more than four classes. AMM 70:4 (1963) 424-426. Gives diagrams with 5 congruent irregular pentagons and with 5 congruent quadrilaterals. Considers problem of finding diagrams that have n-fold rotational symmetry and shows that then n must be a prime. Says he has found an example for n = 7, but doesn't know if examples can be found for all prime n.

Margaret E. Baron. A note on the historical development of logic diagrams: Leibniz, Euler and Venn. MG 53 (No. 384) (May 1969) 113 125. She notes Venn's solutions for n = 4, 5. She gives toothed rectangles for n = 5, 6.

K. M. Caldwell. Multiple set Venn diagrams. MTg 53 (1970) 29. Does n = 4 with rectangles and then uses indents.

A. K. Austin, proposer; Heiko Harborth, solver. Problem E2314 -- Venn again. AMM 78:8 (Oct 1971) 904 & 79:8 (Oct 1972) 907-908. Shows that a diagram for 4 or more sets cannot be formed with translates of a convex set, using simple counting and Euler's formula. (The case of circles is in Yaglom & Yaglom I, pp. 103-104.) Editor gives a solution of G. A. Heuer with 4 congruent rectangles and more complex examples yielding disconnected subsets.

Lynette J. Bowles. Logic diagrams for up to n classes. MG 55 (No. 394) (Dec 1971) 370 373. Following Baron's note, she gives a binary tooth like structure with examples for n = 7, 8.

Vern S. Poythress & Hugo S. Sun. A method to construct convex connected Venn diagrams for any finite number of sets. Pentagon (Spring 1972) 80-83. ??NYS -- cited by Gardner.

S. N. Collings. Further logic diagrams in various dimensions. MG 56 (No. 398) (Dec 1972) 309 310. Extends Bowles.

Branko Grünbaum. Venn diagrams and independent families of sets. MM 48 (1975) 12 22. Considers general case. Substantial survey of different ways to consider the problem. References to earlier literature. Shows one can use 5 identical ellipses, but one cannot use ellipses for n > 5.

B. Grünbaum. The construction of Venn diagrams. CMJ 15 (1984) 238 247. ??NYS.

Allen J. Schwenk. Venn diagram for five sets. MM 57 (1984) 297. Five ovals in a pentagram shape.

A. V. Boyd. Letter: Venn diagram of rectangles. MM 58 (1985) 251. Does n = 5 with rectangles.

W. O. J. Moser & J. Pach. Research Problems in Discrete Geometry. Op. cit. in 6.T. 1986. Prob. 27: On the extension of Venn diagrams. Considers whether a diagram for n classes can be extended to one for n+1 classes.

Mike Humphries. Note 71.11: Venn diagrams using convex sets. MG 71 (No. 455) (Mar 1987) 59. His fourth set is a square; fifth is an octagon.

J. Chris Fisher, E. L. Koh & Branko Grünbaum. Diagrams Venn and how. MM 61 (1988) 36 40. General case done with zig zag lines. References.

Anthony W. F. Edwards. Venn diagrams for many sets. New Scientist 121 (No. 1646) (7 Jan 1989) 51-56. Discusses history, particularly Venn and Carroll, the four set version with ovals and Carroll's four set version where the third and fourth sets are rectangles. Edwards' diagram starts with a square divided into quadrants, then a circle. Fourth set is a two-tooth 'cogwheel' which he relates to a Hamiltonian circuit on the 3-cube. The fifth set is a four-tooth cogwheel, etc. The result is rather pretty. Edwards notes that the circle in the n set diagram meets the 2^n-1 subsets of the n-1 sets other than that given by the circle, hence travelling around the circle gives a sequence of the subsets of n-1 objects and this is the Gray code (though he attributes this to Elisha Gray, the 19C American telephone engineer -- cf 7.M.3). The relationship with the n-cube leads to a partial connection between Edwards' diagram and the lattice of subsets of a set of n things.

New Scientist (11 Feb 1989) 77 has: Drawing the lines -- letters from Michael Lockwood -- describing a version with indented rectangles -- and from Anthony Edwards -- noting some errors in the article.

Ian Stewart. Visions mathématiques: Les dentelures de l'esprit. Pour la Science No. 138 (Apr 1989) 104-109. c= Cogwheels of the mind, IN: Ian Stewart; Another Fine Math You've Got Me Into; Freeman, NY, 1992, chap. 4, pp. 51-64. Exposits Edwards' work with a little more detail about the connection with the Gray code.

A. W. F. Edwards & C. A. B. Smith. New 3-set Venn diagram. Nature 339 (25 May 1989) 263. Notes connection with the family of cosine curves, y = 2^-n cos 2ⁿx on [0, π] and Gray codes and with the family of sine curves, y = 2^-n sin 2ⁿx on [0, 2π] and ordinary binary codes. Applying a similar phase shift to Edwards' diagram leads to diagrams where more than two set boundaries are allowed to meet at a point.

A. W. F. Edwards. Venn diagrams for many sets. Bull. Intern. Statistical Inst., 47th Session, Paris, 1989. Contrib. Papers, Book 1, pp. 311-312.

A. W. F. Edwards. To make a rotatable Edwards map of a Venn diagram. 4pp of instructions and cut-out figures. The author, Gonville and Caius College, Cambridge, CB2 1TA, 21 Feb 1991.

A. W. F. Edwards. Note 75.39: How to iron a hypercube. MG 75 (No. 474) (1991) 433-436. Discusses his diagram and its connection with the n-cube.

Anthony Edwards. Rotatable Venn diagrams. Mathematics Review 2:3 (Feb 1992) 19-21. +  Letter: Venn revisited. Ibid. 3:2 (Nov 1992) 29.