S2T2

Jim Randell 8:37 am on 22 December 2019 Permalink Reply
Tags: by: Christopher Higgins ( 2 ), by: Hugh Bradley ( 2 )   

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  Jim Randell 8:38 am on 22 December 2019 Permalink | Reply

  The clock undergoes a period when the hands are rotating (rotating at a constant rate) much faster that normal. Then, at some point, they begin to slow (slowing at a constant rate) until the hands stop completely, some time after 5pm.

  Presumably this happens when the hands indicate a time of “about 8 days” (say 180 – 204 hours) from the winding time (which is sometime before 12pm).

  I made some assumptions to simplify the programming:

  Having not heard of a “Vienna clock” I assumed the clock was a normal 12 hour timepiece.

  I assumed the clock was wound up sometime between 11am and 12pm, and set off advancing at many times normal speed, but at some point at or before 12pm started slowing. After that the amount advanced by the clock at 12pm, 1pm, 2pm, 3pm, 4pm, 5pm, was sufficient that the clock showed the correct time. So each one was actually displaying a time exactly some multiple of 12 hours in advance of the actual time. The clock then stopped sometime between 5pm and 6pm (actual time).

  The following Python code examines this situation. It runs in 289ms.

  Run: [ @repl.it ]

  from fractions import Fraction as F
  from enigma import irange, Polynomial, nsplit, div, sprintf as f
  
  # format time t as hh:mm:ss.ss
  def hms(t, fmt="{h:d}:{m:02d}:{s:05.2f}"):
    (t, r) = divmod(3600.0 * t, 1)
    (h, m, s) = nsplit(int(t), base=60)
    s += r
    return f(fmt)
  
  # consider number of hours advanced at 12pm
  for h0 in irange(12, 192, step=12):
    # at 1pm
    for h1 in irange(h0 + 1, 193, step=12):
      # at 2pm
      for h2 in irange(h1 + 1, 194, step=12):
        # use the three points to determine the quadratic equation
        # y = a.x^2 + b.x + c
        q = Polynomial.interpolate([(0, h0), (1, h1), (2, h2)])
        if q is None or len(q) < 3: continue
        (c, b, a) = q
  
        # does it achieve a maximum in [5, 6]
        if not(a < 0): continue
        m = F(-b, 2 * a)
        if not(5 < m < 6): continue
  
        # check the times at 3pm, 4pm, 5pm
        hs = [h0, h1, h2]
        for x in (3, 4, 5):
          y = div(q(x), 1)
          if y is not None and y % 12 == x:
            hs.append(y)
        if len(hs) < 6: continue
  
        # find the earliest start time
        s = F(h0, 1 - b)
        if not(-1 < s < 0): continue
  
        # check the total time advanced is in the required range (7.5 - 8.5 days)
        t = q(m) - s
        if not(180 < t < 204): continue
  
        print(f("stop={m} [hs={hs}, start={s}, total={t:.2f}h]", m=hms(m), s=hms(s % 12), t=float(t)))
  

  Solution: The correct time when the clock stopped was 5:35pm.

  ---

  We find a solution, with the number of hours advanced from 12pm at (12pm, 1pm, 2pm, 3pm, 4pm, 5pm) as follows:

  (+12, +73, +122, +159, +184, +197)

  The residues mod 12 of each of the values are:

  (0, 1, 2, 3, 4, 5)

  So we see that the clock would show the correct time on each hour.

  The differences between the successive terms of the sequences are:

  (61, 49, 37, 25, 13)

  Each term decreases by 12 from the previous one, so the clock is slowing at a constant rate, and comes to a stop at x = 67/12, which corresponds to 5:35pm.

  The corresponding quadratic equation is:

  y(x) = –6x² + 67x + 12

  At 12pm the clock is advancing at 67× normal speed (y'(0) = 67), so the maximum possible time after winding would be if immediately after winding, the clock set off at a constant 67× normal speed until 12pm, at which point it began to slow down.

  This places the earliest winding time at 11:49:05.5am.

  If the clock starts advancing at 67× normal speed at this time, then in the first 9.8 seconds it advances to show 12:00, and in the remaining 10m44.8s it advances another 12 hours, so that at 12pm (actual time) the clock shows 12:00.

  In the following hours the slowing clock advances an appropriate number of hours to show the correct time on each hour, until it stops at 5:35pm. In the time since winding it has advanced 199.22 hours, approximately 8.3 days, which is in the range we are expecting.

  If the clock started slowing before 12pm than it would be running at faster than 67× normal rate after winding, and the time of the winding would be closer to 12pm. But the total amount the clock has advanced is at least 197 hours (= 8.2 days), so we still get a viable answer.

  ---

  However, if it is possible that the clock was running at a constant speed until some time after 12pm, then we can get different answers to the published solution.

  For example, if the number of hours advanced at (12pm, 1pm, 2pm, 3pm, 4pm, 5pm) is:

  (+12, +49, +86, +123, +160, +197)

  The the correct time is displayed at the required times, and each value is 37 hours advanced from the previous value, so the clock is advancing at a constant rate of 37× normal speed. The time the clock was wound up was 11:40:00am.

  At (or after) 5pm the clock can be begin to slow and finally run down at whatever time we choose (between 5pm and 6pm).

  For example, if, at 5pm, the clock slows according to the following equation:

  y(x) = –74x² + 777x – 1838

  Then we have y(5) = 197, y'(5) = 37, y'(5.25) = 0, so y achieves a maximum at y(5.25) = 201.625.

  Which means the clock stops at 5:15pm, having advanced 201.96 hours (= 8.4 days). Which provides a viable alternative solution.

  So we do need to make some additional assumptions in order to arrive at a unique solution. The one I’ve highlighted in bold above is probably the key one.

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  • 

    John Crabtree 1:53 am on 23 December 2019 Permalink | Reply

    From 4.00pm to 5.00pm the clock traveled 13 hours
    From 3.00pm to 4.00pm the clock traveled 25 hours
    From 2.00pm to 3.00pm the clock traveled 37 hours
    From 1.00pm to 2.00pm the clock traveled 49 hours
    From 12.00pm to 1.00pm the clock traveled 61 hours
    That is a little over 7.5 days, and so the clock started some time before 12.00pm, and the rate of decrease is correct.

    Let the relative speed at 4.30pm, ie the average of the speeds at 4.00 and 5.00 pm, be 13.
    Then the relative speed at 3.30pm, ie the average of the speeds at 3.00 and 4.00 pm, is 25.
    The clock stopped an additional 13 / (25 – 12) = 13 / 12 hours after 4.30pm.
    And so the clock stopped at 5.35pm.

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