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How often, and when, are the minute & hour hands at any chosen angle?
This is a generalisation of puzzle 4. Remember that the angle in question is that of the smaller sector (internal angle) between the hands on a circular clock face regardless of whether the minute hand is 'behind' or 'ahead' of the hour hand. So:

  • when would the 15th and 16th occurrences of a 35° angle be?
  • does the total number of occasions per day depend on the particular angle?
To run a demo, select an angle from the digit lists below (from 0 - 180°).
Scroll through and click the required numbers which will turn blue.
NB:  reset with thereset buttonbutton before choosing a new angle.


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Answers

  • The 15th time the hands form an angle of 35° is at 7hrs 44 6/11mins (07:44:33) .
    The 16th time is at 8hrs 37 3/11mins     (08:37:16).
  • There are 44 occurrences per day regardless of the chosen angle, except for 0° and 180°, when there are only 22.
Times that the hands are at the chosen 0°:
 

Notes:
  • See the Notes to puzzle 2 regarding the potential errors involved when quoting times to the nearest second.
  • If the question is instead:  "When, after any chosen time, do the minute and hour hands first reach any chosen angle?", see puzzle 9.
  • A related question is:  "Are there any times at which all three hands are mutually at 120°, i.e. equally spaced around the clock face?"
    While it can be shown there are no mathematically exact solutions, a few times come close, such as 02:32:53.
    You can check this out with puzzle 6. How many others can you find?
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Maths
Method 1 (see Basics):
To generalise puzzle 4, where A is the chosen internal angle, the angular difference D = |5.5M - 30H| = A or (360 - A) alternately.
Each occasion can then be found from M = (30H + A + ((360 - 2A) × (1 - Nmod2))) × 2/11, starting from H = 0 and N = 1.

In addition to the single initial interval I1 so calculated, there will then be two different alternating recurrent intervals dependent on the particular angle. However, the recurrent interval between alternate occasions 'IRa' is still 65 5/11 mins.
The Nth occurrence can therefore be calculated from I1 + ((N - 1) × (IRa / 2)) + (((IRa / 2) - (2 × I1)) × (1 - Nmod2)).

For example, if A = 35:
  • initial interval (H = 0, N = 1): I1 = ((30 × 0) + 35 + ((360 - 70) × 0)) × 2/11 = 35 × 2/11, = 6 4/11 mins.
  • so for N = 15:  the time is 6 4/11 + (14 × 32 8/11) + ((32 8/11 - (2 × 6 4/11)) × 0) = 464 6/11, = 7hrs 44mins 33secs† (07:44:33).
  • and for N = 16:  the time is 6 4/11 + (15 × 32 8/11) + ((32 8/11 - (2 × 6 4/11)) × 1) = 517 3/11, = 8hrs 37mins 16secs† (08:37:16).
It will be found that there are two occurrences during most hourly periods but four hours with just one. Which particular hours depends on the angle, but the total will always be 44 regardless of the choice for A.
In the case of 0 & 180°, 44 occasions will still be listed, but each of the 22 unique times is repeated (see the Extra maths for a way to avoid this).
This figure can be found from mins per day / (IRa / 2):  total N = 1440 / 32 8/11, = 44.

NB:  Again, take care to use the precise figures of minutes and fractions of a minute, rather than times rounded to the nearest second.

Method 2 (see Basics):
Since the hour/minute angular difference alternates between A and 360 - A, the cumulative difference C for the Nth occasion will be:
  • for N = 1:  C = A.
  • for N = 2:  C = 360 - A, which may also be written as (180 × 1) + (180 - A).
  • for N = 3:  C = 360 + A,  =  (180 × 2) + A.
  • for N = 4:  C = 720 - A,  =  (180 × 3) + (180 - A), etc...
Like method 1, the alternating A and (180 - A) terms at the end of these expressions may be combined as A + ((180 - 2A) × (1 - Nmod2)). It follows that C = A + 180(N - 1) + ((180 - 2A) × (1 - Nmod2)), and since C = T × 11/120:
T = (A + 180(N - 1) + ((180 - 2A) × (1 - Nmod2))) × 120/11.

For example, if A = 35:
  • for N = 15: T = (35 + (180 × 14) + ((180 - 70) × 0)) × 120/11,  = 27873 secs†,  = 07:44:33.
  • for N = 16: T = (35 + (180 × 15) + ((180 - 70) × 1)) × 120/11,  = 31036 secs†,  = 08:37:16.
Similar to previous puzzles, the total occurrences per day can be predicted by substituting 86400 (secs per day) for T1. To allow for the alternating term one may divide it by two, so 86400 = (A + 180(N - 1) + ((180 - 2A) / 2)) × 120/11.
Solving for N leads to (((86400 × 11/120) - 90) / 180) + 1, but by ignoring the last occasion for the same reason as puzzle 4 results in total N = 44 regardless of the choice for A.
As method 1, the 22 results for angles 0 & 180° will be duplicated. See below for a way to avoid this (the program uses this strategy).

Extra maths

To select only odd-numbered solutions for angles A = 0 & 180:
  • for any sequence N = 1, 2, 3, 4..., (2N - 1) gives the odd numbers 1, 3, 5, 7...
  • note that Amod180 is zero for A = 0 & 180, but some positive number otherwise.
    Dividing by 180 and rounding up (ceiling) the fractional remainder will give 1 in the latter case, but still be zero in the former. Subtracting from 1 will then reverse the result. Call this factor 'f', hence:
    f = 1 - ceiling ((Amod180) / 180), giving 1 for A = 0 & 180, and 0 for all other A.
  • it will now be found that substituting N with the expression N + (f × (N - 1)) results in the required (2N - 1) for A = 0 & 180, but reduces to N for all other A.
† = to the nearest second

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