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When does the angle between the minute & hour hands at any chosen time next occur?
While this puzzle concerns only the the minute/hour hand internal angle, times may be chosen to the nearest second. So:

  • when do the hands next reach the same 120° angle as they assume at, say, 4pm?
  • what is their angle at 16:46:21 and when does this next occur?
To run a demo, select a time from the hrs, mins & secs below. Scroll through and click the required numbers which will turn blue.
NB:  reset with thereset buttonbutton before choosing a new time.


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Answers
  • After 4pm, the next time the minute and hour hands reach a 120° angle is 16:43:38.
  • Their angle at 16:46:21 is 134.925°, and the next time is 17:02:44.
For the demo:
  • At 00:00:00 the min/hr hand internal angle is 0°, and next occurs at 01:05:27.
Notes:
  • See the Notes to puzzle 2 regarding the potential errors involved when quoting times to the nearest second.
    For example, if you enter 01:05:27, as in puzzle 2 for overlapping hands, the precise angle will be found to be 0.025° rather than 0°, so the next position will actually be at 01:05:28 (to the nearest second) rather than 02:10:55.
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Maths
Method 1 (see Basics):
If the minute hand is initially 'behind' the hour hand by angle A (or opposite, when A = 180), it must first catch up before it can then be 'ahead' by the same amount, i.e. it must move 2A plus the hour hand movement over the same period [see puzzle 6 to calculate A].
The minute hand moves 6° and the hour hand ½° per minute, so where 'm' represents the minute hand movement required:
6m = 2A + ½m degrees, hence m = 2A × 2/11 mins.

But if the minute hand is initially 'ahead' by A (or coincident, when A = 0), it must move 360 - 2A plus hour hand movement, i.e. 6m = (360 - 2A) + ½m °,  hence m = (360 - 2A) × 2/11 mins.

These may be combined in the form (2A + ((360 - 4A) × p)) × 2/11 mins, where position factor 'p' is 0 for the minute hand initially behind or opposite, and 1 if ahead or coincident.

For example, if the initial time is 16:46:21:
  • see puzzle 6 to find internal angle A, which is 134.925°.
  • the minute hand is 'ahead' of the hour hand at that time, therefore the time needed to pass until they are next at that angle is:
    (269.85 + ((360 - 539.7) × 1)) × 2/11 = 16 4/11 mins, = 16mins 23secs†.
  • the next occurrence (to the nearest second) of angle 134.925° is thus at 16:46:21 + 00:16:23 = 17:02:44.
The appropriate value of p is easy to decide mentally and by examining or visualising the round clock face, but see the Extra maths below for a purely arithmetic approach.

Method 2 (see Basics):
In puzzle 5 the list of 44 occasions with a chosen angle A was calculated as T = (A + 180(N - 1) + ((180 - 2A) × (1 - Nmod2))) × 120/11. In this puzzle, the initial time T is of course (hrs × 3600) + (mins × 60) + secs, and rounding up T divided by (86400 / 44) will indicate the period underway and appropriate value for N. See puzzle 6 to calculate A.
The next time A occurs will be found by simply adding 1 to N, so next time = (A + 180N + ((180 - 2A) × Nmod2)) × 120/11.

For example, for initial time 16:46:21:
  • T = (16 × 3600) + (46 × 60) + 21 = 60381 secs.
  • N = ceiling 60381 / (86400 / 44) = 31.
  • A = 134.925° (see puzzle 6).
  • therefore, next time = (134.925 + (180 × 31) + ((180 - 269.85) × 31mod2)) × 120/11, = 61364 secs†, = 17:02:44.
See the Extra Maths of puzzle 5 to select only odd-numbered solutions in the cases of A = 0 & 180°.

Extra maths

To determine p for method 1:
The difference 'd' in hour & minute hand positions calculated from d = 5.5M - 30H may be negative (minute hand angle less than hour hand), zero, or positive (minute hand angle greater). Its subsequent expression as the absolute angular difference D = |d| may be less than, equal to, or greater than 180°.
The various combinations of these possibilities are illustrated opposite.
(A = min hand ahead, B = behind, C = coincident, O = opposed).
To distinguish these, the program uses the following tests:		 test diagram
  • Test a (for d -ve, = 0, or +ve):
    The greatest negative value that d can attain is -360°, so adding 360 will produce an all-positive range starting from 0.
    Thus a = round ((d + 360) / 720) gives 0 if d is -ve and 1 if = 0 or +ve.
  • Test b (for D < or >= 180):
    b = round (D / 360) gives 0 for D < 180, and 1 for D >= 180.
  • Test c (for D = 180):
    |D - 180| is zero for D = 180, but a positive number otherwise.
    Thus c = ceiling (|D - 180| / 180) gives 0 for D = 180 and 1 otherwise.
The equation that then leads to the required value of p is p = ((a + b) mod 2) × c, as can be confirmed from the table below. It shows the required value for p depending on d & D and minute vs. hour hand position, the results of tests a, b & c, and subsequent stages of the calculation:

d-ve0+ve
D vs. 180<=><<=>
positionBOACAOB
p to be0011100
test a0001111
test b0110011
test c1011101
a + b0111122
mod 20111100
× c0011100

Can you find a less complicated procedure? See the 'challenge' on the Home page.

† = to the nearest second

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