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How often, and when, would a clock running at the wrong speed 'look right'?
Puzzle 1 on the Home page was the old one about a stopped clock. This last puzzle concerns a running clock but too fast or too slowly.

Assuming it starts correctly at midnight (00:00:00) but its speed is twice normal, it can only next look right at 12:00:00, having in fact advanced to 24:00:00.

  • how often can it look right if running 5 times too fast?
  • what time will it show on the 6th occasion, and at what real time?
  • at speed ratios below a certain threshold there will be no chance to look right during the course of a single day other than at the start. Can you determine that threshold?
  • at what speed would a clock look right on each hour in sequence?
  • if a clock were to 'lose' one second per day, how long would it take before it next looked right?
    The answers may well be unexpected.
In the demos, consider clock 1 to show real time if clock 2 runs fast, but clock 2 to show real time if clock 1 runs slow.
The day clock 2 reaches while clock 1 indicates the passage of a single day (correctly or otherwise) is shown in the results table.

Set the speed ratio below (1.0 - 16.0). Scroll through and click the required numbers which will turn blue, and choose whether clock 2 runs fast or clock 1 runs slow by that ratio.
NB:  reset with thereset buttonbutton before choosing a new ratio.


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Answers
  • At a speed ratio of 5 times faster, the clock looks right on 8 occasions per day.
  • On the 6th occasion it will show 03:00:00, though 3 days later than the actual time of 15:00:00.
  • The threshold at which more than a day must pass before the clock can look right is 1.5.
  • To look right on each hour in sequence the speed ratio must be 13.
  • By losing one second per day, it cannot look right again for just-over 118 years!
For the demo:
  • At a speed ratio of    .
Times and day clocks 1 & 2 look the same:
 

Notes:
  • Clocks are often said to run 'slow' or 'fast' when they have simply been set to start at the wrong time despite running at normal speed. In such cases they can never show the right time, unless they stop or run down sufficiently to momentarily show the right time by chance!
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Maths
After the start, other than for a speed ratio 'R' of 1.0, the clocks cannot look the same until there is an exact 12hr (43200 secs) difference in the times they show. Angle equations are less helpful here than for the other puzzles, but in common with method 2 (see Basics), it is useful to treat time as the total number of seconds passed.

Hence, remembering that the time 'T2' on clock 2 appears to pass faster than time 'T1' on clock 1 (whether wrongly or rightly so in the case of clock 1 running too slowly):
  • T2 = T1 + 43200, or T2 - T1 = 43200.
  • speed ratio R = T2 / T1, so T2 = T1 × R.
  • therefore (T1 × R) - T1 = 43200, and T1 = 43200 / (R - 1).
Generalising to all possible matching positions, where N is the Nth occurrence, it follows that:
T1 = (43200 × (N - 1)) / (R - 1), and T2 = T1 × R.

For example, if R = 5:
  • for N = 1: the term (N - 1) equates to zero, so T1 = 0 and T2 = T1, = 00:00:00 (the start).
  • for N = 6: T1 = (43200 × 5) / 4 = 54000 secs, = 15:00:00 on day 1 (actual time).
    T2 = 54000 × 5 = 270000 secs = 75hrs, = 03:00:00 on day 4 (apparent time).
Note that by putting R = 1 the division term (R - 1) equates to zero, resulting in infinity, i.e. the clocks always show the same time.
Regarding the speed ratio below which more than one day must pass before the clocks appear the same following the start, it would be when gaining or losing 12hrs per day, i.e. R = 1 + (12hrs / 24hrs), = 1.5.

To pre-calculate the total number of matches per day, substitute T1 with seconds in a day, i.e. 86400 = (43200 × (N - 1)) / (R - 1), from which total N = (2 × (R - 1)) + 1.
However, any occurrences at or after 86400 secs are on subsequent days, so the number during a single day will be one fewer so long as the result is then rounded up, i.e: total N = ceiling (2 × (R - 1)).

For example:
  • for R = 1.5: total N = ceiling (2 × (1.5 - 1)), = 1 (at the start).
  • for R = 5: total N = ceiling (2 × (5 - 1)), = 8.
  • for R = 5.1: total N = ceiling (2 × (5.1 - 1)), = 9.
  • for R = 13: total N = ceiling (2 × (13 - 1)), = 24 (on each hour mark in sequence).
Since there are 43200 seconds in a 12hr period, a second gained or lost per day equates to a ratio of 1 1/43200 (≈ 1.000023). By losing 1 second per day, a clock would therefore take 43200 days to next look right, ≈ 118 1/4 years.

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