10.     PHYSICAL RECREATIONS

 

          See also 7.S and 7.Y.

          I will collect here some material on physics toys in general.

 

Christian Ucke.  Physics toys for teaching.  IN: H. Kühnelt, ed.; Interdisciplinary Aspects of Physics Education; proc. of conf. at Altmünster, Austria, 1989; World Scientific, Singapore, 1990, pp. 267-273.  1: Some new and not so well known literature about physics toys gives 12 references.  2: Presenting a database about physics toys.  3. Physics toy experiments with PET bottles.  16 references at the end, some repeating those in part 1.

 

          10.A.  OVERTAKING AND MEETING PROBLEMS

 

          See Tropfke 588.

          Note.  Meeting problems include two pipe cistern problems, 7.H.  Overtaking problems include snail in well problems without end effect, 10.H, and cisterns with one inlet and one outlet, 7.H.  Many of the Indian versions involve gaining or losing wealth rather than covering distance.  Versions going around a circle or an island are related to Conjunction of Planets, 7.P.6, and to problems of clock hands meeting, 10.R.  In the 17, 18 and 19 C, this problem was often discussed in relation to negative numbers as a change in the relative values leads to a negative solution.

 

          NOTATION.  There are five types of meeting (M) and overtaking (O) problems which recur frequently with slight variations.  I have recently converted all problems to this notation and I hope I have done it correctly.  Tropfke 590 gives a more extended classification which includes motion on a right triangle (see 6.BF.5) and on a circle (see 7.P.6, though some occur here) and alternating motion (see 10.H), but doesn't distinguish between problems where times are given and those where rates are given.

          M-(a, b).   Two travellers can cover a route in  a, b  (usually days).  They start at opposite ends at the same time toward each other.  When do they meet?  This is identical to the cistern problem  (a, b)  of 7.H.  Sometimes, the distance apart is given and the point of meeting is also wanted.  If the distance is, say  100, I will then say  "with  D = 100".  This corresponds to asking how much each pipe contributes to a cistern of capacity  D.  Sometimes, one starts later than the other.  If, say the first starts  2  later, I will say  "with the first delayed by  2".  This corresponds to opening one pipe later than the other.  This is the version of Tropfke's  I B a  where times are given.

          MR-(a, b; D).   The same problem except that  a, b  are the actual rates of the two travellers and hence  D  must be given.  This corresponds to a simple form of cistern problem which does not have the characteristic feature of giving the times required to do the entire task.  This is the version of Tropfke's  I B a  with rates given.

          MR-(a, b; c, d; D).   The same problem except that the travellers travel in arithmetic progressions, so this gives: 

          a + a+b + ... + a+(n-1)b  +  c + c+d + ... + c+(n-1)d  =  D.

Hence   MR-(a, 0; c, 0; D)  =  MR-(a, c; D).  The value of  d  is often  0.  One can interpret this as a cistern problem as for  MR-(a, b; D),  but it is even harder to imagine a pipe increasing its rate in arithmetic progression that to imagine a traveller doing so.  (An additional difficulty is that the traveller is usually viewed discretely while a pipe ought to be viewed continuously.)  This is the version of Tropfke's  I B b  with an arithmetic progression specified.

          O-(a, b).   Two travellers start from the same point at rates  a, b,  with the slower starting some time  T  before the faster, or they start at the same time at rates  a, b,  with the slower starting some distance  D  ahead of the other.  I.e. the slower has a headstart of time  T  or distance  D.  When does the faster overtake the slower?  This corresponds to a cistern with rates given, so that  O-(a, b)  with headstart  D, which is the same as  MR-(a, -b; D),  is a cistern problem with one inlet and one outlet.  When  a > b,  then this corresponds to a full cistern of size  D,  inlet rate  a  and outlet rate  b.  The case  a < b  is most easily viewed by negating the amount done, which interchanges inlet and outlet, and taking an empty cistern.  Hound and hare problems are basically of this form, with a headstart of some distance, but usually with rates and distances complicatedly expressed.  This is Tropfke's  I C a.  Sometimes the rates are not given explicitly, so I assume the first has the headstart.

          O-(a, b; c, d).  Two travellers start from the same point at the same time, but in arithmetic progressions.  When do they meet again?  This gives us: 

          a + a+b + ... + a+(n-1)b  =  c + c+d + ... + c+(n-1)d.  Some versions of this are in 7.AF.

If the first has a headstart of time  T,  then we either increase the first  n  by  T  or decrease the second  n  by  T,  depending on which number of days is wanted.  This is Tropfke's  I C b  with an arithmetic progression specified.  Sometimes the first has a headstart of distance  D.  Occasionally it is the second that has the headstart which is denoted by negative values of  T  or  D.

 

          Snail in the well problems without end effect (see 10.H) are special cases of meeting problems, usually  MR-(a, 0; D)  (Tropfke's  I A c).  When there are approaching animals, then it may be  MR(a, b; D)  or  O‑(a, b)  with headstart  D  (Tropfke's  I B c).

 

          Hound and hare problems.  Here one is often only given the ratio of speeds, so one can determine where the hare is caught, but not when.  See:  Chiu Chang Suan Ching;  Zhang Qiujian;  Alcuin;  Fibonacci;  Yang Hui;  BR;  Dell'Abbaco;  Bartoli;  AR;  The Treviso Arithmetic;  Ulrich Wagner;  HB.XI.22;  Calandri, 1491;  Pacioli;  Calandri, c1500;  Tagliente;  Riese;  Apianus;  van Varenbraken;  Cardan;  Buteo;  Gori;  Wingate/Kersey;  Lauremberger;  Les Amusemens;  Euler;  Vyse;  Hutton, c1780?;  Bonnycastle (= Euler);  King;  Hutton, 1798?;  De Morgan, 1831?;  Bourdon;  Hutton-Rutherford;  Family Friend (& Illustrated Boy's Own Treasury);  Anon: Treatise (1850);  Brooks;  Clark.

                    Cases where leaps differ in both time and distance:  Pacioli?;  Apian;  Cardan;  Wingate/Kersey;  Lauremberger;  Euler; Bonnycastle;  King;  Hutton, 1798? (= Lauremberger);  De Morgan, 1831?;  Bourdon (= Lauremberger);  Brooks;  Clark (= Lauremberger); 

 

          General versions.  Newton.

          Versions with geometric progressions.  Chiu Chang Suan Ching;  Chuquet;  Cardan.  See also 7.L.

          Versions with sum of squares.  Simpson.

          Circular versions.  Aryabhata(?);  AR;  Wingate/Kersey;  Vyse;  Pike;  Anon: Treatise (1850);  Perelman.  See also 7.P.6, where problems with more than two travellers in a circle occur.

          Versions using negatives.  Clairaut;  Manning.

 

Chiu Chang Suan Ching.  c‑150.  (See also Vogel's notes on pp. 126‑127.)

   Chap. VI.

Prob. 12, p. 61.  O-(60, 100),  D = 100.  (Mikami 16 gives English.)

Prob. 13, pp. 62‑63.  Slower starts  10  ahead.  After the faster goes  100,  he is  20  ahead.  When did he overtake?

Prob. 14, p. 63.  Hare starts  100  ahead of hound.  Hound runs  250  and is then  30  behind.  How much further for hound to overtake?  (Swetz; Was Pythagoras Chinese?; p. 21, gives this with numbers  50, 125, 30.)

Prob. 16, p. 64.  Slower (the guest) goes  300  per day.  Faster (the host) starts after  _  day, catches the slower and returns after  ¾  of a day.  How fast is the faster?

Prob. 20, p. 66.  Birds can traverse a distance,  M-(7, 9).

Prob. 21, pp. 66‑67.  M-(5, 7)  with first delayed by  2.

   Chap. VII.

Prob. 10, p. 74.  Melon and gourd growing toward each other,  MR‑(7, 10; 90).

Prob. 11, pp. 74‑75.  Rush grows  3, 3/2, 3/4, ....  Sedge grows 1, 2, 4,. ...  When are they equal?  Solution of  2 6/13 = 2.4615  is obtained by linear interpolation between  2  and  3  days.  (Correct solution is  log2 6 = 2.5848.)  (English in Mikami 18.)

Prob. 12, pp. 75‑76.  Two rats separated by a wall  5  thick.  One gnaws  1, 2, 4, ....  The other gnaws  1, 1/2, 1/4, ....  When do they meet?  Solution of  2 2/17 = 2.1176  obtained by linear interpolation between  2  and  3.  (Correct answer is   log2 (2 + Ö6)  =  2.1536,   though Vogel has  8  instead of  6  in the radical.

Prob. 19, p. 79.  Two horses travel  193, 206, 219, 232, ...  and 97, 96 ½, 96, 95 ½, ....  They set out for a city  3000  away.  The faster gets there and then returns.  When does he  meet the slower?  Solution of  15 135/191 = 15.7068  obtained by linear interpolation between  15  and  16.  Correct answer is   (‑227 + Ö147529)/10  =  15.7095.

Zhang Qiujian.  Zhang Qiujian Suan Jing.  Op. cit. in 7.E.  468,  ??NYS.  Analogous to above hare and hound with values  37, 145, 23.  (English in  Sanford 212,  Mikami 41  and  H&S 74.)

Aryabhata.  499.  Chap. II, v. 30-31, pp. 72-74.  (Clark edition: pp. 40-42.)

V. 30.  One man has  a  objects and  b  rupees, another has  c  objects and  d  rupees, and they are equally wealthy.  What is the value of an object?  I.e. solve  ax + b = cx + d.  Though simple, this is the basis of many of the problems in this section, as discussed by Gupta (op. cit. under Bakhshali MS).  Clark gives an example with   a, b, c, d  =  6, 100, 8, 60.

V. 31.  Shukla translates this as describing simple meeting and overtaking problems,  MR-(a, b; D)  and  O-(a, b) with headstart D.  Clark translates it as having two planets separated by  D = s1 - s2  travelling with velocities  v1  and  v2,  so they meet in time   t  =  (s1 - s2) / (v1 ± v2).   This is discussed in Gupta, loc. cit.

Bakhshali MS.  c7C. 

Kaye I 43-44 describes the following types in general terms and indicates that general solutions are stated in the MS.  O-(a, b),  T = t.  O-(a, b; c, d).  O-(a, b; c, 0).  A problem equivalent to  MR(a, b; 2D).  Kaye I 49-52 describes problems of earning and spending which are equivalent to the following.  O-(a, b), headstart  2D.  O‑(a, b)  headstart  D.  Cistern problems with several inlets _ see 7.H.

Kaye III 171, f. 3r, sutra 15  &  Gupta.  General rule for  O-(a, b),  headstart  D.  (Gupta cites Aryabhata as giving the same rule.)

Kaye III 175, f. 4r  &  Gupta.  Examples of the previous with   a, b, D  =  5, 9, 7  and  18, 25, 8*18.  (The second is really an example with the first starting  T = 8  days ahead.)

Kaye III 189, f. 83r.  O-(3/2, 2),  D = 9.

Kaye III 216, ff. 60r-60v, sutra 52  &  Gupta.  Same as f. 3r, sutra 15, except that distance is replaced by wealth.  Example has a man with wealth  30  earning at rate  5/2  and spending at rate  9/3.  This can be considered as  O‑(5/2, 9/3),  D = 30.  This is also very like the simple versions of a snail climbing out of a well and could be considered as  MR(9/3 - 5/2, 0; 30).

Kaye III 217-218, ff. 60v-61v, sutra 53  &  Gupta  &  Hoernle, 1886, p. 146; 1888, p. 44.  Two men earn at rates  a, b,  but the first gives  c  to the second at the beginning.  When are they equally rich?  Examples with   a, b, c  =  5/3, 6/5, 7;   13/6, 3/2, 10.  This is equivalent to  O-(a, b),  D =  2c.  See also: G. R. Kaye;  The Bakhsh_li manuscript;  J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349‑361;  p. 360, Sutra 53 and example, but he has a  +  for a  -  in his formula.

Kaye III 186, f. 31r  &  Gupta.  Another example of the above, with  a, b, c  =  7/4, 5/6, 7.

Kaye III 172, f. 8r  &  Gupta.  O-(a, b; c, 0), T = -e,  i.e.  a + (a+b) + ... + (a+(n‑1)b)  =  c(n+e).  In general this is a quadratic problem _ see below _ but when  e = 0,  a factor of  n  cancels leaving a linear problem.  Example:  O‑(2, 3; 10, 0). 

Kaye III 174-175, ff. 7v & 4r  &  Gupta.  O-(3, 4; 7, 0);  O‑(1, 2; 5, 0).

Kaye III 173, f. 9r  &  Gupta.  O-(1, 1; 10, 0).

Kaye III 176-177, ff. 4.r-5.r, sutra 18  &  Gupta  & in:  G. R. Kaye, The Bakhsh_li manuscript;  J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349‑361; p. 358.  O‑(4, 3; 6, 1);  O-(2, 3; 3, 2)  and  O-(5, 6; 10, 3).  See also 7.AF.

Kaye III 174, f. 4v  &  Gupta.  This is a problem of the same type, but most of it is lost and the scribe seems confused.  Gupta attempts to explain the confusion as due to using the data   a, b; c, d  =  3, 4; 1, 2,   with the rule   n  =  2(c‑a)/(b-d) + 1,   where the scribe takes the absolute values of the differences rather than their signed values.  In this way he gets  n = 3  rather than  n = -1.

Kaye III 173, f. 9v.  Travellers set out at rates  a, b  to a destination  D  away.  The faster, on his return, meets the slower _ when?  This is equivalent to  MR‑(a, b; 2D).  Example with  a, b, D = 1, 6, 70.

Kaye III 190, f. 53v.  MR-(96/18, 27/108; 9).

Kaye I 44-46; III 177-178, ff. 5r-6v.  O-(3, 4; 5, 0),  T = -6,  which leads to a quadratic, but has answer  5.  O-(5, 3; 7, 0),  T = -5  is also given by Hoernle, 1886, pp. 128‑129; 1888, p. 33, and Datta, p. 42.  This also leads to a quadratic, but the answer is  (7 + Ö889)/6  and the MS gives an approximation for the root.  I have recorded that Kaye also has  O-(3, 4; 5, 0)  with no delay, but I can't find this.

Bhaskara I.  629.  Commentary to Aryabhata, chap. II, v. 30-31.  Sanskrit is on pp. 127-132; English version of the examples is on pp. 308-309.

                    V. 30, Ex. 1:  7x + 100  =  9x + 80.

                    V. 30, Ex. 2:  8x + 90  =  12x + 30.

                    V. 30, Ex. 3:  7x + 7  =  2x + 12.

                    V. 30, Ex. 4:  9x + 7  =  3x + 13.

                    V. 30, Ex. 5:  9x - 24  =  2 x + 18.

                    V. 31, Ex. 1:  MR-(3/2, 5/4; 18).

                    V. 31, Ex. 2:  O-(3/2, 2/3),  D = 24.

Anania Schirakatzi (= Ananias of Shirak).  Arithmetical problems.  c640.  Translated by:  P. Sahak Kokian as:  Des Anania von Schirak arithmetische Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919) 112-117.  See 7.E for description.

Prob. 8.  Messengers,  O-(50, 80),  T = 15.

Prob. 16.  One mason working faster than the other:  O‑(140, 218),  T = 39.  [The 140 is misprinted as 218!]

Mahavira.  850.  Chap VI, v. 320‑327, pp. 177‑179.

V. 320:  O-(3, 8; 21, 0).

V. 321:  MR-(5, 3; 216).

V. 323:  O-(4, 8; 10, 2).

V. 325:  O-(5, 8; 45, -8).

V. 327:  O-(9, 13),  D = 100.

Alcuin.  9C.  Prob. 26: Propositio de campo et cursu canis ac fuga leporis.  Hound catching hare, hound goes  9  to hare's  7,  hare has  150  head start.  =  O-(7, 9),  D = 150.  (H&S 72 gives Latin and English.)  The actual rates are not given, only their ratio.

Sridhara.  c900. 

    V. 65‑67(i), ex. 81‑83, pp. 52‑53 & 95.  The verses give rules for various cases.

Ex. 81‑82.  O-(8/(5 ‑ ½),  3),  T =  6 ‑ ¼.

Ex. 83.  Slower goes  2,  faster goes  8  and then returns from  100  away.  When and where do they meet?

    V. 96‑98, ex. 111‑112, pp. 78‑79 & 96.

Ex. 111:  O-(3, 1; 10, 0).

Ex. 112:  O-(a, b; c, 2),  T = 6.  Determine  a, b, c  so that they will meet twice.  Answer assumes  a = 1,  b = 6  and first meeting is at  n = 10,  (which gives  c = 67)  and then asserts there is a second meeting at  n = 18.

al‑Karkhi.  c1010.  Sect. I, no. 5‑9, p. 82.

5:  O-(1, 1; 10, 0).

6:  O-(1, 1; 11, 0),  T = -5.

7:  O-(1, 2; 10, 0).

8:  O-(2, 2; 10, 0).

9:  10 + 15 + 20 + ... + 5(n+2)  =  325.  This could be considered as  MR‑(10, 5; 0, 0; 325).

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  Pp. 103f.  ??NYS _ quoted in Tropfke 593.

No. 4.  O-(6, 9),  T = 4.

No. 5.  O-(1, 1; 30, 0).

No. 6.  M-(5, 7).

Ibn Ezra, Abraham (= ibn Ezra, Abraham ben Meir  = ibn Ezra, Abu Ishaq Ibrahim al‑Majid).  Sefer ha‑Mispar.  c1163.  Translated by Moritz Silberberg as:  Das Buch der Zahl  ein hebräisch-arithmetisches Werk; J. Kauffmann, Frankfurt a. M., 1895.  P. 56.  Brothers meeting,  MR-(17, 19; 100).  (H&S 72 gives English.)  Silberberg's note 121 (p. 109) says a similar problem occurs in Elia Misrachi, c1500.

Fibonacci.  1202.  He has many examples.  I give a selection.

P. 168.  O-(1, 1; 20, 0).

P. 168.  O-(1, 2; 21, 0).

P. 168.  O-(2, 2; 30, 0).

Pp. 168-169.  O-(3, 3; 60, 0).

P. 169.  O-(5, 5; 60, 0).

P. 169.  O-(3, 3; 10, 0).  This has a non‑integral solution.  He computes  n = 5 2/3  days from the equation and then considers the travel on the  6th day to be at constant rates  10  and  18,  with the first starting  5  ahead, so the overtaking is at  5 5/8  days.

Pp. 177‑178:  De duobus serpentibus.  Serpents approaching,  MR‑(1/3 ‑ 1/4,  1/5 ‑ 1/6;  100).

Pp. 179‑180:  De cane et vulpe.  O-(6, 9),  D = 50.

P. 182:  De duabus formicis quorum ua imittatur aliam.  An ant pursuing another.  O‑(1/3 ‑ 1/4,  1/5 ‑ 1/6;  100).

P. 182:  De duabus navibus se se invicem coniungentibus.  Two ships approaching,  M‑(5, 7).

Yang Hui.  Supplements to the Analysis of the Arithmetical Rules in the Nine Sections  (=?? Chiu Chang Suan Fa Tsuan Lei).  1261.  Repeats first problem of the Chiu Chang Suan Ching.

BR.  c1305.

No. 24, pp. 42‑43.  Ship goes  380,  starting  24  days after another and overtakes in  85  days.  How fast does the slower ship go?  This is  O-(a, 380),  T = 24  _ determine  a  such that the solution is  85.

                    General form:  a, b  =  velocities of slower and faster ships;  n  =  time the faster ship sails;  n + T  =  time the slower ship sails, i.e. the slower ship has  T  days headstart or the faster ship is delayed by  T.  This gives  a(n+T) = bn.  The above problem has   b = 380,  T = 24,  n = 85   and asks for the other value, namely  a.  I will denote this by   (a, 380, 24, 85),   etc.

No. 41, pp. 60‑61.  (20, 25, 4, n)  =  O-(20, 25),  T = 4.

No. 42, pp. 60‑61.  (20, b, 4, 16).

No. 43, pp. 62‑63.  (20, 25, d, 16).

No. 44, pp. 62‑63.  (a, 50, 5, 10).

No. 45, pp. 62‑63.  (15, 20, 10, n)  =  O-(15, 20),  T = 10.

No. 46, pp. 64‑65.  MR-(31½, 21; 105).

No. 87, pp. 106‑107.  O-(24, 30),  T = 4.

No. 88, pp. 106‑107.  Hare is  40  leaps ahead but hound's leap is  13/11  of hare's.

No. 94, pp. 114‑117.  MR-(18, 22; 240).

Gherardi.  Liber habaci.  1327?

P. 144.  O-(1, 1; 25, 0).

P. 144.  Couriers,  M-(20, 30),  D = 200.

Munich 14684.  14C.  Prob. VIII & XX, pp. 78 & 81.   Discusses  O‑(1, 1; k, 0).

Dell'Abbaco.  c1370. 

Prob. 91, p. 78 with plate on p. 79 showing hound chasing fox holding a chicken.  Fox is  40  fox‑steps ahead and  3 dog‑steps = 5 fox‑steps  (assuming both steps take the same time).  I have a colour slide of this.

Prob. 108, pp. 89‑91 with plate on p. 90.  M-(8, 5).  Asks for time to meet, but also gives  D = 60.

Prob. 118, pp. 96‑97.  O-(2, 2; 30, 0).  = Fibonacci, p. 168.

Prob. 119, pp. 97.  O-(3, 3; 18, 0).

See Smith, op. cit. in 3.

Lucca 1754.  c1390.  F. 59r, p. 134.  Couriers,  M-(20, 30)  with  D = 200,  though this is not used.

Bartoli.  Memoriale.  c1420.  Prob. 28, f. 78r (= Sesiano, pp. 144 & 149-150.  Fox is  121  (fox-)steps ahead of a dog.  9  dog-steps  =  13  fox-steps.  He computes  9/13  of  121.  Sesiano notes that this assumes the fox stands still and determines how many dog-steps they are apart.  The correct answer, which assumes that both steps take the same time, is that the dog gains  4  fox-steps  for every  9  steps  he makes,  so he has to make  121 · 9/4  =  272 1/4  dog-steps,  which are equal to  121 · 13/4  =  393 1/4  =  121 + 272 1/4  fox-steps.

AR.  c1450.  Several problems with arithmetic progressions which I omit.

Prob. 33, p. 37, 164‑165, 177, 223.  Hound and hare.  Hare  100  ahead and goes  7  for each  10  of the hound.

Prob. 148:  De planetis, pp. 72, 164‑165, 214.  Though described as conjunction by Vogel, this is really just the discussion of the general overtaking problem on a circle.  The text gives a general solution.

The Treviso Arithmetic = Larte de Labbacho.  Op. cit. in 7.H.  1478.

Ff. 54v‑55v (= Swetz, pp. 158‑160).  Messengers,  M-(7, 9)  with  D = 250.  (English is also in  Smith, Source Book I 12  and  Isis 6 (1924) 330.)

Ff. 55v‑56r (= Swetz, pp. 160‑161).  Hare & hound, rates  6  and  10,  starting  150  apart.  (English also in Isis 6 (1924) 330.)

Muscarello.  1478. 

F. 73v, p. 189.  Hare is  70  leaps in front of a hound.  Each hound leap is  7/5  as big as a hare's, but in the same time.  Says the hound catches the hare after  175  leaps.

F. 79v, p. 196.  Couriers,  M(70, 80).

Ff. 81r-81v, pp. 196-197.  Couriers,  O-(1, 2; 15, 0).

Ff. 82r-82v, pp. 198-199.  Couriers,  O-(2, 2; 15, 0).

P. M. Calandri.  c1480.  P. 67.   O-(1, 1; 30, 0);   O-(1, 2; 30, 0);   O-(3, 3; 60, 0).

Ulrich Wagner.  Das Bamberger Rechenbuch, op. cit. in 7.G.1.  1483.

Von Wandern, pp. 112 & 223,  O-(1, 1; 6, 0).

Regel vom Hasen, pp. 113 & 223‑224.  Hare goes  12  for hound's  15  and starts  100  ahead.

Chuquet.  1484.  See also 7.L for a problem with geometric progression.

Prob. 22.  English in FHM 204.  Meeting problem,  M-(7, 9).  He erroneously takes  7  and  9  as rates.

Prob. 98-108, FHM 220-221, are problems involving arithmetic progressions, usually overtaking problems, often with a fractional number of terms.  101 is a cask draining.

Prob. 98.  O-(1, 1; 11, 0).

Prob. 99.  O-(1, 1; 10, 0),  T = -3.

Prob. 105, English in FHM 220-221.  O-(3, 3; 14, 0).  Chuquet says another author, Berthelemy de Romans, gets  8 4/13  instead of  8 1/3.  FHM note that this assumes the speed is constant each day, while Chuquet assumes the speed increases through the day.

Prob. 117.  x + (x+1) + (x+2) + ... + (x+5)  =  30.

Prob. 127.  M-(7, 8).  See FHM 204 for comparison with prob. 22.

Borghi.  Arithmetica.  1484. 

Ff. 109v-109r (1509: f. 92r).  Two ships going between Venice and Padua,  MR‑(5, 8; 2400).  (H&S 74 gives Italian.)

F. 109r (1509: ff. 92r-92v).  O-(1, 1; 16, 0).

HB.XI.22.  1488.  P. 52 (Rath 247).  (Rath says it is similar to Alcuin, but with different numbers.)

Calandri.  Arimethrica.  1491.  F. 97v. 

F. 65r.  O-(3, 3; 60, 0).

F. 66r.  Hare is  3000  hare leaps ahead of a hound, but the hounds step's are  8/5  as long as the hare's and take the same time.

F. 69r.  Ships meeting between Pisa and Genoa.  (3, 5).  Same woodcut as used for cistern problems _ see 7.H.

Pacioli.  Summa.  1494.

Ff. 39r-40r, prob. 1-6  =  the six problems on Fibonacci 168-169.

F. 41r, prob. 13.  Man travelling.  25+x + 2(25+x) + ... + 2x‑1(25+x)  =  300,  i.e.  (2x ‑ 1)(25 + x)  =  300.  Pacioli sets   y = x - 3.   After  3  days, he has gone  196 + 7y  and he travels  224 + 8y  on the  4th day.  Pacioli interpolates both linearly to get   196 + 7y + y(224 + 8y) = 300,  obtaining  x = 3.44341....  I get  x = 3.52567....

Ff. 41r-41v, prob. 14.  100  =  1 + 2 + 3 + ... + a  =  1 + 3 + 5 + ... + 2b-1  =  2 + 4 + 6 + .. + 2c  =  4 + 8 + 12 + ... + 4d.  When should each start so they all arrive at once?  (H&S 73 gives Italian and English and H&S 74 says he gives a problem involving flying about the earth, but I haven't found it.)

F. 41v, prob. 16-18.  These are examples of the general form treated in BR, i.e.  (a, b, T, n)  gives the equation   a(n+T)  =  bn.

  Prob. 16.  (30, 35, 5, n)  =  O-(30, 35),  T = 5.

  Prob. 17.  (32, b, 6, 25).

  Prob. 18.  (a, 35, 4, 20).

F. 41v, prob. 19.  O-(1, 1; 25, 0).

F. 41v-42r, prob. 20.   25n  =  1 + 2 + ... + 2n-2.   He gets  8 73/103 = 8.70874...  by interpolation on the  8-th day.  I get  8.78554....

F. 42r, prob. 21.  M-(4, 5)  with  D = 100.

Ff. 42v-43r, prob. 27.  Hound and hare.  Hare is  60  ahead and  5  hound leaps  =  7  hare leaps.  Says the problem is very unclear.  Assumes the  60  are hare leaps and interprets the relation as saying that  7  hare leaps are the same as  5  hound leaps and leaps take the same time.  Later, he supposes the  60  are hound leaps and converts this to  84  hare leaps.  See Cardan, whose description implies both the lengths and times of the leaps differ.

  Swetz, op. cit. as Treviso Arithmetic in 7.H, p. 244, says this is the first listing of variants of the problem _ but we've seen lots of examples before and many of Pacioli's are taken from Fibonacci.  Swetz gives examples from  Wagner,  Calandri (1491)  and  Köbel (1514).

Calandri.  Aritmetica.  c1500. 

F. 92r, p. 184.  Ships meeting between Livorno and Marseilles _  M(7, 4).

F. 92v, p. 185.  Hare  3000  in front of hound who goes  8  while the hare goes  5.

Blasius.  1513.  F. F.iii.v: Decimasexta regula.  O-(10, 13),  T = 9.

Köbel.  1514.  ??NYS _ given in H&S 73.  O-(10, 13),  T = 9.

Tagliente.  Libro de Abaco.  (1515).  1541. 

Prob. 111, ff. 55v-56r.  Couriers meeting between Rome and Venice _  M-(17, 20).

Prob. 114, ff. 56v-57r.  O-(1, 1; 30, 0).

Prob. 118, f. 58r.  Ships meeting between Venice and Candia _  M‑(3, 5).

Prob. 122, ff. 59r-59v.  Hound chasing goat.  Goat is  50  leaps ahead and hound makes  7  to goat's  5.

Ghaligai.  Practica D'Arithmetica.  1521. 

Prob. 17, f. 64r.  O-(1, 1; 20, 0).

Prob. 18, f. 64r.  M-(4, 5)  with  D = 80.

Tonstall.  De Arte Supputandi.  1522. 

Quest. 35, p. 167.  O-(20, 33),  T = 6.

Quest. 37, pp. 168-169.  Couriers,  M-(3, 5).

Quest. 38, p. 169.  Couriers,  M-(_, ½).

Riese.  Die Coss.  1524.

No. 50, p. 47.  Messengers,  MR-(10, 8; 40).

No. 51, p. 47.  O-(6, 9),  D = 100.

No, 109, p. 54.  Messengers,  MR-(7, 9; 300).

No. 133, p. 59.  Fox is  300  fox leaps ahead of hound.  Each hound leap is  31/20  of a fox leap and they leap at the same rate.

Apianus.  Kauffmanss Rechnung.  1527. 

F. M.v.r.  Couriers between Leipzig and Venice.  M-(18, 24).

F. M.viii.v.  O-(7, 9),  D = 64.

Ff. M.viii.v - N.i.r.  Messengers between Prague and Vienna.  MR-(7, 9; 33).

F. N.i.r.  Hare is 30 leaps ahead of a hound.  Hound makes  8  leaps to hare's  6,  assumed same size.  I.e.  O-(8, 6),  D = 30.

Ff. N.i.r - N.i.v.  Hare is  50  hare leaps ahead of a hound.  Hare makes  4  leaps while hound makes  3,  but 2  hound leaps equal  3  hare leaps in length.  This can be viewed as  O-(8, 9),  where the units are  hare leaps per period in which a hare makes  8  leaps, with a head start of  50  hare leaps.

van Varenbraken.  1532.  ??NYS _ cited by the editor of King, 1795, p. 154.  Hound and hare _  O-(12, 15),  D = 200.

Cardan.  Practica Arithmetice.  1539.  Chap. 66.

Section 7, ff. BB.viii.v - CC.i.r (p. 137).   10n  =  1 + 4/3 + (6/5)(4/3) + (4/3)(6/5)(4/3) + (6/5)(4/3)(6/5)(4/3) + ....

Section 11, ff. CC.ii.v - CC.iii.r (p. 138) (iii  is misprinted  ii).  Hound & hare.  Hare is  60  hound leaps ahead.  Hound makes  63  leaps to hare's  100,  but  61  hound leaps equal  140  hare leaps.  Says Pacioli does it wrong, but I don't see that this is the same as Pacioli's problem??  [The phrasing of the problem is awkward.  H&S 72 gives the Latin and a mistranslation of it.]

Section 12, f. CC.iii.r (p. 139) (iii  is misprinted  ii).  O-(5, 3; 20, 0).

Section 13, ff. CC.iii.r - CC.iii.v (p. 139) (iii  is misprinted  ii).  1 + 2 + 4 + 8 + 16 + ... + 3 + 4 + 6 + 9 + 13 + ...  =  330.  He interpolates between the  7th  and  8th  days to get  7 95/137 = 7.69343,  while the correct(?) answer is  7.83659.  Says Pacioli could not do this.

Section 14, ff. CC.iii.v - CC.iiii.r (p. 139) (iii  is misprinted  ii).  Birds flying around the world.  1 + 2 + 3 + 4 + ... + 1 + 8 + 27 + 64 + ...  =  44310.  Seems to get  20½  days, but  20  is the answer.  (H&S 74 gives Italian and says similar problems are in Pacioli and Stifel.)

Section 15, ff. CC.iiii.r - CC.iiii.v (pp. 139-140).  

          1 + 2 + 4 + 8 + ...   =   2 + 4 + 6 + 8 + ....

Section 66, ff. FF.ii.r - FF.ii.v (p. 155).  (66 is not printed in the Opera Omnia).  O‑(3, 5; 30, 0).

Section 154, ff. MM.iiii.r - MM.iiii.v (p. ??).  An overtaking problem, but with times for the entire route given which can be considered as  M-(10, -7)  with delay of  2,  giving   (n+2)/10  =  n/7.  Described in H&S 73.

Buteo.  Logistica.  1559. 

Prob. 21, p. 219.  A simple hound and hare problem,  O‑(6b/5, b),  D = 1.

Prob. 28, pp. 229-230.  States  O-(1, 1; 22, 0),  but does  O‑(1, 1; 18, 0).

Prob. 29, pp. 230-231.  O-(1, 1; 20, 0).

Prob. 30, p. 231.  O-(2, 2; 15, 0).

Prob. 31, pp. 231-233.  O-(24, -1; 10, 2).  Finds common distance of  199_  by linear interpolation on  11th  day.

Prob. 32, pp. 233-234.  M(3, 2)  with  D = 200  given.

Baker.  Well Spring of Sciences.  1562?  ??check if this in the Graves copy of the 1562/1568 ed.) 

Prob. 3,  1580?: ff. 36v-37r;  1646: pp. 62-63;  1670: pp. 76-77.  O-(1, 1; 8, 0).

Prob. 4,  1580?: ff. 37r-37v;  1646: pp. 63-65;  1670: pp. 77-78.  MR-(1, 1; 2, 2; 200).  He notes that they meet when the first has done  200/3 = 66 2/3,  which is  2/3  on the 12th day.  He assumes they travel at constant speed on each day, so they meet after  11 1/18 = 11.0555....  [18  is misprinted  28  at one point.]  But if the velocity is increasing linearly, so that the distance covered to time  t  is  t(t+1)/2,  then the problem leads to a quadratic with solution  t = 11.0578....

Gori.  Libro di arimetricha.  1571.

F. 72r (p. 78).  Deer is  60  ahead of a dog, which goes  7  for every  4  of the deer, i.e.  O-(4, 7),  D = 60.

F. 72v (p. 78).  Fox is  80  ahead of a dog, which goes  7  for every  5  of the fox, i.e.  O‑(5, 7),  D = 80.

F. 81v (p. 79).  Couriers between Paris and Siena meeting _  M(20, 24)  with  D = 800,  but not used.

F. 82r (p. 79).  Couriers meeting _  MR(40, 50, 1000).

F. 71r (p. 79).  O-(1, 1; 30, 0).

F. 81v (p. 79).  O-(1, 1; 40, 0).

van Halle.  1568.  ??NYS _ cited by the editor of King, 1795, p. 155.  O‑(11, 0; 1, 1).

Io. Baptiste Benedicti (= Giambattista Benedetti).  Diversarum Speculationum Mathematicarum, & Physicarum Liber.  Turin, (1580), Nicolai Bevilaquæ, Turin, 1585; (Venice, 1599).  [Rara 364.  Graves 141.f.16.]  This has a number of overtaking and meeting problems, but he makes diagrams showing the sums of the arithemtic progressions involved.

Theorema CVI, pp. 68-69 (misprinted 70-71).  O-(4, 1; 1, 1)  with diagram showing  1 + 2 + ... + 7 = 4 x 7.

Theoremae CVII & CVIII, pp. 69 (misprinted 71) - 70.  O-(k, 0; 1, 2)  with diagrams showing  1 + 3 + ... + 15  =  8 x 8  and  1 + 3 + ... + 13  =  7 x 7.

Theoremae CIX & CX, pp. 70-71.  O-(k, 0; 2, 2)  with diagrams showing  2 + 4 + ... + 14  =  7 x 8  and  2 + 4 + ... + 16  =  8 x 9.

Theorema CXI, p. 71.  O-(k, 0; 3, 3)  with diagram showing  3 + 6 + ... + 21  =  7 x 12.

Theorema CXII, pp. 71-73.  O-(11, 0; 3, 3),  which does not have an integral solution.  He draws a diagram of  3 + 6 + ... + 21  and tries to make it into a rectangle of side  11,  but sees this does not work.  He then sees the answer is between  6  and  7  days, so he does linear interpolation over this period, getting  6.3  days instead of the correct  6 1/3.  He then uses a simple triangle to show this interpolation process.

Theorema CXIII, pp. 73-74.  O-(a, b)  with first having headstart of time  T  done in general.  Does  O-(20, 25),  T = 8.  Gives a diagram of  a x n+T  =  b x n.  There is an Appendix to this Theorema on pp. 74-75 which applies it to Jupiter overtaking Saturn but I can't make it out.

Theorema CXIIII, pp. 75-77.  M-(9, 11)  starting  D = 400  apart.  Then does the problem in general, giving a diagram which I can't follow.

Theorema CXV, pp. 77-78.  MR-(10, 15; 100),  then  MR-(a, b; D)  in general.

Wingate/Kersey.  1678?. 

Quest. 14, pp. 485-486.  O-(14, 22),  T = 8.

Quest. 15, pp. 486-487.  O-(9, 7)  on a circular island of circumference  36.

Quest. 16, p. 487.  O-(8, 0; 1, 1).

Quest. 17, pp. 487-488.  M-(8, 6; 140).

Quest. 18, pp. 488-489.  M-(a, a + 5/2; 100)  takes time  8.

Quest. 19, pp. 489-490.  M-(11/2, 17/3; 134)  going opposite ways round a circular island.

Quest. 23, pp. 491-492.  Hare is  100  hare-leaps ahead of a hound.  Hare takes five leaps while the hound takes four, but three hound leaps equal four hare leaps.

Edward Cocker.  Arithmetic.  Op. cit. in 7.R.  1678.  Chap. 10, quest. 32.  1678, p. 181;  1715: p. 121;  1787: p. 106.  This problem is attributed to Moor's or More's Arith. cap. 8, qu. 7.  1678 states  O-(40, 50),  T = 3,  giving answers  12  days and  600  miles.  1715 states  O-(40, 50),  T = 3,  giving answers  32  days and  600  miles.  1787 states  O-(48, 50),  T = 3,  giving answers  12  days and  600  miles.  I am surprised at the misprints here.

Peter Lauremberger (Petrus Laurembergus).  Institutiones Arithmeticæ ....  4th ed., Joh. Lud. Gleditsch, Leipzig, 1698.  P. 196, prob. 12.XII.  Fox is  60  hare-leaps ahead of a hound.  She makes  9  leaps while the hound does  6,  but  7  hare-leaps are as long as  3  hound-leaps.

Isaac Newton.  Arithmetica Universalis, 1707.  ??NYS.  English version:  Universal Arithmetic, translated by Mr. Ralphson, with revisions and additions by Mr. Cunn, Colin Maclaurin, James Maguire and Theaker Wilder; W. Johnston, London, 1769.  (De Morgan, in Rara, 652‑653, says there were Latin editions of 1722, 1732, 1761 and Raphson's English editions of 1720 and 1728, ??NYS.)  Resolution of Arithmetical Questions, Problem V, pp. 180‑184.  Begins with  MR(7/2, 8/3; 59)  with second delayed by 1.  Then "The same more generally" does  O(c/f, d/g)  with second having a headstart of distance  e  and either a headstart or delay of time  h  and  MR(c/f, d/g; e)  with second having either a headstart or delay of time  h.  Then does example:  O(13, 1)  with second starting distance  90  ahead, but first delayed by  3  days.  Then repeats original example from general viewpoint.

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790).

Prob. XIX, p. 89 (1790: prob. XXXI, p. 92).  O-(28, -2; 20, 0).

Prob. L, pp. 113-115.  MR-(40, -2; 20, 2; 360).

1790: Prob. LXIII, pp. 114‑116.  MR-(60, -5; 40, 5; 500).

Prob. LI, pp. 115-116 (1790: prob. LXIV, pp. 116‑117).  Overtaking.  8 + 12 + ... + (4+4n)  =  1 + 4 + ... + n2.

Alexis-Claude Clairaut.  Élémens d'algèbre.  1746.  ??NYS _ cited by Tom Henley.  Discusses the relation between overtaking and meeting problems and the use of negative rates or negative headstarts to make them algebraically the same.

Les Amusemens.  1749. 

Prob. 113, p. 255.  O-(2½, ½; 7, 0).

Prob. 114, p. 256.  O-(5, 1; 2, 0).

Prob. 115, p. 257.  Meeting:   4n + n+2 + n+4 + n+6 + n+8  =  104.

Prob. 116, p. 258.  O-(10, 5; 30, 0).

Prob. 117, p. 259.  Cat and mouse,  O-(3, 5),  D = 23.

Euler.  Algebra.  1770.  I.IV.III: Questions for practice.

No. 16, p. 205.  Privateer and prey,  O-(18, 20),  D = 8.  [The numbers  8  and  18  are interchanged in the text.]

No. 25, p. 206.  Hare  50  hare leaps ahead of greyhound.  Hare makes  4  leaps to greyhound's  3,  but hare's leaps are only  _  as long.

Vyse.  Tutor's Guide.  1771? 

Prob. 57, p. 74 & Key p. 99.  O-(22, 32),  T = 4.  Misprint in solution.

Prob. 60, p. 74 & Key P. 100.  MR-(x, x+2½; 135)  with meeting after  8.

Prob. 65, p. 75 & Key p. 101.  MR-(2, 3; 170)  with second delayed by 8.

Prob. 67, p. 75 & Key p. 102.  Hound & hare,  O-(21, 15),  D = 96.  The actual rates are not given, only their ratio, so one can determine where the hare is caught, but not when.

Prob. 12, p. 84 & Key p. 109.  A  and  B  start to circle a wood of circumference  135,  starting at opposite sides and going in the same direction.  A  goes  11/2,  B  goes  17/3.  When do they meet?  O‑(11/2, 17/3),  D = 67½.

Prob. 13, p. 198 & Key p. 241.  Two rowers who can row at  5  set out towards each other at points  34  apart on a river flowing  2½.  Though this appears to belong in Section 10.G, it is simply  MR-(2½, 7½; 34).

Prob. 6, p. 201 & Key p. 245.  Hare starts  5 rods  and  34 sec  before a greyhound and goes at  12 mph,  while the hound goes  20 mph.  Using  320 rods to the mile, this is  O-(16/15, 16/9)  where the slower has a headstart of  T = 34  and  D = 5.

Dodson.  Math. Repository.  1775.

P. 1, Quest. I.  MR-(8, 7; 150)

P. 2, Quest. V.  O-(30, 42),  T = 4.

P. 25, Quest LXIV.  Same as Euler's no. 25.

P. 57, Quest. CX.  MR-(7/2, 8/3; 59)  with second starting one unit of time later.

Pp. 58-59, Quest CXI.  General solution of  O-(a, b)  with one having a headstart and starting before or after the other.  Does  O-(10/3, 5/2),  D = -59,  T = 4;  then  O‑(10/3, 5/2),  D = -59,  T = -4.

P. 178, Quest XXIV.  O-(8, 0; 1, 2).

P. 179, Quest XXVI.  MR-(40, -2; 20, 2; 360).

P. 183, Quest. XXX.  8 + 12 + 16 + ...  =  1 + 4 + 9 + ...

Pp. 191-192, Quest XLII.  1 + 2 + 3 + ...  +  1 + 8 + 27 + ...  =  462.

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping.  Op. cit. in 7.G.2.  [c1780?] 

1804: prob. 68, p. 139.  Hare starts  40  yards before a hound and goes at  10  mph.  The hound doesn't see the hare for  40  seconds, and then goes at  18  mph.

1804: prob. 69, p. 139.  Exeter is  130  miles from London.  A  sets out from Exeter at  8  am going  3  mph;  B  sets out from London at  4  pm going  4  mph.  Where do they meet?

1804: prob. 71, p. 140.  Lincoln is  100  miles from London.  Travellers set out at the same time and meet after seven hours, when they find that  A  has gone  1½  mph faster than  B.  What are their rates?

Bonnycastle.  Algebra.  1782. 

P. 84, no. 5.  MR-(8, 7; 150).

P. 85, no. 19 (1815: p. 107, no. 31).  Same as Euler's no. 25.

Pike.  Arithmetic.  1788.  P. 350, no. 17.  Circle  268  in circumference.  Two men start at ends of a diameter and go in the same direction at rates  22/2  and  34/3.  When and where do they meet?   I. e.  O-(34/3,  22/6),  D = 134.

Eadon.  Repository.  1794. 

P. 80, no. 36.  O-(30, 42),  T = 4.

P. 80, no. 37.  MR-(23, 31; 162).

P. 235, ex. 3.  5 + 9 + 13 + ... + (5 + 13x4).  Find total travel.

John King, ed.  John King  1795  Arithmetical Book.  Published by the editor, who is the great-great-grandson of the 1795 writer, Twickenham, 1995. 

P. 109.  A meeting problem, but the times to meeting and the meeting point are given, so it reduces to:   a + a+2 + a+4 + a+6 + a+8  =  50  =  b + b+3 + b+6 + b+9.

Pp. 117-118.  Hare is  144  hare leaps ahead of a grayhound.  Hare makes  4  leaps while grayhound makes  3,  but grayhound leaps are  3/2  as big.

Thomas Manning.  An Introduction to Arithmetic and Algebra.  2 vols., Nicholson, Lunn & Deighton, Cambridge, 1796 & 1798.  Vol. 1, pp. 208-210.  O-(5/2, 7/2),  T = 2.  Then considers what would happen if the second traveller went slower than the first [more simply, suppose the second traveller had the headstart, i.e.  T  is negative].  This gives a negative solution and he interprets this as that they must have met before the starting time.  He gives a general solution and discussion of the problem.  ??NX.

Hutton.  A Course of Mathematics.  1798? 

Prob. 5,  1833: 210-211;  1857: 214-215.  Hare is  60  hare-leaps ahead of a greyhound.  She makes  9  leaps while the hound does  6,  but  7  hare-leaps are as long as  3  hound-leaps.  = Lauremberger.

Prob. 9,  1833: 213-214;  1857: 217-218.  O-(31½/5, 22½/3),  T = 8.  Then does  O‑(m/t, m'/t')  with first having headstart of  T. 

Prob. 20,  1833: 221;  1857: 225.  M-(8, 7, 150).

Remarks upon Equations of the First Degree,  1833: 224-231;  1857: 228-235,  is an extensive discussion concerning possible negative roots and considers  O-(m, n)  with the second having a headstart of distance  D.  When  m < n,  he says the directions must be reversed.

Robert Goodacre.  Op. cit. in 7.Y.  1804.  Miscellaneous Questions, no. 125, p. 205.  A  goes  4  mi/hr for  7  hrs each day.  B  starts a day later at  5  mi/hr for  8  hours each day, both starting at the same time each morning.  When does  B  overtake?

Augustus De Morgan.  Arithmetic and Algebra.  (1831 ?).  Reprinted as the second, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  Art. 116, pp. 30-31.  Hare is  80  hare-leaps ahead of a greyhound.  Hare makes  3  leaps for every  2  of the hound, but a hound-leap is twice as long as a hare‑leap.  Then considers a hound-leap as  n/m  of a hare-leap.  Takes  n/m = 4/3  and finds a negative solution which he discusses.  Takes  n/m = 3/2  and finds division by zero which he interprets as the hound never catching the hare.

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??) 

Art. 47, prob. 3, pp. 65-66.  Same as Hutton, 1798?, pp. 210-211, with greyhound chasing a fox.  = Lauremberger.

Art. 190, question 6, p. 319.  O-(10, 0; 3, 2).

Augustus De Morgan.  On The Study and Difficulties of Mathematics.  First, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  P. 30 mentions  O-(2, 3),  T = 4  and  O-(a, b) with second delayed by time  T.  Pp. 37-39 discusses the general courier problems  O-(m, n)  with second having headstart of distance  D  and  MR-(m, n; D).  He considers different signs and sizes, getting six cases.

Unger.  Arithmetische Unterhaltungen.  1838.  Pp. 135 & 258, nos. 515 & 516.  MR‑(5/4, 3/5; 57)  with first delayed by time  2½.  The second problem asks what delay of time for the first will make them meet at the half-way point?

Hutton-Rutherford.  A Course of Mathematics.  1841?

Prob. 7,  1857: 81.  Two persons on opposite sides of a wood of circumference  536,  start to walk in the same direction at rates  11, 11_.  How many times has the wood been gone round when they meet?  =  O-(11, 11_),  D = 268.  Answer is the number of times the faster has gone round.

Prob. 23,  1857: 82.  MR-(3, 4; 130),  T = 8.

Prob. 37,  1857: 83.  Hare starts  40 yd  and  40 sec  in front of a hound.  Hare goes  10 mph (=  44/9 yd/sec)  and hound goes  18 mph (= 44/5 yd/sec).  I.e.  O-(44/9, 44/5),  D = 40,  T = 40.

Tate.  Algebra Made Easy.  Op. cit. in 6.BF.3.  1848.  P. 46, no. 15.  Man makes a journey at  4  mph and returns at  3  mph, taking  21  hours in total.  How far did he go?

Family Friend 1 (1849) 122 & 150.  Arithmetical problems _ 1.  "A hare starts  40  yards before a greyhound, and is not perceived by him till she has been up  40  seconds:  she gets away at the rate of  10  miles an hour and the dog pursues her at the rate of  18  miles an hour:  how long will the course last, and what distance will the hare have run?"  = Illustrated Boy's Own Treasury, 1860, Prob. 4, pp. 427 & 431.  = Hutton, c1780?, prob. 68.

Anonymous.  A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools.  3rd ed., 1850.  Op. cit. in 7.H.

Pp. 198-199, no. 117.  Hare is  50  springs ahead of a hound.  Their springs are of equal length, but the hound makes  27  while the hare makes  25.  How many springs does the hare make before being overtaken?  Answer is misprinted  675  instead of  625.

P. 360, no. 47.  = Vyse, prob. 12.

P. 360, no. 48.  See in 7.P.6.

John Radford Young (1799-1885).  Simple Arithmetic, Algebra, and the Elements of Euclid.  IN: The Mathematical Sciences, Orr's Circle of the Sciences series, Orr & Co., London, 1854.  [An apparently earlier version is described in 7.H.]

No. 5, p. 177.  O-(9, 5),  T = 10.

No. 5, p. 216.  A  and  B  travel  90  miles.  A  goes  1  mile per hour faster than  B  and arrives  1  hour before him.  What were their speeds?

No. 8, p. 216.  Same as last with data  150,  3,  8_.

No. 4, p. 228.  MR-(1, 1; 20, -2; 165).

Edward Brooks.  The Normal Mental Arithmetic  A Thorough and Complete Course by Analysis and Induction.  Sower, Potts & Co., Philadelphia, (1858), revised, 1863.  Further revised as:  The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction;  Sower, Potts & Co., Philadelphia, 1873.  Many examples.  I mention just one example.

                    1863 _ pp. 132-13, no. 19;  1873 _ p. 161, no. 18.  "E  takes  60  steps before he is overtaken by  D;  how many steps does  D  take to catch  E,  provided  E  takes  4  steps while  D  takes  3,  and  5  of  D's  equal  8  of  E's,  and how far ahead was  E  when they started?"

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.  No. 4, pp. 166 & 334.  General problem:   ax  =  b (n - x).   Solves for  n ‑ x.

Clark.  Mental Nuts.  1904: no. 62.  The fox and the hound.  Fox is  60  (fox-)leaps in front of a hound.  Hound takes  6  leaps to fox's  9,  but the hound leaps are  7/3  times as long as the fox's.  (= Lauremberger)

Dudeney.  Weekly Dispatch (17 May 1903) 13  &  (14 Jun 1903) 16.

Perelman.  1937.  MCBF.  At the cycle track, prob. 150, pp. 255-256.  Circular track of circumference  170m.  When cyclists are going in opposite directions, they meet every  10  sec;  when going the same direction, the faster passes the slower every  170  sec.

C. Dudley Langford.  Note 1558:  A graphical method of solving problems on "Rate of Work" and similar problems.  MG 25 (No. 267) (Dec 1941) 304-307.  +  Note 2110:  Addition to Note 1558:  "Rate of Work" problems.  MG 34 (No. 307) (Feb 1950) 44.  Uses a graph to show  (a, b)  cistern problems as meeting problems.  Also solves problems  (A, x)  in  B  and  (a, -b),  the latter appearing as an overtaking problem.  The Addition gives a clearer way of viewing  (a, ‑b)  problems as overtaking problems

Gamow & Stern.  1958.  Pp. 9‑10, 59‑63.  Elevator problem.

 

          10.A.1.         CIRCLING AN ARMY

 

          The linear form has an army of length  L.  A rider goes from the rear to the front and then back to the rear, reaching the rear when it has advanced  D.  How far,  d,  does he go?   Here there are insufficient equations to determine all the values (as also occurs in 10.A.3), but the value of  d  can be found as   d  =  L + Ö(L2 + D2).   Other versions of the problem can be solved.

          If the army has a width  W,  and the rider goes across the moving army at each end, the situation is more complex _ see the first example.

J. Gale, proposer;  Joseph Edwards, Jr.  &  Mr. Coultherd, solvers.  Question III.  A Companion to the Gentleman's Diary; ... for the year 1798, pp. 59-60  &  The Gentleman's Mathematical Companion, for the year 1799, pp. 16-17.  Wagoner walking around his wagon and team while it is travelling.  L = 20 yd,  W = 4 yd.  He can walk  V = 4 mph  and he walks  74_ yd  in his circumambulation (hence taking  T = 7/11 min).  A complication arises as to how he goes crossways.  The proposer says he turns at right angles and passes  2 yd  clear at the front and at the back.  (He also says the walker passes  2 yd  clear on each side, but he never gives a width, taking the distance between the two side paths as  4,  which I have accounted for by taking  W = 4.  In some cases, we can account for the  2 yd  at each end by taking  L = 24.)  How fast,  v,  is the team going?  Both solvers get  v = 2 mph, but I find neither is viewing the problem correctly and neither has correctly formulated the problem he has described! 

                    Edwards starts in the middle, 2 in front of the horses, and says the the walker must go 2 to the left, 24 back and 2 to the right to get the the middle in back.  He then says the time required is  28/(V+v).  But this assumes the crossways motion is at the same relative speed as the lengthwise relative motion and seems definitely incorrect to me.  Edwards then says that a similar argument gives the time in the other direction as  28/(V-v).  Setting the sum of these equal to  T  does give  v = 2.

                    Coultherd starts in the middle, 2 behind the wagon, but I will rephrase it to parallel Edwards' solution.  It takes the man  2/V  to go  2  to the left.  During this time, the wagon moves ahead  2v/V,  so he is now only  2 - 2v/V  in front.  When he gets to the back, he only needs to be  2 - 2v/V  behind the wagon when he turns to cross, so that he will be exactly 2 behind at the middle.  So he must make a relative motion of  24 ‑ 4v/V  at the relative velocity  V + v.  Similarly, for the forward trip, he makes relative motion  24 + 4v/V  at relative velocity  V - v.  Adding the times for these to  8/V  for the crossways trips and setting equal to  T,  I get  v = 2.056 mph, whch seems to be the correct answer.  Coultherd forgets to account for the  2 - 2v/V  at the end of his first lengthwise trip and confuses distance travelled at  V  and at  V-v,  which simplifies his algebra to the same final equation as Edwards. 

                    I am rather surprised at the basic errors in both solutions.

Clark.  Mental Nuts.  1904: no. 98;  1916: no. 99.  A West Pointer.  Column is  25  miles long.  Courier goes from the rear to the front and returns to the rear and sees that he is now where the front of the column was when he started.  I.e. his trip take the same time as the time the column moves 25 miles.  How far did he go?  Answer is  60 miles, 1876 feet,  which is correct.  Here  L = 25 = D,  so  d = 60.36.

Loyd.  The courier problem.  Cyclopedia, 1914, pp. 315 & 382.  (= MPSL2, prob. 146, pp. 103 & 167‑168, with solution method provided by Gardner _ Loyd only gives the values "following the rule for solving puzzles of this kind".)  Army  50  miles long.  Rider goes from back to front to back of the army in the time it moves forward  50  miles.  Here  L = 50 = D,  so  d = 120.71  _ Loyd says "a little over 120 miles".  He also extends to the case of a square army.

Abraham.  1933.  Prob. 67 _ The column of troops, pp. 33 & 44 (19 & 116).  Rider circling army _ same as first part of Loyd.

Perelman.  1937.  MCBF.  Reconnaissance at sea, prob. 149 a & b, pp. 253-255.  Squadron moving and a reconnaissance ship is sent ahead.  Part  a  gives distance to go ahead and asks how long it will take.  Part  b  gives time for reconnaissance and asks when the reconnaissance ship will turn back.

McKay.  At Home Tonight.  1940.  Prob. 17: The orderly's ride, pp. 65 & 79.  Same as first part of Loyd with  1  mile army.

William R. Ransom.  Op. cit. in 6.M.  1955.  An army courier, p. 103.  Same as first part of Loyd with  25  mile army, but says it takes a day and asks how fast the courier rides.

Nathan Altshiller Court.  Mathematics in Fun and in Earnest.  Op. cit. in 5.B.  1961.  Prob. m, pp. 189 & 191-192.  Courier going from back to front and then back again at three times the speed of the army.  Where is he when he gets to the back again?

Julius Sumner Miller.  Millergrams.  Ure Smith, Sydney, 1966.  Prob. 25, pp. 25 & 70.  Army is 3 mi long, officer starts at back, goes to front and returns, reaching the back when it has advanced 4 miles.  How far did he go?  L = 3,  D = 4,  giving  d = 8 miles.  He says: "you can get the right answer by erroneous logic", but he doesn't explain how to get the answer!

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Procession, pp. 43, 167 & 189.  = Birtwistle; Calculator Puzzle Book; 1978; prob. 55, pp. 38-39 & 99.  Procession  1½  miles long going at  2  mph.  Marshall starts at head, walks to the back and then forward, reaching his starting point when half the army has passed.  What is his speed?  He continues on the head and returns to the same point.  Where is the end of the procession when he gets back?

 

          10.A.2.         NUMBER OF BUSES MET

 

          New section _ I haven't done much on this yet.  Kraitchik has several examples.

 

Mittenzwey.  1879?  Prob. 82, pp. 18 & 65.  Daily ships crossing the Atlantic in  7  days.

[Richard A. Proctor].  Letters received and short answers.  Knowledge 3 (26 Oct 1883) 264.  Answer to Harry.  Recalls it being posed on by the captain's wife, Mrs Cargill, on the S. S. Australasia, but no date is clear.  Trains going between New York and San Francisco taking  7  days.  How many does one meet on such a trip?  Says he gave the wrong answer!

[Richard A. Proctor].  Editorial gossip.  Knowledge 3 (23 Nov 1883) 318.  Gives a careful answer to the problem stated above.

L. Carroll.  A Tangled Tale.  (1885) = Dover, 1958. 

Knot III, pp. 13‑18, 90‑95.  Circular railway with trains going at different frequencies in the two directions.  How many are met on a round trip in each direction?

Knot V, pp. 27‑28.  Comment on Knot III problem.

Knot VIII, pp. 52‑53, 55‑57, 132‑134.  Buses in both directions pass depot every  15  minutes.  Walker starts from depot at same time as a bus and meets a bus in  12½  minutes.  When is he overtaken by a bus?

Laisant.  Op. cit. in 6.P.1.  1906.  Chap. 49: Du Havre à New-York, pp. 123-125.  Definitely asserts that this problem was posed by Lucas at a meeting 'longtemps déjà'.  Boats leave every noon each way between le Havre and New York and take exactly seven days to make the trip _ how many do they pass?

Pearson.  1907.  Part II, no. 106, pp. 136 & 212‑213.  Tube trains run every  2  minutes.  How many are met in a  30  min journey?  Answer:  30.

Peano.  Giochi.  1924.  Prob. 11, p. 4.  Buses take  7  minutes from the centre to the terminus.  Buses leave from the centre and from the terminus every minute.  How many buses are met in going one way?  Observes that it takes  14  buses to run such a service and so one bus meets the other  13.

McKay.  At Home Tonight.  1940.  Prob. 14: Passing the trains, pp. 65 & 79.  12  hour journey and trains start from the other end every hour.

Sullivan.  Unusual.  1943.  Prob. 24: Back in the days of gasoline, tires, and motorists.  4½  hours from Chicago to Indianapolis, buses leaving every hour [on the hour].

Joseph & Lenore Scott.  Master Mind Brain Teasers.  Op. cit. in 5.E.  1973.  How many buses, pp. 7-8.

 

          10.A.3.         TIMES FROM MEETING TO FINISH GIVEN

 

Simpson.  Algebra.  1745.  Section XI (misprinted IX in 1790), prob. XLVI, pp. 110-111 (1790: prob. LIX, pp. 111‑112).  Travellers set out from each of two cities toward the other, at the same time.  After meeting, they take  4  and  9  hours to finish their journeys.  How long did they take?  He gives a general solution _ if  x  is the time before meeting and  a, b  are the times from meeting to finishing, then  x2 = ab.  [I have seen a 20C version where only the ratio of velocities is asked for _ indeed I used it in one of my puzzle columns, before I knew that the times could be found.]

Dodson.  Math. Repository.  1775.

P. 67, Quest. CXXV.  Travellers set out from London and York at the same time.  When they meet, they observe that  A  had travelled  30  miles more than  B  and that  A  expected to reach York in  4  days and  B  expected to reach London in  9  days.  What is the distance between London and York?

P. 68, Quest. CXXVII.  Travellers set out from London and Lincoln at the same time.  When they meet, they observe that  A  had travelled  20  miles more than  B  and that  A  travelled in  6_  days as much as  B  had gone and  B  expected to get to London in  15  days.  What is the distance from London to Lincoln?

Ozanam‑Montucla.  1778.  Supplement, prob. 42, 1778: 432;  1803: 425;  1814: 360;  1840: 186.  Couriers set out toward one another from cities 60 apart.  After meeting, they take  4  and  6  hours to reach their destinations.  What are their velocities?

Thomas Grainger Hall.  The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges.  Second Edition: Altered and Enlarged.  John W. Parker, London, 1846.  P. 136, ex. 38.  Simpson's problem with cities of London and York and times of 9 and 16.

T. Tate.  Algebra Made Easy.  Op. cit. in 6.BF.3.  1848.  P. 89, no. 4.  Same as Simpson.

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.  No. 132, pp. 306 & 347.  Travellers between London and York reach their destinations  25  &  36  hours after meeting.  How long did each take?

W. W. Rouse Ball.  Elementary Algebra.  CUP, (1890), 2nd ed., 1897.  Ex. 5, p. 231.  A  and  B  are  168 miles  apart.  Trains leave each end for the other starting at the same time.  They meet after  1 hour 52 minutes.  The train from  A  reaches  B  half and hour before the other reaches  A.

Nathan Altshiller Court.  Mathematics in Fun and in Earnest.  Op. cit. in 5.B.  1961.  Prob. b, pp. 188-190.  Trains start toward each other at  7  am;  one takes  8  hours, the other takes  12.  When do they meet?

Shakuntala Devi.  Puzzles to Puzzle You.  Op. cit. in 5.D.1.  1976.  Prob. 37: Walking all the way, pp. 29 & 106-107.  Similar to Simpson, but they start at given times and the time of meeting is given and they get to their destination at the same time.  This is thus the same as Simpson if it is considered with time reversed.  The elapsed times to meeting are  25/12  and  49/12  hours.

Birtwistle.  Calculator Puzzle Book.  1978. 

Prob. 60: Meeting point, pp. 42 & 102.  Travellers start at the same time from opposite ends of a journey.  When they meet, they find that the first has  11 1/5  hours to go, while the other has  17 1/2.  Also the first has travelled  7  miles further than the second.  How long is the journey?

Prob. 71: Scheduled flight, pp. 50-51 & 108.  Planes  A  and  B  start toward each other at  550  mph.  Five minutes later, plane  C  starts from the same place as  A  at  600  mph.  It overtakes  A  and then meets  B  36  minutes later.  When does  A  land?

 

          10.A.4.         THE EARLY COMMUTER

 

          A man is usually met by a car at his local train station, but he arrives  A  minutes early and begins walking home.  The car meets him and picks him up and they arrive home  B  minutes early.  How long was he walking?  The car's trip is  B/2  minutes shorter each way, so the commuter is met  B/2  minutes before his usual time and he has been walking  A - B/2  minutes.  New section _ there must be older examples.

 

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 155, pp. 61 & 151: Catching the postman.  Man driving 10 mph usually overtakes the postman walking 4 mph at the same point every morning.  One morning, the man is four minutes late and he overtakes the postman half a mile beyond the usual point.  Was the postman early or late, and by how much?  (Note that the man is four minutes late starting, not in overtaking.)

Meyer.  Big Fun Book.  1940.  No. 10, pp. 162-163 & 752.  A = 60,  B = 10.

Harold Hart.  The World's Best Puzzles.  Op. cit. in 7.AS.  1943.  The problem of the commuter, pp. 8 & 50.  A = 60,  B = 20.

The Little Puzzle Book.  Op. cit. in 5.D.5.  1955.  Pp. 59-60: An easy problem.  A = 60,  B = 20,  he walks at  4  mph.  How fast does the chauffeur drive?

Nathan Altshiller Court.  Mathematics in Fun and in Earnest.  Op. cit. in 5.B.  1961.  Prob. f, pp. 189 & 191.  A = 60,  B = 16.  The time of arrival is also given, but is not needed.

Liz Allen.  Brain Sharpeners.  Op. cit. in 5.B.  1991.  The Commuter's tale, pp. 87-88 & 136.  A = 60,  B = 10.

 

          10.A.5.         HEAD START PROBLEMS

 

          Doubleday-II gives a typical example.  New section _ I have seen other examples but didn't record them.

          Gardner, in an article:  My ten favorite brainteasers  in Games (collected in Games Big Book of Games, 1984, pp. 130-131) says this is one of his favorite problems.  ??locate

 

Charles Pendlebury.  Arithmetic.  Bell, London, (1886), 30th corrected and expanded printing, 1924.  Section XXV (b): Races and games of skill, pp. 267-268 & answers, p. viii.  Does an example:  A  can give  B  10  yards in  100  yards and  A  can give  C  15  yards in  100  yards.  How much should  B  give  C  in  150  yards?  He gives  11  similar problems.  On pp. 363-364 & answers, part II, p. xix, he gives some further problems, 45-48.

W. W. Rouse Ball.  Elementary Algebra.  CUP, (1890), 2nd ed., 1897. 

Ex. 8, p. 135.  "A  can row a mile in  ¾ of a minute less time than  B.  In a mile race,  B  gets  250 yards head start, and lose by  14 yards."  Determine their times to row a mile, assuming uniform speed.

Prob. 37, pp. 161 & 467,  If  B  gets a  12 sec headstart in a mile race, he loses by  44 yards.  If he gets a  165 yards headstart, he wins by  10 sec.  Determine their times for a mile.

Collins.  Fun with Figures.  1928.  The 100-yard dash, pp. 24-26.  Bill can run  100  yards in  12  sec.  Jack can give Bill a  10  yd headstart and finish even with him.  Bob can give Jack a  5  yd headstart and finish even.  Bob gives Bill a  15  yd headstart and Bill loses  1/20  of a sec in a bad start.  Who wins and by how much?

Doubleday - II.  1971.  Running commentary, pp. 59.  In a  100  yard dash,  A  gives  B  a head start of  25  to make an even race and  B  gives  C  a head start of  20.  How much head start should  A  give  C  to make an even race?

 

          10.A.6.         DOUBLE CROSSING PROBLEMS.

 

          New section.  I recall Loyd and/or Dudeney have versions.  Also, the problems in 10.A with the faster meeting the slower on the return trip are related.

 

Wood.  Oddities.  1927.  Prob. 61: The errand boys, p. 47.  Alan goes from  A  to  B  and back while Bob goes from  B  to  A  and back, both starting at the same time.  They first cross at  720  from  A  and then at  400  from  B.  How far is it between  A  and  B?  Gives a general formula:  If the two distances are  a, b,  then the distance is  3a - b.  He asserts that Alan goes faster, but the ratio of velocities is  a/(2a-b)  which is  9/13  in this case.

 

          10.B.  FLY BETWEEN TRAINS

 

          There are two trains  d  apart, approaching at rates  a, b.  A fly starts at one, flies to the other, then back to the first, then back to the second, etc., flying at rate  c.  How far does he go?

          NOTATION.  We denote this problem by  (a, b, c, d).  For overtaking problems, we let  b  be negative.  I need to check for details.

 

Laisant.  Op. cit. in 6.P.1.  1906.  Chap. 53: Le chien et les deux voyagers, pp. 132-133.  A  is  8  km ahead and goes at  4  km/hr.  B  starts after him at  6  km/hr.  A dog starts at one and runs back and forth between them at  15  km/hr until they meet.  I.e. (4, ‑6, 15, 8).  Notes that the distance the dog travels is independent of where he starts and that the travellers could be meeting rather than overtaking.

Dudeney.  Problem 464: Man and dog.  Strand Mag. (Jul 1919).  ??NX.

Dudeney.  Problem 643: Baxter's dog.  Strand Mag.  (1924?).  ??NX.  A  goes at rate  2  and has an hour's head start.  B  goes at rate  4  and dog starts with him at rate  10.  I.e. (4, ‑2, 10, 2).

G. H. Hardy.  Letter of 5 Jan 1924 to M. Riesz.  In:  M. L. Cartwright.  Manuscripts of Hardy, Littlewood, Marcel Riesz and Titchmarsh.  Bull. London Math. Soc. 14 (1982) 472‑532.  (Letter is on p. 502, where it is identified as Add. MS. a. 275 33, presumably at Trinity College.)  Says it defeated Einstein, Jeans, J. J. Thomson, etc.  Fly between cyclists  (10, 10, 15, 20).  "One thing only is necessary: you must not know the formula for the sum of a geometrical progression.  If you do, you will take  15-20  minutes: if not,  2  seconds."

Ackermann.  1925.  Pp. 116‑117.  Couple walking up a hill.  Their dog, who is twice as fast as they are, runs to the top and back to them continually.  (a, 0, 2a, d) _ he only says the couple start  1/4  of the way up the hill.

Dudeney.  Problem 754: The fly and the motor‑cars.  Strand Mag. (Jun 1925)  ??NX.  (?= PCP 72.)

H&S 53, 1927, says this is 'a modern problem'.

M. Kraitchik.  La Mathématique des Jeux, 1930, op. cit in 4.A.2, chap. 2, prob. 17, p. 30.  (15, 25, 100, 120).  (Identified as from L'Echiquier, 1929, 20, ??NYS.)  I can't find it in his Mathematical Recreations.

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 181, pp. 71 & 156: Little Red Riding Hood.  Little Red Riding Hood with her dog and Grandma set out at the same time to meet: (2, 2, 8, 6).

Dudeney.  PCP.  1932.  Prob. 72: The fly and the motor‑cars, pp. 28 & 86.  = 536; prob. 86: The fly and the cars, pp. 26 & 243.  (50, 100, 150, 300).

Phillips.  Week‑End.  1932.  Time tests of intelligence, no. 38, pp. 21 & 193.  Fly between cyclists.  (10, 15, 20, 60).

Abraham.  1933. 

Prob. 11 _ The fly and the cyclists _ A problem in convergent series?, pp. 6 & 23 (4 & 111).  (10, 10, 15, 20).

Prob. 63 _ The escaping prisoner, pp. 21 & 28 (16 & 115).  Warders going  4  after an escapee going  3  with  3/2  headstart.  A dog runs back and forth at  12.  I.e.  (4, ‑3, 12, 3/2).

Streeter & Hoehn.  Op. cit. in 7.AE.  Vol. 2, 1933, p. 29: Aeroplane dilemma.  Destroyer going  25  overtaking battleship going  20  with headstart of  30.  Plane flies back and forth at  90.  I.e.  (25, ‑20, 90, 30).

Phillips.  Brush.  1936.  Prob. G.21: The busy fly, pp. 20 & 87.  Same as in Week‑End.

J. R. Evans.  The Junior Week‑End Book.  Op. cit. in 6.AF.  1939.  Prob. 37, pp. 265 & 270.  Fly between cyclists,  (10, 10, 15, 20).

McKay.  At Home Tonight.  1940.  Prob. 20: The fluttered pigeon, pp. 66 & 80-81.  Pigeon between walkers _  (3, 3, 21, 30).  Gives solution by adding a GP and the easy solution.

Sullivan.  Unusual.  1943.  Prob. 4: A busy bee.  Bee between motorists _ (10, 10, 15, 20).

L. Lange.  Another encounter with geometric series.  SSM 55 (1955) 472‑476.  Studies the series involved.

William R. Ransom.  Op. cit. in 6.M.  1955.  The bicycles and the fly, pp. 22‑23.  Studies the series.

Eugene Wigner.  In the film:  John von Neumann, MAA, 1966, he relates this as being posed by Max Born to von Neumann, involving a swallow between bicyclists.  He says it was a popular problem in the 1920s.

Paul R. Halmos.  The legend of John von Neumann.  AMM 80 (1973) 382-394.  Gives the fly between two cyclists puzzle and story on pp. 386-387.

David Singmaster.  The squashed fly _ (60, 40, 50, 100)  with fly starting on first train.  Used in several of my series.

On training a fly to fly right.  Los Angeles Times (21 Dec 1987) Section CC Part II.

A very fly braintwister.  Special Holiday Edition of The Daily Telegraph:  The Great British Summer (Aug 1988) 15 & 10.

Flight of fancy.  Focus, No. 2 (Jan 1993) 63 & 98.

The squashed fly.  Games & Puzzles, No. 11 (Feb 1995) 19  &  No. 12 (Mar 1995) 41.

Herbert R. Bailey.  The girl and the fly: A von Neumann Legend.  MS 24 (1991/92) 108-109.  Cites Halmos, but gives a version with a girl walking toward a wall.  Finds the relevant series.

Yuri B. Chernyak & Robert S. Rose.  The Chicken from Minsk.  BasicBooks, NY, 1995. 

Chap. 2, prob. 10: The crazy dog (or the problem that didn't fool John Von Neumann), pp. 16 & 108.  Dog between cyclists; usual solution and some comments.  Refers to MTr (Nov 1991) _ ??NYS.

Chap. 11, prob. 7: The problem that didn't fool Von Neumann, pp. 95 & 183.  Does it by summing one series.

 

          10.C.  LEWIS CARROLL'S MONKEY PROBLEM

 

          A monkey and a barrel of equal weight are on the ends of a rope over a pulley.  The monkey starts to climb the rope _ what happens?

 

Lewis Carroll.  Diary _ entry for 21 Dec 1893.  Ed. by Roger Lancelyn Green.  Cassell, London, 1953, p. 505.

S. D. Collingwood.  The Life of Lewis Carroll.  1898, pp. 317‑318.  ??NYS

S. D. Collingwood & Arthur Brook.  In:  The Lewis Carroll Picture Book, 1899, op. cit. in 5.B, pp. 267‑269 (Collins: 193-194).  Brook writes about another viewpoint and Collingwood discusses it.

Pearson.  1907.  Part II, no. 11: A climbing monkey, pp. 9 & 188.  Asserts that the weight goes up, but the monkey does not!

Dudeney.  Some much‑discussed puzzles.  Op. cit. in 2.  1908.  Cites the diary entry from Collingwood's Life and Brook's comment in the Picture Book.  Says mechanical devices have been built.

Loyd.  Lewis Carroll's monkey puzzle.  Cyclopedia, 1914, pp. 44, 344‑345 (erroneous solution).  (= MPSL2, prob. 1, pp. 1‑2 & 121.)

William F. Rigge.  The climbing monkey.  SSM 17 (1917) 821.  (Refers to L'Astronomie (Jul 1917) ??NYS.)  Asserts that the weight remains fixed and that he made a clockwork climber and demonstrated this.

Wilbert A. Stevens.  The monkey climbs again.  SSM 19 (1919) 815.  Assert's Rigge's climber was too light to overcome friction and that both should ascend together.

William F. Rigge.  The monkey stops climbing.  SSM 20 (1920) 172‑173.  Says he increased the speed of his monkey and that now it and the weight go up together.

Editors, proposers;  E. V. Huntington & L. M. Hoskins, solvers.  Problem 2838.  AMM 27 (1920) 273‑274  &  28 (1921) 399‑402.  Proposal quotes Carroll and cites Collingwood & SSM.

Ackermann.  1925.  Pp. 1‑2.  Says they would go up together, but the rope moves to the side of the monkey and so the weight will rise faster.  Cites Carroll.

Ernest K. Chapin.  Loc. cit. in 5.D.1.  1927. 

P. 83 & Answers p. 6: Balanced swings.  Two equally weighted swings and children connected by ropes over pulleys.  One child starts swinging.  What happens to the other?  Answer says the centrifugal force would pull the other child up.

P. 104 & Answers p. 12: The climbing monkey.  [Unsigned item _ possibly by Loyd Jr??]  Refers to L'Astronomie (Jul 1917), ??NYS.  Gives Carroll's problem.  Suggests making the rope a loop so movement of the rope doesn't cause an imbalance.  Answer says they go up together, but if the inertia of the rope is considered then the monkey will rise faster than the weight.  If the monkey lets go and recatches the rope, then things are again symmetric until one considers the inertia of the rope which will cause the barrel to fall less rapidly and hence the process will cause the barrel to rise.

Jerome S. Meyer.  Fun-to-do.  Op. cit. in 5.C. 1948.  Prob. 6: How good are you in physics?, part 2, pp. 19-20 & 181-182.  Carroll's problem, with no reference to Carroll.  Answer says the weight will go up.

Warren Weaver.  Lewis Carroll: Mathematician.  Op. cit. in 1.  1956.  Mentions the problem.  A. G. Samuelson's letter says a modern version has a mirror at the other end and asks if the monkey can get away from his image.  He says the monkey and the mirror will behave identically so he cannot get away.  Weaver's response is that this is correct, though he was saying that the behaviour of the weight cannot be known unless you know how the monkey climbs.

 

          10.D.  MIRROR PROBLEMS

 

          10.D.1          MIRROR REVERSAL PARADOX

 

          Why does a mirror reverse right and left, but not up and down?  This is a perennial problem in Notes and Queries type columns.

 

Alice Raikes.  Letter.  The Times (22 Jan 1932).  ??NYS _ cited in:  Michael Barsley; The Left Handed Book; (Souvenir, London, 1966);  Pan, 1989, pp. 199‑200.  Quoted in:  Roger Lancelyn Green; Alice _ an excerpt from his:  Lewis Carroll, Bodley Head, 1960;  IN:  Robert Phillips, ed.; Aspects of Alice; (1971;  Gollancz, London, 1972);  Penguin, 1974, pp. 52-53.  Quoted in:  Florence Becker Lennon; Escape through the looking-glass _ an excerpt from her:  Victoria Through the Looking-Glass, 1971;  IN:  Robert Phillips, ibid., pp. 108-109.

                    Miss Raikes was one of Carroll's girl friends.  She relates that Carroll put an orange in her right hand and then asked her to stand in front of a mirror and say which hand the reflection had the orange in.  She said the left hand and Carroll asked her to explain.  She finally said  "If I was on the other side of the glass, wouldn't the orange still be in my right hand?"  Carroll said this was the best answer he had had and later said it was the idea for "Through the Looking Glass".  Green dates this as Aug 1868 and says it took place when Carroll was visiting his uncle Skeffington Lutwidge at his house in Onslow Square, London.

Dudeney.  PCP.  1932.  Prob. 327: Two paradoxes, pp. 112‑113 & 526.   = 536, prob. 526, pp. 216‑217 & 412.

Gardner.  Left and right handedness.  SA (Mar 1958) = 1st Book, chap. 16.

Gardner.  The Ambidextrous Universe.  2nd ed., Pelican (Penguin), 1982.  Pp. 6‑9 & 22‑26 surveys the question and cites three 1970s serious(!) philosophical articles on the question.

Richard L. Gregory.  Mirror reversal.  IN: R. L. Gregory, ed.; The Oxford Companion to the Mind; OUP, 1987, pp. 491-493.  ??NX.  Gives the basic explanation, but seems unhappy with the perceptual aspects.  I would describe it as making heavy weather of a simple problem.

In c1993, there was correspondence in the Answers column of The Sunday Times.  Collected in: The Sunday Times Book of Answers; ed. by Christopher Lloyd, Times Books, London, 1993.  (The column stared in Jan 1993, but 70% of the book material did not appear in the paper.)  Pp. 63-67.  One correspondent said he asked the question in New Scientist almost 20 years ago.  One correspondent clearly states it reverses front and back, not right and left.  Another clearly notes that the appearance of reversing right and left is due to the bilateral symmetry of our bodies, so we turn around to consider the mirror image.

In c1994, there were several letters on the problem in The Guardian's Notes & Queries.  These are reproduced in: Joseph Harker, ed.; Notes & Queries, Vol. 5., Fourth Estate, London, 1994, pp. 178-180.  R. L. Gregory has a slightly confusing letter but adds that he has given the history and solution in his Odd Perceptions (Routledge, 1986, ??NYS) and in the Oxford Companion to the Mind _ see above.

Erwin Brecher.  Surprising Science Puzzles.  Sterling, NY, 1995.  The mirror phenomenon, pp. 16 & 80.  "Why does a mirror reverse only the left and right sides but not up and down?"  Gives a nonsensical answer: "Left and right are directional concepts while top and bottom, or up and down, are postional concepts." and then follows with an unreasonable analogy to walking over the North Pole.

 

          10.D.2.         OTHER MIRROR PROBLEMS

 

Richard A. Proctor.  Our puzzles.  Knowledge 10  (May 1887) 153  &  (Jun 1887) 186-187.  Prob. XXVIII: how to see yourself properly  _  use two mirrors at right angles.  Prob. XXIX: in a fully mirrored room, what do you see when you look into a corner?  _  yourself inverted.  Prob. XXX: in the same room, how many images of yourself do you see?  _  26.

 

          10.D.3          MAGIC MIRRORS

 

          These are oriental (Chinese or perhaps Japanese) polished discs which cast reflections containing a pattern.  I first came across them in one of R. Austin Freeman's detective stories and I was kindly brought one from China a few years ago.  Since about 1991, they have been made in and exported from China and are commercially available.  A fine example, with explanation, is in the Museum of the History of Science in Oxford, but it is not illuminated.  Basically, the pattern is hammered on the disc and this causes molecular changes which remain even when the surface has been made apparently smooth.  Apparently other methods of producing a difference in metallic structure have been used.  Sometimes the pattern is also made in relief on the opposite side of the disc, and sometimes a different pattern is made.

          Peter Rasmussenand Wei Zhang have sent a bundle of material on this, ??NYR.

 

John Timbs.  Things Not Generally Known,  Familiarly Explained.  A Book for Old and Young (spine says  First Series  and a note by a bookdealer on the flyleaf says  2 vol.).  Kent & Co., London, (1857?), 8th ed., 1859.  Chinese magic mirrors, p. 114.  Says the reflected pattern is in relief on the other side of the disc.  He quotes an explanation given by 'Ou-tseu-hing' who lived between 1260 and 1341 and who worked out the process by inspecting a broken mirror.  He says that the disc with the relief pattern is made by casting, in fine copper.  The pattern is then copied by engraving deeply on the smooth side and the removed parts are filled with a rough copper, then the disc is fired, polished and tinned.  The rough copper produces dark areas in the reflection.

R. F. Hutchinson.  The Japanese magic mirror.  Knowledge 10 (Jun 1887) 186.  Says he has managed to fulfill a boyhood longing and obtain one.  Describes the behaviour and asks for an explanation.

J. Parnell.  The Japanese magic mirror.  Knowledge 10 (Jul 1887) 207.  Says he studied it some 20 years earlier and cites:  The Reader (1866);  Nature (Jul 1877)  and  a paper read to the Royal Society by Ayrton & Perry in Dec 1878  _ all ??NYS.  Says the mirror is somewhat convex and the picture lines are slightly flatter, so the reflection of the picture is brighter than of the surrounding area.

R. Austin Freeman.  The Surprising Adventures of Mr. Shuttlebury Cobb.  Story VI: The magic mirror.  The series first appeared in  Pearson's Magazine:  (1 Jun 1913) 438-448,  (15 Jun 1913) 565-574,  (1 Jul 1913) 748-757   and  Red Magazine:  (15 Jul 1913) ??,  (1 Aug 1913) ??,  (15 Aug 1913).  Collected as a book:  Hodder & Stoughton, London, 1927,  with Story VI on pp. 231-281.  "... product of Old Japan, ...."  Says the device on the back is cast as a dark shape with a bright halo.  Says the design is formed by chasing or hammering the lines, which makes them harder than the surrounding bronze (or similar metal), so when polished, they project slightly and produce an image in the reflection.  Says there is an Encyclopedia Britannica article on the subject, but it is not in my 1971 ed.

R. Austin Freeman.  The magic casket.  Pearson's Magazine (Oct 1926) 288-299, ??NYS.  Collected in:  The Magic Casket; Hodder & Stoughton, London, 1927 and reprinted numerous times (I have 5th ptg, 1935), pp. 7-41.  Collected in:  Dr. Thorndyke Omnibus (variously titled); Hodder & Stoughton, London, 1929 (I have a reprint with TP missing), pp. 398-425.  Says the phenomenon was explained by Sylvanus Thompson.  Says the polished lines project slightly and produce dark lines with a bright edges in the reflection.

Will Dexter.  Famous Magic Secrets.  Abbey Library, London, nd [Intro. dated Nov 1955].  P. 56.  Describing a visit to the premises of The Magic Circle, he says:  "Here are Japanese Magic Mirrors _ a whole shelf of them.  Made of a secret bronze alloy, ..., they have a curious property.  ...  Why?  Well now, people have written books to explain this phenomenon, ....  Some other time we'll talk about it...."

 

          10.E.  WHEEL PARADOXES

 

Clark.  Mental Nuts.  1904: no. 44;  1916: no. 46.  The wheel.  "Does the top go faster than the bottom?"  Answer is: "Turning on ground, yes; on shaft, no."

 

          10.E.1.          ARISTOTLE'S WHEEL PARADOX

 

          Wheels of different sizes joined concentrically and rolling on two tracks at different heights.  At first, it appears that they each roll the same distance and hence must have the same circumference!

 

Aristotle (attrib.).  Mechanical Problems.  c-4C?  In:  Aristotle _ Minor Works.  Trans. by W. S. Hett; Loeb Classical Library, 1936, pp. 329‑441.  The wheel paradox is section 24, pp. 386‑395.

Heron (attrib.).  Mechanics (??*).  c2C?  Ed. by L. Nix & W. Schmidt.  Heronis Opera, vol. II, Teubner, Leipzig.  Chap. 7.  ??NYS.  (HGM II 347‑348.)

Cardan.  Opus Novum de Proportionibus Numerorum.  Henricpetrina, Basil, 1570, ??NYS.  = Opera Omnia, vol. IV, pp. 575-576.

Galileo.  Discorsi e Dimostrazione Matematiche intorno à Due Nuove Scienze Attenenti alla Mecanica & i Movimenti Locali.  Elzevirs, Leiden, 1638.  Trans. by S. Drake as:  Two New Sciences; Univ. of Wisconsin Press, 1974;  pp. 28‑34 & 55‑57.  (English also in:  Struik, Source Book, pp. 198‑207.)

Northrop.  Riddles in Mathematics.  1944.  1944: 59-60;  1945: 56‑57;  1961: 63‑64.  Footnote on p. 248 (1945: 229;  1961: 228) only dates it back to Galileo.

Israel Drabkin.  Aristotle's wheel:  notes on the history of a paradox.  Osiris 9 (1950) 162-198.

 

          10.E.2.          ONE WHEEL ROLLING AROUND ANOTHER

 

          How often does a wheel turn when it is rolled around another?  This is a well-known astronomical phenomenon _ the solar day and the sidereal day are different.

 

Clark.  Mental Nuts.  1904: no. 46;  1916: no. 48.  The cog wheels.  "Suppose two equal cog wheels or coins (one stationary), how many turns will the other make revolving around it?"  Answer is: "Two full turns."

Pearson.  1907.  Part II, no. 58: The geared wheels, pp. 58 & 172.  10  tooth wheel turning about a  40  tooth wheel.

Dudeney.  AM.  1917.  Prob. 203: Concerning wheels, pp. 55 & 188.

McKay.  Party Night.  1940.  No. 23, p. 181.  Two equal circles, one rolling around the other.

W. T. Williams & G. H. Savage.  The Penguin Problems Book.  Penguin, 1940.  No. 80: Revolving coins, pp. 46 & 129.  Equal coins, then rolling coin of half the diameter.

Northrop.  Riddles in Mathematics.  1944.  1944: 55-57;  1945: 52‑54;  1961: 60‑62.  Also considers movement of a slab on rollers.

 

          10.E.3.          HUNTER AND SQUIRREL

 

Bubbenhall.  Letter:  A puzzle.  Knowledge 3 (9 Feb 1883) 91, item 719.  "A squirrel is sitting upon a post and a man is standing facing the squirrel, the squirrel presently turns round and the man moves round with it, always keeping face to face.  When the man has been round the post has he been round the squirrel?"  Proctor was editor at the time _ could he have written this letter??

Richard A. Proctor.  Editorial comment.  Knowledge 3 (9 Mar 1883) 141-142.  Says the hunter does go round the squirrel and that the problem is purely verbal.

W. Smith.  Letter:  The squirrel puzzle.  Knowledge 3 (4 May 1883) 268, item 807.  Disagrees with above, but not very coherently.  Proctor's comments do not accept Smith's points.  "In what way does the expression going round an object imply seeing every side of it?  Suppose the man shut his eyes, would that make any difference?  Or, suppose the man stood still and the squirrel turned round, so as to show him every side _ would the stationary man have gone round the squirrel?"

Clark.  Mental Nuts.  1904: no. 15;  1916: no. 21.  The hunter and the squirrel.  Here the squirrel always keeps on the opposite side of the tree from the hunter.  Answer is: "No; he did not.  They travel on parallel lines and do not change their relative position."

Pearson.  1907.  Part I: Round the monkey, p. 126.  Says R. A. Proctor discussed this some years ago in "Knowledge".

William James.  Pragmatism.  NY, 1907.  ??NYS.

Dudeney.  Some much‑discussed puzzles.  Op. cit. in 2.  1908.  'The answer depends entirely on what you mean by "go around."'

Loyd.  Cyclopedia.  1914.  The hunter and the squirrel, p. 61.  c= SLAHP: The hunter and the squirrel, p. 9.  Loyd Jr. says it is a "hundred‑year‑old question".

Collins.  Fun with Figures.  1928.  Monkey doodles business, pp. 232-234.  Monkey on a pole as in Bubbenhall.  "It is really a matter of personal opinion, ...."  Quotes Proctor's comments of 4 May 1883 with some minor changes.

Harriet Ventress Heald.  Op. cit. in 7.Z.  1941.  Prob. 36, p. 17.  Answer says it depends.

Robin Ault.  On going around squirrels in trees.  JRM 10 (1977‑78) 15‑18.  Cites James.  Develops a measure such that if the hunter and the squirrel move in concentric circles of radii  a  and  b,  then the hunter goes  a/(a+b)  around the squirrel and the squirrel goes  b/(a+b)  around the hunter.

 

          10.F.  FLOATING BODY PROBLEMS

 

          Of course, the original floating body problem is Archimedes' testing of Hieron's crown.  I have only included a few examples of this _ it is fairly widely available.

 

Archimedes.  On Floating Bodies, Book I.  In:  T. L. Heath, The Works of Archimedes, ..., op. cit. in 6.AN, 1897 & 1912, pp. 253-262.  On pp. 258-261, Heath describes how Archimedes probably analysed Hieron's crown.

E. J. Dijksterhuis.  Archimedes.  Op. cit. in 6.S.1.  1956.  Pp. 18-21 dicusses the problem, noting that the object was actually a wreath, stating that the oldest known source is Vitruvius (late 1C) and giving several versions of the method thought to have been used by Archimedes.

Isaac Disraeli.  Miscellanies of Literature.  Preface dated 1840 _ my copy is: New edition, revised;  Ward, Lock, London, 1882 (date of publisher's catalogue at end).  P. 211 relates that Charles II, when dining with the Royal Society "on the occasion of constituting them a Royal Society", asked what would happen if one had two pails of water of equal weight and put fish in one of them _ "he wanted to know the reason why that pail with such addition, should not weigh more than the other ...."  This produced numerous confused explanations until one member burst into laughter and denied the fact.  I find the phrasing confusing _ it seems that he wanted to know why the pails remained of equal weight.  However, there is a possible way that the pails would remain of equal weight _ if both pails were full to the brim, then the insertion of fish would cause an equal weight of water to overflow from the pail.  Disraeli is not specific about dates _ there are two basic dates of the beginning of the Society.  It was founded on 28 Nov 1660 and it was chartered on 15 Jul 1662, with a second charter on 22 Apr 1663.  The 1662 date seems most likely.

Dudeney??  Breakfast Table Problems No. 334: Water and ice.  Daily Mail (3  &  4 Feb 1905) both p. 7.  Ice in a full glass of water.  "..., what volume of water will overflow when the ice melts?"

Ackermann.  1925. 

Pp. 62‑63.  Barge in a canal going over a bridge.  How much more weight is on the bridge?

Pp. 94‑95.  Ice in full vessel of water.

Dudeney.  Problem 1060: Up or down?  Strand Mag. (Jun? 1931).  ??NX.  Boat full of iron in a reservoir.

Perelman.  FMP.  c1935?

Which is heavier?, p. 114.  Bucket of water versus equally full bucket with wood floating in it. 

Under water, pp. 199 & 202.  Submerge a balanced balance with stone on one side and iron on the other.

W. A. Bagley.  Paradox Pie.  Op. cit. 6.BN.  1944. 

No. 90: Supporting the ship, p. 60.  Ship in canal going over a bridge.

No. 91: A n'ice question, p. 60.  Ice in glass.

John Henry Cutler.  Dr. Quizzler's Mind Teasers.  Greenberg, NY, 1944.  ??NYS _ excerpted in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun 1992) 47 & 43, prob. 7.  Balance two bowls of water on a scales.  Add some goldfish to one of them.  Do the bowls still balance?

W. T. Williams & G. H. Savage.  The Third Penguin Problems Book.  Penguin, Harmondsworth, 1946.  Prob. 79: Aquatics, pp. 38 & 117.  Boat with iron weight in a bathtub.

J. De Grazia.  Maths is Fun.  Allen & Unwin, London, 1949 (reprinted 1963).  Chap. I, prob. 7, Cobblestones and water level, pp. 12 & 111.

Gamow & Stern.  1958.  The barge in the lock.  Pp. 104‑105.  Barge full of iron in a lock.

H. T. Croft & S. Simons.  Some little naggers.  Eureka 23 (1960) 22 & 36. 

No. 2.  Ice cube in a full glass of water.

No. 3.  Boat on a lake.

David Singmaster.  Any old iron?  Barge full of iron in the lock.  What happens to the boat when the iron is thrown into the lock?

Watertight problem.  The Weekend Telegraph (9 Jun 1990) XXIV  &  (16 Jun 1990) XXIV. 

About 1994, there was some correspondence _ possibly in the Guardian's Notes & Queries column _ about barges in a canal on a viaduct.  Apparently Telford's 1801 Pontcysyllte viaduct on the Shropshire Union Canal at Chirk, near Llangollen, is  1007  ft long, but has a notice restricting the number of barges on it to three, though a barge is about  70  ft long.  Responses indicated that the reason for the restriction may have been wave problems.

 

          10.G.  MOTION IN A CURRENT OR WIND

 

          Simple problems of this type are just variations of meeting problems _ see Vyse and Pike, etc. _ but they only seem to date from the late 18C.

          The comparison of up and down stream versus across and back is the basis of the Michelson-Morley experiment and the Lorentz-Fitzgerald contraction in relativity, so this idea must have been pretty well known by about 1880, but the earliest puzzle example I have is Chapin, 1927.

 

Vyse.  Tutor's Guide.  1771?  Prob. 13, p. 198 & Key p. 241.  Two rowers who can row at  5  set out towards each other at points  34  apart on a river flowing  2½.  Though this appears to belong in Section 10.G, it is simply  MR-(2½, 7½; 34).

Pike.  Arithmetic.  1788.   P. 353, no. 33.  Two approaching rowers, starting  18  apart and normally able to row at rate  4, but the tide is flowing at rate  1½.  I. e.  MR‑(2½, 5½; 18).  (Sanford 218 says this is first published version!)

I imagine this appears in many 19C texts.  I have seen the following. 

T. Tate.  Algebra Made Easy.  Op. cit. in 6.BF.3.  1848.

Pp. 45-46, no. 14.  Rower goes  30  miles down and back in  7  hours.  He can row  2  miles upstream in the time he can row  5  miles downstream.  Find his rates.

P. 86, no. 11.  Rower goes  a  mph with the tide and  b  mph against the tide.  What is the rate of the tide?

Colenso.  Op. cit. in 7.P.1.  1849.

Exercise 47, no. 10, p. 85 & Answers, p. 11.  Rower goes  20  miles and back in  10  hours.  He goes  3  miles downstream in the time he goes  2  miles upstream.  How long does he take each way?

Miscellaneous examples, no. 336, p. xix & Answers, p. 23.  Crew rows  3½  miles down and back up a river in  100  minutes, where the current is  2  miles per hour.  What is their rate in still water?

Edward Brooks.  The Normal Mental Arithmetic  A Thorough and Complete Course by Analysis and Induction.  Sower, Potts & Co., Philadelphia, (1858), revised, 1863.  Further revised as:  The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction;  Sower, Potts & Co., Philadelphia, 1873.  Several examples of the following type.

                    1863 _ p. 136, no. 4;  1873 _ p. 146, no. 4.  Steamboat goes  12  in still water; current is  4.  Goes down river and returns in  6  hours.  How far did it go?

Colenso.  Op. cit. in 7.H.  These are from the 1864? material.  Let  V  be the velocity of the rower and  T  the velocity of the tide.

No. 6, p. 187.  V + T  =  5/3 * (V ‑ T)   and   V + T + ½  =  2 (V ‑ T ‑ ½).

No. 25, pp. 190 & 215.  V = 1/9  mile/minute,  V ‑ T = 1/14  mi/min.  What is  V + T?

Horatio N. Robinson.  New Elementary Algebra: Containing the Rudiments of the Science for Schools and Academies.  Ivison, Blakeman, Taylor & Co., New York, 1875.  Prob. 90, p. 305.  Man can row  15  miles down river in  2½  hours, but requires  7½  to row back.  What are the rates of the rower and of the river?

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.  No. 89, pp. 301 & 347.  Steamboat goes  10  in still water, current is  4,  boat goes down and returns in  10  hours _ how far did it go?

Briggs & Bryan.  The Tutorial Algebra, Part II.  Op. cit. in 7.H.  1898.  Exercises X, prob. 24, pp. 125 & 580.  Stream goes  4.  Man rows up and back and takes  39  minutes longer than in still water.  With a second rower, they can go  3/2  as fast as the single man.  They do the same trip in only  8  minutes longer than in still water.

W. P. Workman.  The Tutorial Arithmetic, op. cit. in 7.H.1.  1902.  Section IX, art. 240, example 3, p. 410 (= 416 in c1928 ed.).  Boat goes  4  mi in  20  min in still water and in  16  min with the tide.  How long against the tide?

Richard von Mises.  c1910.  Described in: George Pólya; Mathematical Methods in Science; (Studies in Mathematics Series, vol. XI; School Mathematics Study Group, 1963); reprinted as New Mathematical Library No. 26, MAA, 1977;  Von Mises' flight triangle, pp. 78-81.  How can one determine the airspeed of a plane from the ground speed when the wind is unknown?  "It was solved by Von Mises some fifty years ago; this I well remember as I heard it from him at that time."

Schubert.  Op. cit. in 7.H.4.  1913.  Section 17, no. 99, pp. 64‑65 & 140.  Steamship covers  60  km in  4  hours going upstream and in  3  hours going downstream.

Loyd.  Cyclopedia.  1914.  Riding against the wind, pp. 199 & 365.  = MPSL2, prob. 49, pp. 34 & 137.  = SLAHP: Wind influence, pp. 38 & 97.  Against and with the wind.

Loyd.  Cyclopedia.  1914.  The Santos‑Dumont puzzle, pp. 202 & 366.  Against and with the wind.

Peano.  Giochi.  1924.  Prob. 14, pp. 4-5.  Two ships travel  6000  miles and return.  The first goes at  8  mph and returns at  12  mph;  the second goes  10  mph both ways.  Which is faster?

Ackermann.  1925.  Pp. 77‑81.  Determine speed of wind using sound echoes.

Ernest K. Chapin.  Loc. cit. in 5.D.1.  1927. 

P. 92 & Answers p. 9.  Flyers go from  A  to  B  and back, one in still air, the other in a steady wind.  Who is faster?  Answer notes that the naive view is that the wind helps and hinders equally, but one only needs to consider a wind as fast as the flyer to see that it is not equal and that for a lesser wind, the flyer is hindered for longer than he is helped.

Prob. 3, p. 98 & answers pp. 10-11: The problem of the swimmers.  Both swim a mile, one up and down stream, the other across and back.  Who is faster?  Answer gives detailed calculation.

Harriet Ventress Heald.  Op. cit. in 7.Z.  1941.  Prob. 48, pp. 21‑22.  Current of  1  mph.  Man rows up in  3  hours & back in  2  hours.  How far did he go?

D. A. Hindman.  Op. cit. in 6.AF.  1955.  Chap. 16, prob. 23: The floating hat, pp. 260‑263.  Hat falls off rower going upstream and is picked up on his return.

Gamow & Stern.  1958.  Boat and bottle.  Pp. 100‑102.  Bottle falls off boat going upstream and is picked up later.

 

          10.H. SNAIL CLIMBING OUT OF WELL

 

          A snail is at the bottom of a well which is  D  deep.  He climbs  A  in the day and slips back  B  in the night.  How long does it take to get out?  The earlier versions had serpents, snakes and lions.  The 'end effect' is that when the snail gets to within  A  of the top, he doesn't slip back.  The earlier versions did not have this clear and thus are just equivalent to meeting problems,  MR-(A, 0; D)  or  MR-(A, B; D)  or  O-(A, B) with headstart  D.  See 10.A:  Bakhshali ff. 60r-60v;  al-Karkhi no. 9(?);  Fibonacci pp. 177-178 & 182. 

          It appears that the  problem grows out of the usage of unit fractions such as  1/2 - 1/3  to specify rates.  At first this just meant  1/6  per day, but then it began to be interpreted as going ahead  1/2  followed by retreating  1/3,  resulting in the 'end effect'.  The earliest to treat the end effect clearly seem to be in c1370. 

          For convenience, let the net gain per day be  G = A - B.  The solution is to take the least  N  such that  NG + A ³ D,  i.e.   N  =  é(D‑A)/Gù,   then interpolate during the daytime of the  (N + 1)‑st day,  getting  (D - NG)/A  of the day time on the  (N + 1)-st day as the time of meeting. 

          When we have approaching animals, going  +a, -b; +c, -d,  set  A = a + c,  B = b + d,  so  G = a - b + c - d.  If we are considering meeting without an end termination such as the cat eating the mouse, and  A ³ G,  then one can have multiple meetings.  The last night on which a meeting occurs during the retrogression is the last  M  such that  MG ³ D,  i.e.   M  =  ëD/Gû.   Thus there will be   2 (M - N) + 1   meetings, though this is reduced by one if  (D - A)/G  is an integer and by one if  D/G  is an integer.  (When both are integers, one reduces by two.)  Simple modifications deal with the cases of negative  B,  say if a snail continues going up at night, but at a different rate, and the case with  A < B. 

          Versions with two approaching animals:  Fibonacci,  Muscarello,  Chuquet,  Borghi,  Pacioli,  Tagliente,  Ghaligai,  Buteo,  Tartaglia.  Chuquet and Buteo treat the end effect clearly and Pacioli and Tagliente almost get it.

          More complex versions:  Mahavira, Wood.

          See Tropfke 588.  Tropke 589 classifies these as  I A c (one person),  I B c (two persons meeting),  I C c (two persons overtaking).

 

Bakhshali MS.  c7C. 

See 10.A for ff. 60r-60v.   v  =  + 5/2 - 9/3,   D = 30.

Kaye I 51; III 222, f. 20v.   v  =  + 1/2 + 1/18 - 1/21,   D = 360.

Kaye I 51; III 225, f. 36v.  Seems to say   v  =  (1/2*3 + 1/3 ‑ 1/4)/(1/2*3)  ‑  (1/2*5)/(3/8),   D = 108,  but the problem is not complete.

Bhaskara I.  629.  Commentary to Aryabhata, chap. II, v. 26-27.  Sanskrit is on pp. 115-122; English version of the examples is on pp. 304-306.  The material of interest is example 4.   v  =  + 1/2 - 1/5,   D = 480.

Chaturveda.  860.  Commentary to Brahma‑sphuta‑siddhanta, chap. XII, section 1, v. 10.  In Colebrooke, pp. 283‑284.   v  =  + 4/5 ‑ 1/60,   D = 76,800,000.

Mahavira.  850.  Chap. V, v. 24‑31, pp. 89‑90.

v. 24:   v  =  (1/5 ‑ 1/9) / (3/7);   D = 4 * 99 2/5.

v. 25:   rate  =  (5/4) / (7/2) ‑ (5/32) / (9/2);   total = 70.

v. 26:   v  =  (3/10) / (11/2) ‑ (2/5) / (7/2);   D = 399/2.

v. 28:   lotus growing in well with outflow and evaporation of water and turtle pulling down lotus;   v  =  (5/2) / (3/3) + (6/5) / (4/3) + (5/2) / (3/2) ‑ (21/4) / (7/2);   D = 960;  problem has  1  for  4/3,  but  4/3  is needed for the given answer.

v. 31:  snake going into hole while growing;  v = (15/2) / (5/14) ‑ (11/4) / (1/4);   D = 768.

Sridhara.  c900.  Ex. 32‑33, pp. 24 & 93.

Ex. 32.   v  =  ½(1+¼)(1‑_)(1+½) / 6(1/5)(1/9)(_)(1+¼)  ‑  2(1‑_) / (1+½);   D = 100.

Ex. 33:   rate  =  (8 ‑ ½) / (1 + _) ‑ ½;   total = 100.

Tabari.  Mift_h al-mu‘_mal_t.  c1075.  Pp. 103f.  ??NYS _ quoted in Tropfke 593.  No. 7.  "A boat comes forward 18 parasangs per day and goes backward 12 parasangs per day.  It comes and goes for 40 days.  How many days does it come and how many days does it go?"  This is not clear _ I wonder if 'days' in the last sentence should be 'parasangs' _ ??

Fibonacci.  1202. 

P. 177: De leone qui erat in puteo.  v = 1/7 ‑ 1/9;  D = 50.  No alternation.  H&S 63 gives Latin and English.  H&S 64 claims that this is an example of day and night alternation, but this is not in the text.

Pp. 177-178: De duobus serpentibus.  Tower is 100.  One comes down  1/3 - 1/4,  other goes up  1/5 - 1/6.  No end effect.

Columbia Algorism.  c1370.  Prob. 67, pp. 88‑89.  Pigeon going down tower.  Rate  =  _ ‑ (_ + ¼),   D = 10 (but Cowley says 50).  These are alternatively day and night, and the end effect is clearly treated to get  112  full days and one more daytime.  (Cowley 399.)

Dell'Abbaco.  c1370.  Prob. 191, pp. 151‑153 with fine plate on p. 152.  Serpent in well of depth  30.  Goes  +2/3  in day and  ‑1/5  at night.  Gives simple answer and then carefully analyses the end effect to get  63  full days plus  9/10  of daytime.  Says that some solve it erroneously.  I have a colour slide of this.

AR.  c1450.  Prob. 65, pp. 47, 177, 224.  Tower  10  high.  Dove flies up  _  in the day and drops back  ¼ + _  at night.  Answer is  120  days, which ignores the end effect.  Should be  112½,  as in the Columbia Algorism.

Muscarello.  1478. 

F. 73r, p. 187-188.  Cat climbing a tower  30  high, going  +½, ‑_.  End effect not treated, so he gets  180  days.

Ff. 74r-74v, pp. 188-189.  Cat at bottom of a tower going  +_, ‑¼.  Mouse at the top going down  ½  and back up  _.  When do they meet?  Again, the end effect is not treated and he gets  240  days.

Tommaso della Gazzaia.  Liber geometriae.  Manuscript C.III.23, Biblioteca Communali di Siena.  15C??  F. 169r.  ??NYS _ quoted in Franci, op. cit. in 3.A, p. 29.  Serpent going  +_ -¼  up a tower  30  high.  End effect ignored.

Chuquet.  1484.  Mentioned on FHM 204 as 'the frog in the well'.  All these treat the end effect clearly.

Prob. 125:  +15, ‑10,  to go  100.  Takes  17  natural days and one 'artificial' day.

Prob. 126:  +15, ‑9,  to go  100.  Takes  15  natural days and  _  of an artificial day.

Prob. 128:  two travellers  100  leagues apart travelling  +12, ‑7  and  +10, ‑6.  Answer:  9  days and  9  nights with  19/22  of a day without the night.

Borghi.  Arithmetica.  1484. 

F. 110r (1509: f. 92v).  Two brothers paying a debt from earnings less expenses:  rate  =  2/3 ‑ 2/5 + 3/4 ‑ 1/2;   total = 700.  (H&S 64 gives Latin.)

Ff. 110r-110v (1509: ff. 92v-92r).  Sparrow hawk at bottom and dove at top of a tower  60  high.  Hawk goes  + _ - ½;  dove goes  + ¾ - _,  approaching in day and retreating at night.  Ignores end effect and gets  240  days.  With end effect, one gets  235  days plus  15/17  of the daytime.

Calandri.  Arimethrica.  1491. 

F. 66r.  Traveller goes  +4  in the day and  -3  in the night to go  20.  Gets  17  days, meaning  16  whole days and a daytime.

F. 71v.  Serpent in well.  +1/7  in the day and  -1/9  in the night to go  50.  He doesn't consider the end effect, so gives  1575  instead of  1572½.)  Nice woodcut, reproduced in Smith, History II 540 and  Rara 48.  Same as Fibonacci, p. 177. 

F. 72r.  Two serpents on a tower of height  100  going  +1/3, -1/4;  +1/5, -1/6,  clearly distinguished as day and night, but he ignores the end effect.  Woodcut showing one serpent (or dragon) at bottom of tower.

F. 72v.  Two ants meeting from  100  away going  +1/3, -1/4;  +1/5, -1/6,  distinguished as day and night, but he ignores the end effect.  Woodcut showing two ants.

F. 73r.  Two ants are at distances  D  and  D + 100  from a pile of grain.  Going  +1/3, ‑1/4;  +1/5, -1/6,  distinguished as day and night, gets them to the pile at the same time.  He ignores the end effect.  Woodcut showing ants by a pile.

F. 73v.  Cat and squirrel in tree of height 26.  They go  +1/2, -1/3;  +1/4, -1/5  clearly distinguished as day and night.  Ignores end effect and gets  120,  but then adds one to get  121  for no clear reason.

Pacioli.  Summa.  1494. 

F. 42r, prob. 22.  Two ants  100  apart approaching at rates  1/3, ‑1/4  and  1/5, -1/6,  clearly stated to be day and night.  He says they approach  7/60  per whole day and then computes  (100 - 7/60) / (7/60)  and adds  1  to get  762 6/7  days.  In the numerator, he seems to have used  88 53/60  instead of  99 53/60.  His process is close to the general solution with end effect, but he should subtract  8/15 = 1/3 + 1/5  instead of  7/60  and he should interpolate on the last day.

Ff. 42r-42v, prob. 23.  Cat & mouse at bottom and top of a growing and shrinking tree originally  60  high.  Mouse descends  1/2  per day and returns  1/6  at night.  Cat goes  +1, -1/4.  Tree grows  +1/4, -1/8  between them.  He says the net gain between them is  23/24  per whole day and computes  (60 - 23/24) / (23/24)  and adds  1  to get  T = 62 14/23.  He asserts the tree has grown  T/8  _ i.e. he is still thinking of overall rates rather than considering the alternation properly.  (Sanford 207‑208.  H&S 64‑65 gives Italian and English.)

F. 42v, prob. 24.  +2, -1  to go  10.  Obtains answer of  10  and says it isn't right and should be  8  whole days and a daytime, which he calls  9  days.  He also does  +3, -2  and gets  7  whole days and a daytime.

Blasius.  1513.  F. F.ii.r: Secunda regula.  +50, -19  to reach  992.  No end effect considered, even though the rates are in the day and in the night.

Tagliente.  Libro de Abaco.  (1515).  1541.  Prob. 121, f. 59r.  Cat and mouse on tree which is  26¾  tall.  Cat goes  +1/2, -1/3 (misprinted  2/3);  mouse goes  +1/4, -1/5.  Seems to say they meet on the  120th  day, but it should be the  121st  day.

Ghaligai.  Practica D'Arithmetica.  1521.  Prob. 19, f. 64v.  Two ants  100  apart going toward a pile of grain.  Further goes  +7, -4;  other goes  +5, -3  during day, night.  How far is the pile if they get there at the same time?  He says they take  100  days, but this ignores the end effect _ they meet at the end of the  99th  daytime.  This is equivalent to a snail going  +2, -1  up a wall of  100,  (H&S 65 gives Italian and English.)

Riese.  Rechnung.  1522.  1544 ed. _ pp. 107‑109;  1574 ed. _ pp. 72v‑73r.  The 1574 ed. calls it Schneckengang.  +4_, ‑3¾,  D = 32.  Treats end effect clearly and says it ws first done correctly by Hansen Conrad, Probierer zu Eissleben (?= assayer at Eissleben).  He discusses how to convince people of the end effect.

Tonstall.  De Arte Supputandi.  1522.  Quest. 36, p. 168.  +70, -15  clearly stated to be day and night, to go  4000,  but he ignores the end effect.

Riese.  Die Coss.  1524.  No. 142, p. 61.  +4¼, ‑3_,  D = 32.  Treats end effect properly.  Says the Nurmbergk Rechenmeister N. Kolberger got it wrong and that Hans Conradtt got it right.

Giovanni Sfortunati.  Nuovo lume.  Venice, 1545.  F. 88r.  ??NYS _ described by Franci, op. cit. in 3.A, p. 41.  Solves a problem with end effect and says that Borghi, Pacioli and Calandri have done it wrong.

Christoff Rudolph.  Künstlich rechnung mit der ziffer und mit den zalpfennigen ...  Auffs new wiederumb fleissig ubersehen und an vil arten gebessert.  Nuremberg, 1553, 1561.  ??NYS _ quoted by:  Grosse; Historische Rechenbücher ...; op. cit. in 7.H under Faulhaber, p. 28.  Von einem schnecken.  Snail climbing  + 7, ‑2  to get out of a well  20  deep.  End effect clearly treated.  [This is a revision of the 1526 ed., ??NYS, which may have the problem?]

                    J. De Grazia; Maths is Fun; p. 12, attributes the problem to Christoff Rudolph (1561).

Buteo.  Logistica.  1559.  Prob 33, pp. 234-237.  Problem of ships with oscillating winds.  They start  20000  apart.  First goes  +1200  per day and  -700  per night.  Second starts on the first night and goes  +1400  per night and  -600  per day.  He first solves without end effect and then end effect, getting  14  days +  12  hours of day +  10 2/7  hours of the night.  (H&S 65 gives Latin & English.)

Gori.  Libro di arimetricha.  F. 73r (p. 80).  Height  50,  rates  +_, ‑ ¼.  Notes end effect, but then forgets to add the last day!

H&S 64 says examples with alternating motion are in:  Fibonacci (1202),  Columbia Algorism (c1300),  Borghi (1484),  Calandri (1491),  Pacioli (1494?),  Tartaglia  and  Riese.  Also that examples with two animals approaching are in:  Fibonacci (two ants),  Borghi (1484, hawk & dove),  Pacioli (1494?, both types)  and  Tartaglia (both types).

Faulhaber.  Op. cit. in 7.H.  1614.  ??NYS _ quoted by Grosse, loc. cit. in 7.H under Faulhaber, p. 120.  No. 12, p. 212.  Worm climbs  +¾, ‑_  up a tree  100  high.  Gives answer with end effect.

Dilworth.  Schoolmaster's Assistant.  1743.  P. 166, no. 91.  +20, -15  to go  150.  No end effect considered so answer is  30  days.

Walkingame.  Tutor's Assistant.  1751.  1777: p. 172, prob. 47;  1860, p. 180, prob. 46.  Snail going  +8, -4  to get up a May pole  20  high.  Answer is  4  days, presumably meaning  3  whole days and a daytime.

Vyse.  Tutor's Guide.  1771?  Prob. 27, pp. 43-44 & Key p. 39.  Same as Walkingame, but answer is done step by step and says 'the fourth Day at Night'.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.  No. 5, pp. 15 & 72.  Snail going  +8, -4  to get up a maypole 20 high.  Treats end effect properly but states the answer is  4  days.

Family Friend 1 (1849) 150 & 178.  Problems, arithmetical puzzles, &c. _ 2.  Snail on wall.  +5, -4  to reach  20.  End effect clearly treated.  = The Illustrated Boy's Own Treasury, 1860, Arithmetical and Geometrical Problems, No. 24, pp. 429 & 433.

Magician's Own Book.  1857.  The industrious frog, p. 234.  + 3, ‑ 2  up a well  30  deep.  End effect treated.  = Boy's Own Conjuring Book, 1860, p. 200.

Charades, Enigmas, and Riddles.  1859?: prob. 24, pp. 59 & 63;  1862?: prob. 569, pp. 107 & 154.  Snail going  +5, -4  up a wall  20  high.  Answer is  16  days, which considers the end effect, but doesn't describe the final daytime well.

Bachet-Labosne.  Problemes.  3rd ed., 1874.  Supp. prob. II, 1884: 182.  +9, -5  to cover  173.  Gives simple answer and then considers end effect.

Lucas.  L'Arithmétique Amusante.  1895.  Prob. XI: La ballade de l'escargot rétrograde, pp. 25-26.  +5, -2  going up a tree of height  9.  Also quoted in Laisant; op. cit. in 6.P.1; 1906; p. 125.

Mr. X.  His Pages.  The Royal Magazine 9:6 (Apr 1903) 544-546.  "A frog was trying to get up a slippery bank twelve feet high.  In the first twelve hours he climbs eight feet, but in the next twelve hours he loses four feet.  How long will he be in reaching the top?"  "Why, having lost his four feet, how could he get to the top at all?"

Clark (1904 & 1916),  Pearson (1907),  Loyd (Cyclopedia, 1914),  Ahrens (A&N, 1918),  Loyd Jr. (SLAHP, 1928)  all have versions.

Wood.  Oddities.  1927.

Prob. 65: A snail's journey, p. 51.  Usual problem with  A, B, H  =  3, 2, 12.

Prob. 66: another snail, pp. 51-52.  A, B, H  =  3, 2, 20,  but the snail has to get to the top and down the other side.  He asserts the level speed of the snail would be  3 + 2 = 5 per day  and hence the downward speed is  5 + 2 = 7 per day.  Hence, at the end of the daylight of the 18th day, he has reached the top and then slides down 2 on the other side in that night.  It then take two more whole days to reach the bottom, making 20 whole days for the trip.

Adams.  Puzzles That Everyone Can Do.  1931.  Prob. 197, pp. 76 & 158: Seaside problem.  Deck-chair man earns £6 when it is sunny but loses £3 when it rains.  The weather alternates, being sunny on the first day.  When is he £60 ahead?

Evelyn August.  The Black-Out Book.  Op. cit. in 5.X.1.  1939.  The man who hoarded petrol, pp. 154 & 215.  Tank holds  8  gallons, man puts in  2  gallons every day and  1  gallon leaks out every night;  when is it full?

 

          10.I.   LIMITED MEANS OF TRANSPORT _ TWO MEN AND A BIKE, ETC.

 

          The men have to get somewhere as quickly as possible and have to share a vehicle which may be left to be picked up or may return to pick up others.

 

Laisant.  Op. cit. in 6.P.1.  1906. 

Chap. 51: Deux cyclistes pour une bicyclette, pp. 127-129.  Graphic solution, assuming walking speeds and riding speeds are equal, but notes one can deal with the more general problem with a little more mathematics.

Chap. 52: La voiture insuffisante, pp. 129-132.  Two couples, but the car can only carry two persons besides the driver.  Graphic solution, again assuming walking speeds and driving speeds are equal, but making an estimate for calculational convenience.  He also states the exact solution.

Loyd.  Tandem puzzle.  Cyclopedia, 1914, pp. 322 & 382 (erroneous solution).  = MPSL2, prob. 123, pp. 88 & 160‑161, with solution by Gardner.  Three men and a tandem bike.

Loyd Jr.  SLAHP.  1928.  A tandem for three, pp. 52 & 104.  Like Loyd's but with different data.

Dudeney.  PCP.  1932.  Prob. 75: A question of transport, pp. 29 & 136.  = 536, prob. 89, pp. 27 & 244.  Twelve soldiers and a taxi which can take four of them.

Gaston Boucheny.  Curiositiés & Récréations Mathématiques.  Larousse, Paris, 1939, pp. 77‑78.  Two men and a bike.

R. L. Goodstein.  Note 1797:  Transport problems.  MG 29 (No. 283) (Feb 1945) 16‑17.  Graphical technique to solve general problem of a company of men and a lorry.

William R. Ransom.  Op. cit. in 6.M.  1955.  A ride and walk problem, pp. 108‑109.  Same as Dudeney.

Karl Menninger.  Mathematics in Your World.  Op. cit. in 7.X.  1954??  A bit of 'hitch hiking', pp. 100‑101.  Three men and a motorcyclist who can carry one passenger.  Graphical method, but he neglects to consider that the passengers can be dropped off before the goal to walk the remaining distance while the cyclist returns to pick up the others.

Doubleday - II.  1971.  Dead heat, pp. 79-80.  Two men and a pony, but the pony is assumed to stay where it is left, so this is like two men and a bike.  If the pony could go back to meet the second traveller, then this would be two men and a motorcyclist.

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.  Men and a vehicle, pp. 282-285.  Uses appropriate variables to make the equations more symmetric and hence easily solvable, even with different speeds.

 

          10.J.   RESISTOR NETWORKS

 

          Find the resistance between two points in some network of unit resistors.

 

E. E. Brooks & A. W. Poyser.  Magnetism and Electricity.  Longmans, Green & Co., London, 1920.  Pp. 277‑279.  ??NYS.

Anon.  An electrical problem.  Eureka 14 (1951) 17.  Infinite square lattice of one ohm resistors along edges.  Asks for resistance from  (0, 0)  to  (0, 1)  and to  (1, 1).  Solution to be in No. 15, but isn't there.

Huseyin Demir, proposer;  C. W. Trigg, solver.  Problem 407 _ Resistance in a cube.  MM 33 (1959‑60) 225‑226  &  34 (1960‑61) 115‑116.  All three inequivalent resistances for a cubical network are found, i.e. the resistances between points which are distance  1, 2, 3  apart.  Trigg says  distance = 1 and 3  cases are in Brooks & Poyser.

Gardner.  SA (Dec 1958) = 2nd Book, p. 22.  Gives cube solution and cites Trigg's reference to Brooks & Poyser.

B. van der Pol & H. Bremmer.  Operational Calculus _ Based on the Two‑sided Laplace Integral.  2nd ed., Cambridge Univ. Press, 1959.  ??NYS.  'Very last section' obtains formulae for the resistance  R(m, n)  in the infinite plane network between  (0, 0)  and  (m, n).  These are in terms of Chebyshev polynomials or Bessel functions.

Albert A. Mullin & Derek Zave, independent proposers;  A. A. Jagers, solver.  Problem E2620 _ Symmetrical networks with one‑ohm resistors.  AMM 83 (1976) 740  &  85 (1978) 117‑118.  All regular polyhedra and the  n‑cube, but only between furthest vertices.  (Editorial note says the cubical case first occurs in Coxeter's Regular Polytopes, but I can't find it??  It also erroneously says Gardner gives other solutions.)

D. C. Morley, proposer;  Friend H. Kierstead Jr., solver.  Problem 529 _ Hyper‑resistance.  JRM 9 (1976‑77) 211  &  10 (1977‑78) 223‑224.  4‑cube, resistance between points  3  apart.

David Singmaster, proposer;  Brian Barwell, solver.  Problem 879 _ Hyper‑resistance II.  JRM 12 (1979‑80) 220  &  13 (1980‑81) 229‑230.  n‑cube, furthest vertices.  What happens as  n  goes to infinity?

David Singmaster, proposer;  B. C. Rennie, partial solver.  Problem 79‑16 _ Resistances in an  n‑dimensional cube.  SIAM Review 21 (1979) 559  &  22 (1980) 504‑508.  In an  n‑cube, what are the resistances  R(n, i)  between points  i  apart?  Results are only known for  i = 1, 2, 3, n‑2, n‑1, n.  Other solvers considered the infinite  n‑cubical lattice and obtained a general result in terms of an  n‑fold integral, including  R(n, 1) = 1/n.

P. Taylor & C. Feather.  Problems drive 1981.  Eureka 44 (Spring 1984) 13‑15 & 71.  No. 1.  Find all resistances for regular polyhedral networks.

P. E. Trier.  An electrical resistance network and its mathematical undercurrents.  Bull. Inst. Math. Appl. 21:3/4 (Mar/Apr 1985) 58‑60.  Obtains a simple form for  R(m, n)  (as defined under van der Pol) which involves a double integral.  He evaluates this explicitly for small  m, n.  The integral is the same as that of the other solvers of my problem in SIAM Review.

P. E. Trier.  An electrical network _ some further undercurrents.  Ibid. 22:1/2 (Jan/Feb 1986) 30‑31.  Letter making a correction to the above and citing several earlier works (McCrea & Whipple, 1940;  Scraton, 1964;  Hammersley, 1966 _ all ??NYS) and extensions:  an explicit form for  R(m, m),  the asymptotic value of  R(m, m)  and extensions to three and  n  dimensions.

 

          10.K.  PROBLEM OF THE DATE LINE

 

          A man who circles the earth gains or loses a day. 

          I include some related problems here.  See also 6.AF.

 

E. John Holmyard.  Alchemy.  Penguin, 1957, p. 119, says Roger Bacon (1214-1294 (or 1292)) foresaw circumnavigation, but doesn't indicate if he recognized the date problem.

Nicolas Oresme.  Traitié de l'espere.  c1350.  ??NYS _ described in:  Cora E. Lutz; A fourteenth‑century argument for an international date line; Yale University Library Gazette 47 (1973) 125‑131.  Chap. 39.  Three men.  One circles the world eastward in  12  days, another westward and the third stays at home.  He computes their effective day lengths.

                    Lutz describes the occurrence of the problem in other of Oresme's writings.

                    Quaestiones supra speram (c1355),  where the travellers take  25  days and Oresme suggests "one ought to assign a definite place where a change of the name of the day would be made".

                    His French translation of Aristotle's  De caelo et mundo  as:  Traitié du ciel et du monde (1377),  where they take  9  days.

Kalendrier des Bergers; 1493.  = The Kalendayr of the Shyppars (in a Scottish dialect); 1503.  = The Shepherds' Kalendar; R. Pynson, London, 1506.  ??NYS.  Described in:  E. G. R. Taylor; The Mathematical Practioners of Tudor & Stuart England; (1954); CUP for the Inst. of Navigation, 1970; pp. 11‑12 & 311.  Three friends, one stays put, others circle the earth in opposite directions.  When they meet, they disagree on what day it is.

Antonio Pigafetta.  Magellan's Voyage: A Narrative Account of the First Circumnavigation ....  Translated and edited by R. A. Skelton, 1969.  Vol. I, pp. 147‑148.  ??NYS _ quoted by Lutz.  When they reached Cape Verde in 1522, a landing party was told "it was Thursday, at which they were much amazed, for to us it was Wednesday, and we knew not how we had fallen into error."

Cardan.  Practica Arithmetice.  1539.  Chap. 66, section 34, ff. DD.iiii.v ‑ DD.v.r (p. 145).  Discusses ship which circles world three times to the west, and also mentions going east.  (H&S 11 gives Latin.  Sanford 214 thinks he was first to note the problem.)

Cardan.  De Rerum Varietate.  1557, ??NYS.  = Opera Omnia, vol. III, pp. 239-240.  Liber XVII.  Navis qua orbem cir_ambivit.  Describes Magellan's circumnavigation, giving dates and saying they had lost a day.

van Etten.  1624.  Prob. 91 (96), part IV (7), p. 141 (231‑232).  "How can two twins, who are born at the same time and who die together, have seen a different number of days?"  The English edition comments on a Christian, a Jew and a Saracen having their Sabbaths on the same day.

"A Lover of the Mathematics."  A Mathematical Miscellany in Four Parts.  2nd ed., S. Fuller, Dublin, 1735.  The First Part is:  An Essay towards the Probable Solution of the Forty five Surprising PARADOXES, in GORDON's Geography, so the following must have appeared in Gordon.

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 73o22', 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.  In:  The Lewis Carroll Picture Book, 1899, op. cit. in 5.B, pp. 4‑5 (Collins: 11-12).  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.

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.

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?

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 frquency of the oscillation is  Ö(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.

 

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!

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. 20o, 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.

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. 

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 straightforwad 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 Ö(R/g).  The shortest time is  π Ö{(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  Ö{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.

 

          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.  Pp. 7 & 54 and Plate I.  ??NX.

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.

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.

Charles W. Sherburne, 3409 Patton Ave., San Pedro, California, 90731, USA, has published material claiming that the rattleback demonstrates the failure of Newton's laws and that it is the shape of Noah's Ark!  He claims to have a patent D210,947 or D.E.S. 210,947 on his version of it. but the current US patent numbers are close to 6,000,000, so I suspect he has a registered design.

 

          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.

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 explantion 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.

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.

Jules Leopold.  At Ease!  Op. cit. in 4.A.2.  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?

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.

Proverbs 11:1.  "A false balance is abomination to the Lord:  But a just weight is his delight."

Proverbs 16:11.  "A just weight and balance are the Lord's:  All the weights of the bag are his work."

Proverbs 20:10.  "Divers weights, and divers measures,  Both of them are alike abomination to the Lord."

Proverbs 20:23.  "Divers weights are an abomination unto the Lord:  And a false balance is not good."

Aristotle.  Mechanical Questions.  c-340.  ??NYS _ cited by van Etten.

Pappus.  Collection.  Book 3.  c320.  ??NYS - cited by Leybourn.

Muhammed.  Koran.  c630.  Translated by J. M. Rodwell, Everyman's Library, J. M. Dent, 1909.  Sura LXXXIII _ Those who stint: 1-3.  "Woe to those who STINT the measure:  Who when they take by measure from others, exact the full;  But when they mete to them or weigh to them, minish _".  (I saw the following version on a British Museum label, erroneously attributed to Sura LXXX:  "Woe be unto those who give short measure or weight.")

Cardan.  De Subtilitate.  1550, Liber I, ??NYS.  = Opera Omnia, vol. IV, p. 370: Modus faciendi libr_, que pondera rerum maiora quàm sunt[?? _ nearly obliterated in the text I have seen] ostendat.  Describes a scale with arm divided  11 : 12.

John Wecker.  Op. cit. in 7.L.3.  (1582), 1660.  Book XVI _ Of the Secrets of Sciences: Chap. 20 _ Of Secrets in Arithmetick: Fraud in Balances where things heavier shall seem to be lighter, p. 293.  Says such fraud is mentioned by Aristotle.

van Etten.  1624.  Prob. 54 (49), pp. 49‑50 (73‑74).  Mentions Aristotle's mechanical questions and cites Archimedes' law of the lever.  Discusses example with arm lengths  12 : 11

Leybourn.  Pleasure with Profit.  1694.  Tract. IV, pp. 2-3.  Cites Solomon and Pappus' Collections, Book 3.  Discusses arm lengths  11 : 10.

Ozanam.  1694.  Prob. 4, 1696: 275-276 & fig. 131, plate 46.  Prob. 4 & fig. 26, plate 14, 1708: 351‑352.  Vol. II, prob. 7, 1725: 339‑340 & fig. 131, plate 46.  Vol. II, prob. 3, 1778: 4-5;  1803: 4-6;  1814: 3-5;  1840: 196-197.  Construct a balance which is correct when empty, but gives dishonest weight.  This can be detected by interchanging the contents of the two pans.  Hutton notes that the true weight is the geometric mean of the two weights so obtained, and that this is close to the average of these two values.  Illustrates with weights  16  and  14.  The figure is just a picture of a balance and is not informative _ the same figure is also cited for various sets of weights.

Vyse.  Tutor's Guide.  1771?  Prob. 1, p. 316 & Key p. 356.  Cheese weighs  76  in one pan and  56  in the other.  States the general rule with no explanation.

Jackson.  Rational Amusement.  1821.  Curious Arithmetical Questions.  No. 8, pp. 16 & 72.  Cheese weighs  16  on one side and  9  on the other.  Says the answer is the mean proportional.  = Illustrated Boy's Own Treasury, 1860, prob. 12, pp. 428 & 431. 

Julia de Fontenelle.  Nouveau Manuel Complet de Physique Amusante ou Nouvelles Récréations Physiques ....  Nouvelle Édition, Revue, ..., Par M. F. Malepeyre.  Librairie Encyclopédique de Roret, Paris, 1850.  Pp. 407-408 & fig. 146 on plate 4 (text erroneously says V): Balance trompeuse.  A bit like Ozanam, but doesn't indicate the true weight is the geometric mean.  Figure copied from Ozanam, 1725.

Magician's Own Book.  1857.  The false scales, p. 251.  Cheese weighs  9  on one side and  16  on the other.  Says the true weight is the mean proportional, hence  12  here.  = Book of 500 Puzzles, 1859, p. 65.  = Boy's Own Conjuring Book, 1860, p. 223.  Almost identical to Jackson.

Hoffmann.  1893.  Chap. IV, no. 94: The false scales, pp. 169 & 229.  On one side a cheese weighs  9  and on the other it weighs  16.  Answer notes that the true weight is always the geometric mean.  Almost identical to Jackson and Magician's Own Book.

Clark.  Mental Nuts.  1904: no. 66;  1916: no. 84.  The grocer puzzled.  Weights of  8  and  18.  Answer says to solve  8 : x :: x : 18.

Briggs & Bryan.  The Tutorial Algebra _ Part II.  Op. cit. in 7.H.  1898.  Exercises X, pp. 125 & 580.  Weighing one way gains an extra  11%  profit, but weighing the other way gives no profit at all.  What is the legitimate profit?

Pearson.  1907.  Part II, no. 72, pp. 128 & 205.  Same problem as Hoffmann.  Answer says to take the square root of  9 x 16.

Ernest K. Chapin.  Loc. cit. in 5.D.1.  1927.  Prob. 1, p. 98 & Answers p. 10: The druggist's balance.  One arm is longer than the other, but the shorter is weighted to give balance when the scales are empty.  He uses the two sides equally _ does he gain or lose (or come out even)?  If the lengths are  L  and  l,  then balancing against weight  W  will result in  WL/l  and  Wl/L  equally often and the arithmetic mean of these is greater than their geometric mean of  W,  so the druggist is losing.  Answer only does the example with  L = 2l.

Loyd Jr.  SLAHP.  1928.  The jeweler's puzzle, pp. 21‑22 & 90.  More complex version.

Kraitchik.  La Mathématique des Jeux.  Op. cit. in 4.A.2.  1930.  Chap. II, prob. 26, p. 34.  = Mathematical Recreations; op. cit. in 4.A.2; 1943; Chap. 2, prob. 54, pp. 41‑42.  Merchant has a false balance.  He weighs out two lots by using first one side, then the other.  Is this fair on average?

 

          10.Q.  PUSH A BICYCLE PEDAL

 

          Holding a bicycle upright, with the pedals vertical, push the bottom pedal backward.  What happens?

 

Pearson.  1907.  Part II, no. 17: A cycle surprise, pp. 14 & 189.

Ernest K. Chapin.  Loc. cit. in 5.D.1.  1927.  P. 99 & Answers p. 11.  Pull the bottom pedal forward.  What happens?  What is the locus of the pedal in ordinary travel?  Answer says it is a cycloid but it is actually a curtate cycloid.

W. A. Bagley.  Paradox Pie.  Op. cit. in 6.BN.  1944.  No. 60: Another cycling problem, p. 46.  Uses a tricycle _ which simplifies the experimental process.

David E. Daykin.  The bicycle problem.  MM 45:1 (Jan 1972) 1.  Short analysis of this 'old problem'.  No references.

 

          10.R.  CLOCK HAND PROBLEMS

 

          These are questions as to when the hands can meet, be opposite, be interchangeable, etc.  There are also questions with fast and slow clocks.  Many examples are in 19C arithmetic and algebra books and in Loyd, Dudeney, etc.  I am somewhat surprised that my earliest examples are 1678? and 1725, as clocks with minute hands appeared in the late 16C.  The problems here are somewhat related to conjunction problems _ see 7.P.6. 

          See also 5.AC for digital clocks, which are a combinatorial problem.

          These are related to Section 7.P.6.

 

Wingate/Kersey.  1678?.  Quest. 20, p. 490.  Clock with an hour hand and a day hand, which goes round once every 30 days.  They are together.  When are they together again?  In 30 days, the faster hand must pass the slower 59 times, so the time between coincidences is  30/59  of a day.

Ozanam.  1725.  Prob. 11, question 3, 1725: 76‑77.  Prob. 2, 1778: 75-76;  1803: 77-78;  1814: 69;  1840: 37.  When are the hands together?  1725 does it as a geometric progression, like Achilles and the tortoise.  1778 adds the idea that there are  11  overtakings in  12  hours, but this does not appear in the later eds.

Les Amusemens.  1749.  Prob. 122, p. 264.  When are clock hands together?

Vyse.  Tutor's Guide.  1771?  Prob. 7, p. 317 & Key pp. 357-358.  When are the hands together between  5  and  6  o'clock?

Dodson.  Math. Repository.  1775.

P. 147, Quest. CCXXXIV.  When are hour and minute hands together?

P. 147, Quest. CCXXXV.  When are hour, minute and second hands together?

Charles Hutton.  A Complete Treatise on Practical Arithmetic and Book-keeping.  Op. cit. in 7.G.2.  [c1780?]  1804: prob. 35, p. 136.  When are watch hands together between  4  and  5?

Pike.  Arithmetic.  1788.  P. 352, no. 27.  When are the clock hands together next after noon?

Bonnycastle.  Algebra.  1782.

P. 86, no. 22.  When are the hands next together after  12:00?

P. 201, no. 1 (1815: p. 226, no. 1).  When are the hands together between  8:00  and  9:00?

Eadon.  Repository.  1794.  P. 195, no. 13.  When do the hands meet after noon?

Hutton.  A Course of Mathematics.  1798?  Prob. 36,  1833: 223;  1857: 227.  When do the hands next meet after noon?

Robert Goodacre.  Op. cit. in 7.Y.  1804.  Miscellaneous Questions, no. 128, p. 205.  When are clock hands next together after  12:00?

Louis Pierre Marie Bourdon (1779-1854).  Élémens d'Algèbre.  7th ed., Bachelier, Paris, 1834.  (The 6th ed., was 1831, but I haven't yet looked for the early editions _ ??)  Art. 57, prob. 10, p. 85.  When are clock hands together?

Augustus De Morgan.  Examples of the Processes of Arithmetic and Algebra.  Third, separately paged, part of: Library of Useful Knowledge _ Mathematics I, Baldwin & Craddock, London, 1836.  P. 90.  How much time elapses between times when the hands are together?  Suppose one hand revolves in  a  hours, the other in  b  hours _ how long between conjunctions?  In the latter case, how long does it take the faster hand to gain  p/q  of a whole revolution on the other?

Hutton-Rutherford.  A Course of Mathematics.  1841?

Prob. 11,  1857: 81.  When are clock hands next together after 12:00?

Prob. 19,  1857: 82.  When are clock hands together between 5 and 6 o'clock?

T. Tate.  Algebra Made Easy.  Op. cit. in 6.BF.3.  1848.

Pp. 46-47, no. 19.  When are the hands together between  2:00  and  3:00?

P. 46, no. 20.  Same between  3:00  and  4:00?

Anonymous.  A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools.  3rd ed., 1850.  Op. cit. in 7.H.

P. 356, no. 4.  When are hands together between  5  and  6?

P. 360, no. 46.  When are hands together between  1  and  2?

Dana P. Colburn.  Arithmetic and Its Applications.  H, Cowperthwait & Co., Philadelphia, 1856.  Miscellaneous Examples.

No. 12, p. 359.  When are hands together after  12:00?

No. 13, p. 359.  When are hands opposite after  12:00?

No. 32, p. 361.  When is the hour hand as much after  12  as the minute hand is before  1?

Colenso.  Op. cit. in 7.H.  These are from the 1864? material.

No. 3, p. 186.  When are hour and minute hands at right angles, or coincident, between  5:00  and  6:00?

No. 21, pp. 190 & 215.  When are hour and minute hands  162o  apart between  11:00  and  12:00?

William J. Milne.  The Inductive Algebra ....  1881.  Op. cit. in 7.E.

No. 95, p. 140.  When are hands together after  2:00?

No. 96, pp. 140 & 332.  When are hands together after  5:00?

No. 97, pp. 141 & 332.  When are hands together after  8:00?

No. 98, pp. 141 & 332.  When are hands at  180o  after  4:00?

No. 99, pp. 141 & 332.  When are hands at  180o  after  5:00?

No. 100, pp. 141 & 332.  When are hands at  90o  after  6:00?

No. 101, pp. 141 & 332.  When are hands at  90o  after  8:30?

No. 104, pp. 303 & 347.  When are hands at  180o  after  11:00?

Ernest K. Chapin.  Loc. cit. in 5.D.1.  1927.  Prob. 5, p. 89 & Answers p. 8.  Can the three hands make equal angles?  [Last line of the answer has slipped into the next column.]

Perelman.  1937.  MCBF.  Interchanging the hands of a clock, prob. 141, pp. 235-238.  Says the following was posed to Einstein by his friend A. Moszkowski:  at what times can one exchange the hour and minute hands and get a valid position of the hands?  There are  143  solutions, of which  11  are the positions where the hands coincide, which Perelman discusses as prob. 142, pp. 238-239.

Sullivan.  Unusual.  1943.  Prob. 1: Watch and see.  How many places do hands meet?

Pierre Berloquin.  The Garden of the Sphinx, op. cit. in 5.N.  1981.

Prob. 11: What coincidences?, pp. 10 & 91.  When are the hour, minute and second hands together?

Prob. 12: Meetings on the dot, pp. 10 & 92.  At what other times are the three hands closest together?

 

          10.S.   WALKING IN THE RAIN

 

W. A. Bagley.  Paradox Pie.  Op. cit. in 6.BN.  1944.  No. 93: Better take a brolly, p. 61.  Says the faster you go, the wetter you get.

A. Sutcliffe.  Note 2271:  A walk in the rain.  MG 41 (1957) 271‑272.

C. O. Tuckey.  Note 2384:  A walk in the rain [Note 2721].  MG 43 (No. 344) (May 1959) 124‑125.

M. Scott.  Nature.  ??NYS _ reported in This England 1965-1968, p. 70.  "When walking into the rain one should lower the head and walk as fast as possible.  When the rain is coming from behind one should either walk forward leaning backwards, or backwards leaning forwards, at a deliberate pace."

David E. Bell.  Note 60.21:  Walk or run in the rain?  MG 60 (No. 413) (Oct 1976) 206-208.  "... keep pace with the wind if it is from behind; otherwise, run for it."

D. R. Brown.  Answer to question.  The Guardian (2 Apr 1993) 13.  Cites Bell.

Yuri B. Chernyak & Robert S. Rose.  The Chicken from Minsk.  BasicBooks, NY, 1995.

Chap. 4, prob. 5: Rainy day on the carousel, pp. 27 & 113.  How to hold your umbrella when riding on a carousel in vertical rain.

Chap. 4, prob. 9: The May Day parade, pp. 31 & 116-117.  How to point a cannon on a truck so that vertical rain does not touch the sides of the barrel.

Chap. 4, prob. 11: Boris and the wet basketball _ reference frames and fluxes, pp. 33 & 118-120.  Sliding a sphere in the rain, it gets wetter faster, the faster it goes.  However the total wetness is reduced by going faster because it takes less time to get home.

Chap. 4, prob. 12: Chicken feed, pp. 34 & 120.  Boris is wheeling a wheelbarrow with a bucket of chicken feed in it when it starts to rain vertically.  Assuming the bucket is level, should he run?  The flux of rain entering the bucket is constant, so running reduces the time in the rain and hence the amount of rain which gets into the bucket.

Chap. 4, prob. 13: Fluxes and conservation laws (or it always helps to run in the rain), pp. 34 & 121-122.  Is there a contradiction between the rate results in prob. 11 and prob. 13?  No, because the effective cross sections behave differently.

 

          10.T.  CENTRIFUGAL PUZZLES

 

Fred Swithenbank.  UK Patent 11,801 _ An Improved Toy or Puzzle, applicable also as an Advertising Medium.  Applied 21 May 1913;  completed 21 Nov 1913;  accepted 1 Jan 1914.  2pp + 1p diagrams.  Shows two ball version and four ball round version.

I have acquired an example in wood with a celluloid(?) top with the following printed on the sides.   'Zebra' Grate Polish.  Get one ball in each hole at the same time.   'Brasso' Metal Polish.  Patent No. 11801/13.  I have seen others of similar date advertising:  "Swan" Pens / "Swan" Inks;  Hoffmann Roller Bearings.  Both have the same instruction and patent no.

Slocum says that he has the same Zebra/Brasso version and another version from R. Journet, called Spoophem, with the same patent number and that a round version "The Balls in the Hole Puzzle" is in Gamage's 1913 catalogue.

James Dalgety.  R. Journet & Company.  A Brief History of the Company & its Puzzles.  Published by the author, North Barrow, Somerset, 1989.  P. 13 mentions the Spoophem puzzle, patented in 1913.

Slocum.  Compendium.  Shows 4 ball Centrifugal Puzzle from Johnson Smith 1919 & 1929 catalogues.  Shows 2 ball Spoophem Puzzle from the latter catalogue.

Western Puzzle Works, 1926 Catalogue.  No. 72: Four Ball Puzzle.

William R. Maxwell.  US Patent 1,765,019 _ Ball puzzle device.  Filed 6 Apr 1929;  patented 17 Jun 1930.  2pp + 1 p diagrams.  Shows 2-ball and 4-ball version.

L. Davenport & Co. and Maskelyne's Mysteries.  New Amusement Guide [a novelties catalogue].  Davenport's, London, nd [Davenport's identifies this as being in the period 1938-1942], p. 6, shows a Boat Puzzle, which is a 2 ball version..

 

          10.U.  SHORTEST ROUTE VIA A WALL, ETC.

 

          I have just added this.  I imagine there are ancient versions of this problem.  See also 6.M and 6.BF.3 for some problems which use the same reflection principle.  Basically, this will cover two-dimensional cases where the reflection process is very physical.

 

Ozanam‑Montucla.  1778.  Prob. 24 & fig. 36, plate 5, 1778: 305;  1803: 300-301;  1814: 256;  1840: 130.  Run from  A  to  B,  touching the wall CD.

Ozanam‑Montucla.  1778.  Vol. II.  Prob. 35: Du jeu de billard & figs. 34 & 35, plate 7, 1778: 58-62;  1803: 63-66;  1814: 52-55;  1840: 222-224.  Use cue ball to hit another after hitting one or two cushions.

Birtwistle.  Math. Puzzles & Perplexities.  1971.  Playground circuit, pp. 142-143.  In a rectangular playground with a point marked in it, what is the shortest route from the point to all four walls and back to the point?  He draws it and says it's a parallelogram, but doesn't see that the length is twice the diagonal of the rectangle, independently of the starting point.

 

          10.V.  PICK UP PUZZLES  =  PLUCK IT

 

          I have recently added these.  The puzzle comprises a ball which is about half into a hole and the object is to remove it without moving the hole.  It can be done with the fingers, but one can also use the Venturi effect!

 

Western Puzzle Works, 1926 Catalogue.  No. 11: Pick up ball.  Negro head.

 

          10.W. PUZZLE VESSELS

 

          New section.  S&B 140-141 gives an outline of the ancient history.  Vases which fill from the bottom are among the Phoenician items found in Cyprus in the 1870s and some are among the Treasure of Curium in the Cesnola Collection at the Metropolitan Museum of Art, NY.  These date from early second millennium BC.  There are examples or descriptions of such jugs in -5C Greece, in the works of Hero and Philo of Alexandria in 1C, and in 13C France.  From the 16C, they are common throughout Europe and China, and the bottom filling 'Cadogan' teapot goes back to perhaps c1000 in China.  I have now looked at some of the literature and have realised that this topic covers a much larger range of types than I had initially thought.  In Banu Musa, there are about a hundred types.  Some of these are common in later works, e.g. van Etten, Ozanam, etc. have casks which pour different liquids from the same spout.  I don't know if I will try to include all these later versions.

          A Tantalus or temperance cup has a siphon such that if it is filled above a certain point, the siphon drains the cup, usually into the drinker's lap!

          In response to an exchange on NOBNET in Feb 1999, Peter Rasmussen kindly sent a page and a half of bibliography on Chinese puzzle vessels.  A number of his entries are auction, sale or collection catalogues which apparently only show one or two items, so I won't list them here, but I have included the more general books at the end of this section.

          Norman Sandfield has also sent a copy of his bibliographical and historical material, 10pp, dated 27 Jan 2000, mostly recent literature on Chinese vessels, together with a version of this section.

 

Joseph Veach Noble.  Some trick Greek vases.  Proc. Amer. Philosophical Soc. 112:6 (Dec 1968) 371-378.  He describes and illustrates a number of types of trick vases from Athens:  false bottoms (-4C);  dribble vases (which dribble wine over the guest) (-5C);  a covered drinking cup which fills from the bottom like a Cadogan teapot (‑5C);  vases with concealed contents which pour when the host releases fingers from holes (-5C).   In each case he cites museums and extended descriptions.  One reference gives a list of covered cups.

Jasper Maskelyne.  White Magic.  Stanley Paul, London, nd [c1938], p. 110.  Discusses puzzle pitchers, saying  "On the site of a Greek temple at Athens, excavators a few years ago discovered a magic pitcher which was famous in Greek legend in trials for witchcraft."  He describes the standard hollow handled pitcher.

Walter Gibson.  Secrets of the Great Magicians.  (Grosset & Dunlap, 1967);  Collins, Glasgow, 1976.  Describes several ancient devices based on siphons. 

Pp. 13-14: The tomb of Belus.  This was encountered by Xerxes, c‑480, and was a vessel which maintained its level when liquid was poured into it.

Pp. 18-19: The ever-full fountain.  This was described by Hero and is a vessel which maintains its level when water is drawn from it.

Pp. 95-96: The Bundar boat.  This is a device which has intermittent flow which was used by Indian magicians, apparently already in use when Westerners got to India.

John Timbs.  Things Not Generally Known,  Familiarly Explained.  A Book for Old and Young (spine says  First Series  and a note by a bookdealer on the flyleaf says  2 vol.).  Kent & Co., London, (1857?), 8th ed., 1859.  Hydrostatic wonders, p. 111, mentions some classical examples as:  "The magic cup of Tantalus, which he could never drink, though the beverage rose to his lips; the fountain in the island of Andros, which discharged wine for seven days, and water during the rest of the year; the fountain of oile which burnt out to welcome the return of Augustus from the Sicilian war; the empty urns which, at the annual feast of Bacchus, filled themselves with wine, to the astonishment of the assembled strangers; the glass tomb of Belus, which, after being emptied by Xerxes, would never again be filled; the weeping statues of the ancients; and the weeping virgin of modern times, whose tears were uncourteously stopped by Peter the Great when he discovered the trick; and the perpetual lamps of the magic temples, _ were all the obvious effects of hydrostatical pressure.  _ North British Review, No. 5."

Banu Musa  = Ban_ M_s_ bin Sh_kir  (Sons of Moses),  but largely the work of Ahmad  = Abu-l-Hasa Ahmad ibn M_sa.  Kit_b al-Hiyal.  c870.  Translated and annotated by Donald R. Hill as:  The Book of Ingenious Devices; Reidel, 1979.  Describes 103 devices, most of which are trick vessels, as well as fountains, etc.  E.g. Model 1:  "We wish to explain how a beaker is made in which a quantity of wine is poured, and if [a measure] of wine or water is added to it all its contents are discharged."  This is a Tantalus cup.  Model 4:  "We wish to make a jar with an open outlet: if water is poured into it nothing issues from the outlet, and if pouring is stopped the water issues from the outlet, and if pouring is resumed [discharge] ceases again, and if pouring is stopped the water discharges, and so on continuously."  Models 12 - 15 have finger holes under the handle allowing the pourer to produce various effects.

Sandfield says the earliest Chinese puzzle vessels are in the Xian Museum and are dated to either the Song (951-960) or Northen Song (960-1127 or 1279).  (My Chinese chronology has Song being 960-1279, with Northern Song being 960-1127 and Southern Song being 1127-1279.)

Al-Jazari  = Bad_‘al-Zam_n Ab_ al-‘Izz Isma‘il ibn al-Razz_z al-Jazar_.  Kit_b f_ ‘rifat al-hiyal al-handasiyya.  c1204.  Translated and annotated by:  Donald R. Hill as:  The Book of Knowledge of Ingenious Mechanical Devices; Reidel, Dordrecht, 1974.  Introduction, pp. 3‑12, and Conclusion, pp. 279-280, include comparisons with other works.  Not much of interest to us, except for the following.

Category II, chapter 5, pp. 110-114, 221, 256-257, 272.  "It is a pitcher for wine which is used in carousals, into which water and wines of [different] colours are poured; it has a valve from which each colour is drawn separately."

Category III, chapter 1, pp. 127-129, 259, 272.  "A pitcher from which hot water, cold water and mixed water is poured."

In the Historical Museum at Cologne is the 'goblet of Albertus Magnus', which has a false bottom which allowed him to introduce antimony and make the drink emetic.  He lived 1193-1280, so this might date from 1250.  Described and illustrated in:  Edwin A. Dawes; The Great Illusionists; Chartwell Books, Secaucus, New Jersey, 1979, p. 21.

Ashmolean Museum, Oxford, item 1921-202, presently in Room 4, in the case devoted to Decorative Techniques, item 14.  Late 13C, different than the Exeter jug below.  Described, with photo and photo of a cut-away model, in Crossley, below.  The same item is illustrated by a drawing in the following entry.

Jeremy Haslam.  Medieval Pottery in Britain.  Shire, Aylesbury, (1978);  2nd ed., 1984.  On pp. 19, under: Some regional types; Oxford region; Thirteenth and fourteenth centuries, he mentions, among characteristic items of the Oxford region in the late 13C, "'puzzle‑jugs' decorated with applied scales, strips of red-firing clay and stags' heads (13, 10)".  Fig. 13 on p. 47, Oxford, Fourteenth- and Fifteenth-Century Vessels, item 10, is "Puzzle-jug, glazed green, with app. scales, red strips, face masks and deer head over spout."  The drawing is by the author from the above mentioned item in the Ashmolean Museum, Oxford.

The Royal Albert Museum, Queen Street, Exeter, EX4 3RX, has a notable example of a puzzle jug, c1300, from SW France or the Saintonge region of west France, described as the finest piece of imported medieval pottery in England.  A postcard of it is available from the Museum.  Crossley, below, says it is late 13C from the Saintonge and was excavated in Exeter in 1899.  He gives a cross-sectional drawing and cites a 1988 excavation report.  In fact, the puzzle aspect is quite simple _ the upper level is connected to the base via the handle which leaves room for three levels of decoration, showing some apparently unclothed bishops inside, then some ladies leaning out of windows and then some musicians serenading outside!  It is depicted and described in: John Allan & Simon Timbs; Treasures of Ancient Devon; Devon Books (Devon County Council), Tiverton, 1996, pp. 34-35. 

Cardan.  De Rerum Varietate.  1557, ??NYS.  = Opera Omnia, vol. III, pp. 252.  Liber XVIII.  Urcei qui non se mergunt.  Apparently a vase which one fills with water and then pours wine out of.

Prévost.  Clever and Pleasant Inventions.  (1584), 1998.  Pp. 41-44.  A Tantalus cup.

John Bate.  The Mysteries of Nature and Art.  In foure severall parts.  The first of Water works.  The second of Fire works.  The third of Drawing, Washing, Limming, Painting, and Engraving.  The fourth of sundry Experiments.  (Ralph Mab, London, 1634.)  The second Edition; with many additions into every part.  Ralph Mabb, London, 1635.  (3rd ed., Andrew Crooke, London, 1654.)  [BCB 20-22.  Toole Stott 81-83.  HPL [Bate] RBC has 2nd & 3rd eds.]  The first part has several examples, notably the following.

                    P. 2.  How to make a conceited pot, which being filled with water, will of it selfe runne all out; but not being filled will not run out.

                    P. 3.  Another conceited Pot out of which being first filled with wine and water, you may drinke pure wine apart, or faire water apart, or els both together.

Louis L. Lipski.  Dated English Delftware: Tin-glazed Earthenware, 1600-1800.  Sotheby Publications, London  &  Harper & Row, Scranton, Pennsylvania,  1984.  ??NYS _ cited and quoted by Sandfield.  Their earliest upright puzzle jug is no. 1009, dated to 1653.

Edgar Gorer & J. F. Blacker.  Chinese Porcelain and Hard Stones, Illustrated ....  Quaritch, London, 1911.  Vol. 1, plate 135.  ??NYS _ cited and quoted by Sandfield.  Two illustrated examples of puzzle jugs, i.e. 'how-to-pour puzzles', dated from the Kang-he (= K'ang Hsi) period, 1662-1722.  ??check dates.

Ozanam.  1694.  Probs. 14 & 26: 1696: 286 & 294 & figs. 137 & 138, plate 47 & fig 146, plate 49;  1708: 362-363 & 370-371 & figs. 33 & 34, plate 15 & fig. 44, plate 18.  Probs. 18 & 30: 1725: vol. 2: 389-390 & 404 & figs. 137 & 138, plate 47 & fig. 146, plate 49.  Probs. 4 & 5: 1790: vol. 4: 33-35 & figs. 5-7, plate 1;  1803: 34-36 & figs. 5-7, plate 1;  1840: 613‑614.  Three forms of Tantalus cups.  In 1790, it is called Tantalus and a figure of Tantalus is put in the cup so that when the liquid approaches him, it runs out.  Some of these are such that they pour out when the cup is tilted.

Patrick J. Donnelly.  Blanc de Chine: The Porcelain of Tehua in Fukien.  London, 1969.  Pp. 95, 345, Appendix 5: N86 & N69.  ??NYS _ cited and quoted by Sandfield.  Describes some tantalus cups, where the siphon is concealed in a figure, dated 1675-1725.  The T'ao-ya says the figure represents Lu Hung-chien (= Lu Hong Jian), a 16C author on porcelain whose name was Hsiang Yuan-pien Tzu-ching and that the vessel is called a 'quiet mind' dish.  Augustus the Strong (of Saxony?) had an example before 1721.  There is an example in the Princessehof Museuem, Leeuwarden, The Netherlands.  Other versions mentioned in Sandfield have naked men or women in the cups.

The Victoria and Albert Museum, London, has an example from the K'ang Hsi period (1622-1722).  ??NYS _ cited and quoted by Sandfield.  ??check dates.

Alberti.  1747.  Art 31, pp. ?? (131-132) & fig. 57 on plate XVI, opp. p. ?? (130).  Vase pours different liquors.

Tissandier.  Récréations Scientifiques.  1880?;  2nd ed., 1881, pp. 327-328 describes 'vases trompeurs', with illustrations on pp. 324-325.  Says they were popular in the 18C and earlier and the illustrations are of examples in the Musée de la Manufacture de Sèvres.  Not in 5th ed., 1888.

Tissandier.  Popular Scientific Recreations.  1890?  Pp. 65-67, with illustrations on pp. 67-69, describes "vases of Tantalus" _ cups which cannot be filled too full _ and then gives the material from the 2nd French ed., 1881.

The Ashmolean Museum, Oxford, Room 47 has a fine example of a Lambeth puzzle jug dated 1745.

The Rijksmuseum, Amsterdam, Room 255 has a fine example of a Delft puzzle jug dated 1768.

Henry Pierre Fourest.  Delftware: Faience Production at Delft.  (Translated from La Faiences de Delft.)  Rizzoli, NY, 1980.  P. 136, no. 130.  ??NYS - cited and quoted by Sandfield.  Shows a Delft puzzle jug from second half of 18C.  In a footnote, the author says: "This very old common form was copied from the Chinese for the East India Company."

Badcock.  Philosophical Recreations, or, Winter Amusements.  [1820].  Pp. 130-131, no. 195 & Frontispiece fig. 10: Hydraulic Experiment called Tantalus's Cup.

The Boy's Own Book.  1828: 446.  Tantalus's cup.  With statue of Tanatalus inside.

An exhibit in the Bramah Tea & Coffee Museum, London, says Mrs Cadogan brought the first examples, of what became Cadogan teapots, to England about 1830.

Young Man's Book.  1839.  Pp. 172-173.  Tantalus's Cup.

Robert Crossley.  The circulatory systems of puzzle jugs.  English Ceramic Circle Transactions 15:1 (1993) 73-98, front cover & plate V.  Starts with medieval examples from late 13C _ see above.  Says the traditional jug with sucking spouts is first known from late 14C England, though it probably derives from Italy but no early examples are known there.  Classifies puzzle jugs into 10 groups, subdivided into 20 types and into 40 variations.  Some of these are only known from one example, while others were commercially produced and many examples survive.  64 B&W illustrations, 4 colour illustrations on the cover and plate V.  Many of the illustrations show the circulation systems by cut-away drawings or photos of cut-away models, followed by photos of actual jugs.  72 references, some citing several sources or items.  The Glaisher collection at the Fitzwilliam Museum, Cambridge, provided the most examples _ 13. 

                    In the same issue of the journal, pp. 45, 48 & plate IIb show 6 examples of Cadogan teapots.

The Glaisher Collection of pottery in the Fitzwilliam Museum, Trumpington Street, Cambridge, CB2 1RB; tel: 01223-332900, has the largest number of puzzle vessels on display that I know of.  I found 22 on display, including one specially made for Glaisher.  James Whitbread Lee Glaisher (1848-1928) was a mathematician of some note at Trinity College, Cambridge.  His collection included over 3000 items.  Many of the items on display are described in Crossley's article, but Crossley mentions six items of the Glaisher Collection which are not on display.  I have prepared a list of the items on display and the further six items.

                    The collection includes several 'tygs', which are large cups with several handles.  Some of these are specifically described as puzzle vessels.  In some other cases one cannot see if a tyg is a puzzle tyg or not, but these cases are all included in Crossley.  [Crossley, pp. 92-93.]

                    The collection includes some 'fuddling cups' which are multiple interconnected small cups which either spill on you if you don't use them correctly or cause you to drink several cupfuls instead of one _ when this is not expected, you get fuddled!  One example has ten cups in a triangular array.  [Crossley, pp. 91-92] mentions these briefly.

 

J. F. Blacker.  Chats on Oriental China.  T. Fisher Unwin, Ltd., London, 1908, 406 pp.  See pp. 272‑273.  Cadogan pots.  (??NYS _ from Rasmussen.)

R. L. Hobson.  Chinese Pottery and Porcelain.  Funk & Wagnalls, New York, 1915.  Reprinted by Dover, 1976.  See pp. 276, 278.  Tantalus cups, Cadogan pots.  (??NYS _ from Rasmussen.)

R. L. Hobson & A. L. Hetherington.  The Art of the Chinese Potter from the Han Dynasty to the end of the Ming.  Benn, London  & Knopf, New York,  1923, 20[sic ??] pp.  Reprinted as: The Art of the Chinese Potter: An Illustrated Survey, Dover, 1982, 137 pp.  See Plates 128 & 149.  Tantalus cups.  Cadogan pots.  (??NYS _ from Rasmussen.)

R. L. Hobson; Bernard Rackham & William King.  Chinese Ceramics in Private Collections.  Halton & Truscott Smith, London, 1931, 201 pp.  See pp. 109‑110.  Tantalus cups.  (??NYS _ from Rasmussen.)

D. F. Lunsingh Scheurleer.  Chinese Export Porcelain: Chine de Commande.  Pitman, New York, 1974, 256 pp.  See pp. 94‑95, 215, plate 105.  Puzzle jugs.  (??NYS _ from Rasmussen.)

Miriam Godofsky.  The Cadogan pot.  The Wedgewoodian (Oct 1982) 141-142.  ??NYS _ cited and quoted by Sandfield.  Three illustrations and mentions of several in museums.

C. J. A. Jörg.  Interaction in Ceramics: Oriental Porcelain & Delftware.  Hong Kong Museum of Art, Hong Kong, 1984, 218 pp.  See pp. 78‑79 and 162‑163, nos. 36 and 115.  Puzzle jugs.  (??NYS _ from Rasmussen.)

Lynn Pan.  True to Form: A Celebration of the Art of the Chinese Craftsman.  FormAsia Books Ltd., Hong Kong, 1995, 148 pp.  See p. 18.  Cadogan pots.  (??NYS _ from Rasmussen.)

Fang Jing Pei.  Treasures of the Chinese Scholar.  Weatherhill, New York, 1997, 165 pp.  See pp. 116‑117.  Cadogan pots.  (??NYS _ from Rasmussen.)

Rik van Grol.  Puzzling China.  CFF 47 (1998??) 27-29.  ??NYR _ cited and quoted by Sandfield. 

Franz de Vreugd.  Oriental puzzle vessels.  CFF 49 (Jun 1999) 18-20.  ??NYR_ cited by Sandfield.  12 puzzle vessels in colour.

 

          10.X.  HOW FAR DOES A PHONOGRAPH NEEDLE TRAVEL?

 

          New section.  I have just seen a recent version of this and decided it ought to be entered.  There must be examples back to the early part of this century.

 

Meyer.  Big Fun Book.  1940.  No. 2, pp. 173 & 755.

The Diagram Group.  The Family Book of Puzzles.  The Leisure Circle Ltd., Wembley, Middlesex, 1984.  Problem 95, with Solution at the back of the book. 

 

          10.Y.  DOUBLE CONE ROLLING UPHILL

 

Leybourn.  Pleasure with Profit.  1694.  Tract. III, pp. 12-13: "A Mechanical Paradox: or, A New and Diverting Experiment.  Whereby a Heavy Body shall by its own Weight move up a sloping Ascent.  Written by J. P."  Nice drawing.  No indication of who J. P. is and he is not one of the publishers or the additional author.

Andrew Q. Morton.  Science in the 18th Century.  The King George III Collection.  Science Museum, London, 1993.  P. 33 shows the example in the George III Gallery and says it is c1750 and that the idea was invented at the end of the 17C and was popular in 18C lectures on mechanics.

Henk J. M. Bos.  Descriptive Catalogue  Mechanical Instruments in the Utrecht University Museum.  Utrecht University Museum, 1968, pp. 35-37.  Describes several examples, saying they are described in the classic experimental mechanics texts of the early 18C, citing 's Gravesande, Desaguliers, Nollet, Musschenbroek.  Item M 5 was bought in 1755. 

                    The Museum also has two oscillatory versions where the cone seems to roll over a peak and down the other side, then back again, ....  Item M 7 is first mentioned in an inventory of 1816.

Ozanam-Montucla.  1778.  Vol. II, prob. 26:  1790: 45-46 & figs. 22-24, plate 5;  1803: 49-50 & figs. 22-24, plate 5;  1840: 216-217.

Catel.  Kunst-Cabinet.  1790.  Der bergangehende Kegel, p. 12 & fig. 4 on plate I. 

Bestemeier.  1801.  Item 40 _ Der Berggehende Kegel.  Copies most of Catel.  Diagram is copied from Catel, but less well done.

Gardner D. Hiscox.  Mechanical Appliances  Mechanical Movements  and  Novelties of Construction.  A second volume to accompany his previous Mechanical Movements, Powers and Devices.  Norman W. Henley Publishing Co, NY, (1904), 2nd ed., 1910.  Item 931, p. 372, is an attempt to use this device as a perpetual motion by having the rails diverging where it goes uphiil and parallel where it goes downhill in alternate sections.  Patented in 1829!

Magician's Own Book (UK version).  1871.  A double cone ascending a slope by its own weight, pp. 136-137.  The picture of the cone looks more like an octahedron!

 

          10.Z.  THE WOBBLER

 

          This is formed by slotting two discs on radial lines and fitting them together.  When the distance between centres is  Ö2  times the radius, then the centre of gravity at two obvious positions is at the same height and the object rolls rather smoothly with a 'wobbly' motion.  The basic result is that the centre of gravity stays at constant height as it rolls.  Similar shapes are included here.

 

Gardner D. Hiscox.  Mechanical Appliances  Mechanical Movements  and  Novelties of Construction.  A second volume to accompany his previous Mechanical Movements, Powers and Devices.  Norman W. Henley Publishing Co, NY, (1904), 2nd ed., 1910.  Item 355: The pantanemone, p. 144.  Take a disc and cut it along a diameter.  Twist one half by 90o about the diameter perpendicular to the cut, so the planes of the semi-circles are at right angles, looking  bit like the Wobbler.  Mount the pieces on an axle at 45o to each plane _ in practice this requires some guy wires between the pieces.  He asserts that this rotates in a wind from any direction, except perpendicular to the axle, so it can be used as a stationary windmill, giving 60 days more work per year than conventional windmills.

 

A. T. Stewart.  Two-circle roller.  Amer. J. Physics 34:2 (Feb 1966) 166-167.  Shows the basic result and gets the height of the centre of gravity for any spacing.  Shows examples made from slotted plastic discs.  I learned of Stewart's work in 1992 from Mike Berry.  In a letter of 27 Oct 1992 to Berry, he says the problem is difficult and little known.  It took him "a dozen pages of cumbersome trigonometry".  He offered $5 to anyone who could derive the result on one page and only two people ever collected.

Anthea Alley.  Rocking Toy  1969.  This is an example of the wobbler made from 48" diameter disks, 1" thick, with holes about 12" diameter in the middle.  This is shown, with no text, in a photograph on p. 108 of:  Jasia Reichardt, ed.; Play Orbit [catalogue of an exhibition at the ICA, London, and elsewhere in 1969-1970]; Studio International, 1969.

Paul Schatz.  Rhythmusforschung und Technik.  Verlag Freies Geistesleben, Stuttgart, 1975.  In this he considers a number of related ideas, but his object _ the Oloid _ has the distance between centres being the radius.  This does not roll smoothly on the level, but he would roll it down a slope _ see Müller.

Paul Schatz.  Swiss patent 500,000.  (??NYS _ cited in Müller.)

Georg Müller, ed.  Phänomena _ Eine Dokumentation zur Ausstellung über Phänomene und Rätsel der Umwelt an der Seepromenade Zürichhorn, 12 Mai - 4 November 1984.  Zürcher Forum, Zürich, 1984.  P. 79 shows the Oloid rolling down a slope.

David Singmaster and Frederick Flowerday.  The wobbler.  Eureka 50 (1990) 74-78.  Flowerday developed the idea in the 1980s, but I can't recall if he developed it from Schatz's work.  This paper proves the basic result and discusses unsolved problems such as the path of the centre of gravity.  For the Oloid, the distance between the two contact points is of constant length.  Stewart's paper was not known when this was written.

Christian Ucke & Hans-Joachim Schlichting.  Wobbler, Torkler oder Zwei-Scheiben-Roller.  Physik in unserer Zeit 25:3 (1994) 127-128.  Short description, mentioning Flowerday and citing Stewart and Schatz.  Says they are sold in Germany and Switzerland as Wobbler or Go-On.  Shows a version using slotted plastic discs called Rondi.  Instructions on how to make one.

Christoph Engelhardt & Christian Ucke.  Zwei-Scheiben-Roller.  Preprint of 3 May 1994, to appear in Mathematisch-Naturwissenschaftlicher Unterricht (1995).  Basic description and derivation of the basic result.  Shows this also works for elliptical discs!  Attempts to find the path of the centre of gravity.  Singmaster & Flowerday was not known when this was written.

 

          10.AA.         NON-REGULAR DICE

 

          This deals with determining the probability of the various faces of a die which is not a regular polyhedron.  The immediate approach is a simple geometric model _ the probability of a face should be proportional to the solid angle subtended by that face viewed from the centroid.  However, this fails to agree with reality and a number of authors have attempted to explain the real situation by more complex modellings of the physical situation.

 

Scott Beach.  Musicdotes.  Ten Speed Press, Berkeley, California, 1997, p. 77.  Says Jeremiah Clarke (c1674-1707), the organist of St. Paul's Cathedral and a composer best known for the Trumpet Voluntary (properly the Prince of Denmark's March, long credited to Purcell) became enamoured of a lady above his station and was so despondent that he decided to commit suicide.  Being somewhat indecisive, he threw a coin to determine whether to hang himself or drown himself.  It landed on the ground and stuck on edge!  Failing to recognise this clear sign, he went home and shot himself!  (The text is given as a Gleaning: A loss of certainty, submitted by me, in MG 66 (No. 436) (1982) 154.)

J. D. Roberts.  A theory of biased dice.  Eureka 18 (1955) 8-11.  Deals with slightly non-cubical or slightly weighted dice.  He changes the lengths by  ε  and ignores terms of order higher than first order.  He uses the simple geometric theory.

L. E. Maistrov.  Probability Theory.  A Historical Sketch.  Academic, 1974.  ??NYS _ Heilbronner says he measured ancient dice at Moscow and Leningrad, finding them quite irregular _ the worst cases having ratios of edge lengths as great as 1.2 and 1.3.

R. Gibbs, et al.  ??  MM 51 (1978) 308-??.  Shows one can load a pair of dice so each value has equal probability of occurring.  I.e. each of  2, 3, ..., 12  has probability  1/11.  ??NYS.

Scot Morris.  The Book of Strange Facts and Useless Information.  Doubleday, 1979, p. 105.  Says  6  is the most common face to appear on an ordinary die because the markings are indentations in the material, making the six side the lightest and hence most likely to come up.  He says that this was first noticed by ESP researchers who initially thought it was an ESP effect.  The effect is quite small and requires a large number of trials to be observable.  (I asked Scot Morris for the source of this information _ he couldn't recall but suspected it came from Martin Gardner.  Can anyone provide the source?)

Frank Budden.  Note 64.17: Throwing non-cubical dice.  Math. Gaz. 64 (No. 429) (Oct 1980) 196-198.  He had a stock of 15mm square rod and cut it to varying lengths.  His student then threw these many times to obtain experimental values for the probability of side versus end.

David Singmaster.  Theoretical probabilities for a cuboidal die.  Math. Gaz. 65 (No. 433) (Oct 1981) 208-210.  Gives the simple geometric approach and compares the predictions with the experimental values obtained by Budden's students and finds they differ widely.

                    Correspondence with Frank Budden led to his applying the theory to a coin and this gives probabilities of landing on edge of  8.1%  for a UK 10p coin and  7.4%  for a US quarter.

Trevor Truran.  Playroom: The problem of the five-sided die.  The Gamer 2 (Sep/Oct 1981) 16 & 4 (Jan/Feb 1982) 32.  Presents Pete Fayers' question about a fair five-sided die and responses, including mine.  This considered a square pyramid and wanted to determine the shape which would be fair.

Eugene M. Levin.  Experiments with loaded dice.  Am. J. Physics 51:2 (1983) 149-152.  Studies loaded cubes.  Seeks for formulae using the activation energies, i.e. the energies required to roll from a face to an adjacent face, and inserts them into an exponential.  One of his formulae shows fair agreement with experiment.

E. Heilbronner.  Crooked dice.  JRM 17:3 (1984-5) 177-183.  He considers cuboidal dice.  He says he could find no earlier material on the problem in the literature.  He did extensive experiments, a la Budden.  He gives two formulae for the probabilities using somewhat physical concepts.  Taking  r  as the ratio of the variable length to the length of the other two edges, he thinks the experimental data looks like a bit of the normal distribution and tries a formulae of the form  exp(-ar2).  He then tries other formulae, based on the heights of the centres of gravity, finding that if  R  is the ratio of the energies required to tilt from one side to another, then  exp(-aR)  gives a good fit.

Frank H. Berkshire.  The 'stochastic' dynamics of coins and irregular dice.  Typescript of his presentation to BAAS meeting at Strathclyde, 1985.  Notes that a small change in  r  near the cubical case, i.e.  r = 1,  gives a change about  3.4  times as great in the probabilities.  Observes that the probability of a coin landing on edge depends greatly on how one starts it - e.g. standing it on edge and spinning it makes it much more likely that it will end up on edge.  Says professional dice have edge  3/4"  with tolerance of  1/5000 " and that the pips are filled flush to the surface with paint of the same density as the cube.  Further, the edges are true, rather than rounded as for ordinary dice.  These carry a serial number and a casino monogram and are regularly changed.  Describes various methods of making crooked dice, citing Frank Garcia; Marked Cards and Loaded Dice; Prentice Hall, 1962, and John Scarne; Scarne on Dice; Stackpole Books, 1974.  Studies cuboidal dice, citing Budden and Singmaster.  Develops a dynamical model based on the potential wells about each face.  This fits Budden's data reasonably well, especially for small values of  r.  But for a cylinder, it essentially reduces to the simple geometric model.  He then develops a more complicated dynamical model which gives the probability of a 10p coin landing on edge as about  10-8.

David Singmaster.  On cuboidal dice.  Written in response to the cited article by Heilbronner and submitted to JRM in 1986 but never used.  The experimental data of Budden and Heilbronner are compared and found to agree.  The geometric formula and Heilbronner's empirical formulae are compared and it is found that Heilbronner's second formulae gives the best fit so far.

                    I had a letter in response from Heilbronner at some point, but it is buried in my office.

Joseph B. Keller.  The probability of heads.  AMM 93:3 (Mar 1986) 191-197.  Considers the dynamics of a thin coin and shows that if the initial values of velocity and angular velocity are large, then the probability of one side approaches  1/2.  One can estimate the initial velocity from the amount of bounce _ he finds about  8 ft/sec.  Persi Diaconis examined coins with a stroboscope to determine values of the angular velocity, getting an average of  76π rad/sec.  He considers other devices, e.g. roulette wheels, and cites earlier dynamically work on these lines.

Frank H. Berkshire.  The die is cast.  Chaotic dynamics for gamblers.  Copy of his OHP's for a talk, Jun 1987.  Similar to his 1985 talk. 

J. M. Sharpey-Schafer.  Letter: On edge.  The Guardian (20 Jul 1989) 31.  An OU course asks students to toss a coin 100 times and verify that the distribution is about 50 : 50.  He tried it a 1000 times and the coin once landed on edge.

D. Kershaw.  Letter: Spin probables.  The Guardian (10 Aug 1989) 29.  Responding to the previous letter, he says the probability that a spun coin will land on edge is zero, but this does not mean it is impossible.

A. W. Rowe.  Letter.  The Guardian (17 Aug 1989) ??  Asserts that saying the probability of landing on edge is zero admits 'to using an over-simplified mathematics model'.

K. Robin McLean.  Dungeons, dragons and dice.  MG 74 (No. 469) (Oct 1990) 243-256.  Considers isohedral polyhedra and shows that there are 18 basic types and two infinite sets, namely the duals of the 5 regular and 13 Archimedean solids and the sets of prisms and antiprisms.  Then notes that unbiased dice can be made in other shapes, e.g. triangular prisms, but that the probabilities are not obvious, citing Budden and Singmaster, and describing how the probabilities can change with differing throwing processes.

Joe Keller, in an email of 24 Feb 1992 says Frederick Mosteller experimented with cylinders landing on edge 'some time ago', probably in the early 1970s.  He cut up an old broom handle and had students throw them.  He proposed the basic geometric theory.  Keller says Persi Diaconis proposed the cuboidal problem to him c1976.  Keller developed a theory based on energy loses in rolling about edges.  Diaconis made some cuboidal die and students threw them each 1000 times.  The experimental results differed both from Diaconis' theory (presumably the geometric theory) and Keller's theory. 

Hermann Bondi.  The drop of a cylinder.  European J. Physics.  1994??  Considers a cylindrical die, e.g. a coin.  Considers the process in three cases: inelastic, perfectly rough planes;  smooth plane, for which an intermediate case gives the geometric probabilities;  imperfectly elastic impacts. 

In late 1996 through early 1997, there was considerable interest in this topic on NOBNET due to James Dalgety and Dick Hess describing the problem for a cubo-octahedron.  I gave some of the above information in reply.

 

          10.AB.         BICYCLE TRACK PROBLEMS.

 

          There are three different problems involved here. 

          First, does the front wheel wear out more rapidly than the rear one?  Why?

          Second, ignoring the first point, can you determine from bicycle tracks which one is front and which is rear?

          Third, can you determine which way the bicycle was going? 

This section was inspired by running across several modern items and recalling Doyle's article.

 

I have recently read that the front wheel of a bicycle wears out faster than the rear wheel because the front wheel travels further _ on a curve, front wheels travel in an arc of larger radius than rear wheels, and even in fairly straight travel, the front wheel oscillates a bit to each side of the line of travel.  In addition, front wheels are often turned when the vehicle is at rest.  However, I cannot relocate my source of this, though I recall that the answer simply said the front wheel travels further with no further explanation.

Yuri B. Chernyak & Robert M. Rose.  The Chicken from Minsk.  BasicBooks (HarperCollins), NY, 1995.  Asks why the front tires on a car wear out faster than the rear tires and says that proper turning requires the front wheels to turn by different amounts and that this, with some other undiscussed reasons, leads to the front wheels being set slightly out of parallel, which causes the extra wear.  The solution concludes: "The perfect suspension, which would turn the wheels at exactly the proper angles, has yet to be devised." 

(On the other hand, a cross-country cyclist recently told me that his rear tire wears out faster.)

 

From the fact that the front wheel makes a more sinuous path, or that it is the outer track on a curve, or that the rear track goes over the front track, or, perhaps, that it makes a shallower track, we can tell which of the two tracks is the front wheel, but Doyle, below, does not refer to this point.

 

A. Conan Doyle.  The Adventure of the Priory School.  In this, Holmes says he can tell which way a bicycle was going from its tracks on a pathway. 

A. Conan Doyle.  The truth about Sherlock Holmes.  The National Weekly (= Collier's Weekly) (29 Dec 1923).  Reprinted in: The Final Adventures of Sherlock Holmes; ed. by Peter Haining; W. H. Allen, London, 1981, pp. 27-40 in the PB ed., esp. p. 38.  See also: The Uncollected Sherlock Holmes; ed. by Richard Lancelyn Green; Penguin, 1983, pp. 305-315, esp. pp. 313-314, which gives a longer version of the article that appeared as Sidelights on Sherlock Holmes; Strand Mag. (Jan 1924) and is basically a part of a chapter in Doyle's autobiography which he was writing in 1923.

                    Doyle recalls the bicycle track episode in "The Adventure of the Priory School" and says that a number of letters objected to this, so he went out and tried it, finding that he couldn't tell on the level, but that "on an undulating moor the wheels make a much deeper impression uphill and a more shallow one downhill; so Holmes was justified of his wisdom after all." 

Ruth Thomson  &  Judy Hindley.  Tracking & Trailing.  Usborne Spy Guides, 1978, Usborne, London, pp. 44-45.  This says: "The front wheel of a cycle makes a loopy track as the cyclist turns it from side to side to keep his balance.  As he goes faster he turns it less, so the loops are flatter.  The narrow end of the loops point in the direction where the cyclist is heading."  When I first read this, I thought that one could tell the direction from the fact that the loops get flatter as the cycle goes downhill, but the track going uphill will look similar - the cycle travels faster at the bottom then at the top.  I am not convinced that 'the narrow end of the loops' works _ see my analysis below.

Joseph D. E. Konhauser, Dan Velleman & Stan Wagon.  Which Way Did the Bicycle Go? ...and Other Intriguing Mathematical Mysteries.  MAA, Dolciani Math. Expos. 18, 1996, prob. 1, pp. 1 & 63-64.  This is a careful treatment of determining which way the bicycle was going from the geometry of the tracks in general, but I have found there is a much simpler solution in ordinary cases. 

                    Consider when the bicycle is going essentially straight and begins to turn.  Both wheels move off the straight route onto curves, so the front wheel will have gone a bit further (namely the distance between the axles) along the straight route than the rear one did, so the outer track, which is made by the front wheel, has a short straight section at the beginning of the turn.  When the bicycle completes its turn and both wheels are now going straight, the front wheel is the same distance ahead, so the rear wheel makes a bit of a straight track before meeting the track of the front wheel.  So the inner track, made by the rear wheel, has a short straight section at the end of the turn.  Knowing this, one can tell which way the cycle was going from examination of one end of a turn, provided the track is distinct enough.

 

          10.AC.         ROBERVAL'S BALANCE.

 

          This is a mechanism commonly used in pan balances but if one extends part of it outward, then it exhibits the paradoxical behaviour that the position of a weight doesn't affect the equilibrium, apparently in violation of the law of the lever.  Imagine a rectangle with pivoted corners.  Let the long edges be horizontal and the short edges be vertical.  Attach the midpoints of the long edges to an upright, so these can pivot.  As the rectangle pivots the short ends will remain vertical.  Now attach horizontal rods to these ends.  As the rectangle pivots, these remain horizontal.  If you hang equal weights on these rods, the whole thing balances, regardless of where the weights are positioned on these rods.

 

Nouvelle maniere de Balance inventée par M. de Roberval, Professor Royal des Mathématiques dans l'Université de Paris.  Journal des Sçavans (10 Feb 1670).  ??NYS - cited and described in: Henk J. M. Bos; Descriptive Catalogue  Mechanical Instruments in the Utrecht University Museum; Utrecht University Museum, 1968, pp. 37-38.

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