Sources page probability recreations

Part I, no. 8, pp. 8-9. Two men born & dying together, but living different numbers of days. Gives various explanations, e.g. differing calendars, living in the Arctic and going around the earth.

Part I, no. 9, p. 9. Two nearby places have their dates different. Again, this can be due to different sabbaths: Christians on Sunday, Graecians on Monday, Persians on Tuesday, Assyrians on Wednesday, Egyptians on Thursday, Turks on Friday, Jews on Saturday. Better, Macao and the Philippines differ by a day since one was colonized from the west and the other from the east.

Ozanam Montucla. 1778. Vol. III, prob. 5, 1778: 32-33; 1803: 34-35; 1814: 33-34; 1840: 415-416. Two men, born & dying together, but one living one day, or even two days, more than the other.

Philip Breslaw (attrib.). Breslaw's Last Legacy. 1784? Op. cit. in 6.AF. 1795: pp. 36-38.

Geographical Paradoxes.

Paradox IV, p. 36. How can there be two persons who were born and died at the same time and place, but one lived several months longer than the other?

Paradox VI, pp. 37-38. A man sees one day which is the longest day, the shortest day and a day whose day and night are equal. Answer: he crosses the equator on 22 Jun.

Carlile. Collection. 1793.

Prob. CXIV, p. 68. Two children who were born and died at the same time, "yet one was several months older than the other." One lived within the Arctic Circle, at 73^o22', where days are three months long!

Prob. CXV, p. 68. Two sailors meet and find their calendars are off by a day. They have gone to New Zealand, one eastward and the other westward.

Jackson. Rational Amusement. 1821. Geographical Paradoxes.

No. 6, pp. 35-36 & 102-103. Two nearby places in Asia whose reckonings differ by a day. Solution is either Christians and Jews, whose sabbaths are a day different, or Macao and the Philippines, which differ by a day since the Portuguese came to Macao from the west and the Spanish came to the Philippines from the east.

No. 8, pp. 36 & 103-104. Children born and dying at the same times, but one lives longer than the other. Solution is either due to difference between lunar and solar calendars or due to one sailing round the world.

No. 47, pp. 47 & 114. How can Christians, Jews and Turks all have their sabbath on the same day. Christian sails east and the Turk west.

Endless Amusement II. 1826? Pp. 74-75: "How Two Men may be born on the same Day, die at the same Moment, and yet one may have lived a Day, or even two Days more than the other." Notes that this can give three Thursdays in one week.

Carroll. The Rectory Umbrella. c1845?? Difficulties: No. 1. In: The Rectory Umbrella and Mischmasch; (Cassell, London, 1932), Dover, 1971, pp. 31 33. Also in: Carroll-Collingwood, pp. 4 5 (Collins: 11-12). Carroll-Gardner, pp. 81-82, says he published a letter on this in the Illustrated London News of 18 Apr 1857. In 1860, he later gave this as a lecture to the Ashmolean Society at Oxford.

Warren Weaver. Lewis Carroll: Mathematician. Op. cit. in 1. 1956. Mentions Carroll's interest in the problem.

J. Fisher. Where does the day begin? The Magic of Lewis Carroll. Op. cit. in 1, pp. 22 24. This discusses several of Carroll's versions of the problem.

Vinot. 1860. Art. CXXIV: La semaine des trois jeudis, p. 145. How to have three Thursdays in a week.

Jules Verne. Le tour du monde en quatre-vingts jours (Around the World in Eighty Days). 1873. Features the gain of a day by going eastward.

In 1883 and 1884, the Rome and Washington Conferences for the Purpose of Fixing a Prime-Meridian and a Universal Day adopted the Greenwich Meridian and the basic idea of time zones, which implied the acceptance of the International Date Line. The Philippines, having been colonized from the New World, had to skip a new to conform with its Asian neighbours, but I don't know when this happened. When Alaska was purchased in 1867, it had to drop 12 days to convert to the Gregorian calendar and then had to have one eight-day week to conform with the rest of the New World.

Clark. Mental Nuts. 1897, no. 89. Where is the day lost? Travel from San Francisco to Pekin at the speed of the sun and one finds it is a day later. Answer says "New day begins at 180th meridian, which is midway in the Pacific Ocean."

A. M. W. Downing. Where the day changes. Knowledge 23 (May 1900) 100-101. Observes that the precise location of the Date Line has not yet been fixed. Gives four versions on a map, the last two of which only differ at a few islands.

F. &. V. Meynell. The Week End Book. Op. cit. in 7.E. 1924. 2nd. ed., prob. eight, pp. 276 277; 5th? ed., prob. twelve, p. 410. What happens if you go around the world from west to east in 24 hours, or in less than 24 hours? Suggests that in the latter case, you get back before you started!

McKay. Party Night. 1940. No. 36, p. 184. Two airmen circle the earth in opposite directions, both taking 14 days. Which gets back home first? Answer is that the eastbound one gets back two days sooner because the 14 days are considered as days viewed by the airmen.

A. P. Herbert. Codd's Last Case and Other Misleading Cases. Methuen, London, 1952. Chap. 14: In re Earl of Munsey: Stewer v. Cobley -- The missing day case, pp. 72-83. (This probably appeared in Punch, about 7 Dec 1949.) Reprinted in: More Uncommon Law; Methuen, 1982, pp. 74-83. Lord Munsey left property to his great-nephew "if he has attained the age of 21 before the date of my death", otherwise the property went to Lord Munsey's brother. The great-nephew's birthday was 2 May. On 1 May, Lord Munsey was on a cruise around the world and the ship crossed the Dateline at about noon, going westward, so the Captain declared that the next day would be 3 May at midnight, but Lord Munsey expired just after midnight. So who inherits?

Jonathan Always. Puzzles for Puzzlers. Tandem, London, 1971. Prob. 2: Another birthday poser, pp. 11 & 60. Four persons, born on different dates, lived to be fifty years of age, but never saw their fiftieth birthday. One was born on 29 Feb. "Another crossed the International Date-line travelling eastward, so that he gained a day and never arrived at his fiftieth birthday at all however long he lived." This confuses me -- crossing the Dateline goes back a day, so it would seem his fiftieth birthday would simply be delayed by at most a day. Always continues: "The third crossed the International Date-line travelling westwards on the eve of his fiftieth birthday, thus losing a day, a died a few hours later." I think he has confused things. If a man travels eastwards across the Dateline on the day before his 50th birthday, then he goes back a day and hasn't yet arrived at his 50th birthday, though he is now 50 years old. If he dies before midnight, then that fits the problem. If another man travels westwards across the Dateline at midnight on the eve of his fiftieth birthday, he goes ahead two days and really loses a day, namely his fiftieth birthday, which he never saw. The fourth person was blind!

Shakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 138: The Sabbath day, pp. 86 & 134. In order for a Moslem, a Jew and a Christian to have their Sabbath at the same time, send the Moslem around the world to the west and the Christian around the world to the east. When they meet again they will all have the same Sabbath day!

David Singmaster. Letter [on the International Date Line]. Notes and Queries column, The Guardian, section 2 (20 Dec 1995) 7. Reprinted in Guardian Weekly (7 Jan 1996) 24. Sketches the history from Oresme onward and notes the anomalies that Alaska and the Philippines essentially crossed the Date Line.

Ed Barbeau. After Math. Wall & Emerson, Toronto, 1995. Double Christmas, pp. 153-154. Straightforward problem illustrating that flying east from Australia, one may return to the previous day. Here the heroine gets 36 hours of Christmas, in two disconnected parts! [Indeed if one just crosses the date line from west to east at midnight, one gets 48 hours of the same day.]

10.L. FALLING DOWN A HOLE THROUGH THE EARTH
The angular frequency of the oscillation is (g/R) where g is the acceleration of gravity at the surface and R is the radius of the earth. Taking a mean radius of 3956.665 mi and g = 32.16 ft/sec² gives half-period of 42.20 min = 42 min 12 sec.
Hesiod. The Theogony. c-700. IN: The Homeric Hymns and Homerica, translated by H. G. Evelyn White; Harvard Univ. Press, 1959, p. 131, lines 724-725. ??NYS -- information sent by Andrew Simoson [email of 25 Feb 2003]. This claims that a brazen anvil dropped from the Earth's surface will reach Tartarus in nine days.

Plutarch. Plutarch's Moralia, vol. XII, translated by H. Cherniss & W. C. Helmbold. Harvard Univ. Press, 1957, pp. 65-67. ??NYS -- cited by Simoson. Plutarch notes that if the centre of the earth is at a person's navel, then both his head and his feet are pointing up. Simoson quotes Plutarch as: "Not that ... masses ... falling through the ... earth stop when they arrive at the centre, though nothing encounter or support them; and, if in their downward motion the impetus should carry them past the centre, they swing back again and return of themselves?"

Galileo. Dialogo ... sopra i due Massimi Sistemi del Mondo ... (Dialogue Concerning the Two Chief World Systems). Gio. Batista Landini, Florence, 1632. Translated by Stillman Drake; Univ. of Calif. Press, Berkeley, 1953; pp. 22 23, 135 136, 227 & 236. Asserts the object will oscillate. No mention of air resistance.

van Etten. 1624. Prob. 91 (88), part II (2), p. 139 (220). Says a millstone dropped down such a hole at 1 mile per minute will take more than 2½ days to reach the centre, where "it would hang in the air".

Ozanam. 1694. Prob. 7, corollary 3, 1696: 218; Prob. 7, Remark, 1708: 312. Prob. 7, part 8, corollary 3, 1725: vol. 2, 151-152. Considers falling down a well to the centre of the earth but uses a hypothetical constant value of g. Then considers a tube through the earth and says the object will oscillate, but air resistance will slow it down to rest at the centre.

Euler. A Physical Dissertation on Sound. 1727? ??NYS -- described by Truesdell in his Introduction to Euler's Algebra, p. xiv. Annex announces the solution of the problem of oscillation through a hole in the earth.

Euler. Algebra. 1770. I.III.X.501, p. 163. How far would an object fall in a hour under constant g as at the earth's surface? 39272 miles!

Euler. Letters to a German Princess. ??NYS -- Simoson cites an Arno Press, 1975, reprint of the 1833 ed. Vol. 1, letter L, pp. 178-182. Simoson quotes Euler as: "You will remember how Voltaire used to laugh at the idea of a hole reaching to the centre of the earth, ... but there is no harm is supposing it, in order to discover what would be the end results."

Ozanam Montucla. 1778. Vol. IV, prob. 9, 1778: 41-42; 1803: 42-43; 1814: 34-35; 1840: 616-617. First finds the time to reach the centre if g is constant, namely 19  minutes. Then considers that gravity will decrease and quotes a result of Newton to find the time to the centre is 21' 5" 12"'.

John Baines, proposer; Wm. Rutherford; N. J. Andrew & George Duckett; independent solvers. Question (23). The Enigmatical Entertainer and Mathematical Associate for the Year 1830; .... Sherwood & Co., London. No. III, 1829. This has two separate parts with separate pagination. The second part is The Mathematical Associate and the problem is on pp. 36-37 in the Answers to the Questions Proposed Last Year. "If a hole were bored through the earth, parallel to the equator, in lat. 20^o, and a heavy body let fall into it from the surface, it is required to determine its velocity at any point of its descent, taking into account the variation of gravity, but abstracting all resistance."

Lewis Carroll. Alice in Wonderland. Macmillan, London, 1865. Chap. I, pp. 27-28 in Gardner's Annotated Alice, below. "I wonder if I shall fall right through the earth!"

Lewis Carroll. Sylvie and Bruno Concluded. Macmillan, London, 1893. Chap. 7, pp. 96 112, esp. pp. 106 108. Discusses trains using straight holes, not through the centre. Cf Carroll-Gardner, pp. 7-8, where Gardner notes that Carroll frequently uses the value 42 which is the half-period in minutes.

Martin Gardner. The Annotated Alice. Revised ed., Penguin, 1970. Chap. I, note 4 (to the line given above), pp. 27 28. Describes Carroll's interest in the problem. Says it interested Plutarch, Bacon and Voltaire and that it had been resolved by Galileo (see above). Gardner also cites C. Flammarion, Strand Mag. 38 (1909) 348; ??NYS.

Clark. Mental Nuts. 1897, no. 53. Matters of gravity. "Suppose you drill a hole through the earth and drop an iron ball in it, where will the ball go?" Answer is "Centre of earth."

Pearson. 1907. Part II, no. 79: Dropped through the globe, pp. 130 & 207. Says it will oscillate, but air friction will cause it to come to rest at the centre.

Ackermann. 1925. Pp. 60 61. Similar to Pearson.

Collins. Fun with Figures. 1928. A hole through the earth, p. 203. Says it will oscillate like a pendulum and if air is present, it will slow down and stop at the centre.

W. A. Bagley. Puzzle Pie. Op. cit. in 5.D.5. 1944. No. 46: Down Under, pp. 51-53. Various discussions of what happens to a person or a cannonball falling through the earth. Seems to think a man would turn over so that he would be rightside up at the other side?? Says the oscillation will continue with diminishing periods (presumably meaning amplitude) until the person is stuck at the centre. A cannonball would burn up from friction.

R. E. Green. A problem & H. Martyn Cundy. A solution. Classroom Note 178: Quicker round the bend! MG 52 (No. 382) (Dec 1968) 376 380. Green notes the well known fact that the time to fall through a straight frictionless hole is independent of its length -- about 42 minutes. He asks what path gives the least time? Cundy says it is a straightforward application of the calculus of variations. He finds a solution for θ as a function of r, in terms of R, the radius of the earth, and m, the distance of closest approach of the curve to the centre. Also m/R = 1   2a/π, where 2a is the central angle between the ends of the tube. The straight through time is π (R/g). The shortest time is π {(R² m²)/Rg}.

K. E. Bullen. The earth and mathematics. MG 54 (No. 390) (Dec 1970) 352 353. A riposte to Classroom Note 178, pointing out Saigey's result of c1890, that g increases as you start down a hole because the interior of the earth is denser than the surface. More recent theoretical and practical work indicates g is essentially constant for at least the first 2000 km.

H. Lindgren. Classroom Note 250: Quicker round the bend (Classroom Note 178). MG 55 (No. 393) (Jun 1971) 319 321. Shows the optimal curve is a hypocycloid and rephrases the time required. If d is the surface distance between the ends and C is the earth's circumference, the minimal time is {d(C d)/Rg}. He cites 1953 and 1954 papers which treat the problem in general.

Erwin Brecher. Surprising Science Puzzles. Sterling, NY, 1995. Hole through the earth, pp. 20 & 88. Asks a number of straightforward questions and then asks whether a ball will take more or less time to fall through a hole in the moon. He says it will take about 53 min on the moon -- I get 54.14 min.

In 2000 or 2001, Tim Rowett asked me the following. If an apple could be dropped from a point on the earth's orbit, but only attracted by the sun, how long would it take to reach the sun? This is complicated by gravity varying and I found it quite awkward to do, having to make some approximations to get a time of 64.4 days. After seeing Simoson's article, below, I asked him if his work would deal with the problem and he pointed out that the formula for falling to the centre of the earth considered as a point mass (p. 349, case 3) can be used if one adapts the parameters appropriately. After a little conversion, this is identical to the formula I obtained. He computes 64.57 days, or 63.89 days to the surface of the sun, the difference being due to his using more accurate values for the astronomical constants.

Andrew J. Simoson. The gravity of Hades. MM 75:5 (Dec 2002) 335-350. Considers the problem for several models of the earth. Cites Plutarch, Halley, Euler and discusses models from Greek mythology, Dante, Hooke (a multi-layer, onion-like, earth), Halley (a hollow earth), etc. He finds the time to the centre is about 21.2 min. He finds the time to the centre for a constant acceleration model is about 19.0 min -- cf Ozanam and Ozanam-Montucla. For the best known results on the earth's mass distribution, he finds about 19.2 min. If all the earth's mass is at the centre, he gets about 15.0 min and observes this is the model of the the earth's mass distribution which gives the fastest time (p. 349, case 3). He then poses a problem of falling down the z-axis to a galaxy in the x-y plane, as Satan may have done when cast out of heaven, thereby estimating the Miltonian distance between Heaven and Hell.

10.M. CELTS = RATTLEBACKS
When rotated, these objects stop and then start turning in the opposite direction. The word 'celt', with a soft 'c', so it sounds like 'selt', means a stone hand axe, chisel or similar primitive implement -- see 1910 below. I once heard that the phenomenon was discovered by anthropologists examining handaxes and that they used the spin as a form of classification.
The OED entry for Celt is long and not definitive. The word 'celte' appears in the Vulgate translation of the Bible and is understood to mean some kind of tool, but others feel it is a miscopying -- it seems to be 'certe' in some manuscripts. By 1700, it was considered a proper Latin word and was adopted by British archaeologists for primitive tools.

Chambers's Encyclopedia. Revised edition, W. & R. Chambers, London & Edinburgh, 1885. Vol. II, p. 711. About a column on celts. "CELT (Lat. celtis (?), a chisel), the name by which certain weapons or implements of the early inhabitants of Western Europe are known among archaeologists."

G. T. Walker. J. Walker says his investigations occur in old books on rotational mechanics in the chapter on asymmetrical tops. ??NYS

G. T. Walker. On a curious dynamical property of celts. Proc. Camb. Philos. Soc. 8 (1892/95) 305 306. (Meeting of 13 May 1895.) ??NX

G. T. Walker. On a dynamical top. Quarterly J. Pure & App. Mathematics 28 (1896) 175 184. (& diagrams??) ??NX

Harold Crabtree. An Elementary Treatment of the Theory of Spinning Tops and Gyroscopic Motion. Longmans, Green & Co., London, 1909, ??NX; 2nd ed, 1914. Pp. 7 & 54 and Plate I. (I have a citation to these pages in the 1st ed and the material is on the same pages in the 2nd ed. The Preface to the 2nd ed says changes occur elsewhere.) Says the term comes from Latin, celtis, a chisel, and uses it for the stone axes in general. Shows and discusses three examples. The first is an ordinary axe. The second is a two-way version -- when spun in either direction, it will reverse. The third is a one-way version -- when spun in one direction, it continues, but when spun in the other direction, it reverses. Cites Walker, 1896. The plate has photos of the three examples.

Encyclopedia Britannica. 11th ed., 1910. Entry for Celt. ??NYS -- quoted by Bürger, below. "CELT, a word in common use among British and French archeologists to describe the hatchets, adzes or chisels of chipped or shaped stone used by primitive man. The word is variously derived from the Welsh celit, a flintstone (that being the material of which the weapons are chiefly made, though celts of basalt, felstone and jade are found); from being supposed to be the implement peculiar to the Celtic peoples; or from a Low Latin word celtis, a chisel. The last derivation is more probably correct."

Andrew Gray. Treatise of Gyrostatics and Rotational Motion. (1918); Dover, 1959. ??NYS.

Charles W. Sherburne (3409 Patton Ave., San Pedro, California, 90731, USA). US Design 210,947 -- Scientific demonstration toy. Filed: 12 Nov 1995; patented: 7 May 1968. 1p. This simply says it 'shows my new design' and there is no indication of the skew curvature of the bottom. Sherburne has published material claiming that the rattleback demonstrates the failure of Newton's laws and that it is the shape of Noah's Ark!

Karl Magnus. The stability of rotations of a non-symmetrical body on a horizontal surface. Festschrift Szabo, Berlin, 1971, pp. 19-23. ??NYS -- cited by Bürger. This determines which direction of rotation is stable. I don't know if it deals with 'both-way reversing' examples.

Jearl Walker. The mysterious "rattleback": a stone that spins in one direction and then reverses. SA 241:4 (Oct 1979) 144 150. Reprinted with extra Note and recent references in: Jearl Walker; Roundabout -- The Physics of Rotation in the Everyday World; Freeman, NY, 1985; Chap. 6, pp. 33 38 & 66. Cites Crabtree and G. T. Walker. Discusses work of Nicholas A. Wheeler and of A. D. Moore.

Allan J. Boardman. The mysterious celt. Fine Woodworking (Jul/Aug 1985) 68 69. Describes how to make celts.

Hermann Bondi. The rigid body dynamics of unidirectional spin. Proc. Royal Soc. London A 405 (1986) 265 274. Analyses the dynamics and shows that the phenomenon occurs even without friction. Only cites G. T. Walker, Quarterly J. The Cavendish Laboratory has made a fine steel model with adjustable weights which Bondi has seen make five reversals.

Wolfgang Bürger. A Celtic rocking top. English version of leaflet to accompany the plastic version of the celt distributed by Nixdorf Computers. Nd [probably late 1980s]. Cites Walker, 1896, and quotes Encyclopedia Britannica, 11th ed., 1910, for the term as quoted above. He conjectures that the spinning property may have been discovered by an archaeologist and he gives a myth that such spinning was used by ancient priests to determine guilt or innocence. (Frame-ups were common even then.) Until recent realization that these objects were man-made, they were the subject of superstitions throughout the world. He gives a short discussion of the physics/geometry involved and says that since 1980, nine scientific papers have tried to analyse the motion and that it was the subject of a German Jugend forscht (Young Researchers) prize winning project in 1985.

10.M.1. TIPPEE TOPS
The physics of this is hard and I will only give some general articles.
Helene Sperl. German Patent 63261 -- Wendekreisel. Patented: 7 Oct 1891; published: 12 Jul 1892. 1p + 1p diagrams. Several slightly diferent shapes. Diagram is reproduced by Kuypers & Ucke, below.

Harold Crabtree. An Elementary Treatment of the Theory of Spinning Tops and Gyroscopic Motion. Longmans, Green & Co., London, (1909, ??NX); 2nd ed, 1914. Pp. 4-5 describes some rising tops. One of these is a loaded sphere, which seems to have the same inverting properties as the Tippee Top. Appendix IV: The rising of a spinning top, pp. 145-158, is new in the 2nd ed and discusses the loaded sphere on pp. 155-156.

Gwen White. Antique Toys and Their Background. Op. cit. in 5.A. 1971. P. 45. "An interesting little top known as a 'tippe top' came in 1953 .... It was a great commercial success for the British Indoor Pastimes Company."

D. G. Parkyn. The inverting top. MG 40 (No. 334) (Dec 1956) 260 265. Cites 4 papers in 1952.

Leslie Daiken. Children's Toys Throughout the Ages. Spring Books, London, 1963. [This may be a reprint of an earlier publication??] P. 38. "... the most recent craze, invented by a Swede. Made from plastic, and known as the Tippy Tap [sic], this type will turn upside down and spin on its head!"

Jearl Walker. Amateur Scientist columns. SA (Oct 1979) -- op. cit. in 10.M and The physics of spinning tops, including some far out ones. SA 244:3 (Mar 1981) 134 142. Reprinted with extra notes in: Roundabout, op. cit. in 10.M, chap. 6 & 7, pp. 33 44 & 66, esp. pp. 37 44. 15 references on p. 66, not including Parkyn.

Wolfgang Bürger. Elementary dynamics of simple mechanical toys. Mitteil. der Ges. f. Angew. Math. und Mechanik 2 (Jul 1980) 21-60. (Reproduced in: Spielzeug-Physik; Bericht Nr. 98, Akademie für Lehrerfortbildung Dillingen, 1986, pp. 159-199.) Pp. 49 52 (= 188-191) discusses the Tippe Top, noting that Fraülein Sperl's explanation in her patent 'was far from being correct.' Says the dynamics is difficult and cites 1952 & 1978 articles on it. J. L. Synge asserted that friction was not necessary for the turning over motion, but Bürger shows it is essential.

Friedhelm Kuypers & Christian Ucke. Steh' auf Kreisel! Physik in unserer Zeit 25:5 (Sep 1994) cover & 214-215. The German names are Stehaufkreisel and Kippkreisel. Describes Sperl's patent and modern work on the mechanics involved -- "it is not so easy as we first believed".

10.N. SHIP'S LADDER IN RISING TIDE
Water is touching the bottom rung of a rope ladder hanging over the side of a ship. The tide is rising at a known rate. How many rungs will be covered after some time?
Phillips. Brush. 1936. Prob. O.1: The ship's ladder, pp. 49 & 106.

Haldeman-Julius. 1937. No. 42: The Queen Mary problem, pp. 7 & 22.

Morley Adams. The Children's Puzzle Book. Faber, London, 1940. Prob. 174: The ship's ladder, pp. 55 & 78.

Shirley Cunningham. The Pocket Entertainer. Blakiston Co., Philadelphia, and Pocket Books, NY, 1942. Prob. VI: The rope ladder, pp. 72 & 222.

Harold Hart. The World's Best Puzzles. Op. cit. in 7.AS. 1943. The rope ladder, pp. 31 & 59.

Leopold. At Ease! 1943. Of time and tide, pp. 9-10 & 195.

Sullivan. Unusual. 1943. Prob. 3: Time the tide.

Leeming. Op. cit. in 5.E. 1946. Chap. 3, prob. 40: Rising tide, pp. 36 & 162.

10.O. ERRONEOUS AVERAGING OF VELOCITIES
See also 7.Y which involves erroneous averaging of unit costs. There must be earlier examples than I have here.
H. A. Ripley. How Good a Detective Are You? Frederick A. Stokes, NY, 1934, prob. 42: Class day. Average 10 mph and 50 mph over the same distance.

Dr. Th. Wolff. Die lächelnde Sphinx. Academia Verlagsbuchhandlung, Prague, 1937. Prob. 17, pp. 190 & 201. Going and returning at 100 km/hr versus going at 120 and returning at 80.

Harriet Ventress Heald. Op. cit. in 7.Z. 1941. Prob. 31, pp. 15 16. Man goes 30 mph for a mile. How fast must he go for a second mile in order to average 60 mph overall?

Sullivan. Unusual. 1943. Prob. 22: Don't get caught trying it. If you are going 60 mph, how much faster do you have to go to save a minute on each mile?

E. P. Northrop. Riddles in Mathematics. 1944. 1944: 11-13; 1945: 10 12; 1961: 20-22. Same as Heald with rates 15, 30. Relates to airplane going with and against the wind.

Leeming. Op. cit. in 5.E. 1946. Chap. 5, prob. 20: Sixty miles per hour, pp. 59 60 & 178. Does 30 mph for 10 miles. How fast for the next 10 miles to average 60 mph?

Sullivan. Unusual. 1947. Prob. 31: A problem without a title. Same as Heald.

Birtwistle. Calculator Puzzle Book. 1978. Prob. 40: Motoring problem, pp. 30 & 87-88. Same as Heald with rates 15, 30.

10.P. FALSE BALANCE
Solomon (or The Preacher). c-960.